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2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.1

October 25, 2024October 23, 2024 by Prasanna

Students can Download Basic Maths Exercise 19.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.1

Part-A

2nd PUC Basic Maths Application of Derivatives Ex 19.1 Two Marks Questions and Answers

Question 1.
The displacement ‘s’ of a particle at time ‘t’ is given by S = 4t³ – 6t² + t – 7. Find the velocity and acceleration when t = 2 sec.
Answer:
Given S = 4t₃ – 6t + t – 7
Velocity = v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 12t² -12t + 1
At t = 2secs, v = 12(2)² – 12(2) + 1 = 48 – 24 + 1 = 25 units/sec.
At t = 2 sec, acceleration = 24.2 – 12 = 48 – 12 = 36 units/sec².

Question 2.
If S = 5t² + 4t – 8. Find the initial velocity and acceleration, (s = displacement, t = time).
Answer:
Given s = 5t² + 4t – 8
V = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 10t + 4 dt
Initial velocity is velocity when t = 0
i.e., = 10.0 + 4 = 4 units/sec.
Acceleration = \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 10 units /sec².

Question 3.
A stone thrown vertically upward rises ‘s’ ft. in ‘t’ sec. where s = 80t – 16t². What its velocity after 2 sec.? Find the acceleration?
Answer:
Given S = 80t -16t²
v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 80 – 32t
At t=2 sec, v = 80 – 32 (2)
= 80 – 64 = 16 ft./sec.
Acceleration = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = -22 ft/sec² .

Question 4.
A body is thrown vertically upwards its distance S feet is’t’ sec. is given by S = 5 + 12t – t². Find the greatest highest by the body.
Answer:
Given s = 5 + 12t – t²
v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 12 – 2t dt
Maximum height ⇒ Kinetic energy = 0 ⇒ v = 0 ⇒ 12 – 2t = 0 ⇒ t = 6 sec
∴ Greatest height = s = 5 + 12.6 – 6²
= 5 + 72 – 36 = 77 – 36 = 41 feet.

Question 5.
If v = \(\sqrt{s^{2}+1}\) prove that acceleration is ‘S’ (V = velocity, S = displacement).
Answer:
Given v = \(\sqrt{s^{2}+1}\)
Acceleration = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = \(\frac{1}{2 \sqrt{s^{2}+1}} \cdot 2 s \frac{d s}{d t}=\frac{1}{2 v}\) . 2 . s . v . s units / sec².

Question 6.
If S = at³ + bt. Find a and b given that when t = 3 velocity is ‘O’ and the acceleration is 14 unit. (S = displacement, t = time).
Answer:
Given s = at³ + bt; v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 3at²+b
Acceleration = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 6at
When v = 0 then 3at² + b = 0 ⇒ 27a + b = 0.
t = 3 sec
When acceleration = 14 then 14 = 6a.3, b = -27a = \(-\frac{27.7}{9}\) = -21
F = 3 sec. a = \(\frac{14}{18}=\frac{7}{9}\)
∴ a = \(\frac { 7 }{ 9 }\) and b = -21.

Question 7.
When the brakes are applied to moving car, the car travels a distance ‘s’ ft. in ‘t’ see given by s = 8t – 6t² when does the car stop?
Answer:
Given s = 8t – 6t²
v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 8 – 12t car stops when v = 0
∴ 8 – 12t = 0 ⇒ t = \(\frac{8}{12}=\frac{2}{3}\) sec.

Part-B

2nd PUC Basic Maths Application of Derivatives Ex 19.1 Three Marks Questions and Answers.

Question 1.
The radius of sphere is increasing at the rate of 0.5 mt/sec. Find the rate of increase of its surface area and volume after 3 sec.
Answer:
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 0.5 , t = 3 sec, \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = ? \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = ?
surface area = s = 4πr^{2

\(\frac{d s}{d t}\) = 4π . 2r . \(\frac{d r}{d t}\)}
= 4π × 2 × 1.5 × 0.5
= 6π m²/sec
dr = 0.5 × dt
⇒ r = 0.5 t
= 0.5 × 3
= 1.5

Question 2.
The surface area of a spherical bubble is increasing at the rate of a 0.8cm² / sec. Find at what rate is its volume increasing when r = .25cm [r = radius of the sphere].
Answer:
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 0.8 cm² / sec. r = 2.5 cm, \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = ?
s = 4πr²
\(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 4π . 2r × \(\frac{\mathrm{dr}}{\mathrm{dt}}\)
0.8 = 4π × 2 × 2.5. \(\frac{\mathrm{dr}}{\mathrm{dt}}\) ⇒ \(\frac{0.8}{8 \pi \times 2.5}=\frac{0.1}{2.5 \pi}\)
v = \(\frac { 4 }{ 3 }\)πr³
\(\frac{d v}{d t}=\frac{4}{3} \pi 3 r^{2} \cdot \frac{d r}{d t}=4 \pi \times(2.5)^{2} \times \frac{0.1}{2.5 \pi}=1 \mathrm{cc} / \mathrm{sec}\)

Question 3.
A spherical balloon is being inflated at the rate 35cc/sec. Find the rate at which the surface area of the balloon increases when its diameter is 14cm.
Answer:
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 35cc /sec 2r = 14 ⇒ r = 7, \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = ?, \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = ?
v = \(\frac { 4 }{ 3 }\)πr³ s = 4πr²
\(\frac{d V}{d t}=\frac{4}{3} \pi 3 r^{2} \frac{d r}{d t} \quad \frac{d S}{d t}=4 \pi 2 r \frac{d r}{d t}\)
35 = 4π . 7² \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 4π × 2 × 7 × \(\frac{5}{28 \pi}\)
\(\frac{d r}{d t}=\frac{35}{196 \pi}=\frac{5}{28 \pi}=10 \mathrm{cm}^{2} / \mathrm{sec}\)

Question 4.
The radius of a circular plate is increasing at the rate of \(\frac{2}{3 \pi}\) cm/sec. Find the rate of change of its area when the radius is 6cm.
Answer:
Given \(\frac{d r}{d t}=\frac{2}{3 \pi} r=6 \mathrm{cm} \frac{d A}{d t}=?\)
A = πr²
2nd PUC Basic Maths Question Bank Chapter 19 Indefinite Integrals Ex 19.1 - 1

Question 5.
A circular patch of oil spreads on water the area growing at the rate of 16cm²/min. How fast are radius and the circumference increasing when the diameter is 12cm.
Answer:
Given \(\frac{\mathrm{dA}}{\mathrm{dt}}\) = 16cm²/min, d = 2r = 12cm, \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = ? \(\frac{\mathrm{dc}}{\mathrm{dt}}\) = ? r = 6cm
A = πr² c = 2πr
\(\frac{\mathrm{dA}}{\mathrm{dt}}\) = π . 2r . \(\frac{\mathrm{dr}}{\mathrm{dt}}\) \(\frac{\mathrm{dc}}{\mathrm{dt}}\) = 2π \(\frac{\mathrm{dr}}{\mathrm{dt}}\)
16 = 2π .6 . \(\frac{\mathrm{dr}}{\mathrm{dt}}\) 2π . \(\frac{4}{3 \pi}=\frac{8}{3}\)cm / min
⇒ \(\frac{d r}{d t}=\frac{16}{12 \pi}=\frac{4}{3 \pi} \mathrm{cm} / \mathrm{min}\)

Question 6.
A stone is dropped into a pond waved in the form of circles are generated and the radius of the outer most ripple increases at the rate 2 inches/sec. How fast is the area increasing when the (a) radius is 5 inches (b) after 5 sec.?
Answer:
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 2inch/sec, r = 5 inch, \(\frac{\mathrm{dA}}{\mathrm{dt}}\) = ?
(a) A = πr²
\(\frac{\mathrm{dA}}{\mathrm{dt}}\) = π2r. \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = π × 2 × 5 × 2 = 20π sq. inches / sec.

(b) After 5 sec, dr = 2dt
r = 2t
⇒ when t = 5, r = 10 inches
\(\frac{\mathrm{dA}}{\mathrm{dt}}\) = π . 2r \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = π × 2 × 20 × 2 = 40π square inches/sec.

Question 7.
The side of an equilateral triangle is increasing at the rate \(\sqrt{3}\) cm./sec. Find the rate at which its area is increasing when its side is 2 meters.
Answer:
Given \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = \(\sqrt{3}\) cm/sec., x = 2 meters, \(\frac{\mathrm{dA}}{\mathrm{dt}}\) = ?
Area of equilateral Δ^le = A = \(\frac{\sqrt{3}}{4}\) x²
\(\frac{d A}{d t}=\frac{\sqrt{3}}{4} \cdot 2 x \frac{d x}{d t}\)
= \(\frac{\sqrt{3}}{4}\) . 2. 200. \(\sqrt{3}\) = 300cm² / sec.

Question 8.
Water is being poured at the rate of 30 mt³/min. into a cylindrical vessel whose base is a circle of radius 3 mt. Find the rate at which the level of water is rising?
Answer:
Given r = 3mts, \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 30 m³/min, \(\frac{\mathrm{dh}}{\mathrm{dt}}\) = ?
V = πr²h, r = constant
\(\frac{\mathrm{dv}}{\mathrm{dt}}\) = π . (3)² . \(\frac{\mathrm{dh}}{\mathrm{dt}}\)
30 = 9π \(\frac{\mathrm{dh}}{\mathrm{dt}}\) ⇒ \(\frac{\mathrm{dh}}{\mathrm{dt}}\) = \(\frac{10}{3 \pi}\) meter/min.

Question 9.
Sand is being dropped at the rate of 10 mt³/sec. into a conical pile. If the height of the pile twice the radius of the base, at what rate is the height to the pile is increasing when the sand in the pile is 8mt high.
Given
\(\frac{d v}{d t}\) = 10m³/sec, h = 2r, h = 8, \(\frac{d h}{d t}\) = ?
r = \(\frac{\mathrm{h}}{2}\) \(\mathrm{v}=\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{h}=\frac{1}{3} \pi \cdot\left(\frac{\mathrm{h}}{2}\right)^{2} \cdot \mathrm{h}=\frac{1}{3} \pi \frac{\mathrm{h}^{3}}{4}\)
\(\mathrm{v}=\frac{1}{12} \pi \mathrm{h}^{3} \Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\pi}{12} \cdot 3 \mathrm{h}^{2} \frac{\mathrm{dh}}{\mathrm{dt}} ; \quad 10=\frac{\pi}{4} \cdot 8^{2} \cdot \frac{\mathrm{dh}}{\mathrm{dt}}\)
\(\frac{d h}{d t}=\frac{40}{64 \pi}=\frac{5}{8 \pi} \mathrm{m} / \mathrm{sec}\)

Question 10.
A ladder of 15ft. long leans against a smooth vertical wall. If the top slides downwards at the rate of 2ft sec. Find how fast the lower and is moving when the lower end is 12ft. from the wall.
Answer:
2nd PUC Basic Maths Question Bank Chapter 19 Indefinite Integrals Ex 19.1 - 2
Given \(\frac{\mathrm{d} y}{\mathrm{dt}}\) = 2ft /sec, x = 12
From fig, x² + y² = 15²
12² + y² = 15²
y² = 15² – 12²
y = \(\sqrt{225-144}\)
y = \(\sqrt{81}\) = 9
x² + y² = 15² ⇒ 2x \(\frac{\mathrm{dx}}{\mathrm{dt}}\) + 2y\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = 0; 2.12.\(\frac{\mathrm{dx}}{\mathrm{dt}}\) + 2 .9 . 2 = 0
⇒ \(\frac{d x}{d t}=\frac{-36}{24}=\frac{-3}{2} f t / \sec\)

Question 11.
An edge of a variable cube is increasing at the rate of 10cm/sec. How fast the volume and also its surface area is increasing when the edge is 5cm long.
Answer:
Given \(\frac{d x}{d t}\) = 10 cm/sec, x = 5 cm, \(\frac{d v}{d t}\) = ? \(\frac{d s}{d t}\) = ?
(i) V = x³
\(\frac{d v}{d t}\) = 3x² \(\frac{d x}{d t}\) = 3 × (5²) × 10 = 750 cm³/sec.

(ii) S = 6x² .
\(\frac{d s}{d t}\) = 12 × .\(\frac{d x}{d t}\) = 12.5 .10 = 600 cm²/sec.

Question 12.
A man 6ft. tall is moving directly away from a lamp post of height 10ft. above the ground. If he is moving at the rate 3ft./sec. Find the rate at which the length of his shadow is increasing and also the tip of his shadow is moving?
Answer:
2nd PUC Basic Maths Question Bank Chapter 19 Indefinite Integrals Ex 19.1 - 3
Let the shadow be x & y be the distance the man walks
Given \(\frac{d y}{d t}\) = 3ft/sec From a similar Δ^les we have
\(\frac{6}{10}=\frac{x}{x+y}\)
6x + 6y = 10x
6y = 4x ⇒ 3y = 2x
⇒ \(3 \frac{d y}{d t}=2 \frac{d x}{d t} \Rightarrow 2 \frac{d x}{d t}=3.3=9\)
∴ the shadow is increasing at the rate.
∴ \(\frac{d x}{d t}=\frac{9}{2}\) ft/sec & the tip of the shadow moves is
\(\frac{d x}{d t}+\frac{d y}{d t}=\frac{9+6}{2}=\frac{15}{2} \mathrm{ft} / \mathrm{sec}\)

Question 13.
The height of circular cone is 30 cm. and it is constant. The radius of the base is increasing at the rate of 0.25cm/sec. Find the rate of increase of volume of the cone when the radius of base is 10cm.
Answer:
dr
Given h = 30 cm, \(\frac{d r}{d t}\) = 0.25cm/sec. r = 10cm. dt
V = \(\frac { 1 }{ 3 }\) πr²h
\(\frac{d v}{d t}\) = \(\frac{\pi}{3}\)h.2r. \(\frac{d r}{d t}\) = π. \(\frac { 30 }{ 3 }\) . 20.(0.25) = 50π cm² / sec

Question 14.
The volume of a spherical ball in increasing at the rate 4πcc/sec. Find the rate of increase of the radius of the ball when the volume is 288πCC.
Answer:
Given V = 288π C.C., \(\frac{d v}{d t}\) = 4πcc/ sec \(\frac{d r}{d t}\) = ?
V = \(\frac { 4 }{ 3 }\) πr³
2nd PUC Basic Maths Question Bank Chapter 19 Indefinite Integrals Ex 19.1 - 4
r = 6cm
\(\frac{d v}{d t}=\frac{4}{3} \pi \cdot 3 r^{2} \frac{d r}{d t} 4 \pi=\frac{4}{3} \pi \times 3 \times 36 \times \frac{d r}{d t} \Rightarrow \frac{d r}{d t}=\frac{1}{36} \mathrm{cm} / \mathrm{sec}\)

Question 15.
A drop of ink spreads over a blotting paper so that the circumferences of the blot is 4πcm and it changes 3cm/sec. Find the rate of increase of its radius and also find the rate of increase of its area?
Answer:
Given c = 4π, \(\frac{d C}{d t}\) = 3cm / sec \(\frac{d A}{d t}\) = ? \(\frac{d r}{d t}\) = ?
Circumference = c = 2πr
4π = 2πr ⇒ r = 2
Again
C = 2πr & A = πr²
\(\frac{d c}{d t}\) = 2π . \(\frac{d r}{d t}\) \(\frac{d A}{d t}\) = π . 2r. \(\frac{d r}{d t}\)
3 = 2π . \(\frac{d r}{d t}\) = π . 2. 2. \(\frac{3}{2 \pi}\)
⇒ \(\frac{d r}{d t}\) = \(\frac{3}{2 \pi}\) cm/ sec \(\frac{d A}{d t}\) = 6cm² / sec

Question 16.
A circular plate of metal is heated so that its radius increase at the rate of O.lmm/min. At what rate is the [plate’s area increasing when the radius is 25cm [1cm = 10mm].
Answer:
Given \(\frac{d r}{d t}\) = 0.1 mm/min, r = 25 cm, \(\frac{d A}{d t}\) A = πr²
\(\frac{d A}{d t}\) = π. 2r . \(\frac{d r}{d t}\) = π . 2. /250 (0.1) = 50πmm² /min.

Question 17.
The surface area of a spherical soap bubble increasing at the rate of 0.6cm²/sec. Find the rate at which its volume is increasing when its radius is 3cm.
Answer:
Given \(\frac{ds}{d t}\) = 0.6 cm² / sec, r = 3cm, \(\frac{d v}{d t}\) = ?
s = 4 πr² &
\(\frac{d s}{d t}\) = 4π. 2r. \(\frac{d r}{d t}\)
0.6 = 4π × 3 × 3 × \(\frac{d r}{d t}\)
∴ \(\frac{d r}{d t}=\frac{0.6}{6 \times 4 \pi}=\frac{0.1}{4 \pi} \mathrm{cm} / \mathrm{sec}\)
\(\frac { 1 }{ 40 }\) πcm/sec.
v = \(\frac { 4 }{ 3 }\) πr³
\(\frac{d v}{d t}=\frac{4}{3} \pi \cdot 3 r^{2} \frac{d r}{d t}\)
= 4πr² \(\frac{d r}{d t}\)
= 4π(3)² . \(\frac{0.1}{4 \pi}\)
= 0.9 cm³ / sec.

Question 18.
A rod 13 feet long slides with it end A and B as two straight lines at right angles which meet at ‘O’. If A is moved away from O with a uniform speed at 4ft./sec., find the speed of the end B move when A is 5 feet from O.
Answer:
2nd PUC Basic Maths Question Bank Chapter 19 Indefinite Integrals Ex 19.1 - 5
From fig we have
x² + y² = 13²
y² = 13² – 15² = 144
y = \(\sqrt{144}\) = 12
Als0
x² + y² = 13² ⇒ 2x \(\frac{d x}{d t}\) + 2y\(\frac{d y}{d t}\) = 0
\(5 \times 4=-12 \frac{\mathrm{dy}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=-\frac{20}{12}=\frac{-5}{3}=\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{-5}{3} \mathrm{ft.} / \mathrm{sec}\)

Question 19.
A street lamp is hung 12 feet above a straight horizontal floor on which a man of 5 feet is walking how fast his shadow lengthening when he is walking away from the lamp post at the rate of 175ft./min.
Answer:
2nd PUC Basic Maths Question Bank Chapter 19 Indefinite Integrals Ex 19.1 - 6
Let the shadow be y & the distance from the man walks is x.
Given \(\frac{d x}{d t}\) = 175 ft/min.
From figure we have
\(\frac{12}{5}=\frac{x+y}{y}\) ⇒ 12y = 5x + 5y ⇒ 7y = 5x
⇒ \(\frac{7 \mathrm{dy}}{\mathrm{dt}}=5 \frac{\mathrm{dx}}{\mathrm{dt}}\)
∴ the shadow is lengthening ⇒ \(\frac{d y}{d t}=\frac{5}{7} \times 175\) = 125 ft/ min.

Question 20.
Find a point on the parabola y² = 4x at which the ordinate increases at twice the rate of the abscissa [Ordinate = y, abscissa = x].
Answer:
Given y² = 4x diff. w.r.t. x
2y \(\frac{d y}{d x}\) = 4 \(\frac{d x}{d t}\)
Also given \(\frac{d y}{d x}\) = 2. \(\frac{d x}{d t}\) ⇒ 2y .2 \(\frac{d x}{d t}\) = 4. \(\frac{d x}{d t}\)
⇒ y = 1 ⇒ 1² = 4x
⇒ x = \(\frac { 1 }{ 4 }\)
∴ the point on the parabola is (\(\frac { 1 }{ 4 }\), 1 ).

2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2

October 25, 2024October 23, 2024 by Prasanna

Students can Download Maths Chapter 6 Application of Derivatives Ex 6.2 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2

2nd PUC Maths Application of Derivatives NCERT Text Book Questions and Answers Ex 6.2

Question 1.
Find the intervals in which the function f given by f (x) = 3x + 17 is strictly increasing in R
Answer:
f (x) = 3x + 17
f’ (x) = 3 > 0 , ∀ x ∈ R
hence f(x) is strictly increasing on R.

Question 2.
Show that the function given by f (x) = e^2xis strightly increasing on R
Answer:
f’ (x) = 2 e^2x > 0 ∀ x ∈ R
hence f (x) is strictly increasing on R.

Question 3.
Show that the function given by f(x) = sin x is
(a) strictly increasing in \((0, \pi / 2)\)
(b) strictly decreasing in \((\pi / 2, \pi)\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.1

Question 4.
Find the intervals in which the function f given by f (x) = 2x² – 3x is
(a) strictly increasing
(b) strictly decreasing
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.2

Question 5.
Find the intervals in which the function f given by f (x) = 2x² – 3x² – 36x +7 is
(a) strictly increasing
(b) strictly decreasing
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.3

Question 6.
Find the intervals in which the following functions are strictly increasing or decreasing:
(a) x² + 2x – 5
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.5

(b) 10 – 6x – 2x²Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.6

(d) 6 – 9x – x²Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.8

(e) (x + 1)³(x -3)³
Answer:
f (x) = (x + 1)³ (x – 3)³f’ (x) = (x + 1)³x 3(x – 3)² + (x – 3)³ x 3 (x + 1)²= (x + 1)² (x – 3)² [3x + 3 + 3x – 9]
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.9

Question 7.
Show that \(y=\log (1+x)-\frac{2 x}{2+x}, x>-1\) an increasing function of x throughout its domain.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.10

Question 8.
Find the values of x for which y = [x(x – 2)]²is an increasing function.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.11

Question 9.
Prove that \(y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta\) is an a increasing function of θ in \(\left[0, \frac{\pi}{2}\right]\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.13

Question 10.
Prove that the logarithmic function is strictly increasing on (0, ∞).
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.14

Question 11.
Prove that the function f given by f (x) = x² – x + 1 is neither strictly increasing nor strictly decreasing on (- 1, 1).
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.15

Question 12.
Which of the following function are strictly decreases on \((0, \pi / 2)\)
(a) cos x
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.16

(b) f(x) = cos 2x
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.17

(d) tan x
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.20

Question 13.
On which of the following intervals is the function f given by f (x) = x¹⁰⁰ + sin x – 1 strictly decreasing ?
(A) (0,1)
(B) \(\left(\frac{\pi}{2}, \pi\right)\)
(C) \(\left(0, \frac{\pi}{2}\right)\)
(D) None of these
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.21

Question 14.
Find the least value of a such that the function f given by f (x) = x² + ax + 1 is strictly increasing on (1, 2).
Answer:
f (x) = x² + ax + 1
f’ (x) = 2x + a
f (x) is increasing if f’ (x) > 0
2x + a > 0 is x > – a/2
2x > -a – a < 2x ⇒ a > – 2x
since x ∈ (1,2) a > -2
The least value is -2.

Question 15.
Let I be any interval disjoint from (-1, 1). Prove that the function f given by \(f(x)=x+\frac{1}{x}\) is strictly increasing on I.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.22

Question 16.
Prove that the function f given by f (x) = log sin x is strictly increasing on
\(\left(0, \frac{\pi}{2}\right)\) and strictly increasing on \(\left(0, \frac{\pi}{2}\right)\)and strictly decreasing on \(\left(\frac{\pi}{2}, \pi\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.23

Question 17.
Prove that the function f given by f (x) = log cos x is strictly decreasing on
\(\left(0, \frac{\pi}{2}\right)\)and strictly increasing on \(\left(\frac{\pi}{2}, \pi\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.24

Question 18.
Prove that the function given by f (x) = x³ – 3x² + 3x – 100 is increasing in R.
Answer:
f (x) = x³ – 3x² + 3x
f'(x) = 3x² – 6x + 3
= 3 (x – 2)²f'(x)>0 ∀ x ∈ R
∴ function is continuous on R

Question 19.
The interval in which y = x² e^-x is increasing is
(A) (- ∞ , ∞)
(B) ( – 2, 0)
(C) (2, ∞)
(D) (0,2).
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.2.25

2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1

October 25, 2024October 23, 2024 by Prasanna

Students can Download Basic Maths Exercise 5.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1

Part – A

2nd PUC Basic Maths Partial Fractions Ex 5.1 Two or Three Marks Questions and Answers

Question 1.
Express the following as a sum of polunomial and proper rational fraction ; \(\frac{x^{2}+x+1}{x^{2}-1}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1 - 1
Question 2.
\(\frac{3 x^{2}-4 x+7}{x+7}\)
Answer:

Question 3.
\(\frac{x^{2}-1}{x^{2}+1}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1 - 3

Question 4.
\(\frac{5 x^{2}}{x^{2}+4 x+3}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1 - 4

Question 5.
\(\frac{x^{4}}{(x-1)\left(x^{2}+1\right)}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1 - 5

Question 6.
\(\frac{4 x^{3}-2 x^{2}+3 x+1}{2 x^{2}+4 x-1}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1 - 6

Question 7.
\(\frac{4 x^{2}-4 x-1}{2 x-1}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1 - 7

Question 8.
\(\frac{x^{3}+7}{x^{2}-2 x+1}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.1 - 8

2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2

October 25, 2024October 23, 2024 by Prasanna

Students can Download Basic Maths Exercise 4.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2

Part – A

2nd PUC Basic Maths Binomial Theorem Ex 4.2 Two Marks Questions and Answers

1. Find
Question (i).
The 5^th term in \(\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{8}\)
Answer:
\(\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{8}\) compare with (x + a)ⁿ
⇒ x → \(\frac{4 x}{5}\), a → \(\frac{4 x}{5}\) n → 8,
To find ^th term put r = 4
T_r+ = ⁿC_r.x^{n – r}.a^r
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 1
T₅ = ⁸C₄.(2²).2^-4
⁸C₄.2^8-4 = ⁸C₄.2⁴ = 1120

Question (ii).
The 8^thterm in \(\left(\frac{a}{2}-\frac{3}{b}\right)^{10}\)
Answer:
\(\left(\frac{a}{2}-\frac{3}{b}\right)^{10}\) compare with (x+a)ⁿ
⇒ x → \(\frac{a}{5}\), a → \(\frac{2}{b}\) n = 10,
To find 8^thterm put r = 7
T_r+ = ⁿC_r.x^{n – r}.a^r
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 2

Question (iii).
The 6^th term in (√x – √y)¹⁷Answer:
Compare (√x – √y)¹⁷ with (x + a)ⁿ
x → √x a → -√y and n = 17
To find 6^th term put r = 5
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 3

Question (iv).
The 7^th term in (3x² – \( \frac{y}{3}\) )⁹Answer:
Here x → 3x²a → \(-\frac{y}{3}\) nn = 9
Put r = 6
T₆₊₁=⁹C₆ (3x²)^9-6. ( \(-\frac{y}{3}\) )⁶

Question (v).
the 10^th term in \(\left(\frac{a}{b}-\frac{2 b}{a^{2}}\right)^{12}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 5

Question (vi).
the 11^th term in \(\left(x+\frac{1}{\sqrt{x}}\right)^{14}\)
Answer:
Here x = x a = \(\frac{1}{\sqrt{x}}\) ,n = 14 and
put r = 10
T₁₀₊₁ = ¹⁴C₁₀ .x^{14 – 10} . \(\left(\frac{1}{\sqrt{x}}\right)^{10}\) ¹⁴C₄.x⁴ \(\frac{1}{x^{5}}\) = \(\frac{1001}{x}\)

Part – B

2nd PUC Basic Maths Binomial Theorem Ex 4.2 Three Marks Questions and Answers

Question 2.
Find the middele term in the expansion of

Question (i).
\(\left(x-\frac{1}{2 y}\right)^{10}\)
Answer:
\(\left(x-\frac{1}{2 y}\right)^{10}\)
Here n = even i.e 10 ∴ we have only one middle term \(\frac{\mathrm{n}}{2}+1=\frac{10}{2}+1\) = 6^th term
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 6

Question (ii).
\(\left(\frac{a}{x}+b x\right)^{12}\)
Answer:
\(\left(\frac{a}{x}+b x\right)^{12}\)
Here
n = 12 (even)
∴ We have only one middle term
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 7

Question (iii).
\(\left(\frac{2 a}{3}-\frac{3}{3 a}\right)^{6}\)
Answer:
Here n = 6 (even)
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 8

Question (iv).
\(\left(3 x-\frac{1}{6} x^{3}\right)^{8}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 9

Question (v).
\(\left(\frac{a}{3}+\frac{b}{3}\right)^{8}\)
Answer:
Here n = 8(even)
∴ middle term \frac{\mathrm{n}}{2} + 1 = 4 + 1 = 5^th
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 10

3. Find the middle terms in the expansion of

Question (i).
Find the middle term in the expansion of
\(\left(3 x-\frac{2}{x^{2}}\right)^{15}\)
Answer:
Here n = 15 and odd, so we have two middle terms i.e, \(\frac{n+1}{2}=\frac{15+1}{2}=8^{t h}\) and 8 + 1 = 9^th terms to find 8 ^thterm to find 8 ^th term pur r = 7
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 11

Question (ii)
\(\left(\frac{x}{2}+\frac{3}{x^{2}}\right)^{19}\)
Answer:
Here n = 19 (odd)∴ We have two middle terms \(\frac{n+1}{2}=\frac{20}{2}\) 10^th and 10 + 1 =11^th term To find 10^th term put r = 9
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 12
Question (iii)
\(\left(2 x^{2}+\frac{1}{\sqrt{x}}\right)^{11}\)
Answer:
Here n = 11 (odd)
∴ the two middle terms are \(\frac{13+1}{2}\) = 7^th and 6 + 1 = 7^th terms
To find 6^th term put r = 5

