mesin 128

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

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:

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:

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

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:

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:

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 ).

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^2x is 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:

Question 4.
Find the intervals in which the function f given by f (x) = 2x² – 3x is
(a) strictly increasing
(b) strictly decreasing
Answer:

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:

Question 6.
Find the intervals in which the following functions are strictly increasing or decreasing:
(a) x² + 2x – 5
Answer:

(b) 10 – 6x – 2x²
Answer:

(c) -2x³ – 9x² – 12x + 1
Answer:

(d) 6 – 9x – x²
Answer:

(e) (x + 1)³ (x -3)³
Answer:
f (x) = (x + 1)³ (x – 3)³
f’ (x) = (x + 1)³ x 3(x – 3)² + (x – 3)³ x 3 (x + 1)²
= (x + 1)² (x – 3)² [3x + 3 + 3x – 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:

Question 8.
Find the values of x for which y = [x(x – 2)]² is an increasing function.
Answer:

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:

Question 10.
Prove that the logarithmic function is strictly increasing on (0, ∞).
Answer:

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:

Question 12.
Which of the following function are strictly decreases on \((0, \pi / 2)\)
(a) cos x
Answer:

(b) f(x) = cos 2x
Answer:

(c) cos 3x
Answer:

(d) tan x
Answer:

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:

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:

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:

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:

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)²
f'(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:

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:

Question 2.
\(\frac{3 x^{2}-4 x+7}{x+7}\)
Answer:

Question 3.
\(\frac{x^{2}-1}{x^{2}+1}\)
Answer:

Question 4.
\(\frac{5 x^{2}}{x^{2}+4 x+3}\)
Answer:

Question 5.
\(\frac{x^{4}}{(x-1)\left(x^{2}+1\right)}\)
Answer:

Question 6.
\(\frac{4 x^{3}-2 x^{2}+3 x+1}{2 x^{2}+4 x-1}\)
Answer:

Question 7.
\(\frac{4 x^{2}-4 x-1}{2 x-1}\)
Answer:

Question 8.
\(\frac{x^{3}+7}{x^{2}-2 x+1}\)
Answer:

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

T₅ = ⁸C₄.(2²).2^-4
8C₄.2^8-4 = ⁸C₄.2⁴ = 1120

Question (ii).
The 8^th term 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^th term put r = 7
T_r+ = ⁿC_r.x^{n – r}.a^r

Question (iii).
The 6^th term in (√x – √y)¹⁷
Answer:
Compare (√x – √y)¹⁷ with (x + a)ⁿ
x → √x a → -√y and n = 17
To find 6^th term put r = 5

Question (iv).
The 7^th term in (3x² – \( \frac{y}{3}\) )⁹
Answer:
Here x → 3x² a → \(-\frac{y}{3}\) 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:

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

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

Question (iii).
\(\left(\frac{2 a}{3}-\frac{3}{3 a}\right)^{6}\)
Answer:
Here n = 6 (even)

Question (iv).
\(\left(3 x-\frac{1}{6} x^{3}\right)^{8}\)
Answer:

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

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

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

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.

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

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

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

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:

Question (iv).
\(\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{9}\)
Answer:

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).
\(\left( \sqrt { 2 } \frac { 2 }{ { x }^{ 2 } } \right)\)²⁰
Answer:

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

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: