2nd PUC

Expert Teachers at KSEEBSolutions.com has created Karnataka 2nd PUC Physics Question Bank with Answers Solutions, Notes, Guide Pdf Free Download of 2nd PUC Physics Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2026-27 in English Medium and Kannada Medium are part of 2nd PUC Question Bank with Answers. Here KSEEBSolutions.com has given the Department of Pre University Education (PUE) Karnataka State Board NCERT Syllabus 2nd Year PUC Physics Question Bank with Answers Pdf.

Students can also read 2nd PUC Physics Model Question Papers with Answers hope will definitely help for your board exams.

Karnataka 2nd PUC Physics Question Bank with Answers

• Chapter 1 Electric Charges and Fields
• Chapter 2 Electrostatic Potential and Capacitance
• Chapter 3 Current Electricity
• Chapter 4 Moving Charges and Magnetism
• Chapter 5 Magnetism and Matter
• Chapter 6 Electromagnetic Induction
• Chapter 7 Alternating Current
• Chapter 8 Electromagnetic Waves
• Chapter 9 Ray Optics and Optical Instruments
• Chapter 10 Wave Optics
• Chapter 11 Dual Nature of Radiation and Matter
• Chapter 12 Atoms
• Chapter 13 Nuclei
• Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits
• Chapter 15 Communication Systems

Karnataka 2nd PUC Physics Question Bank with Answers

• Chapter 1 Electric Charges and Fields Question Bank
• Chapter 2 Electrostatic Potential and Capacitance Question Bank
• Chapter 3 Current Electricity Question Bank
• Chapter 4 Moving Charges and Magnetism Question Bank
• Chapter 5 Magnetism and Matter Question Bank
• Chapter 6 Electromagnetic Induction Question Bank
• Chapter 7 Alternating Current Question Bank
• Chapter 8 Electromagnetic Waves Question Bank
• Chapter 9 Ray Optics and Optical Instruments Question Bank
• Chapter 10 Wave Optics Question Bank
• Chapter 11 Dual Nature of Radiation and Matter Question Bank
• Chapter 12 Atoms Question Bank
• Chapter 13 Nuclei Question Bank
• Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits Question Bank
• Chapter 15 Communication Systems Question Bank

Karnataka 2nd PUC Physics Syllabus and Marking Scheme

Karnataka 2nd PUC Physics Blue Print of Model Question Paper

Karnataka 2nd PUC Physics Question Paper Design

Time: 3 Hours 15 Minutes (of which 15 minutes for reading the question paper).
Maximum Marks: 70

The weightage of the distribution of marks over different dimensions of the question paper is as follows:

A. Weightage to Objectives:

Objective	Weightage	Marks
Knowledge	40%	43/105
Understanding	30%	31/105
Application	20%	21/105
Skill	10%	10/105

B. Weightage to Content/Subject Units:

C. Weightage to forms of Questions:

Note:

1. Questions in IV Main must be set from Unit I to V.
2. Questions in V Main must be set from Unit VI to X.
3. Questions in VI Main must be set such that one Numerical Problem is from every 2 successive units.

D. Weightage to the Level of Difficulty:

Level	Weightage	Marks
Easy	40%	43/105
Average	40%	42/105
Difficult	20%	20/105

General Instructions

• Questions should be clear, unambiguous, understandable and free from grammatical errors.
• Questions which are based on the same concept, law, fact etc. and which generate the same answer should not be repeated under different forms (VSA, SA, LA and NP).
• Questions must be set based on the blowup syllabus only.

2nd PUC Physics Practical Exam Syllabus

List of Experiments

1. To find resistance of a given wire using metre bridge and hence determine the specific resistance of its material.
2. To determine resistance per cm of a given wire .by plotting a graph of potential difference versus current.
3. To verify the laws of combination (series/ parallel) of resistances using a metre bridge.
4. To compare the EMFs of two given primary cells using potentiometer.
5. To determine the internal resistance of given primary cell using potentiometer.
6. To determine resistance of a galvanometer by half-deflection method and to find its.figure of merit.
7. To convert the given galvanometer (of known resistance of figure of merit) into an ammeter and voltmeter of desired range and to verify the same.
8. To find the frequency of the ac mains with a sonometer.
9. To find the value of v for different values of u in case of a concave mirror and to find the focal ‘ length.
10. To find the focal length of a convex mirror, using a convex lens.
11. To find the focal length of a convex lens by plotting graphs between u and v or between 1/uand 1/v.
12. To find the focal length of a concave lens, using a convex lens.
13. To determine angle of minimum deviation for a given prism by plotting a graph between the angle of incidence and the angle of deviation.
14. To determine the refractive index of a glass slab using a travelling microscope.
15. To find the refractive index of a liquid by using
    • Concave mirror
    • Convex lens and plane mirror.
16. To draw the I-V characteristics curves of a p-n junction in forward bias and reverse bias.
17. To draw the characteristics curve of a Zener diode and to determine its reverse break down voltage.
18. To study the characteristics of a common- emitter NPN or PNP transistor and to find out the values of current and voltage gains.

2nd PUC Physics Practical Exam Marking Scheme

General Instructions

• Duration of practical examination: 2 hours.
• Maximum marks allotted: 30 marks.
• At least TEN (10) different experiments have to be set in the practical Examination.

Scheme of Evaluation

A. Weightage of Marks

S.No.	Particulars	Marks
I	Performing the Experiment	20
II	Viva – Voce	4
III	Practical Record	6
	Total	30

B. Distribution of Marks

I. Performing the Experiment

1. Writing the principle of the experiment (2 Marks)
2. Writing the formula and explaining the terms (2 Marks)
3. Writing the diagram/figure/circuit with labelling (At least two parts) (2 Marks)
4. Writing the tabular column/observation pattern (2 Marks)
5. Constructing the experimental set up/ circuit (3 Marks)
6. Performing the experiment and entering the readings into the tabular column/Observation pattern (4 Marks)
7. Substitution and calculation/plotting the graph and calculation (3 Marks)
8. Result with a unit (2 Marks)
9. Total 20 marks

Note for S.No.6

• At least three (3) trials have to be taken in case of finding mean value.
• At least six (6) readings have to be taken in case of plotting the graph.

II. Viva – Voce

1. Four questions must be asked and each question carries 1 mark.
2. The questions in the viva- voce should be simple, direct and related to the experiment being performed by the student.

III. Practical Record

1. If the student has performed and recorded 13 experiments or more (6 Marks)
   (91% to 100% of the experiments prescribed for the practical examination or more)
2. If the student has performed and recorded 11 or 12 experiments. (5 Marks)
   (81% to 90% of the experiments prescribed for the practical examination)
3. If the student has performed and recorded 10 experiments. (4 Marks)
   (71% to 80% of the experiments prescribed for the practical examination)
4. If the student has performed and recorded below 10 and above 5 experiments. (3 Marks)
   (41% to 70% of the experiments prescribed for the practical examination)
5. If die student has performed and recorded 5 or less than 5 experiments. (0 Marks)
   (40% & below 40% Of the experiments prescribed for the practical examination)

We hope the given Karnataka 2nd PUC Class 12 Physics Question Bank with Answers Solutions, Notes, Guide Pdf Free Download of 2nd PUC Physics Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2026-2027 in English Medium and Kannada Medium will help you.

If you have any queries regarding Karnataka State Board NCERT Syllabus 2nd Year PUC Class 12 Physics Question Bank with Answers Pdf, drop a comment below and we will get back to you at the earliest.

Students can Download Maths Chapter 7 Integrals Ex 7.1 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 7 Integrals Ex 7.1

2nd PUC Maths Integrals NCERT Text Book Questions and Answers Ex 7.1

Find an anti derivative (of integral) of the following functions by the method of inspection.

Question 1.
sin 2x
Answer:

Question 2.
cos 3x
Answer:

Question 3.
e^2x
Answer:

Question 4.
(ax + b)²
Answer:

Question 5.
sin 2x – 4 e^3x

Find the following integrals in Exercises 6 to 20:

Question 6.
(4e^3x + 1)dx
Answer:

Question 7.
\(\int x^{2}\left(1-\frac{1}{x^{2}}\right) d x\)
Answer:

Question 8.
∫(ax² + bx+c)dx
Answer:

Question 9.
∫(2x²+e^x)dx
Answer:

Question 10.
\(\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} d x\)
Answer:

Question 11.
\(\int \frac{x^{3}+5 x^{2}-4}{x^{2}} d x\)
Answer:

Question 12.
\(\int \frac{x^{3}+3 x+4}{\sqrt{x}} d x\)
Answer:

Question 13.
\(\int \frac{x^{3}-x^{2}+x-1}{x-1} d x\)
Answer:

Question 14.
\(\int(1 – x) \sqrt{x} d x\)
Answer:

Question 15.
\(\int \sqrt{x}\left(3 x^{2}+2 x+3\right) d x\)
Answer:

Question 16.
∫(2x – 3cosx + e^x)dx
Answer:

Question 17.
\(\int\left(2 x^{2}-3 \sin x+5 \sqrt{x}\right) d x\)
Answer:

Question 18.
∫sec x(sec x + tan x)dx
Answer:
∫sec x (sec x + tan x) dx
= ∫(sec²x+sec x tan x) dx
= tan x + sec x + C

Question 19.
\(\int \frac { \sec ^{ 2 } x }{ { cosec }^{ 2 }x } dx\)
Answer:

Question 20.
\(\int \frac{2-3 \sin x}{\cos ^{2} x} d x\)
Answer:
∫(2sec² x – 3sec x tan x)dx
= 2 tan x – 3 sec x + C

Choose the correct answer in Exercises 21 and 22.

Question 21.
The anti derivative of \(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\) equals
(A) \(\frac{1}{3} x^{\frac{1}{3}}+2 x^{\frac{1}{2}}+C\)
(B) \(\frac{2}{3} x^{\frac{2}{3}}+\frac{1}{2} x^{2}+C\)
(C) \(\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C\)
(D) \(\frac{3}{2} x^{\frac{3}{2}}+\frac{1}{2} x^{\frac{1}{2}}+C\)
Answer:

Question 22.
\(\frac{d}{d x} f(x)=4 x^{3} \cdot \frac{3}{x^{4}}\) such that f(2) Then f(x) is ………
(A)\(x^{4}+\frac{1}{x^{3}}-\frac{129}{8}\)
(B)\(x^{3}+\frac{1}{x^{4}}+\frac{129}{8}\)
(C)\(x^{4}+\frac{1}{x^{3}}+\frac{129}{8}\)
(D)\(x^{3}+\frac{1}{x^{4}}-\frac{129}{8}\)
Answer:

Students can Download Maths Chapter 6 Application of Derivatives Miscellaneous Exercise 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 Miscellaneous Exercise

Question 1.
Using differentials, find the approximate value of each of the following

(a) \(\left(\frac{17}{81}\right)^{1 / 4}\)
Answer:

(b) (33)^-1/5
Answer:

Question 2.
Show that the function given by log has maximum at \(f(x)=\frac{\log x}{x}\) has maximum at x = e
Answer:

Question 3.
The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base ?
Answer:

Question 4.
Find the equation of the normal to curve y² = 4x which passes through the point (1, 2).
Answer:

Question 5.
Show that the normal at any point 6, to the curve x = a cos θ +a θ sin θ ,y = a sin θ –
a θ cos θ is at a constant distance from the origin.
Answer:

∴ equation of normal is
y – (a sin θ – a θ cos θ)
= – cot θ (x – a cos θ – a θ sin θ)
y + x cot θ= a cot θ cos θ+a θ cot θ sin θ – a θ cos θ
\(y+\frac{x \cos \theta}{\sin \theta}=\frac{a \cos ^{2} \theta}{\sin \theta}+\frac{a \sin ^{2} \theta}{\sin \theta}\)
x cos θ a cos²θ, a sin²θ
normal form is x cos θ+ y sin θ = p when p is the normal from the normal to the given curve is at a consistent distance ‘a’ from the origin.

Question 6.
Find the intervals in which the function f given by
\(f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}\)
(i) increasing
(ii) decreasing
Answer:

Question 7.
Find the intervals in which the function f given by
x³ + -1/x³, x ≠ 0
(i) increasing
(ii) decreasing
Answer:

Question 8.
Find the maximum area of an isosceles triangle inscribed in the ellipse
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) with its vertex at one end of the major axis.
Answer:

Question 9.
‘A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m³. If building of tank costs ₹ 70 per sq metres for the base and ₹ 45 per square metre for sides. What is the cost of least expensive tank?
Answer:

Question 10.
The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.
Answer:

Question 11.
A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.
Answer:

Question 12.
A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the maximum length of the hypotenuse is
\(\left(a^{\frac{3}{2}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\)
Answer:

Question 13.
Find the points at which the function f given by f (x) = (x – 2)⁴ (x + 1)³ has
(i) local maxima
(ii) local minima
(iii) point of inflexion
Answer:
f (x) = (x – 2)⁴ (x + 1)³
f’ (x) = (x – 2)⁴ 3 (x + 1)² + (x + 1)³ 4(x – 2)³
= (x – 2)³ (x + 1) [3 (x -2) + 4 (x + 1)]
= (x – 2)³ (x + 1)² [3x – 6 + 4x + 4]
= (x – 2)³ (x + 1)² [7x – 2]
f’ (x) = 0 ⇒ x = 2, x = -1, x = 2/7
f’ (x)=(x – 2)³ (x+1)² x 7 + (x – 2)³ (7x – 2) 2 (x +1) + (x + 1)² (7x – 2)3 (x – 2)²
here second demivalive test fails as f” (x) = 0 then we apply first demivalive test
At x = 2 f'(x) changes from negative to positive.
∴ function has minimum value and the local minimum value is f (2) = 0
At x = -1
f’ (x) does not change (positive to negative)
∴ f^x has neither maximum value nor minimum value.
∴ x = -1 is the point of inflection at x = 2/7
f’ (x) change from positive to negative
hence function has been maximum at x = 2/7 and local maximum value is