Question (iv).
\(\left(\sqrt{x}-\frac{4}{x^{2}}\right)^{11}\)
Answer:
Here n = 11(odd) we have two middle term
i.e, \( \frac{1+1}{2}\) = 6^th and 6 + 1 = 7^th terms
To find 6^th term put r = 5.
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 14

Question (v).
\(\left(\sqrt{x}-\frac{3}{x^{2}}\right)^{13}\)
Answer:
\(\left(\sqrt{x}-\frac{3}{x^{2}}\right)^{13}\)
Here n = 13 (odd)
∴ the two middle terms are \(\frac{13+1}{2}\) = 7^th and 7 + 1 = 8^th terms
To find 7^th term put r = 6
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 16

Part – C

2nd PUC Basic Maths Binomial Theorem Ex 4.2 Five Marks Questions and Answers

4.
Question (i).
Find the coefficent of xⁿ in \(\left(x+\frac{2}{x^{2}}\right)^{17}\)
Answer:
Here x = x a = \(\frac{2}{x^{2}}\) and n = 17
T_r+1 = ⁿC_r . x ^n-r.a^r
= ¹⁷C_r .x ^17-r.( \(\frac{2}{x^{2}}\) )^r = ¹⁷C_r . 2^r.x ^17-r-2r
= ¹⁷C_r 2 ^r.x ^17-3r
To find coefficient of x¹¹ equate the power of x to 11
⇒ 17 – 3r = 11 ⇒ 17 – 11 = 3r
⇒ 3r = 6 ⇒ r = 2
T₂₊₁ = ¹⁷C₂ . 2².x¹¹
∴ Coefficient of x” is ¹⁷C₂ .2² = \(\frac{17 \times 16 \times 2}{2 \times 1}\) = 544

Question (ii).
Y³ in \(\left(7 y^{2}-\frac{2}{y}\right)^{12}\)
Answer:
Here x =7y², a = \(-\frac{2}{y}\) and n = 12
T_r+1 = 12C^r(7Y²)^12-r.\(\left(\frac{-2}{y}\right)^{r}\)
= ¹²C_r.7^12-r .y^24-2r. y^-r(-2)^r
T_r+1 = ¹²C_r.7^12-r.(-2)^r.y^24-3r
To find the coefficient of y³ equate the power of y ro 3
i.e., 24 – 3r = 3 ⇒ 21 = 3r ⇒ r= 7
∴T₇₊₁ = ¹²C₇ .7^12-7. (-2)⁷ y³
= ^-12C₇ .7⁵ 2⁷ . y³
∴Coefficient of y³ is ^-12C₇.7⁵.2⁷

Question (iii).
x¹¹ in \(\left(\sqrt{x}-\frac{2}{x}\right)^{17}\)
Answer:
lere, x → √x, a →\(\frac{-2}{x}\) and n = 17
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 17
To find the coefficient of x2 equate the power of x to -11
∴ \(\frac{17-3 r}{2}\) = -11 ⇒ 17 – 3r = – 22
⇒ 17+22 = 3r ⇒ 39 = 3.r ⇒ r = 13
T₁₃₊₁ = ¹⁷C₁₃(-2)¹³.x^-11
Coefficient of x^-11 is ^-17C₁₃.2¹³

Question (iv).
X¹⁸ in \(\left(x^{2}-\frac{6}{x}\right)^{15}\)
lere, x → x², a →\(\frac{-6}{x}\) and r = 15
T_r+1 = ¹⁵C_r.(x²)^15-r\(\left(\frac{-6}{x}\right)^{r}\)
T_r+1 = ¹⁵C_r.x^30-2r.(-6)^r.x^-r
= ¹⁵C_r.(-6)^r.x^30-2r-r
= ¹⁵C_r(-6)^r.x^30-3r.
To find the coefficient of x¹⁸,equate the power of x to 18
∴ 30 – 3r = 18 ⇒ 30 – 18 = 3r ⇒ 3r = 12 ⇒ r = 4
T₄₊₁ = ¹⁵C₄(-6) ⁴x¹⁸
∴ Coefficient of x¹⁸ is ¹⁵C₄. (6)⁴

Question (v).
X^-2 in \(\left(x+\frac{1}{x^{2}}\right)^{17}\)
Answer:
Here, x → x, a → \(\frac{1}{x^{2}}\) and n = 17
∴ T_r+1 = ¹⁷C_r.x^17-r\(\left(\frac{1}{x^{2}}\right)^{r}\)
T_r+1 = ¹⁷C_r.x^17-r-2r
= ¹⁷C_rx^17-3r
To find the coefficient of x^-2,equate the power of x to -2
17 – 3r = -2 ⇒ 17 + 2 = 3r ⇒ 3r ⇒ r = \(\frac{19}{3}\)
Since r is a fraction the coefficient of x^-2 is 0.

Question (vi).
X⁵ in \(\left(x+\frac{1}{x^{2}}\right)^{17}\)
Answer:
Here, x → x, a → \(\frac{1}{x^{2}}\) and n = 17
T_r+1 = ¹⁷C_r.x^17-r\(\left(\frac{1}{x^{2}}\right)^{r}\)
= ¹⁷C_rx^17-3r
To find the coefficient of x⁵, equate the power of x to 5
∴ 17 – 3r = 5
12 = 23 ⇒ r = 4
T₄₊₁ = T₅ = ¹⁷C₄.x⁵
∴ the coefficient of x⁵ is ¹⁷C₄

Question (vii).
X¹⁸ in \(\left(x^{2}+\frac{3 a}{x}\right)^{15}\)
Answer:
Here x → x² a → \(\frac{3 \mathrm{a}}{\mathrm{x}}\) , n=15
∴ T_r+1 = ¹⁵C_r.(x²)^15-r \(\left(\frac{3 a}{x}\right)^{r}\)
= ¹⁵C_rx^30-2r.(3a)^rx^r
=¹⁵C_r .3^r.a^r.x^30-3r
To find the coefficient of x¹⁸, we get 30 – 3r = 18
12 = 3r ⇒ r =4
∴ T₄₊₁ = ¹⁵C₄.3⁴.a⁴.x¹⁸
∴ coefficient of x¹⁸ is ¹⁵C₄.(3a)⁴

5. Find the term independent of x in

Question (i).
\(\left(\frac{4 x^{2}}{3}+\frac{3}{2 x}\right)^{9}\)
Here x → \(\frac{4 x^{2}}{3}\) , a = \(\frac{3}{2 x}\) and n = 9
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 18
To find the term independent of x, equate the power of x to zero.
e., 18 – 3r = 0 ⇒ r = 6

Question (ii).
\(\left(x^{3}-\frac{3}{x^{2}}\right)^{15}\)
Answer:
Here x → x³, a = \(\frac{-3}{x^{2}}\) and n = 15
T_r+1 = ¹⁵C_r.(x³)^15-r.\(\left(\frac{-3}{x^{2}}\right)^{r}\)
= ¹⁵C_r.x^45-r.x^-2r (-3)^r
= ¹⁵C_r.(-3)^r.x^45-5r
To find the term independent of x we have 45 – 5r = O
:. 45 = 5r ⇒ r = 9
T₉₊₁ = ¹⁵C₉.(-3)⁹.x⁰
T₁₀ = ^-15C₉.(3)⁹ is the term independent of x.

Question (iii).
\(\left(\sqrt{x}+\frac{1}{3 x^{2}}\right)^{10}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 20

Question (iv).
\(\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{9}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 21

Question (v).
\(\left(3 x-\frac{2}{x^{2}}\right)^{15}\)
Answer:
Here x → 3x a → \(\frac{-2}{x^{2}}\) and n =15
T_r+1 = ¹⁵C_r.(3x)^15-r.\(\left(\frac{-2}{x^{2}}\right)^{r}\)
= ¹⁵C_r.3^15-r.(-2)^r . x^15-r
We have x^15-3r. = x⁰ ⇒ 15 = 3r ⇒ r = 5
T = ¹⁵C₅.3^15-5.(-2)⁵ = ^-15C₅ .3¹⁰ 2⁵
∴ The term independent of x is ^-15C₅.3¹⁰.2⁵

Question (vi).
\(\left(x^{2}-\frac{2}{x^{3}}\right)^{5}\)
Answer:
Here x → x² a → \(\frac{-2}{x^{3}}\) = 5_r(x²)^{5 – r} . \(\left(\frac{-2}{x^{3}}\right)^{r}\)= 5_r.x^10-2r.(-2)^r
= 5C_r(-2)^r.x^10-5r
We have 10 – 5r = 0 ⇒ r = 2
T₂₊₁ = T₃ = ⁵C₂(-2)².x⁰ = 4. \(\frac { 5.4}{ 2.1 }\) = 40
∴ The term independent of x is 40

Question (vii).
\(\left(x-\frac{1}{x^{2}}\right)^{21}\)
Answer:
Here x →x and a → \(\frac{-1}{x^{2}}\) and n= 21
T_r+1 = ²¹C_r.(x)^21-r. \(\left(\frac{-1}{x^{2}}\right)^{r}\)
= ²¹C_r.x^21-3r.(-1)^r
We have x⁰= x^21-3r ⇒21 = 3r ⇒ r = 7
∴ T₇₊₁ = T₈ = ²¹C₇ (-1)⁷.x⁰ = ^-21C₇
∴ The term independent of x is -21C₇

Question (viii).
\(\left( \sqrt { 2 } \frac { 2 }{ { x }^{ 2 } } \right)\)<sup>20</sup>
Answer:
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 22

Part – D

2nd PUC Basic Maths Binomial Theorem Ex 4.2 Four Marks Questions and Answers

6. Use binomial theorem to evaluate upt 4 decimals place

Questoin (i)
(102)⁶
(102)⁶ = (100 + 2)⁶
= (100)⁶ + ⁶C₁(1 0O)⁵. 2 + ⁶C₂(100)⁴.2²+ ⁶C₃(100)³.2³ + ⁶C₄(100)² 2⁴ + ⁶C₅.100.2⁵ + ⁶C₆2⁶
= 1000000000000 + 120000000000 + 6000000000 + 160000000 + 2400000 + 19200 + 64
= 1,126,162,419,264

Question (ii).
(98)⁴
Answer:
(98)⁴ = (100 – 2 )⁴
= (100) ⁴– ⁴C₁ (100)³.2 + ⁴C₂(100)².2²– ⁴C₃(100).2³ + ⁴C₄.2⁴
= 100000000 – 8000000 + 240000 – 3200 + 16
= 92236816

Question (iii).
(1.0005)⁴
Answer:
(1.0005)⁴ = (1 + 0.0005)⁴
= 1⁴ + ⁴C₁(O.0005) + ⁴C₂(0.0005)²+ ⁴C₃(0.0005)³ + ⁴C₄(0.0005)⁴
1 + 0.002 + 0.0000015 + …………….
= 1.00200150 ≈ 1.0020

Question (iv).
(0.99)⁴
Answer:
(0.99)⁴ = (1- 0.01)⁴ = ⁴C₀(0.01) – ⁴C₁ (0.01) + ⁴C₂(0.O1)² – ⁴C₃(0.01)³ + ⁴C₄(0.01)⁴
= 1 – 0.04 + 0.0006 – 0.000004 + 0.00000001
= 0.96059601 ≈ 0.9606

Question 7.
The first three terms in (1 + ax)ⁿ where n is a positive integer are 1, 6x, 16x². Find the vaIue
Answer:
Given . T₁ = lin (1 + ax)ⁿ; T₂ = 6x
ⁿC_{1 .}ax = 6x
nax = 6x
⇒ na = 6 ⇒ a = \(\frac{6}{n}\)
and T₃ = 16x²
\(\frac{n(n-1)}{2}\) a²x² = 16x²

Question 8.
In the expansion of (3 + kx)⁹ the x² and x³ are equal. Find k.
Answer:
Given , T_r+1= ⁹C_r.3^9-r.(kx) ^r
= ⁹C_r.3^9-r.k^rx^r
Coefficient of x² ⇒ x² ⇒ r = 2
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 27
36x 3⁷.k² x x²
Coefficient of x³ ⇒ x³ = x^r⇒ r = 3

T₃₊₁ = T₄ = ⁹₃.3^9-3.k₃.x₃ = ⁹C₃.3⁶
k₃.x₃ = 84.3⁶k₃.x₃
84 x 3⁶k₂= 36 x 3₇ x k²

Question 9.
Find the ratio of the coefficient of x⁴ in the two expansions (1+x)⁷ and (1+x)¹⁰?
Answer:
2nd PUC Basic Maths Question Bank Chapter 4 Binomial Theorem Ex 4.2 - 26

KSEEB Solutions for Class 8 Maths Chapter 11 Tribhujagala Sarvasamate Ex 11.7

October 25, 2024October 23, 2024 by Prasanna

Students can Download Maths Chapter 11 Tribhujagala Sarvasamate Ex 11.7 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths in Kannada helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

KSEEB Solutions for Class 8 Maths Chapter 11 Tribhujagala Sarvasamate Ex 11.7

KSEEB Solutions for Class 8 Maths Chapter 11 Tribhujagala Sarvasamate Ex 11.7 1

KSEEB Solutions for Class 8 Maths Chapter 11 Tribhujagala Sarvasamate Ex 11.7 3

1st PUC Economics Question Bank Chapter 4 Poverty

October 21, 2024October 18, 2024 by Prasanna

Students can Download Economics Chapter 4 Poverty Questions and Answers, Notes Pdf, 1st PUC Economics Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Economics Question Bank Chapter 4 Poverty

1st PUC Economics Poverty TextBook Questions and Answers

Question 1.
Why caloric – based norm is not adequate to identify the poor?
Answer:
The government uses Monthly Per Capita Expenditure (MPCE) as proxy for income of households to identify the poor. Poverty line is estimated by the monetary value (per capita expenditure) of the minimum calorie intake that was estimated at 2400 calories for a rural person and 2100 for a person in the urban area. But this calorie based norm is not adequate to identify the poor duo to following reasons:

1. This mechanism groups all the poor together and does not differentiate between the very poor and the other poor which makes it ’ difficult to identify who amongst them needs help the most.

2. Economists question, the basis of taking expenditure on food and a few select items as proxy for income.

3. This norm does not take into account the other factors associated with poverty such as accessibility to basic education, health care, drinking water, and sanitation.

4. This norm does not take social factors such as illiteracy, lack of access to resources, discrimination or lack of civil and political freedoms into consideration.

Question 2.
What is meant by the ‘Food for work programme’?
Answer:
‘Food for work’ programme was started in the 1970’s to raise the standard of living of poor. These poverty alleviation programme aimed to raise income and employment for the poor through the creation of incremental assets and by means of work generation.

Question 3.
Why are employment generation programmes important in poverty alleviation in india?
Answer:
In India, twin problems exist i.e., poverty and unemployment. Poverty alleviation has been one of the guiding principles of the planning process in India. Poverty can effectively be eradicated only when the poor start contributing to growth by their active involvement in the growth process.

This can only be achieved by launching various employment schemes. Following points discussed the importance of Employment Generation Programmes to eradicate poverty.

1. Nexus between Unemployment and Poverty:
There exists a deep nexus between unemployment and poverty. If employment opportunities are generated, then more people will be employed leading to rise in their income which in turn will reduce poverty.

2. Availability of Basic Facilities:
With the rise in employment opportunities, income increase and poor people are able to get access to education, health facilities, proper sanitation, etc.

3. Creation of Assets:
The Employment Generation Programmes aim at creation of assets like water harvesting, irrigation facilities, construction of roads, construction of dams, etc. All these assets help in the social and economic development of the rural areas and hence eradication of poverty.

4. Creation of skills
An essential element of employment generation programmes is the formation of human capital by imparting skills to the unskilled labourers through training. This alleviation programmes like Prime Minister’s Rozgar Yojana, Swarna Jayanti Shahari Rogzar Yojana, National Food for Work Programme, Annapurna came into existence.

Question 4.
How can creation of income earning assets address the problem of poverty?
Answer:
By creating income earning assets, we can generate employment opportunities through which poor can raise their income which ultimately helpful in improving standard of living. Thereby, it address the problem of poverty.

Question 5.
The three dimensional attack on poverty adopted by the government has not succeded in poverty alleviation in India. Comment.
Answer:
Poverty alleviation has always been accepted as one of the major objectives of planned development process in India but even after vast spending on poverty alleviation programmes, the government has not succeeded in poverty alleviation in India.

Despite various strategies to alleviate poverty, problems like hunger, malnorishment illiteracy, and lack of basic amenities are prevalent in India. None of the poverty alleviation strategies resulted in any radical change in the ownership of assets, process of production and improvement of basic amenities to the needy.

Due to unequal distribution of assets, the benefits from poverty alleviation programmes have not actually reached the poor. The amount of resources allocated for the poverty alleviation programmes is not sufficient when we take the magnitude of poverty into consideration.

The implementation of the poverty alleviation programmes is the responsibility of government and bank officials who are ill motivated, inadequately trained, corruption prone and vulnerable to pressure from local elites.

The resources are thus used inefficiently. Government policies have also failed to address the various issues related to poverty due to non – participation of local level institutions in programme implementation. It is evident that high grouwth alone is not sufficient to reduce poverty without the active participation of the people.

Further, it is necessary to identify poverty stricken areas and provide infrastructure such as schools, roads, power, telecom, IT services, training institution s, etc. Institutional weaknesses abound and implementation failures are the biggest reasons that these programmes not succeeded.

Question 6.
What programme has the Government adopted to help the elderly people and poor and destitute women?
Answer:
National Social Assistance Programme’ is one of the programme started by Government to help the elderly people and poor and destitute women this programme targets elderly people, widows and the poor and destitute women who are alone and have no one to take care of them. Under this programme, these targeted people are given pension to sustain their livelihood.

Question 7
Is there any relationship between unemployment and poverty? Explain.
Answer:
There exists a deep nexus between unemployment and poverty. Unemployment or underemployment and the casual and intermittent nature of work in both rural and urban areas drives unemployed people who do not have resources to make their ends meet into indebtedness and poverty.

If employment opportunities are generated, then more people will be employed leading to rise in their income which in turn will reduce poverty.

Due to unemployment, income of the people is reduced to a large extent and they are unable to get access to education, health facilities, proper sanitation, etc. This causes poor quality of living and hence poor human capital and skills which in turn lead to poverty making a vicious circle of poverty.

Question 8.
Suppose you are from a poor family and you wish to get help from the Government to set up a petty shop under which scheme will you apply for assistance and why?
Answer:
For setting up a petty shop, I would apply for financial assistance under the programme prime Minister’s Rozgar Yojana’ (PMRY). Under this programme, an unemployed educated person from low-income family in rural and urban areas can set up any kind of enterprise that can generate employment.

Question 9.
Illustrate the difference between rural and urban poverty. It is correct to say that poverty has shifted from rural to urban areas? Use the trends in poverty ratio to support your answer?
Answer:

Rural Poverty	Urban Poverty
1. Open unemployment under & disguised unemployment are found simultaneously in rural areas	Open unemployment is generally found
2. It is difficult to differentiate between open employment & under employment in rural areas	One can differentiate between open and disguised unemployment
3. As the population increases dependent on agriculture increase. This leads to further rise in seasonal, open, under & distinguished unemployment in rural areas.	In urban areas the main reason for increase in open unemployment is increase in education, health & other facilities
4. No facility of education, employment, social welfare etc. that lead to poverty among the masses	Good facility of education, social welfare etc., that make the standard of living of urban people high
5. Rural people posses few assets	Urban people posses many assets
6. Malnutrition among rural people is high	Generally malnutrition is not found.

Question 10.
Suppose you are a resident of a village, suggest a few measures to tackle the problem of poverty.
Answer:
Following measures have been taken by the Government to remove poverty under five year plan:
1. Integrated rural development programme:
With a view to remove poverty in rural areas and making provision for full employment under this programme, attempts are being made to provide more employment by developing agriculture, animal husbandry, fisheries, small scale, and cottage industries etc.,

2. Jawahar Rozsar Yojana:
It was launched in 1989. Its aim was to provide employment to at least one member of a IPUC Economics INDIAN ECONOMIC DEVELOPMENT rural poor family for 50 to 100 days in a year.

3. Jawahar Gram Samriddhi Yojana (JGSY):
It was launched on 1st April 1999. It has two main objectives:

Creation of durable productive community assets at the village level.
Generation of supplementary employment for the unemployed poor in rural areas.

4. Swarnaiavanthi Gram Swaroiear Yoiana (SGST):
It was started on 1^st April 1999. Its aims are:

Focussed approach to poverty alleviation.
Capitalising advantages of group lending
Overcoming the problems.

5. Employment Assurance Scheme:
The programme is presently being implemented in all rural blocks.

Creation of additional wage employment opportunities during the period of acute shortage for the rural poor living below the poverty line.
Creation of durable community, social and economic assets for sustained economic development.

6. Pradhan Mantri Gramodava Yojana (PMGY):
It was introduced in 2000-2001 focussing on village development in five critical areas, health, primary education, drinking water, housing and rural roads with the objectives of improving the quality of life or people in the rural areas.

7. Sampoorna Grameen Roisar Yojana (SGRY):
The scheme aims at providing wage employment in rural areas as also food, security, creation durable, community, social and economic assets.

8. Grameena Roisar Guarantee Scheme:
It was launched in February 2005 to provide work for 100 days a year in rural areas.

1st PUC Economics Poverty Additional Questions and Answers

1st PUC Economics Poverty Very Short Answer Type Questions

Question 1.
Give the meaning of Poverty?
Answer:
Poverty is a situation in which, section of the people in a country are unable to get the minimum basic needs (like food, clothing, housing, education and health facilities)

Question 2.
What makes the poor physically weak?
Answer:

Malnutrition
111 health & disability
makes the poor physically weak.

Question 3.
Mention two kinds of Poverty?
Answer:
The two kinds of poverty are

Absolute poverty
Relative poverty

Question 4.
On what basis the poor are identified?
Answer:
The poor people are identified on the basis of their occupation and ownership of assets. In both rural and urban areas, the people who are unable to get even minimum income from their present occupation to lead their life are considered as poor.

Question 5.
Who are called churning poor?
Answer:
The churning poor are those who regularly move in and out of poverty, ex: small farmers, seasonal workers

Question 6.
Define poverty line as per Planning Commission?
Answer:
Poverty line as per planning commission is measured on the basis of minimum nutrition calories required for rural and urban population. It is estimated that a person is said to be poor if he is unable to consume 2400 calories per day in rural areas and 2100 calories per day in case of urban areas.

Question 7.
What do you mean by ‘Head Count Ratio’
Answer:
The number of poor estimated as the proportion of people below the poverty line is known as ‘Head Count Ratio’.

Question 8.
Name the National Programme which provides social security to aged and widows?
Answer:
National SocialAssistance Programme.

Question 9.
Write the meaning of Absolute Poverty?
Answer:
Absolute poverty is one of the kinds of poverty where consumption and expenditure of a person is not sufficient to maintain a minimum acceptable level of living, according to national standard, which is expressed in terms of food grains.

Question 10.
What is Relative Poverty?
Answer:
Relative Poverty refers to poverty which is measured in relation to levels of income of individuals. A section of people whose level of income is low when compared to other sections of people.

Question 11.
Who discussed the concept of poverty line first in India?
Answer:
‘Dadabhai Naroji’ was the first to discuss the concept of poverty line India.

Question 12.
Name the programme which guarantees minimum of 100 days of employment in rural areas?
Answer:
‘Mahatma Gandhi National Rural Employment Guarantee’ (MGNREGP).

Question 13.
Suppose you are from a poor family of a rural area and you wish to get help from the Government to set up a petty shop. Under which scheme will you apply for assistance.
Answer:
‘SwamajayanthiGramSwarojgarYojana’.

Question 14.
Mention the states which have 70% of India’s poor people?
Answer:

Orissa
Bihar
West Bengal
Uttar Pradesh
Madhya Pradesh.

Question 15.
What programmes has the Government adopted to help the elderly people and poor and destitute women?
Answer:
National social assistance programme help the elderly people and poor and destitute women. This programme in Karnataka named as ‘Sandhya Suraksha Yojana’.

Question 16.
Name the programme introduced by the Government to provide insurance to the head of the family of rural landless households?
Answer:
AamAadmi Bima Yojana.

Question 17.
Name the programme introduced by the Government of Karnataka to provide health insurance?
Answer:
‘Yashaswini Yojana’.

Question 18.
Name the programme introduced by the Government of Karnataka to provide financial support to female child of poor family.
Answer:
‘Bhagya LakshmiYojana’.

Question 19.
Expand‘VAMBAY’?
Answer:
‘Valmiki Ambedkar Awas Yojana’.

Question 20.
Name the states which are poorest in India?
Answer:

Bihar and
Orissa

Question 21.
Expand IAY?
Answer:
Indira Awas Yojana.

Question 22.
Expand PMRY?
Answer:
‘PradhanaManthriRojgar Yojana’.

Question 23.
Who are chronic poor?
Answer:
These are the person who always remains as poor for a long period of time, ex: casual workers, agricultural labourer’s, ragpickers, beggars etc.,

Question 24.
Who are called Transient Poor?
Answer:
The transient poor are those people who are rich most of the time but may sometimes have a pitch of bad luck.

Question 25.
Expand SGRY?
Answer:
‘SampoomaGrameena Rojgar Yojana’.

Question 26.
Expand SGSY?
Answer:
‘Swamajayanthi Gram Swarozgar Yojana’ (SGSY).

Question 27.
Expand SJSRY?
Answer:
‘Swarna jayanthi Shahari Rozgar Yojana’ (SJSRY).

1st PUC Economics Poverty Short Answer Type Questions

Question 1.
Write the important features of poorest households?
Answer:
The important features of poorest households:

The poor people possess few assets and live in huts but the poorest of them do not even. have such huts also.
They will not have capacity to eat a square meal a day.
Starvation, hunger, lack of basic literacy and skills.

Question 2.
Name the categories of Poverty?
Answer:
The categories of poverty are :

Chronic poor
Transient poor
Never poor or non-poor.

Question 3.
Name the factors which are to be considered to develop poverty line other than income and assets?
Answer:

Accessibility to basic education
Health care
Drinking water and sanitation have to be considered to develop poverty line.

Question 4.
Do you think the growth oriented approach is successful in reducing poverty? why?
Answer:
The growth oriented approach has not succeeded in reducing poverty. This is mainly because of rapid growth rate of population and lack of proper implementation of land reforms.

Question 5.
How are poor people are identified?
Answer:
The poor people are identified on the basis of their occupation and ownership of assets. In both rural and urban areas, the people who are unable to get even minimum income from their present occupation to lead their life are considered as poor.

Question 6.
How do you calculate ‘Head Count Ratio’?
Answer:
The HeadCount Ratio is calculated with following formula.
HCL = Number of people living below the poverty line/Total population of the country.

Question 7.
What are the approaches to reduce poverty in India?
Answer:

Growth oriented approach
Income and employment generation approach
Providing minimum basic needs to the people approach.

Question 8.
State any two self employment programmes?
Answer:
The major self employment programmes are :

Swamajayanthi Gram Swarozgar Yojana (SGSY)
Swamajayanthi ShahariRozgar Yojana (SJBRY)
Pradhana Manthri Rozgar Yojana (PMRY).

Question 9.
Mention any two.national programmes to generate wage employment?
Answer:
The major national programmes to generate wage employment are as follows :

National Food for Work Programme (NFWP)
Sampooma Grameena Rozgar Yojana (SGRY)
Mahathma Gandhi National Rural Employment Guarantee Programme (MGNREGP).

Question 10.
Mention the programmes aimed at improving food and nutritional status of poor?
Answer:
There are programmes introduced by the Government to improve the food and nutritional status of poor. They are as follows:

Public distribution system (PDS)
Integrated child development scheme (ICDS)
Mid-day meals scheme.

Question 11.
Mention any four causes for poverty?
Answer:
The main causes for poverty are :

Unemployment
Population explosion
Inflation
Exploitation under British rule

Question 12.
State the programme introduced to provide basic infrastructure in rural areas?
Answer:
Programmes like:

Pradhan Mantri Gram Sadak Yojana
Basava Vasathi Yojana (Karnataka)
Public Distribution System
Integrated Child Development Scheme.

Question 13.
Illustrate the difference between rural and urban poverty. Is it correct to say that poverty has shifted from rural to urban areas?
Answer:
The difference between rural and urban poverty is the nature of poverty. In rural areas, poor people are those who are landless agricultural labourers, small and marginal farmers. While in urban areas, poor people are those who are unemployed, under employed or employed in low productivity occupation with low wages.
Poverty Ratio:
1st PUC Economics Question Bank Chapter 4 Poverty img2
Estimates source:
planning commission estimates (Uniform Reference Period).

Question 14.
Who are poor according to Tendulkar Committee?
Answer:
According to Tendulkar Committee Report (2009-10), the poverty line is defined on the basis of per capita consumption expenditure of Rs.672.80 in rural areas and Rs.859.60 in urban areas.

1st PUC Economics Poverty Long Answer Type Questions

Question 1.
Briefly explain the causes of Poverty?
Answer:
The causes of Poverty:
1. Exploitation under British rule :
Britishers ruthlessessly exploited our country during their colonial rule. They destroyed our traditional cottage and small-scale industries to encourage their modem industries.

2. Economic Inequalities:
In India inequalities in the distribution of wealth and income are also responsible for poverty. A large section of the people are forced to remain under poverty due to concentration of wealth and income in the hands of few people.

3. Low resource base:
The most important factor which is responsible for poverty in India is the low resource base of poor both in rural and urban area. At present a large section of rural poor are having very little land or no land at all. The income from this smallholdings is not sufficient to meet their basic needs.