Question 14.
Find the absolute maximum and minimum values of the function f given by
f (x) = cos² x + sin x, x ∈ [o,π]
Answer:

Question 15.
Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is \(\frac{4 r}{3}\)
Answer:
Let the radius of sphere be ‘r’

Question 16.
Let f be a function defined on [a, b] such that f’ (x) > 0, for all xe (a, b). Then prove that f is an increasing function on (a, b).
Answer:
Let x₁, x₂ be any two real number [a, b] such that x₁ <  x₂, then f (x) satisfies the conditions of L.M.V theorem. ∋ C ∈  (a, b)

Question 17.
Show that the height of the cylinder of maximum volume that can be inscribed in sphere of radius R is \(\frac{2 \mathbf{R}}{\sqrt{3}}\). Also find the maximum volume.
Answer:
Let r be the radius of the cylinder and height 2x

Question 18.
Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle a is one-third that of the cone and α the greatest volume of cylinder is \(\frac{4}{27} \pi \mathrm{h}^{3}\)
Answer:
Height of cone – h
Radius of cone – r
Height of cylinder – y
Radius of cylinder – x



Choose the correct answer in the Exercises from 19 to 24.

Question 19.
A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of
(A) 1 m³/h
(B) 0.1 m³h
(C) 1.1 m³/h
(D) 0.5 m³/h
Answer:

Question 20.
The slope of the tangent to the curve x = t² + 3t – 8, y = 2t² – 2t – 5 at the point (2,-1) is
(A) \(\frac{22}{7}\)
(B) \(\frac{6}{7}\)
(C) \(\frac{7}{6}\)
(D)\(\frac{-6}{7}\)
Answer:

Question 21.
The line y = mx + 1 is a tangent to the curve y² = 4x if the value of m is  ………..
(A) 1
(B) 2
(C) 3
(D) \(\frac{1}{2}\)
Answer:

Question 22.
The normal at the point (1,1) on the curve 2y + x² = 3 is ……………….
(A) x + y = 0
(B) x – y = 0
(C) x + y = 1
(D) x – y = 1
Answer:

Question 23.
The normal to the curve x² = 4y passing (1,2) is
(A) x + y = 3
(B) x – y = 3
(C) x + y = 1
(D) x – y = 1
Answer:

Question 24.
The points on the curve 9y² = x³, where the normal to the curve makes equal intercepts with the axes are ………………
(A) \(\left(4, \pm \frac{8}{3}\right)\)
(B) \(\left(4, \frac{-8}{3}\right)\)
(C) \(\left(4,+\frac{3}{8}\right)\)
(D) \(\left(\pm 4, \frac{8}{3}\right)\)
Answer:

2nd PUC Maths Application of Derivatives Miscellaneous Exercise Additional Questions and Answers

Question 1.
If the length of the three sides of a trapezium other than the base is 10cm, find the area of the trapezium when it is maximum (CBSE 2010)
Answer:

Question 2.
Find the intervals in which the function sinx + cos x x ∈ [0,π] is (1) strictly increasing (2) strictly decreasing. (CBSE 2010)
Answer:

Question 3.
If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, find the apps error in. Calculating its surface area, (CBSE 2011)
Answer:

Question 4.
Find the relationship between a and b so that the function of defined by
\(f(x)=\left\{\begin{array}{l}{a x+1, \text { If } x \leq 3 \text { is continous at } x=3} \\{b x+3 \text { if } x>3} \end{array}\right.\)
Answer:

Question 5.
An open box with a square box is to be made out of a given quantity of sheet of area a². Show that the maximum volume of the box is \(a^{3} / 6 \sqrt{3}\) (PUC 2011)
Answer:

Question 6.
Show that the rectangle of maximum perimeter which can be insenibed in a circle of radius a is a square of side \(\sqrt{2}\) a. (CBSE 2008)
Answer:

Question 7.
Discuss the continuity of
\(f(x)=\left\{\begin{array}{cl}{\frac{1-\cos x}{x^{2}}} & {x \neq 0} \\{1} & {x=0} \end{array}\right.\) (KPUC 2008)
Answer:

Students can Download Basic Maths Exercise 18.3 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 18 Differential Calculus Ex 18.3

Part – A

2nd PUC Basic Maths Differential Calculus Ex 18.3 One or Two Marks Questions and Answers

Question 1.
3x² + 4y² = 10
Answer:
Given 3x² + 4y² = 10
Diff w.r.t x
6x + 8y \(\frac{d y}{d x}\) = 0
\(\Rightarrow \quad \frac{d y}{d x}=\frac{-8 y}{6 x}=\frac{-4 y}{3 x}\)

Question 2.
\(\sqrt{x}+\sqrt{y}=3\)
Answer:

Question 3.
y² = 4ax.
Answer:
Given y² = 4ax.
Differentiate with respect to x
2y \(\frac{d y}{d x}\) = 4a.1 ⇒ \(\frac{d y}{d x}\) = \(\frac{4 a}{2 y}=\frac{2 a}{y}\)

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

Question 5.
x² = 4ay
Answer:
Given x² = 4ay
Differentiate with respect to x, 2x = 4a \(\frac{d y}{d x}\) ⇒ \(\frac{d y}{d x}\) = \(\frac{2 x}{4 a}=\frac{x}{2 a}\)

Question 6.
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Answer:

Question 7.
x³ + y³ = 3axy
Answer:
Given x³ + y³ = 3axy
3x² + 3y² \(\frac{d y}{d x}\) = 3a \(\left(x \cdot \frac{d y}{d x}+y \cdot 1\right)\)
\(\frac{d y}{d x}\) (3y² – 3ax) = 3ay – 3x² = \(\frac{d y}{d x}=\frac{a y-x^{2}}{y^{2}-a x}\)

Question 8.
x – y = 0
Answer:
Given x – y = 0
Differentiate with respect to x,
1 – \(\frac{d y}{d x}\) = 0 ⇒ \(\frac{d y}{d x}\) = 1

Question 9.
x² – y² = a²
Answer:
Given x² – y² = a²
Differentiate with respect to x we get,
2x – 2y. \(\frac{d y}{d x}\) = 0 ⇒ \(\frac{d y}{d x}\) = \(\frac{2 x}{2 y}=\frac{x}{y}\)

Question 10.
x + \(\sqrt{x y}\) = x².
Answer:

Part-B

2nd PUC Basic Maths Differential Calculus Ex 18.3 Three marks Questions and Answers

Question 1.
log(xy) = x² + y²
Answer:
Given log(xy) = x² + y²
log x + log y = x² + y² differentiate w.r.t x

Question 2.
2^x + 2^y = 2^x+y
Answer:
Given 2^x + 2^y = 2^x+y
Differentiate w.r.t. x we get
2^x log 2 + 2^y log 2 \(\frac{d y}{d x}\)