4. Unemployment:
Unemployment and underemployment are much more wide spread problem in India. The degree of unemployment among the poor is very high.

5. Rapid growth in population :
The single most important reason of poverty in India is rapid growth population. This is responsible for low level of per capita income and consumption.

6. Inflationary pressures:
Inflationary rise in the prices of food grains and other essential commodities further intensifies the situation of poverty.

7. Vicious circle of poverty:
At present India is facing the challenge of vicious circle of poverty. Here poverty is both the cause and the effect.

8. Social factors:
In India, social factors are also causing poverty. Illiteracy, ignorance, backwardness, inadequancy of social structure, narrow outlook affecting the quality of life of the people and their employability.

Question 2.
The three dimensional attack on poverty adopted by the Government has not succeeded in poverty alleviation in India. Comment.
Answer:
The three dimensional approach of economic growth, employment generation and alleviating poverty could not achieve the desired result. Although there has been a reduction in the percentage of absolute poor in some of the states still the poor people lack basic amenities, literacy and nourishment.
This is because of:

Unequal distribution of land and other assets among rich and poor farmers.
Improper implementation of poverty alleviation programmes by ill-motivated and inadequately trained bureaucrats further worsened the situation.
Corruption along with inclination towards interest of elites led to an efficient and misallocation of scarce resources.

Question 3.
Suppose you are a resident of a village, suggest a few measures to tackle the problem of poverty?
Answer:
Being a resident of a village, I would suggest the following measures to tackle the problem of poverty:

Identification of poor
Generating employment opportunities for the identified poor
Free access to education and health care facilities.
Establishment of small scale industries.
Re-distribution of income-earning assets.
Encouraging poor for their active participation.
Organising training camps and night classes for importing vocation training to unskilled labourers.
Advancing financial and technical assistance to establish small enterprises.
Upgradation of Agricultural practices to raise productivity.
Enforcement of measures to check population growth.
Development of infrastructure.
Motivating the poor to acquire skills information and knowledge.

Question 4.
What are the various policies and programmes towards poverty alleviation of the Government?
Answer:
The second five year plan (1956-61) has pointed out that the benefits of economic development must accure more and more to the relatively less privileged classes of society.
The Government’s approach to poverty reduction is of three dimensions:
1. Growth oriented approach:
It is based on the expectation that the effects of economic growth that is rapid increase in the gross domestic product and per capita income of a nation would spread to all sections of society. Rapid industrial development and transformation of agriculture through Green Revolution in select regions would benefit underdeveloped regions and more backward sections of the community.

2. Food for Work:
In 1970, Food for work programme was started. The policymakers started thinking that incomes and employment for the poor could be increased through the creation of incremental assets and by means of work generation through specific poverty alleviation programmes. Under self-employment programmes, financial assistance is given to families or individuals.

3. Provision of minimum basic amenities :
Through public expenditure on social consumption needs such as education, health, water supply and sanitation, people’s standard of living could be improved. Programmes under this approach are expected to supplement the consumption of the poor, create employment opportunities and improvement in health and education.

For this purpose ‘Pradhana Manthi Gram Saraj Yojana, Pradhana Mantri Gramodaya, Yojana, Valmiki Ambedkar Awas Yojana and National Social Assistance Programme’ were started.

Question 5.
Write some measures to remove poverty in India?
Answer:
Following steps should be taken to remove poverty in India.
1. Creation of employment opportunities:
There should be greater encouragement to small scale industries in rural areas, poultry farming, dairy farming and piggeries should be developed.

2. Population Control:
Population growth is a hindrance in our coming the problem of population. It should be checked. Various family planning methods should be adopted by the people. The people should be educated so that they understand the charm of a small sized family.

3. Economic development:
There should be improvement in agricultural and industries. Their production should be increased.

4. Provision of minimum needs :
The Government should provide water, housing, sanitation, other facilities. The public distribution system should be regulated properly. So that the poor people may get essential commodities at cheaper rates.

5. Removal of economic inequalities:
The Government should encourage small industries and agriculture. It should give incentive to industries in rural areas. The taxation policy should be progressive. The money collected by the Government should be spent on the welfare of poor people.

Question 6.
How is the poverty line determined?
Answer:
In simple words the poverty line is defined as the per capita expenditure at which the average calorie in take was 2400 calories for a person in rural areas and 2100 calories for urban areas. Most of the studies use data collected by national sample survey organisation on various aspects including consumption expenditure and region.

1. In 1999-2000 the poverty line was defined for rural areas as consumption worth Rs.328 per month and for urban areas it was Rs.454.

2. In 1973-74 more than 321 million people were below the poverty line. In 1999-2000 this has come down to about 260 million.

3. In 1973,74, about 55 percent of the total population were below the poverty line. In 1999-2000 it has fallen to 26 percent. In rural areas, about 75 percent of the total poor are below the poverty line.

Question 7.
Briefly explain the three dimensional attack on poverty adopted by the Government?
Answer:
The Government’s approach to poverty reduction is of three dimensions:
1. The first one is the growth-oriented approach. It is based on the expectation that the effects of economic growth – rapid increase in gross domestic product and per capita income would spread to all sections of society and will trickle down to poorer sections.

This was the major focus of planning in the 1950’s and early 1960s. It was felt that rapid industrial development and transformation of agriculture through green revolution in select (regions would benefit the underdeveloped regions).

2. The second approach has been initiated from the third five-year plan (1961 -1966) that suggest that incomes and employment for the poor could be raised through the creation of incremental assets and by means of work generation. This could be achieved through specific poverty alleviation programme. The example based on this approach are :

a. Food for work programme:
Minimum food was ensured to workers and their families instead of money under this programme.

b. Rural Employment Generation Programme:
It aims at creating self employment opportunities in rural areas and small towns.

c. Prime Minister’s Rozgar Yojana:
The educated unemployed from low income families in rural and urban areas can get financial help to setup any kind of enterprise that generate employment under this programme.

d. Swama Jayanthi Shasri Rozgar Yojana:
It mainly aims at creating employment opportunity both of self-employment and wage employment in urban areas.

e. National Food for work programme and Sampooma Grameen Rojgar Yojana:
These programmes aim at generating wage employment for the poor unskilled people living in rural areas.

f. In August 2005, the parliament has passed a new Act to provide guaranteed wage employment to every household whose adult volunteer to do unskilled manual work for a minimum of 100 days in a year.

3. The third approach to tackle poverty is to provide minimum basic amenities to the people. Programmes under this approach are expected to supplement the consumption of the poor, create employment opportunities and bring about improvement in health and education.

Three major programmes that aim at improving the food and nutritional status of the poor are public distribution system, integrated child development scheme and mid-day meal scheme.

Question 8.
Critically examine the growth oriented approach to poverty reduction?
Answer:
Growth oriented approach to poverty reduction:
It is the first Government’s approach to poverty reduction. This approach was based on the expectation that the effects of economic growth would spread to all sections of society and trickle down to the poor section also. This approach was the major focus of planning in the 1950’s and early 1960’s.

It was felt that rapid industrial development and transformation of agriculture through green revolution would benefit the under developed regions and the more backward sections of the community. But this approach was failure. The benefits of economic growth has not trickled down to the poor.

The green revolution exacerbated the disparities regionally and between large and small farmers. There was unwillingness and inability to redistributed lands.

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling

October 21, 2024October 18, 2024 by Prasanna

Students can Download Maths Chapter 10 Data Handling Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 5 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 5 Maths Chapter 10 Data Handling

KSEEB Class 5 Maths Data Handling Ex 10.1

1. Observe the pictograph and answer the questions

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 2

Number of Science books read Six
Number of Kannada books read Four
Number of Adventure books read Twelve
Number of novels read Eight
Total number of books read Thirty

2. The following pictograph shows the number of cars in 5 villages of a taluk.

Observe the pictograph and answer the following questions:

Which village has the maximum number of cars? Village 3
Which village has the minimum number of cars? Village 1
What is the total number of cars in five villages? 150 cars
How many more cars are there in village 3 than in village 5? Twenty-five
Which two villages have the same number of cars? Village 2 & 4

Data Handling Additional Questions and Answers

1. A survey of sale of cars made by an agency for a particular month is an follows:

Question
Represent this data through a pictograph.
Answer:
Scale = 5 cars.

Answer the following questions:

Question
In which week were the maximum number of cars sold?
Answer:
First week

Question
In which week were the minimum number of cars sold?
Answer:
Second week

Question
What is the difference in the number of cars sold in the third and fourth week?
Answer:
Five years

Question
What is the total number of cars sold in the month?
Answer:
110 cars

Question
Arrange the number of cars sold in the descending order along with the week.
Answer:
Example:
35, 30, 25, 20
First Week

Question 3.
Ramu had 5 pens. 3 erasers, 6 books, 2 pencils and one sharpener in his bag.
Answer:
Write the data in tubular form and represent the same through a pictograph.

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 5

4. The number of bulbs manufactured in a week by a factory is given below. Draw a pictograph to represent the data. (Choose a suitable scale)

Bar Graphs

I. Read the bar graph and answer the given questions:

Question
What is the information given in the bar graph?
Answer:
Days against Earnings of flourist in rupees

Question
Mention the scale taken.
Answer:
50 flourist = 1 cm

Question
Mention the earnings on each day.
Answer:
First day – 300
Second day – 500
Third day – 200
Fourth day – 400
Fifth day – 350

Question
What is the total earnings in five days?
Answer:
1750 flourist in Rupees

Question
What is the difference in the amount earned on the second day and the fifth day?
Answer:
150 flourist in Rupees

Question
Arrange the amount earned on each day in the descending order.
Answer:
500,400,350,300,200, 2nd day 4th day 5th day ! day 3rd day

2. Study the bar graph and answer the questions given.

Question
What is the information given in the bar graph?
Answer:
Saplings planted by a farmer against week

Question
Mention the scale taken.
Answer:
100 Plants = 1 week

Question
Mention the number of saplings planted in each week.
Answer:
First week – 700
Second week – 900
Third week – 500
Fourth week – 800
Fifth week – 600

Question
What is the total number of saplings planted in the five weeks?
Answer:
3500 saplings

Question
In which week did the farmer plant the maximum number of saplings?
Answer:
Second week

Question
In which week did he plant the minimum number of saplings?
Answer:
Third week

3. A survey of 120 school students was done to find the activity they prefer to do in their free time.

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 10

Draw a bar graph to illustrate the above data.
Scale: 1 cm= 5 students

Question
Which activity is preferred by most of the students other than playing?
Answer:

4. The number of belts sold by a shopkeeper on six consecutive days of a week is as follows:

Draw a bar graph to represent the data.
Scale : 1 cm = 5 belts

Question
What is the total number of belts sold in 6 days?
Answer:
175

5. The number of hours spent by a student on different activities in a day is given below.

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 14

Represent the above data by a bar graph.
Scale: 1 cm = 1 hour

6. The following table shows the number of bicycles manufactured in a factory during the year 2005 – 2009. Illustrate the data using a bar graph.
(choose a suitable scale)

KSEEB Solutions for Class 5 Maths Chapter 10 Data Handling 16

Question a.
In which year was the maximum number of bicycles manufactured?
Answer:
2007

Question b.
In which year was the minimum number of bicycle manufactured?
Answer:
2008

1st PUC Maths Model Question Paper 1 with Answers

October 21, 2024October 18, 2024 by Prasanna

Students can Download 1st PUC Maths Model Question Paper 1 with Answers, Karnataka 1st PUC Maths Model Question Paper with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Maths Model Question Paper 1 with Answers

Time : 3 hrs 15 min
Max. Marks : 100

Instructions:
1. The question paper has five parts A, B, C, D and E and answer all parts,
2. Part-A carries 10 marks, Part-B carries 20 marks, Part-C carries 30 marks, Part-D carries 20 marks, Part-E carries 10 marks.

Part – A

Answer any TEN questions : (10 × 1 = 10)

Question 1.
Given that the numbers of subsets of a set A is 16. Find the number of elements of A.
Answer:
Let n(A) = m
n [P(A)] = 16 ⇒ 2^m = 16 ⇒ m= 4
n(A)= 4

Question 2.
If tan x = \(\frac { 3 }{ 4 }\) and x lies in the third quadrant, find sin X.
Answer:
tan x = \(\frac{3}{4}=\frac{-3}{-4}\)
1st PUC Maths Model Question Paper 1 with answers - 1
ordinate = -3, abscissa = 4, distance = 5 sin x = \(\frac { -3 }{ 5 }\)

Question 3.
Find the modulus of \(\frac{1+i}{1-i}\)
Answer:
\(\frac{1+i}{1-i}=\frac{(1+i)^{2}}{1-i}=\frac{2 i}{2}\) = i Modulus of i= 1

Question 4.
Find ‘n’ if ⁿC₇ = ⁿC₆
Answer:
ⁿC_n-7 = ⁿC₆ ⇒ n – 7 = 6 ⇒ n = 13.

Question 5.
Find 20th term of GP. \(\frac{5}{2}, \frac{5}{4}, \frac{5}{8}—-\)
Answer:
1st PUC Maths Model Question Paper 1 with answers - 2

Question 6.
Find the distance between 3r + 4y +5= 0 and 6x + 8y +2=0.
Answer:
3x + 4y + 5 = 0) ………….. (1)
3x + 4y + 1 = 0 …………. (2)
Required distance \(\left|\frac{5-1}{\sqrt{9+16}}\right|=\frac{4}{5}\) units

Question 7.
1st PUC Maths Model Question Paper 1 with Answers - 3
Answer:
\(\lim _{x \rightarrow 0+}\) f(x) = \(\lim _{x \rightarrow 0} \frac{x}{x}=\frac{1}{1}=2\) ∴ f(x) in Discontinuous

Question 8.
Write the negation of For all a, b ∈ I, a – b ∈ I’.
Answer:
“There exists a, b ∈ I, such that a – b € I’
or ‘∃ a, b ∈ I, a – b ∉ I’.

Question 9.
A letter is chosen at random from the word “ASSASINATION”. Find the probability that letter is vowel.
Answer:
No of ways of selecting one vowel out of six vowels (3A’S, 21’s, 10’s) = ⁶C₁ = 6.
P(1 vowel) = \(\frac{^{6} \mathrm{C}_{1}}{^{13} \mathrm{C}_{1}}=\frac{6}{13}\)

Question 10.
Let A = {2,3,4} and R be a relation on A defined by
R={(x,y)|x,y ∈ A,x divides y}, find ‘R’.
Answer:
R= {(2, 2), (2, 4), (3, 3), (4,4)}.

Part – B

Answer any TEN questions : (10 × 2 = 20)

Question 11.
If A and B are two disjoimt sets and n(A) = 15 and n(B) = 10 find n(A ∪ B), n(A ∩ B)
Answer:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
n(A ∩ B) = 0, n(A ∪ B) = 15 + 10 = 25.