Question 3.
x^y = y^x.
Answer:
Given x^y = y^x., taking log^m both sides
y log x = x log y differentiate
Both sides w.r.t x

Question 4.
sin xy = cos(x + y).
Answer:
Given sin xy = cos(x + y), diff w.r.t x.
cos(xy) \(\left[\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\right]\)
\(\frac{d y}{d x}\) [sin(x + y) + xcos (xy)]
= -sin(x + y) – y cos (xy)
\(\frac{d y}{d x}=\frac{-[\sin (x+y)+\cos x y]}{(\sin (x+y)+x \cos x y)}\)

Question 5.
y = 4^x+y
Answer:
Given y = 4^x+y, diff. w r.t. x
\(\frac{d y}{d x}\) = 4^x+y log 4(1 + \(\frac{d y}{d x}\)) = 4^x+y
\(\frac{d y}{d x}\)(1 – 4^x+y log 4) = 4^x+y log 4
∴ \(\frac{d y}{d x}=\frac{4^{x+y} \cdot \log 4}{1-4^{x+y} \cdot \log 4}\)

Part-C

2nd PUC Basic Maths Differential Calculus Ex 18.3 Five Marks Questions and Answers.

Question 1.
If \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}\) = a, Prove that x . \(\frac{d y}{d x}\) = y.
Answer:

Question 2.
If x^y = e^{y – x}, show that \(\frac{d y}{d x}\) = \(\frac{2-\log x}{(1-\log x)^{2}}\)
Answer:
Given x^y = e^{y – x} . Taking log both sides
y log x = (y – x)log e^e
x = y (1 – log x) ∵ log e^e = 1

Question 3.
If cos y = x cos(a + y). show that \(\frac{d y}{d x}\) = \(\frac{\cos ^{2}(a+y)}{\sin a}\)
Answer:

Question 4.
If e^x = y^x show that \(\frac{d y}{d x}\) = \(\frac{(\log y)^{2}}{\log y-1}\)
Answer:
Given e^x = y^x Taking log^m both sides
y log e^e = x log y
y = x log y differentiate w.r.t x

Question 5.
If e^x+y = xy show that \(\frac{d y}{d x}\) = \(\frac{y(1-x)}{x(y-1)}\)
Answer:
Given ye^x+y = xy
Taking log m both sides
(x + y) loge = log(xy)
x+y = log x + log y diff w.r.t x

Question 6.
If y^x = x^y show that \(\frac{d y}{d x}\) = \(\frac{y(y=x \log y)}{x(x-y \log x)}\)
Answer:
Given y^x = x^y, Taking log^m both sides
x log y = y log x, diff w.r.t x

Students can Download Maths Chapter 6 Application of Derivatives Ex 6.5 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 6.5

2nd PUC Maths Application of Derivatives NCERT Text Book Questions and Answers Ex 6.5

Question 1.
Find the maximum and minimum values, if any, of the following functions given by

(i) f(x) = (2x – 1)² + 3
Answer:
f (x) = (2x – 1)² + 3
For all values of x, f (x) > 3 (2x – 1)² + 3 > 3
∴ minimum value is 3 when 2x -1 = 0 is x = \(\frac{1}{2}\) however the function has no maximum values as f (x) → ∞ as |x| ∞

(ii) f (x) = 9x² + 12x + 2
Answer:
f (x) = 9x² + 12x + 2

(iii) f (x) = – (x – 1)² + 10
Answer:
f (x) = – (x – 1)² + 10
f(x)= 10 – (x – 1)²
10 – (x – 1)² < 10
f(x) has a maximum value when x – 1 = 0,x= 1.
how ever f (x) has no minimum value.

(iv) g(x) = x³ + 1
Answer:
g(x) = x³ + 1 as x → ∞ g (x) → ∞ and a
s -+ – ∞, g (x) ⇒ – ∞
∴ g (x) has neither minimum value nor maximum value.

Question 2.
Find the maximum and minimum values, if any, of the following functions given by

(i) f (x) = |x + 2| – 1
Answer:
f (x) = |x + 2| – 1
f (x) = |x + 2| – 1 ≥ -1
minimum value is – 1 when x + 2 = 0, x = -2
however it has no maximum value.

(ii) g(x) = – | x + 1| + 3
Answer:
g(x) = – | x + 1| + 3 = 3,-1 x + 1|
g (x) < 3 v x + 1 = 0
∴ max. value is 3
when x = – 1
how ever no minimum value.

(iii) h(x) = sin (2x) + 5
Answer:
h(x) = sin (2x) + 5
maximum value of sin 2x = 1 and minimum value is -1
f(x) = 1 + 5 is 1 + 5
∴ max. h (x) = 6 and
min. h (x) = 4.

(iv) f (x) = |sin 4x + 3|
Answer:
f (x) = |sin 4x + 3|
-1 < sin 4x < 1 ⇒ 3 -1 < sin 4x + 3 < + 1 + 3 + 2′< sin 4 x + 3 < 4
2 < | sin 4x + 3 | < 4
f (x) > 2 and f (x) < 4
min. f (x) = 2 when sin 4x + 3 = 0
max f (x) = 4 when sin 4x + 3 = 0
∴ minimum value is 2 at sin 4x = -1
maximum value is 4 at sin 4x = 1

(v) h(x) = x + 1, x ∈ (- 1, 1)
Ans:
h(x) = x + 1, x ∈ (-1,1)
given that x ∈ (-1, 1)
ie. -1 < x < 1
-1 + 1< x + 1 < 1 + 10 < x + 1< 2
∴ x + 1 > 0 or x + 1 < 2
x + 1 > 0 so no minimum value
x + 1 < 2, so no maximum value.

Question 3.
Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:
(i) f (x) = x²
Answer:
f (x) = x²
f’ (x) = 2x, f’ (x) = 0
⇒ 2x = 0, x = 0
f'(x)= 2
f'(x)> 0, hence f (x) has minimum value at x = 0
and the minimum value is (0)²= 0.

(ii) g(x) = x³ – 3x
Answer:

(iii) h(x) = sin x + cos x, o < x< \(\frac{\pi}{2}\)
Answer:

(iv) f (x) = sin x – cos x, 0 < x < 2π
Answer:

(v) f (x) = x³ – 6x² + 9x + 15
Answer:

(vi) \(g(x)=\frac{x}{2}+\frac{2}{x}, x>0\)
Answer:

(vii) \(g(x)=\frac{1}{x^{2}+2}\)
Answer:

(viii) \(f(x)=x \sqrt{1-x}, x>0\)
Answer:

Question 4.
Prove that the following functions do not have maxima or minima :

(i) f (x) = e^x
Answer:
f (x) = e^x
f’ (x) = e^x
f'(x) > 0 ∀ x ∈ R
hence function has no critical point There is no point at which the function is maximum or minimum.

(ii) g(x) = log x, x > 0
Answer:
g (x) = log x, x > 0
g'(x) = \(\frac{1}{x}\), where x > 0
hence the function has no critical point
∴ There is no point at which the function is maximum or minimum.

(iii) h (x) = x³ + x¹ + x +1
Answer:
h (x) = x³+ x²+ x +1
h’ (x) = 3x² + 2x + 1
h’ (x) = 0 ⇒ 3x² + 2x + 1 = 0,
x has no real value, hence there is no critical point.
∴ For no point the function has max. or min. value.

Question 5.
Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i) f(x) = x³,x ∈ [-2,2]
Answer:
f (x) = x³,x ∈ [- 2,2]
f’ (x) = 0 ⇒ 3x² = 0 ⇒ x = 0 for finding the absolute maximum and absolute minimum, we have to evaluate f (0), f (2), f (-2)
F (0) = (0)³ = 0, F (2) = (2)³ = 8,
F (-2) = (-2)³ = – 8
Absolute maximum = 8 and
Absolute minimum = -8
∴ maximum at 2 is 8 and minimum at -2 is – 8.

(ii) f (x) = sin x + cos x , x ∈ [0, π]
Answer:

(iii) \(f(x)=4 x-\frac{1}{2} x^{2}, x \in[-2,9 / 2]\)
Answer:

(iv) f (x) = (x – 1)² + 3, x ∈ [-3,1]
Answer:
f (x) = (x – 1)² + 3, x ∈ [-3,1]
f'(x) = 2 (x – 1)
f’ (x) = 0 ⇒ (x – 1) = 0
⇒ x = 1 we will evaluate f (-3) and f (1)
f(-3) = (-3 – 1)² + 3= 16 + 3 = 19
f (1) = (0)² + 3 = 3
Absolute maximum at x = -3 is 19 and
Absolute minimum x = 1 is 3.

Question 6.
Find the maximum profit that a company can make, if the profit function is given by
p(x) = 41+24x – 18x²
Answer:

Question 7.
Find both the maximum value and the minimum value of
3x⁴ – 8x³ + 12x² – 48x + 25 on the interval [0, 3].
Answer:
f (x) = 3x⁴ – 8x³ + 12x² – 48 x + 25
f’ (x) = 12x³ – 24x² + 24x – 48
_f‘(x) = 0 ⇒ 12 (x³ – 2x² + 2x -4) = 0
⇒ 12 (x² (x – 2) + 2 (x – 2))
⇒ 12 ( (x – 2) (x² + 2) ) = 0 (x – 2)(x² + 2) = 0
⇒ x = 2 but x² + 2 ≠ 0
The points are f (0), f (2), f (3)
f (x) = 3x⁴ – 8x³ + 12x² – 48x + 25
f (0) = 25
f (2) = 3 (16) – 8 (8) + 12 (4) – 48 (2) + 25 = 48 – 64 + 48 – 96 + 25 = -39
f(3) = 3 (3⁴) – 8 (3³) + 12 (3²) – 48 (3) + 25 = 243 – 216 + 108 – 144 + 25
376 – 360 = 16
∴ maximum of f (x) at x = 0 is 25
minimum of f (x) at x = 2 is – 39.

Question 8.
At what points in the interval [0,2π] does the function sin 2x attain its maximum value?
Answer:

Question 9.
What is the maximum value of the function sin x + cos x on \(\left[0, \frac{2}{\pi}\right]\)
Answer:

Question 10.
Find the maximum value of 2x³ – 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [-3, -1]
Answer:
f (x) = 2x³ – 24x +107
f’ (x) = 6x² – 24 ⇒ x² = 4, x = ± 2 + 2 s [1,3]
∴ we have to evaluate
f (1). f (2), f (3)
f (1) = 2 (1)³ – 24 x 1 + 107 = 85
f (2) = 2 (2)³– 24 x 2+ 107 = 75
f(3) = 2(3)³-24 x 3 + 107 = 89
maximum at x = 3 and miximum value is 89 minimum at x = 2, is 75
Now the interval is [-3, -1]
∴ points are -3, -2, -1
f (-3) = 2 (-3)³ – 24 (-3) + 107 = 125
f (-2) = 2 (-2)³ – 24 (-2) + 107
= 2 (-8+ 48 + 107 = 139 = 139
f (-1) = 2 (-1) 3-24 (-1) + 107 = 129
maximum at -2 is 139
minimum at (-3) is 125.

Question 11.
It is given that at x = 1, the function x⁴ – 62x² + ax + 9 attains its maximum value, on the interval [0,2]. Find the value of a.
Answer:
f (x) = x⁴ – 62x² + ax + 9
f'(x) = 4x³ – 124x + a
at x = 1, the function has maximum value
∴ f'(1) = 0 ⇒ 4 (1)² -124 (1) + a = 0, a = 120
∴ when a = 120, the function attains maximum value.

Question 12.
Find the maximum and minimum values of x + sin 2x on [0,2π]
Answer:

Question 13.
Find two numbers whose sum is 24 and whose product is as large as possible.
Answer:

Question 14.
Find two positive numbers x and y such that x + y = 60 and xy³ is maximum.
Answer:
x + y = 60 ⇒ y = 60 – x

Question 15.
Find two positive numbers x and y such that their sum is 35 and the product x² y⁵ is a maximum.
Answer:

Question 16.
Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Answer:

Question 17.
A square piece of tin of side 18 cm ¡s to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.
Answer:


x = 9 is not possible, it so side = 0
volume is maximum when the sides are 12,12,3
volume =12 x 12 x 3 = 432 cm³
The volume is maximum when the side of the square to be cut off is 3cm.

Question 18.
A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum ?
Answer:
The sides of the box are x, 45 – 24, 45 – 2x

Question 19.
Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Answer:
Let the sides of rectangle be x and y and radius of the circle is r

hence rectangle becomes square.
∴ The area is maximum when the rectangle is a square.

Question 20.
Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.
Answer:

Question 21.
Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?
Answer:

Question 22.
A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?
Answer:

Question 23.
Prove that the volume of the largest cone that can be inscribed in a sphere of radius
R is \(\frac{8}{27} \)of the volume of the sphere.
Answer:

Question 24.
Show that the right circular cone of least curved surface and given volume has an altitude equal to \(\sqrt{2}\) time the radius of the base.
Answer:

Question 25.
Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is \(\tan ^{-1} \sqrt{2}\).
Answer:

Question 26.
Show that semi-vertical angle of right circular cone of given surface area and maximum volume is \(\sin ^{-1}\left(\frac{1}{3}\right)\)
Answer:

Choose the correct answer in the Exercises 27 and 29.