Question 12.
If U = {x:r s≤10, r ∈ N} A = {x : x ∈ N, x is prime) B = {x : x ∈ N, x is even} write A ∩ B’ in roster form.
Answer:
U = {1, 2, 3 … 10), A = {2, 3, 5, 7}
B = {2, 4, 6, 8, 10} B¹ = {3, 5, 7,9}
[A ∩ B¹ = {3,5,7)]

Question 13.
If A × B = {(a,1) (a,2) (a,3) (b,1) (b,2) (b,3)}, find the sets A and B and hence fimnnd B × A
Answer:
A = {a, b}, B = {1, 2, 3}
B × A= {(1, a) (1, b) (2, a) (2, b) (3, a) (3, b)}

Question 14.
The difference between two acute angles of a right angled triangle is to \(\frac{3 \pi}{10}\) radiAnswer: Express the angles in degrees.
Answer:
Let A and B be acute angles
Given A + B = \(\frac{\pi}{2}\) = 90° : A – B = \(\frac{3 \pi}{10}\) = 54° ⇒ A = \(\frac{2 \pi}{5}\) and B = \(\frac{\pi}{5} \) ⇒ A = 72 and B = 18°.

Question 15.
Find sin \(\frac{x}{2}\) if tan x = \(-\frac{4}{3}\)and x lies in second quadrant,
Answer:
tan x = \(\frac{-4}{3}=\frac{4}{-3}\) ⇒ ∴ cos x = \(\frac{-3}{5}\)
1st PUC Maths Model Question Paper 1 with Answers - 5

Question 16.
\(\lim _{x \rightarrow 3} \frac{x-3}{x^{2}-5 x+6}\)
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 6

Question 17.
CV = 60 σ = 21
Answer:

Question 18.
Write the inverse, converse of ‘If a parallelogram is a square, then it is rhombus’.
Answer:
Inverse: If a parallelogram is not a square then it is not a rhombus
Converse: If a parallelogram is a rhombus then it is a square.

Question 19.
On her vacations Veena visits four cities A, B, C and D in random order. What is the probability that she visits A before B?
Answer:
n (S) = 24
P(visiting A before B) = \(\frac{12}{24}\) = 1/2

Question 20.
In a triangle ABC with vertex A(2,3), B(4, -1) and C(1,2). Find the Length of the alitude from vertex A.
Answer:
Equation BC = x + y – 3 = 0
Length of the alitude from A (2, 3) = \(\left|\frac{2+3-3}{\sqrt{2}}\right|=\left|\frac{2}{\sqrt{2}}\right|=\sqrt{2}\) = √2.

Question 21.
Represent the complex number z = 1 + i in polar form.
Answer:
r = √2 and θ = \(\frac{\pi}{4}\) Polar form is √2 \(\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\)

Question 22.
Obtain all pairs of consecutive odd natural numbers such that in each pair both are more than 50 and their sum is less than 120.
Answer:
Taking the pair as x, x + 2,
we get x > 50, 2x + 2 < 120 ⇒ x < 59 and writing the required pairs are (51, 53), (53, 55), (55, 57), (57, 59).

Question 23.
A line cuts off equal intercepts on the co – ordinate axes. Find the angle made by the line with the positive x-axis.
Answer:
Slope =-1 ⇒ tanθ = -1
angle made = 135°.

Question 24.
If the origin is the centroid of the triangle PQR with vertices P (2a, 4, 6) Q(-4, 3b, -10) and R(8, 14, 2c) then find the values of a, b, c.
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 9

Part-C

Answer any EIGHT of the following questions : (10 × 3 = 30)

Question 25.
Out of a group of 200 students (who know at least one language), 100 students know English, 80 students know Kannada, 70 students know Hindi. If 40 students know all the three languages, find the number of students who know exactly two languages.
Answer:
writing n(E)= 100, n(K)= 80, n(H)= 70 .
n(E ∪ K ∪ H) = 200, n (E ∩ K N ∩ H) = 40
We know that n(E ∪ K ∪ H) = n(E) + n (K) + n(H) –n (E ∩ K)
– n (K ∩ H) -n (H ∩ E) + n (E ∩ K ∩ H)
∴ n (E ∩ K) + n (K ∩ H) + n (H ∩ E) = 90

Question 26.
Let R: Z → Z be a relation defined by R = {(a,b) : a, b, ∈ Z, a – b ∈ z}. Show that
(i) ∀ a ∈ Z, (a, a) ∈ R.
(ii) (a, b) ∈ R ⇒ (b,a) ∈ R
(iii) (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R
Answer:
∀ a ∈ Z, (a, a) ∈ R since a – a = 0 ∈ Z
(a, b) ∈ R ⇒ a – b ∈ Z ⇒ ∴ b – a ∈ Z ⇒ (b, a) ∈ R
(a, b) ∈ R, (b, c) ∈ R ⇒ a – b ∈ Z, b – C ∈ Z
∴ a – b + b – c ∈ Z ⇒ (a, c) ∈ Z

Question 27.
Prove that (cos x + cos y)² + (sin x – sin y)= 4cos² \(\left(\frac{x+y}{2}\right)\)
Answer:
LHS = cos²x + cos² y + 2 cos x cos y + sin²x + sin²y – 2sin x sin y
= 1 + 1 + 2 (cos x cos y – sin x sin y) = 2 [1 + cos (x + y)] = 4cos² \(\left(\frac{x+y}{2}\right)\)

Question 28.
Solve: √2x² + x + √2 = 0.
Answer:
√2x² + x + √2 = 0; Here a = √2, b = 1,c = √2
∴ D = 5² – 4ac = (1)² – 4(√2)(√2) = 1 – 8 = -7
1st PUC Maths Model Question Paper 1 with Answers - 10

Question 29.
How many letters words with or with out meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.
(i) 4 letters are used at a time,
(ü) all letters are used at a time
(iii) all letters are used but first letter is a vowel?
Answer:
There are 6 letters in the word MONDAY. So, the total number of words is equal to the number of arrangements of these letters taken four at a time.
The number of such arrangements = ⁶P₄ = \(\frac{6 !}{(6-4) !}=\frac{6 !}{2 !}=\frac{6.5 .4 .3 .2 .1}{2.1}\) = 360
Hence, total number of words = 360
(ii) Total number of arrangements in this case = ⁶P₆ = 6! = 720
(ii) Total number of arrangements when all letters are used but the first letter is a vowel
= 2 × ⁵P₅ = 2 × 5! = 2 × 5 × 4 × 3 × 2 × 1 = 240.

Question 30.
If x + iy = \(\frac{2+i}{2-i}\) then prove that x² + y²
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 11

Question 31.
Find the term independent of x in the expansion of \(\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{6}\)
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 12
The term will be independent of x if the index of x is zero, i.e., 12 – 3r= 0. Thus, r=4
Hence 5th term is independent of x and is given by (-1)⁴ ⁶C₄ \(\frac{(3)^{6-8}}{(2)^{6-4}}=\frac{5}{12}\)

Question 32.
8, A₁, A₂, A₃, 24
Answer:
T_n = a + (n − 1)d
24 = 8 + (5 – 1)d
16 = 4d ⇒ d = 4
3 Am’s are 12, 16, 20.

Question 33.
(i) At least one man.
Answer:
1 man and I woman
or 2
men and 0 women
= ²C₁ × ²C₁ + ²C₂ × ²C₀ = 2 × 2 + 1 × 1 = 4 + 1 = 5.

(ii) At most one man
Answer:
1 men and 1 women
or
0 men and 2 women
= ²C₁ × ²C₁ + ²C₀ × ²C₂ = 2 × 2 + 1 × 1 = 4 + 1 = 5.

Question 34.
Find the derivative of the function ‘cos x’ w.r.t ‘x’ from first principle
Answer:
Let f ‘(x) = \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
1st PUC Maths Model Question Paper 1 with Answers - 13

Question 35.
A parabola with vertex at origin has its focus at the centre of x² + y² – 10x + 9 = 0 Find the direction its directrix and latus rectum.
Answer:
The centre of the circle = (5,0).
The equation y² = 20x ⇒ y² = 4ax ⇒ a = 5
directrix is x = -5, LR = 20.

Question 36.
If an A.P. if m^th term is n and the n^th term is m, where m ≠ n, find the p^th term
Answer:
We have
a_m = a + (m – 1) d = n, …… (1)
and a_n = a + (n – 1) d = m …….. (2)
Solving (1) and (2), we get
(m, n) d = n – m, or d = -1,
and a = n + m – 1 ……. (3)
Therefore a_p = a + (p – 1)d ……. (4)
= n + m – 1 + (p – 1) (-1)= n + m – p
Hence, the p^th term is n + m – p.

Question 37.
Verify by the method of contradiction that √2 is irrational
Answer:
If possible Let √2 is rational
∴ √2 = \(\frac{p}{q}\), p, q ∈ z, q ≠ 0
We assume that p and q donot have any common factor
p = √2q ⇒ p = 2q²
p² is a multiple of 2 ⇒ ∴ p is a multiple of 2
∴ p = 2k, k ∈ z ⇒ p² = 4k
2q² = 4k² ⇒ q² = 2k²
q² = 2k²
q² is a multiple of 2 ⇒ ∴ q is a multiple of z
∴ p and q are both multiple of 2 and hence has a common factor z which is a contradiction
∴ our assumption is wron
∴ √2 is irrational

Question 38.
Two students Anil and Sunil a appear in the examination The prpbability that Anil will qualify in the examination is 0.05 and that Sunil will qualify is 0.10. The probability that both will qualify in the examination is 0.02. Find the probability that Anil and Sunil will not qualify in the examination.
Answer:
Let A, B denote the events that Anil, Sunil qualify in the exam
P (A) = 0.05, P(B) = 0.1, P (A ∩ B) = 0.02
P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.05 + 0.1 – 0.02 = 0.13
P(A’ ∩ B’) = 1 – P (A ∪ B) = 1 – 0.13 = 0.87.

Part – D

Answer any Six questions : (6 × 5 = 30)

Question 39.
Define signum function. Draw the graph of the signum function. Also write its domain and range.
Answer:
Singum function :
The function f:R → R defined by
1st PUC Maths Model Question Paper 1 with Answers - 14

is called the signum function. The domain of the signum function is R and the range is the set {-1, 0, 1}.

Question 40.
Prove that \(\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\) (q being in radians) and hence show that \(\lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta}=1\)
Answer:
Consider a circle with centre O and radius ‘r’ Mark two points A and B on the circle so that

Join AB. Draw BM ⊥ OA
From the figure it is clear that.
Area of ∆OAB < Area of sector OAB < sector of ∆OAC …. (1)
Area of ∆OAB = \(\frac { 1 }{ 2 }\) OA.BM
[In ∆OBM. Sin θ \(\frac{\mathrm{BM}}{r}\) ⇒ BM = r Sin θ ]
1st PUC Maths Model Question Paper 1 with Answers - 17
∴ Area of ∆OAB = \(\frac { 1 }{ 2 }\) r r. sin θ = \(\frac { 1 }{ 2 }\)r² sin θ
Area of sector OAB = \(\frac { 1 }{ 2 }\) r²θ
Area of ∆OAC = \(\frac { 1 }{ 2 }\) OA.AC
= \(\frac { 1 }{ 2 }\)r² tan θ [In ∆OAC, tan θ = \(\frac{A C}{O A}=\frac{A C}{r}\) ⇒ AC = r tan θ]
∴ (1) becomes
1st PUC Maths Model Question Paper 1 with Answers - 18

Question 41.
1² + 2² + 3² + ………… + n² = \(\frac{n(n+1)(2 n+1)}{6}\)
Answer:
Let P(n) : 1² + 2² + 3² + ………… + n² = \(\frac{n(n+1)(2 n+1)}{6}\)

step 1 : Prove that P(1) is true
when n= 1; L.H.S = 1² = 1
R.H.S = \(\frac{1(1+1)(2+1)}{6}\)
∴ L.H.S = R.H.S ⇒ ∴ P(1) is true

step 2: Assume that P(m) is true
i.e., 1² + 2² + 3² + ………… + m² = \(\frac{m(m+1)(2 m+1)}{6}\) …… (i)

step 3: Prove that P(m + 1) is true
i.e., 1² + 2² + 3² + ………… + m²+ (m+1)² = \(\frac{(m+1)(m+2)(2 m+3)}{6}\)
L.H.S = [1² + 2² + 3² + ………… + m²] + (m+1)²
1st PUC Maths Model Question Paper 1 with Answers - 20
= R.H.S ⇒ ∴ P(m+1) is true
Conclusion: P(1) is true ⇒ P(m) is true ⇒ P(m+1) is true
∴ By principle of mathematical induction, the result is true for all natural numbers ‘n’.

Question 42.
A group consists of 7 boys and 5 girls. Find the number of ways in which a team of 5 members can be selected so as to have atleast one boy and one girl.
Answer:
Number of ways of selecting
1 B and 4 G = ⁷C₁ × ⁵C₄ = 35
2 B and 3G = ⁷C₂ × ⁵C₃ = 210
3 B and 2 G = ⁷C₃ × ⁵C₂ = 350
4. B and 1 G = ⁷C₄ ×⁵C₁ = 175
Total number of selections = 770.

Question 43.
State and prove binomial theorem for positive integers.
Answer:
Statement: “If’n’ is a +ve integer then
(x + a)ⁿ = ⁿC₀ xⁿ a⁰ + ⁿC₁ x^n-1 a¹ + ⁿC₂ x^n-2a² + ………….. + ⁿC_n x⁰ aⁿ.

Proof : (By Mathematical induction):
Let p(n) : (x + a)ⁿ = ⁿC₀ xⁿ a⁰ + ⁿC₁ x^n-1 a¹ + ⁿC₂ x^n-2a² + ………….. + ⁿC_n x⁰ aⁿ.