Question 27.
The point on the curve x² = 2y which is nearest to the point (0, 5) is
(A) (2,\(\sqrt{2}\),4)
(B) (2, \(\sqrt{2}\), 0)
(C) (0,0)
(D) (2,2)
Answer:

Question 28.
For all real values of x, the minimum value of \(\frac{1-x+x^{2}}{1+x+x^{2}}\) is
(A) 0
(B) 1
(C) 3
(D) 1/3
Answer:

Question 29.
The maximum value of [x(x – 1) + 1]^1/3< x < 1 is 0 ≤ x ≤ 1 is
Answer:

Students can Download Basic Maths Exercise 19.2 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 19 Application of Derivatives Ex 19.2

Part-A

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

Question 1.
Find whether the following functions are increasing or decreasing or neither.
(i) f(x) = x⁴ – 8x³ + 22x² – 24x + 5 at x = 0, -2
(ii) f(x) = 4x³ – 15x² + 12x – 2 at x = 1,-1
(iii) f(x) = (x – 1)(x – 2)² at x = 1,3.
Answer:
(i) Given
f(x) = x⁴ – 8x³ + 22x² – 24x + 5 at x = 0, -2
f'(x) = 4x³ – 24x² + 44x – 24
At x = 0, f'(0) = -24 < 0 ∴ f(x) is decreasing at x = 0
At x = -2, f'(-2) = 4(-2)³ – 24(-2)² + 44 (-2) – 24 < 0
= 32 – 96 – 88 – 24 <0 (negative)
∴ f(x) is decreasing at x = -2

(ii) Given
f(x) = 4x³ – 15x² + 12x – 2 at x = 1,-1
f'(x) = 12x² – 30x + 12
At x = 1, f'(x) = 12 – 30 + 12 = -6 < 0
∴ f(x) is increasing at x = 1
At x = -1, f'(-1) = 12(-1)² – 30 (-1) + 12 = 54 > 0
∴ f(x) is increasing at x = -1

(iii) Given
f(x] = (x – 1) (x – 2)² at x = 1, 3
f'(x) = (x – 1) (2 (x – 2} + (x – 2)^2>)
At x = 1, f'(1) = 0 + (-1)² = 1 > 0
∴ f(x) is increasing at x = 1
At x = 3, f'(3) = 2 × 2(1] + 1 = 5 > 0
∴ f(x) is increasing at x = 3.

Question 2.
Find the value of x (Interval) for which the function is increasing or decreasing.
(i) f(x) = 2x³ – 15x² – 84x + 7
(ii) f(x) = x⁴ – 2x³ + 1
(iii) f(x) = x³ – 3x² + 3x – 100
(iv) f(x) = 2x² – 96x + 5
(v) f(x) = 10 – 6x – 2x²
(vi) f(x) = 2x³ + 9x² + 12x + 20
Answer:
(i) Given
f(x) = 2x³ – 15x² – 84x + 7
f'(x) = 6x² – 30x – 84
= 6(x² – 5x – 14) = 6(x – 7] (x + 2)
f(x) is increasing if f'(x) > 0
(x – 7) (x + 2) > 0
Case – 1
x – 7 > 0 & x + 2 > 0
x > 7 & x > -2
x > 7 ⇒ (7, ∞)

Case – 2:
x – 7 < 0 & x + 2 < 0
x < 7 & x < -2
x < -2 ⇒ (-∞ , -2) U (7,∞)
∴ Interval is (-∞ , -2) U (7,∞)
f(x) is decreasing if f'(x) < 0
(x – 7) (x + 2)
x – 7 < 0 & x + 2 > 0

Case -1:
x < 7 & x > -2
∴ -2 < x < 7

Case – 2 x – 7 > 0 & x + 2 < 0
x > 7 & x < -2 (not possible)

(ii) Given f(x) = x⁴ – 2x³ + 1
f'(x) = 4x³ – 6x²
= 2x² (2x – 3) Here 2x² > 0
f(x) is increasing if f'(x) > 0
⇒ 2x – 3 > 0
x > \(\frac { 3 }{ 2 }\) ⇒ (\(\frac { 3 }{ 2 }\), ∞) is the interval
f(x) is decreasing if f ‘(x) < 0
2x – 3 < 0
x < \(\frac { 3 }{ 2 }\) (-∞, \(\frac { 3 }{ 2 }\)) is the interval`

(iii)
Given f(x) = x³ – 3x² + 3x – 100
f'(x) = 3x² – 6x + 3
= 3(x² -2x + 1)
= 3(x – 1)²
f(x) is increasing
f ‘(x) > o ⇒ (x – 1)² > 0
increasing for all
∵ f'(x) > 0 it will not be decreasing

(iv) Given f(x) = 2x² – 96x + 5
f ‘(x) = 4x – 96 = 4 (x – 24)
f(x) is increasing if f’ (x) > 0
x – 24 > 0
x > 24
f(x) is decreasing if f'(x) < 0
x – 24 < 0
x < 24

(v) f(x) = 10 – 6x – 2x²
f'(x) = -6 -4x = -(6 + 4x) = -(6 + 4x)
f(x) is increasing if -(6 + 4x) > 0
⇒ 6 + 4x < 0
⇒ 4x < -6
⇒ x > –\(\frac { 3 }{ 2 }\)
f(x) is decreasing : if -(6 + 4x) < 0 ⇒ 6 + 4x > 0 ⇒ 4x > -6 ⇒ x > –\(\frac { 3 }{ 2 }\)

(vi) Given f(x) = 2x³ + 9x² + 12x + 20
f'(x) = 6x² + 18x + 12
= 6(x² + 3x + 2)
= 6(x + 2) (x + 1)
f(x) is increasing if f ‘(x) > 0
(x + 1) (x + 2) > 0

Case 1:
x + 2 > 0 and x + 1 > 0
x > -2 and x > -1
⇒ x > -1 ⇒ (-1, ∞)

Case 2:
x + 2 < o & x + 1 < 0
x < -2 & x < -1
x < -2 ⇒ (-∞, -2)
f(x) is decreasing if f ‘(x) < 0
(x + 2) (x + 1) < 0

Case 1:
x + 2 < 0 and x + 1 > 0
x < -2 and x > -1
Not possible

Case – 2:
x + 2 > 0 and x + 1 < 0
x > -2 and x < -1
x > -2 and x < -1 ⇒ -2 < x < -1.

Students can Download Basic Maths Exercise 19.3 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 19.3

Part- A

2nd PUC Basic Maths Application of Derivatives Ex 19.3 Two or Three Marks Questions with Answers.

Question 1.
Find the maximum and minimum value of the following function.
(i) f(x) = x³ – 3x
(ii) f(x) = x³ – 6x² + 9x + 15(0 ≤ x ≤ 6)
(iii) f(x) = x⁴ – 62x² + 120x + 9
(iv) f(x) = 2x³ – 3x² – 12x + 12
(v) f(x) = 2x³ – 3x² – 36x + 10
(vi) f(x) = 9x² + 12x + 2
(vii) f(x) = 2x³ – 15x² + 36x + 10
(viii) f(x) = 2x³ – 21x² + 36x – 20
(ix) f(x) = 2x³ – 15x² + 36x + 10
(x) f(x) = 12x⁵ – 45x⁴ + 40x³ + 6
Answer:
(i) Given f(x) = x³ – 3x …..(1)
f'(x) = 3x² – 3 = 3(x² – 1) = 3(x – 1) (x + 1) …..(2)
f'(x) = 0 ⇒ x = ±1
f”(x) = 6x – (3)
At x = 1, f”(1) = 6 > 0, f(x) is minimum at x = 1
& minimum value is f(1) = 1 – 3 = -2
At x = -1, f”(-1) = -6 < 0 f(x) is maximum at x = -1
Maximum value is f(-1) = -1 + 3 = 2

(ii) f(x) = x³ – 6x² + 9x + 15 (0 ≤ x ≤ 6) – (1)
f'(x) = 3x² – 12x + 9
= 3(x² – 4x + 3) = 3 (x – 3) (x + 1] = 0
f “(x) = 6x – 12
f'(x) = 0 ⇒ x = 1 or 3
At x = 1 f”(1) = 6 – 12 = -6 < 0 the function is maximum at x = 1.
And maximum value is f (1) = 1 – 6 + 9 + 15 = 19
At x = 3, f”(3) = 6 × 3 – 12 = 18 -12 = 6 > 0, f is minimum at x = 3
And minimum value is f(3) = 3³ – 6.3² + 9.3 + 15 = 27 – 54 + 27 + 15 = 15.

(iii) Given f(x) = x⁴ – 62x² + 120x + 9 …..(1)
f”(x) = 4x³ – 124x + 120 ….(2)
f”(x) = 12x² – 124 …..(3)
for a function to be maximum or minimum of f'(x) = 0.
⇒ x³ – 31x + 30 = 0 here x = 1 is a root

⇒ x² + x – 30 = 0 ⇒ (x + 6)(x – 5) = 0 ⇒ x = 5 – 6
Put x = 1 in (3] we get f”(x) = (12 – 124) < 0
f(x) att-ains maximum at = 1 & maximum at x = 1 & max value is
f(1) = 1 – 62 + 120 + 9 = 68.
At x = 5, f”(5) = 12(5)² 124 = 300 – 124 > 0 f(x) attains
minimum at x = 5, & minimum value is f(5) = 625 – 1550 + 600 + 9 = – 316
At x = – 6, f “(-6) = 12 (-6)² – 124 = 432 – 124 > 0
f(x) attains minimum at x = -6 & minimum value is
f (-6) = (-6)⁴ – 62(-6)² + 120 (-6) + 9
= 1296 – 2232 – 720 + 9 = -1647.

(iv) f(x) = 2x³ – 3x² – 12x + 12 ….(1)
f'(x) = 6x² – 6x – 12 = 6 (x² – x – 2) …(2)
f'(x) = 12x – 6 ….(3)
for a function to be maximum or minimum f ‘(x) = 0 ⇒ (x – 2) (x + 1) = 0
⇒ x = 2 or – 1
Put x = (-1) in (3) we get f “(-1) = -12 – 6 = -18 < 0
f(x) att-ains maximum at x = -1 & maximum value is
f(-1) = 2 (-1)3 – 3 (-1)2 – 12(-1) + 12 = 19
Put x = 2 in(3) f”(2) = 24 – 6 = 18 > 0
f(x) attains minimum at x = 2 & minimum value is
f(2) = 2 (2)³ – 3(2)² – 12(2) + 12 = 6- ^-24+ >^=8

(v) Given f(x) = 2x³ – 3x² – 36x + 10 ….(1)
f ‘(x) = 6x² – 6x – 36 = 6 (x² – x – 6) …..(2)
f “(x) = 12x – 6 ……(3)
For a function to be maximum of minimum f ‘(x) = 0
⇒ (x² – x – 6) = 0 => (x – 3) (x + 2) = 0
⇒ x = 3 or – 2
Put x = 3 in equation (3) we get
f “(3) = 36 – 6 = 30 >0
⇒ f(x) attains minimum at x = 3
Minimum value is f(3) = 2(3)³ – 3 (3²) – 36 (3) + 10
f(3) = 54 – 27 – 108 + 10 = -71
Put x = -2 in equation (3) we get
f”(-2) = -24 – 6 = – 30 < 0 ⇒ f (x) attains maximum at x = -2
Maximum value is f(-2) = 2(-2)³ -3(-2)² – 36 (-2] + 10
f(-2) -16-12 + 72 + 10 = 54

(vi) Given f(x) = 9x² + 12x + 2 ….(1)
f'(x) = 18x + 12 …(2)
f”(x) = 18 > 0 ……(3)
⇒ f(x) attains minimum
f'(x) = 0 ⇒ 18x+ 12 = 0 ⇒ x = \(-\frac{2}{3}\)
& f” \(\left(-\frac{2}{3}\right)\) 18 > 0 ⇒ f(x) is minimum & the minimum value is
f\(\left(-\frac{2}{3}\right)\) = 9\(\left(\frac{4}{9}\right)\) + 12 \(\left(-\frac{2}{3}\right)\) + 2
= 4 – 8 + 2 = -2

(vii) f(x) = 2x³ – 15x² + 36x + 10 …….(1)
f ‘(x) = 6x² – 30x + 36 = 6 (x² – 5x + 6) ……..(2)
f”(x) = 12x – 30 …..(3)
f'(x) = 0 ⇒ x² -5x + 6 = 0 ⇒ (x – 3)(x – 2) = 0 ⇒ x = 3 or 2
when x = 3 f “(x) = 12x – 30
f “(3) = 36 – 30 = 6 > 0 ⇒ f(x) has minimum
Minimum value is f(3) = 2(3)³ – 15(3)² + 36(3) + 10
f(3) = 54 – 135 + 108 + 10 = 37
when x = 2, f”(2) = 24 – 30 = -6 < 0 ⇒ f(x) has maximum
maximum value is f(2) = 2(2)³ – 15(2) + 36(2) + 10
f(2) = 16 – 60 + 72 + 10 = 38.

(viii) Given f(x) = 2x³ – 21x² + 36x – 20 ….. (1)
f'(x) = 6x² – 42x + 36 …… (2)
= 6(x² – 7x + 6)
f'(x) = 6 (x – 1) (x – 6) = 0 ⇒ x = 1 or 6
f”(x) = 12x – 42 … (3)
when x = l,f “(1) = 12 – 42 = -30 < 0 ⇒ f(x) is maximum
maximum value is f(1) = 2 – 21 + 36 – 20 = -3
when x = 6, f “(6) = 72 – 42 = 30 > 0 ⇒ f(x) is minimum
minimum value is f(6) = 2(6)3³ – 21 (6)² + 36(6) – 20
f(6) = 432 – 756 + 216 – 20 = -128

(ix) Given f(x) = 12x⁵ – 45x⁴ + 40 x³ + 6 ….(1)
f'(x) = 60x⁴ – 180x³ + 120x² ….(2)
= 60x² (x² – 3x + 2)
= 60x² (x – 1) (x – 2)
f ‘(x) = 0 ⇒ 60x² (x – 1) (x – 2) = 0 ⇒ x = 0, 1, 2
f “(x) = 60 (4x³ – 6x + 4x) ……. (3)
when x = 0, f”(x) = 0 ⇒ f(x) has neither maximum nor minimum
when x = 1, f”(x) = -1 < 0 ⇒ f(x) has a maximum & maximum value is
f(1) = 12 – 45 + 40 + 6 = 13
When x =2 f”(x) = 4 > 0 ∴ f(x) has a minimum
minimum value is f(2) = 12(32) – 45(16) + 40(8) + 6
f(2) = 384 – 720 + 320 + 6 = -10.