Step1 : Prove that P(1) is true
when n = 1, L.H.S = (x + a)¹ = x + a; RHS = ¹c₀ x¹ a⁰ + ¹c₁ x⁰ a¹ = 1.x.1 + 1.1.a = x+a
L.H.S. = R.H.S. ⇒ ∴ P(1) is true

Step 2: Assume that P(m) is true
i.e., (x + a)^m = ^mc₀ x^m a⁰ + ^mc₁ x^m-1 a¹ + ^mc₂ x^m-2 a² + ………….+ ^mc^mx⁰a^m …….(i)

Step 3 : Prove that P(m + 1)^m+1 is true
i.e. (x + a)^m+1 = ^m+1c₀ x^m+1 a⁰ + ^m+1c₁ x^m a¹ + ^m+1c₂ x^m-1 a² + …….. + ^m+1c_m+1 x⁰ a^m+1
Multiply both sides of equation (i) by (x+a)
∴ (x + a)^m (x + a) = (x + a) [^mc₀ x^m a⁰ + ^mc₁ x^m-1 a¹ + ^mc₂ x^m-2 a² + ………….+ ^mc^mx⁰a^m]
(x + a)^m+1 = x [^mc₀ x^m a⁰ + ^mc₁ x^m-1 a¹ + ^mc₂ x^m-2 a² + ………….+ ^mc^mx⁰a^m] + a [^mc₀ x^m a⁰ + ^mc₁ x^m-1 a¹ + ^mc² x^m-2 a² + ………….+ ^mc^mx⁰a^m]
(x + a)^{m + 1} = ^mc₀x^m+1 a⁰ + ^mc₁x^ma¹ + ^mc₂ x^m-1 a² + ……. + ^mc_m x⁰ a^m+1
= ^mc₀x^m+1a⁰ + (^mc₁ + ^mc₀)x^m a¹ + (^mc₂ + ^mc₁) x^m-1 a² + ………… + (^mc_m + ^mc_m-1) x¹ a^m + ^mc^m x⁰ a^m+1

(x + a)^m+1 = ^m+1c₀ x^m+1 a⁰ + ^m+1c₁ x^m a¹ + ^m+1c₂ x^m-1 a² + …….. + ^m+1c_m+1 x⁰ a^m+1
[∴ ^mc_m = ^m+1c₀
^mc₁ + ^mc₀ = ^m+1c₁
^mc₂ + ^mc₁ = ^m+1c₂
^m+1c_m]
⇒ ∴ P(m + 1) is true
Conclusion: P(1) is true, P(m) is true ⇒ P(m+1) is true
∴ By principle of mathematical induction the result is true for all natural numbers n.

Notes: 1. The number of terms in the expansion of (x + a)ⁿ is n + 1.
2. The Gen term of binomial expansion is T_r+1 = ⁿc_r x^n-ra^r.

Question 44.
Derive an expression for the coordinates of a point that divides the line joining the points A(x₁, y₁, z₁,) and B (x₂, y₂, z₂.) internally in the ratio m : n. Hence, find the coordinates of the midpoint of AB where A = (1, 2, 3) and B = (5, 6, 7).
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 21
Let the two given points be P (x₁, y₁, z₁,) and Q (x₂, y₂, z₂). Let the point R (x, y, z) divide PQ in the given ratio m : n internally, Draw PL,
QM and RN perpendicular to the XY-plane. Obviously PL ∥ RN ∥ QM and feet of these perpendiculars lie in a XY-plane. The points L, Mand N will lie on a line which is the intersection of the plane containing PL, RN and QM with the XY-Plane. Through the point R draw a line ST parallel to the line LM. Line ST will intersect the line LP externally at the point S and the line MQ at T, as shown in Fig 12.5.

Also note that quadrilaterals LNRS and NMTR are parallelograms.
The triangles PSR and QTR are similar. Therefore,
1st PUC Maths Model Question Paper 1 with Answers - 22

Similarly, by drawing perpendiculars to the XZ and YZ-planes, we get

Hence, the coordinates of the point R which divides the line segment joining two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂,) internally in the ratio m: n are

Question 45.
Derive a formula for the angle between two lines with slopes m¹ and m². Hence find the slopes of the lines which make an angle \(\frac{\pi}{4}\) with the line x – 2y + 5 = 0.
Answer:
m₁ = tan θ₁, m₂ = tan θ₂
1st PUC Maths Model Question Paper 1 with Answers - 26

Question 46.
Prove that \(\frac{\cos 4 x+\cos 2 x+\cos 3 x}{\sin 4 x+\sin 2 x+\sin 3 x}=\cot 3 x\)
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 27

Question 47.
Solve graphically 2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6
Answer:
2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6 …..(1)
Points A(2, 0) and B(0,4) lie on 2x + y = 4
Plot the points and join them to get line AB
(0,0) does not satisfy 2x + y ≥ 4
⇒ Half plane given by 2x + y ≥ 4 is away from origin …..(2)
Points C(3, 0) and D(0, 3) lie on x + y= 3
Plot the points and join them to get line CD.
(0,0) satisfies x + y ≤ 3 ….. (3)
⇒ Half plane given by x + y ≤ 3 is towards origin
Points C(3, 0) and E(0, -2) lie on 2x – 3y = 6
1st PUC Maths Model Question Paper 1 with Answers - 28
Plot the points and join them to get line CE.
(0,0) satisfies 2x – 3y ≤ 6
⇒ Half plane given by 2x – 3y ≤ 6 is towards origin
From (2), (3), (4) common region shown shaded in figure represents solution of (1).

Question 48.
Find the mean deviation about the mean for the following data.
1st PUC Maths Model Question Paper 1 with Answers - 29
Answer:

Part – E

Answer any ONE questions : (1 × 10 = 10)

Question 49.
(a) To cos (A + B) = cos x. cos y – sin x sin y and hence find cos 75°.
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 32
(i) becomes
∴ cos (A + B)= cos A.cos B – sin B
cos 75° = cos (45 +30)
= cos 45 cos 30 – sin 45 sin 30

(b) Find the sum to n terms of the series 1² + (1² + 2²) + (1² + 2² + 3²) + ……..
Answer:
nth term T_n = 1² + (1² + 2²) + (1² + 2² + 3²) + ……. + n² = \(\frac{n(n+1)(2 n+1)}{6}=\frac{2 n^{3}+3 n^{2}+n}{6}\)
∴ T_n = \(\frac{2 n^{3}+3 n^{2}+n}{6}\)
Sum to n terms, S_n = ΣT_n
= \(\frac{1}{6}\) [2Σn³ +3Σn² + Σn]
1st PUC Maths Model Question Paper 1 with Answers - 34
∴ S_n = \(\frac{n(n+1)^{2}(n+2)}{12}\)

Question 50.
(a) An elipse is the set of all ninte in a plane tha e of whose distance from two fived points in the need of all points in a plane the sun of whose distance points in the plane is a constant.
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 35
Let F₁ and F₂ be the foci and O be the mid point of the line segment F₁F₂ Let O be the origin and the line from 0 through F₂ be the positive x-axis and that through F₁ as the negative x-axis. Let, the line through O perpendicular to the x-axis be the y-axis. Let the coordinates of F₁ be (-c, 0) and F₂ be (c, 0).

Let P(x, y) be any point on the ellipse such that the sum of the distances from P to the two foci be 2a
i.e., PF₁ + PF₂ = 2a. …(1)
Using the distance formula, we have
\(\sqrt{(x+c)^{2}+y^{2}}+\sqrt{(x-c)^{2}+y^{2}}=2 a\)
i.e., \(\sqrt{(x+c)^{2}+y^{2}}=2 a-\sqrt{(x-c)^{2}+y^{2}}\)
Squaring both sides, we get
(x + c)² + y² = 4a² – 4a \(\sqrt{(x-c)^{2}+y^{2}}\) + (x – c)² + y²
which on simplification gives
\(\sqrt{(x-c)^{2}+y^{2}}=a-\frac{c}{a} x\)
Squaring again and simplifying, we get
1st PUC Maths Model Question Paper 1 with Answers - 36

(b) \(y=\frac{x^{5}-\cos x}{\sin x}\)
Answer:
1st PUC Maths Model Question Paper 1 with Answers - 37

KSEEB Solutions for Class 8 Maths Chapter 2 Bijoktigalu Ex 2.2

October 21, 2024October 18, 2024 by Prasanna

Students can Download Maths Chapter 2 Bijoktigalu Ex 2.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths in Kannada helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

KSEEB Solutions for Class 8 Maths Chapter 2 Bijoktigalu Ex 2.2

KSEEB Solutions for Class 8 Maths Chapter 2 Bijoktigalu Ex 2.2 1

KSEEB Solutions for Class 8 Maths Chapter 2 Bijoktigalu Ex 2.2 3

2nd PUC Biology Model Question Paper 3 with Answers

October 21, 2024October 18, 2024 by Prasanna

Students can Download 2nd PUC Biology Model Question Paper 3 with Answers, Karnataka 2nd PUC Biology Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Model Question Paper 3 with Answers

Time: 3.15 Hours
Max Marks: 70

General Instructions:

This question paper consists of four parts A, B, C and D. Part D consists of two parts, Section – I and Section – II.
All the parts are Compulsory.
Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.

Part – A

I. Answer the following questions in One Word or One Sentence each : ( 10 × 1 = 10 )

Question 1.
Name the site of fertilization in female human beings.
Answer:
Upper limb (Ampullary – isthmic junction) of fallopian tube.

Question 2.
Define adaptive radiation
Answer:
It is the evolution of different animals in different direction from a common ancestor adapting to different ecological niches. It is also called as divergent evolution.

Question 3.
Name the organism that causes typhoid.
Answer:
Salmonella typhi.

Question 4.
What is green revolution?
Answer:
The dramatic increase in food production due to improvement of high yielding and disease resistant crop varieties is called green revolution.

Question 5.
Name the microbe that causes big holes in the Swiss cheese.
Answer:
Propioni bacterium sharmanii

Question 6.
Why does the logistic growth curve becomes S shaped?
Answer:
In sigmoid growth unlimited resources are rare, population growth is, therefore becomes stable due to environmental resistance.

Question 7.
Breast feeding during the initial period of infant growth is recommended. Give reason?
Answer:
The yellowish milk (colostrum) secreted from mammary glands during the initial days contains antibodies (IgA) to boost the development of immune system of the child.

Question 8.
Give an example for chemical carcinogen that causes cancer.
Answer:
Alcohol or tobacco or coal tar or coal gas or asbestos cement etc

Question 9.
What is totipotency?
Answer:
The capacity to generate a whole plant from any cell or explant is called totipotency

Question 10.
How do birds cope with temporary fluctuations in their environmental conditions?
Answer:
They migrate to hospitable areas for food and breeding purposes during unfavourable conditions in their habitat.

Part – B

II. Answer any FIVE of the following questions in 3 – 5 sentences each, wherever applicable : ( 5 × 2 = 10 )

Question 11.
Name the plant that shows unusual flowering phenomenon in hilly tracks of Karnataka, Kerala and Tamilnad.
Answer:
Strobilanthus kuluthiana ( Neelakuranji)

Question 12.
What is Biochemical Oxygen Demand (BOD)? Mention its significance.
Answer:
BOD refers to the amount of oxygen that would be consumed if all the organic matter in one liter of water were oxidized by bacteria.

Significance

The BOD test measures the rate of uptake of oxygen by micro-organisms in a sample of water.
Indirectly BOD is a measure of the organic matter present in the water.
The greater the BOD of waste water more is its polluting potential.

Question 13.
Differentiate test cross and out cross.
Answer:
Test cross
It is a cross between F₁hybrid and homozygous recessive parent.

Out cross
It is a cross between F₁ hybrid and homozygous dominant parent.

Question 14.
Give two examples for natural selection in organisms.
Answer:

Industrial melanism in Biston betularia
Antibiotic resistance in bacteria

Question 15.
What are single cell proteins? Mention their importance?
Answer:
Dried biomass of a single species of microbes that can be used as protein source in the diet is known a single cell protein (SCP).

Single cell proteins (SCP) are alternative source of proteins used to feed the ever increasing population with protein rich diet to overcome the problems of protein malnutrition.

Question 16.
Why is India said to have greater ecosystem diversity than Norway?
Answer:
India with diverse ecosystems such as rain forests, coral reefs, wet lands, coastal areas, mangroves, estuaries and deserts has greater ecosystem diversity than Norway.

Question 17.
Draw a diagram of sparged tank bioreactor and label the parts.
Answer:
2nd PUC Biology Model Question Paper 3 with Answers 1

Question 18.
Explain the mechanism of action of restriction endonucleases.
Answer:
Restriction endonucleases recognize a specific base sequence of a DNA molecule.
They cut double stranded DNA at different sites. This generates protruding (51 or 31) ends. Such ends are called cohesive or sticky ends. These ends can join with single stranded sequences of other DNA having complementary base pairs.
E.g. EcoRI, Bam H1
2nd PUC Biology Model Question Paper 3 with Answers 2
Cuts symmetrically placed around the line of symmetry

Part – C

III. Answer any FIVE of the following Questions in 40-80 words each, wherever applicable. ( 5 × 3 = 15 )

Question 19.
Describe the different phases of life cycle.
Answer:
Juvenile phase
The period of growth from birth till individual attains sexual maturity is called Juvenile phase. It is known as vegetative phase in plants.

Reproductive phase
The period of growth after Juvenile or vegetative phase during which individual reaches sexual maturity and capable of producing gametes is called reproductive phase. In human beings reproductive phase begins on the onset of puberty usually after 13 or 14 years. In plants flowering marks the beginning of reproductive phase.

Senescent phase
The phase begins after reproductive phase and marks the end of growth cycle in organisms.

Question 20.
Give a brief account of Human Origin and evolution.
Answer:

Human evolution states that humans developed from primate or ape like ancestors.
Human origin occurred in Central Asia and Africa, China, Java and India (Shivalik hills).
The first mammals were shrew like terrestrial, insectivores or rat like creatures.
The insectivorous mammals namely tree shrews gave rise to the primitive primates called prosomians which include tarsiers, lemurs and lories and anthropoids (marmosets, baboons, monkeys, apes and man).
Modem man Homo sapiens belongs to class mammalia, order primata and suborder anthropoidea.
Dryopithecus africanus is regarded as common ancestor of man and apes.
It is lived about 20 – 25 million years ago (Miocene).
Dryopithecus gave rise to Rama pithecus; he appeared 14-15 million years ago.
The fossil of Rama pithecus was discovered from Pliocene rocks of Shivalik hills of India by Edward Lewis.
Rama pithecus gave rise to Australo pithecus after a gap of 9 – 10 million years.
Rama pithecus the first ape man commonly called southern ape .He appeared about 5 million years ago (early Pleistocene).
Homo habillus lived in early Pleistocene about 2 million years ago.
He was the first tool maker or handy man.
Homo erectus is the direct ancestor of modern man.
Homo erectus evolved from homo habillus about 1-7 million years ago in middle Pleistocene.
Homo erectus include Java ape man, pieking man and Heidelberg man.
The Neanderthal man existed in the late Pleistocene he was roughly equal to the modem man.
Cro-Magnon is regarded as direct and most recent ancestor of the living modem man.
He lived about 34.0000 years ago in Holocene epoch.
The cranial capacity was about 1650cc and considered more intelligent than the man of today.
The modern man Homo sapiens arose in Africa about 25.0000 years ago in Holocene epoch.
It is believed that the man of today appeared about 11.000 or 10.000 years ago in the region around Caspian and Mediterranean seas from their members migrated west wards, eastwards and south wards respectively changing into present day white or Caucasoid, Monogoloid and black or Negroid races.
Modern man underwent cultural evolution. Cave art developed about 18.000 years ago.
Agriculture came around 10.000 years back and human settlement started.

Question 21.
List the various public health measures as a safeguard against infectious diseases?
Answer:
Following are the public health measures, considered safe against infectious diseases.

Maintenance of personal and public hygiene like proper disposal of waste and excreta through proper sanitation.
Provision of safe water supply, periodic cleaning and disinfection of water reservoirs, pools, and tanks, etc.
Standard practices of hygiene in public health catering.
Avoiding stagnation of water in and around residential areas to prevent breeding of mosquitoes and vectors.
Spraying of insecticides like DDT in ditches, drainage areas, swamps, etc., to prevent mosquito breeding.
Measures against vector borne diseases like dengue and chikungunya, include fixing of wire meshes to doors and windows to prevent the entry of mosquitoes.
Regular health camps, immunization (Vaccination) programmes to prevent the spread of infectious diseases.

Question 22.
What is food chain? Describe the types of food chain.
Answer:
The transfer of food energy from producers to decomposers through a series of organisms with repeated eating and being eaten is called food chain.
1. Grazing food chain
The food chain begins with producers (green plants) at the first trophic level. It involves grazing Of grass by herbivores and flow of energy occurs from smaller organisms (herbivores) to larger animals (consumers). It is based on prey – predator relationship and is therefore, called predator food chain.
Grass → Deer → Tiger

2. Parasitic food chain
The food chain begins with producers and flow of energy goes from larger organisms (host) to smaller organisms (parasites) as parasites draw nourishment from the host. This is based on host – parasite relationship.
2nd PUC Biology Model Question Paper 3 with Answers 4

3. Detritus or saprophytic food chain
The food chain begins with dead and decaying organic matter (detritus). Which is acted upon by decomposers or saprophytes and energy from decomposers goes to detrivores.
Leaf litter → Earth worm → bird → hawk

Question 23.
Define the terms
(a) Emasculation
Answer:
The removal of anthers from the flower bud of a bisexual flower before the anther dehisces is called emasculation.

(b) Bagging
Answer:
The process of covering the emasculated flowers with a bag of suitable size to prevent contamination with unwanted pollen is called bagging.

Question 24.
What is transformation? Explain any two methods of introduction of recombinant DNA in to host organism.
Answer:
Transformation is the process by which plasmid or DNA can be introduced into a cell.

1. Gene gun or particle gun or Biolistics method
It is a popular and widely used direct gene transfer method for delivering foreign genes into any cells and tissues or even intact seedlings.

The foreign DNA is coated or precipitated onto the surface of minute gold or tungsten particles (1-3 pm).
It is bombarded or shot onto the target tissue or cells using the gene gun or micro projectile gun or shot gun.with a device much like a particle gun. Hence the term biolistics
The bombarded cells or tissues are cultured on selection medium to regenerate plants from the transformed cells.

2. Micro injection
In this technique a solution of DNA is directly injected in to the host nucleus using a fine micro capillary pipette or micro syringe, under a phase contrast microscope to aid vision.

Question 25.
Define trophic level. Show schematically the different tropic levels in a food chain.
Answer:
The amount of energy available at each step in food chain is called Trophic level.
2nd PUC Biology Model Question Paper 3 with Answers 5

Question 26.
What is pedigree? Show diagrammatic representation of pedigree for sickle cell anemia.
Answer:
Pedigree is a chart of graphic representation of record of inheritance of a trait through several generations in a family
2nd PUC Biology Model Question Paper 3 with Answers 6

Part – D

Section – I

IV. Answer any FOUR of the following questions on 200 – 250 words each, wherever applicable . ( 4 × 5 = 20 )

Question 27.
What are assisted reproductive technologies? Describe the various reproductive technologies to assist an infertile Couple to have children.
Answer:
The couple can be assisted to have children through certain special techniques known as assisted reproductive technologies (ART).
1. In vitro fertilization and embryo transfer (IVF-ET)

In this technique egg cells are fertilized by sperm (usually 100,000 sperm / ml) outside the body in almost similar conditions as that in the body.
In this process Ova from the wife or donor female and sperm from husband or donor male are allowed to fuse in shallow containers called Petri dishes. (Made of glass or plastic resins) Under laboratory condition.
The fertilized egg (zygote) is then transferred to the patient’s uterus this called embryo transfer (E.T.), Where she provides suitable conditions for the development of embryo.
The baby born by this technique is called test tube baby.

2. ZIFT ( Zygote intra fallopian transfer)
Zygote or early embryo up to eight blastomeres is transferred into the fallopian tube.

3. Gamete intra fallopian transfer (GIFT)
Transfer of an ovum collected from a donor to fallopian tube of another female who cannot produce ova, but provide suitable conditions for fertilization and further development up to parturition.

4. Intra uterine transfer (IUT)
Embryo with more than eight blastomeres is transferred in to the uterus.

5. Intra cytoplasmic sperm injection (ICSI)
The sperm is directly injected into the cytoplasm of the ovum to form an embryo in the laboratory and then embryo transfer is carried out.

6. Artificial insemination (AI)

This method is used in cases where infertility is due to the inability of the male partner to inseminate the female or due to very low sperm counts in the ejaculates.
In this method, the semen collected from the husband or a healthy donor is artificially introduced into the vagina or into the uterus (IUI – Intra uterine insemination).

Question 28.
What is dihybrid cross? State and explain Mendel’s second law of independent assortment.
Answer:
When a cross is made between two parents differing in two pairs of contrasting characters like round yellow and wrinkled green is called dihybrid cross.

Based on the results obtained from dihybrid cross, Mendel postulated his second law “The law of independent assortment”. It states that, when two pairs of contrasting characters are brought into the hybrid they behave independently of each other without contamination. The presence of one character is not contaminated with the other character. These characters are assorted independently of one another during the formation of gametes.
Mendel conducted dihybrid cross for which he selected two characters at a time.

2nd PUC Biology Model Question Paper 3 with Answers 7
Self crossed (F2)

Question 29.
What is meant by semi conservative replication? How did Meselson and Stahl prove it experimentally?
Answer:

2nd PUC Biology Model Question Paper 3 with Answers 8
It is a process by which DNA produces daughter DNA molecules, which are exact copies of the original DNA. In each new DNA molecule, one strand is old (original) while the other is newly formed. Hence, Watson and Crick described this method as semi-conservative replication.

Meselson and Stahl conducted experiments on E.coli to prove that DNA replication is semi conservative.

They grew E.coli cells in a culture medium containing radioactive isotopes N¹⁵ (NH₄cl).
The ¹⁵N was incorporated into the newly synthesized DNA and other nitrogen containing compounds.
Bacteria labeled with ¹⁵N are transferred into a medium containing normal ¹⁴NH₄cl.
The analysis was done after every generation at regular intervals.
The DNA samples were separated by centrifugation on CSCl₂ gradient.
The hybrid DNA separated from fresh generations possesses one heavier strand contributed by parental with ¹⁵N isotope which is radioactive and lighter strand is newly synthesized with ¹⁴N isotope, is non-radioactive.
The DNA of the second generation showed that in one DNA molecule one strand was radioactive (¹⁵N) and other was non radioactive (¹⁴N), Whereas both the strands of other DNA molecule were non-radioactive (¹⁴N).
It is clear that the DNA of the first generation is intermediate it contains both ¹⁵N and ¹⁴N.
In the second generation there were two types of DNA one heavier DNA with ¹⁵N and ¹⁴N isotopes while the lighter DNA with ¹⁴N and ¹⁴N, the ratios of two types of DNA was 50:50.
In the third generation the ratio of intermediate (¹⁵N and ¹⁴N) was reduced from 50% to 25% while the lighter fraction (¹⁴N and ¹⁴N) increased from (50% – 75%).
In the 4^th generation the percentage of intermediate still reduced to 12.5% while that of lighter fraction increased from 75% to 87.5%.
These experiments prove that DNA shows semi conservative replication.

Question 30.
What is water pollution? Write short notes on the following with respect to water pollution.
a) Eutrophication
b) Biomagnifications
Answer:
Any undesirable change in the physical, chemical and biological characteristics of water, which makes water unfit for human use and harmfully affect aquatic organisms is called water pollution.

a) Eutrophication
It the process of nutrient enrichment of water bodies and subsequent loss of species diversity like fishes.
Harmful effects

Excess nutrients cause algal bloom, which may cover the whole surface of water body and release toxins.
It causes oxygen deficiency in water that leads, to the death of aquatic animals like fishes.

b) Biomagnifications
It refers to increase in concentration of toxic substances at successive trophic levels in a food chain. Example: Biomagnifications of DDT in an aquatic food chain.

Harmful Effect
High concentration of DDT disturbs calcium metabolism in birds, which causes thinning of egg shell and their premature breaking, causing decline in bird’s population.

Question 31.
Explain the structure of an anatropous ovule with a neat labeled diagram.
Answer:
2nd PUC Biology Model Question Paper 3 with Answers 9
Ovule is the female gametophyte develops inside the ovary from a cushion like structure called placenta. It is attached to placenta by a stalk known as funicle. The nutritive tissue enclosed inside the ovule is the nucellus. Nucellus is surrounded by outer and inner integuments. These integu-ments leave a small pore at the anterior end called micropyle. Anterior end of the ovule is micropyle chalazal end is the posterior part.

Question 32.
Distinguish DNA and RNA with respect to their structure or Chemistry and function.
Answer:

DNA	RNA
1. Present mainly in the chromatin the nucleus.	1. Most of RNA is present in the cell of cytoplasm (90%) and a little (10%) in the nucleolus
2. Normally double stranded and rarely single stranded.	2. Normally single stranded and rarely double stranded.
3. It contains deoxyribose sugar.	3. It has ribose sugar.
4. It contains bases like Adenine, Guanine, cytosine and Thymine.	4. It contains bases like, Adenine, Guanine cytosine and uracil.
5. DNA acts as template for its Synthesis.	5. RNA does not act as a template for its synthesis.

Section – II

Answer any THREE of the following questions in 200 – 250 words each, wherever applicable. ( 3 × 5 = 15 )

Question 33.
What is biocontrol? Name the principle behind biological method of pest control. Mention examples of biocontrol agents and their function.
Answer:
It is the use of micro organisms to control or eliminate insect pests. The micro organisms employed in biological control are called bio control agents.

Principle
It is based on prey – predator relationship.

Examples
1. Ladybird and Dragon flies useful to get rid of aphids and mosquitoes.

2. Bacillus thuringiensis (Bt) is used to control butterfly caterpillar. Spores available in sachets are mixed with water and sprayed on plants, eaten by insect larva, toxin released in gut kills larvae.
Example: Bt toxin genes are introduced into cotton plants and made resistant to insect pests such as cotton boll worms, stem borer, aphids and beetles.

3. Nucleo polyhedrovirus ( NPV) is a virus suitable for narrow spectrum insecticide applications. It has no negative impacts on plants, mammals, birds, fish or target insects. It is suitable for overall integrated pest Management programme (IPM) in ecologically sensitive areas.

Question 34.
What is logistic growth? Explain sigmoid curve with a suitable diagram.
Answer:
The rate of population growth slows as the population size approaches carrying capacity, leveling to a constant level.
2nd PUC Biology Model Question Paper 3 with Answers 10
Logistic growth curve or S-shaped growth curve (sigmoid growth curve) is characteristic of all higher animals including man and microbes like yeast cells growing in a natural environment.

Logistic growth is a pattern of growth in which the population density of few individuals introduced into the new habitat increases slowly initially, in a positive acceleration phase (lag phase) then increases rapidly, approaching an exponential growth rate (log phase) due to abundance of food, favourable environmental conditions such a climate, few predators and low levels of disease. Then declines in a negative acceleration phase.

This decline reflects increasing environmental resistance at higher population densities. This type of population growth is termed density-dependent, since growth rate depends on the numbers present in the population. The point of stabilization, or zero growth rates, is termed the saturation value or carrying capacity (K) of the environment for that organism.

When population growth rate is plotted graphically against time S shaped or sigmoid curve is obtained. This type of population growth is also called Verhulst- pearl logistic growth.
The mathematical equation for sigmoid growth is
dN / dt = rN(N – K) / k
Where
N = Population size at a given time
t = Time,
r = intrinsic rate of natural increase
k = Saturation value or carrying capacity for that organism in that environment.

Question 35.
What is animal breeding? Explain the controlled methods of animal breeding and mention their significance.
Answer:
The production of new varieties of animals which are superior to their parents through selective breeding or mating is called animal breeding.
There are two control breeding methods

Artificial insemination
Multiple ovulation embryo transfer technology (MOET)

Artificial insemination
Semen of superior male is collected and-injected unto the reproductive tract of selected female.

Significance:

It helps to overcome several problems of normal mating.
Semen of a desirable bull can be inseminated into cows.
Semen can be stored by cryopreservation to use at later stages and can be easily transported to distant places.
Semen of single ejaculation can be used to inseminate large number of cows.

Multiple Ovulatuion Embryo transfer technology (MOET)
Technique for herd improvement by successful production of hybrids

Hormones (FSH) are administered to the cow for inducing follicular maturation and super ovulation.
Cow produces 6-8 eggs instead of one egg and is either mated with, superior bull or artificially inseminated.
Fertilized egg at 8-32 cell stage are recovered non-surgically & transferred to surrogate mother.

Significance:

It is practiced in cattle, sheep, rabbits, buffaloes, etc.
Genetically superior animals can be produced.
It solves the problems of infertility.
A single female can donate many ovules.
Preservation and transport of embryos is convenient.

Question 36.
What is recombinant DNA technology? Draw a schematic representation of various steps involved in recombinant DNA technology.
Answer:
The branch of biotechnology that deals with altering the characters of an organism by introducing genes selected from desired organism is called genetic engineering or gene cloning or gene splicing.

2nd PUC Biology Model Question Paper 3 with Answers 11

Question 37.
What is gametogenesis? Mention the types. Explain spermatogenesis with a schematic representation.
Answer:
The process of formation of haploid male and female gametes in gonads is called gametogenesis.
It is of two types

Spermatogenesis
Oogenesis

2nd PUC Biology Model Question Paper 3 with Answers 12
1. Multiplication phase (mitosis)
Each testis is composed of numerous seminiferous tubules. Epithelial cells lining these tubules are called germinal epithelium. At the time of gametogenesis some cells of the epithelium become active, detach from the layer and start dividing mitotically to produce large number of primordial germ cells (2n), in turn primordial germ cells undergo repeated mitotic divisions to form sperm mother cells or spermatogonial cells which are diploid in nature.

2. Growth phase
Some of the spermatogonial cells (2n) stop dividing and enter into the seminiferous tubule where they grow in size by accumulating cytoplasm and duplication of DNA; sertoli or nurse cells provide necessary nutrients for the growing sperm mother cells, now sperm mother cells are termed as primary spermatocytes.

3. Maturation phase (meiosis)
Each primary spermatocyte (2n) enters into first meiotic division and forms two cells with haploid number of chromosomes called secondary spermatocytes. The secondary spermatocytes undergo second meiotic division to give rise to four haploid spermatids. The spermatids are unspecialized cells and do not undergo further division; all the four spermatids are transformed into four haploid sperms. This process of transformation of spermatids into mature and motile sperms is called spermiogenesis or spermateliosis.