Question 2.
The sum of two natural numbers is 48. Find the numbers when their product is maximum.
Answer:
Let the two numbers be x & y.
Given x + y = 48 & product: = xy where y 48 – x.
Let p = xy = x (48 – x) = 48x – x².
\(\frac{d p}{d x}\) = 48 – 2x
\(\frac{d p}{d x}\) = 0 ⇒ 48 – 2x = 0 x = 24
\(\frac{\mathrm{d}^{2} \mathrm{p}}{\mathrm{d} \mathrm{x}^{2}}\) = -2 < 0 ⇒ product is maximum
x = 24 ⇒ y = 48 – 24 = 24
⇒ the two numbers are 24, 24.

Question 3.
Find two positive numbers whose sum is 14 and the sum of whose square is minimum.
Answer:
Let the two numbers be x and y
Given x + y = 14 & S = x² + y² where y = 14 – x
∴ S = x² + (14 – x)² = x² + 14² + x² – 28x = 2x² – 28x + 14²
\(\frac{d s}{d x}\) = 4x – 28 → (1) \(\frac{d s}{d x}\) = 0 ⇒ 4x – 28 = 0 ⇒ x = 7
\(\frac{\mathrm{d}^{2} \mathrm{s}}{\mathrm{d} \mathrm{x}^{2}}\) = 4 > 0, sum is minimum.
∴ y = 14 – x = 14 – 7 = 7
∴ the two positive number are 7 & 7.

Question 4.
Find two positive numbers whose sum is 30 and the sum of their cubes is minimum.
Answer:
Let the two numbers be x & y.
Given x + y = 30 & S = x³ + y³ where y = 30 – x
S = x³ + (30 – x)³ = x³ + (30)³ – x³ – 2700x + 90x²
\(\frac{d s}{d x}\) = – 2700 + 180x
\(\frac{d s}{d x}\) = 0 ⇒ x = \(\frac{2700}{180}\) = 15
\(\frac{\mathrm{d}^{2} \mathrm{s}}{\mathrm{d} \mathrm{x}^{2}}\) = 180 > 0 ⇒ sum of cubes is minimum & y = 30 – 15 = 15
∴ two positive number are 15 & 15.

Question 5.
The product of two natural numbers is 64. Find the numbers is their sum is minimum
Answer:
Let the two numbers be x & y
Given xy = 64 ⇒ y = \(\frac{64}{x}\)
Let s = x + y = x + \(\frac{64}{x}\).
\(\frac{d s}{d x}\) = 1 – \(\frac{64}{x^{2}}, \frac{d s}{d x}\) = 0 ⇒ x² = 64 ⇒ x = ±8
\(\frac{d^{2} s}{d x^{2}}=+\frac{128}{x^{3}}\)
When x = 8, \(\frac{d^{2} s}{d x^{2}}=\frac{128}{8^{3}}\) > 0 ⇒ s is minimum
When x = -8, \(\frac{d^{2} s}{d x^{2}}=\frac{128}{8^{3}}\) < 0 ⇒ s is maximum
The two numbers are 8 & 8.

Students can Download Basic Maths Exercise 6.1 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 6 Mathematical Logic Ex 6.1

Part – A

2nd PUC Basic Maths Mathematical Logic Ex 6.1 One Mark Questions and Answers

Question 1.
Symbolise the following propositions:
(1) 3x = 9 and x<7
(ii) 33 + 11 ≠ 3 or 8 – 6 = 2
(iii) If two numbers and equal then their squares are not equal.
(iv) If oxygen is a gas then gold is a compound
(v) y + 4 ≠ 4 ore is not a vowel
Answer:
(i) Let p:3x = 9, q = x < 7
Given in symbols is p ∧ q

(ii) Let p:33 = 11 = 3, q= 8 – 6 = 2
Given is ~p ∨ q

(iii) Let p: Two numbers are equal, q: Squares are equal then given is p → ~q

(iv) Let p: Oxygen is a gas
q: Gold is a compound
Then given is p → q

(v) P: y + 4 = 4, q: e is a vowel given proposition is ~p v ~q

Part – B

2nd PUC Basic Maths Mathematical Logic Ex 6.1 Two or Three Marks Questions and Answers

Question 1.
if p, q and r are propositions with truth values F, T and F respectively, then find the truth values of the following compound propositions:
(i) (~p → q) ∨ r
(ii) (p ∧ ~q) → r
(iii) p → (q → r)
(iv) ~(p → q) ∨~(p ↔ q)
(v) (p ∧ q) ∨ ~ r
(vi) ~(p ∨ r) → ~q
Answers:
(i) (~p → q)∨ r
(~F →T) ∨ r
(T →T) ∨ F
T ∨ F
= T

(ii) (p ∧~q) → r
(F∧ ~T) → F
(F ∧F) → F
F →F
= T

(iii) p → (q → r)
F →(T →F)
F →(T →F)
= T

(iv) ~(p → q) ∨~(p↔ q)
~(F →T) ∨ ~(F↔T)
~T ∨ ~F
F ∨ T = T

(v) (p ∧ q) ∨ ~ r
(F ∧ T) ∨ ~ r
F ∨ T
= T

(vi) ~(P ∨ r) → ~ q
~(F ∨ F) ∨ ~ T
~F → ~ T
T → F
= F

Question 2.
Answer:
(1) If the compound proposition “(p → q) ∧ (p ∧ r)” is false, then find the truth values of p, q and r.
(ii) If the compound proposition p→ (q ∨ r) is false, then find the truth values of p, q and r.
(iii) If the compound proposition p → (~q ∨ r) is false, then find the truth values of p, q and r.
(iv) If the truth value of the propositions (p ∧ q) → (r ∨ ~s) is false, then find the truth values of p, q, rand s.

Answers:
(i) Given (p → q) ∧ (p ∧ r) is false
(a) Case 1: p → q is true & p ∧ r is false
p is T q is T & p is T &ris F
p = T, q = T, r = F
Case 2(a): p = F, q = T p ∧ r= F → p = F, r = F
p = F, q = T, r = f

(b): (p →q) is F & par is true
T → F = F
T ∧ T is T
P=T, q = F, r=T

Case 3: (p → q) is F & (p ∧ r) is false
T → F = F
F ∧ F = F
F ∧ T = F
F ∧ F = F .
∴ p = T, q = F, r= F.

(ii) Given p → (q ∨ r) is false
T → F = F
∴ p = T & q ∨ r is false =
F ∨ F= F
∴ p = T, q = F & r= F

(iii) Given p → (q ∨ r) is false
Then T → F= F
∴ P = T, ~ q ∨ r= F
F ∨ F = F
∴ q = T, q = T, r = F.

(iv) Given (p ^ q) → (r ∨ ~s) is false
We know that T → F = F
∴ p∧q = T and r ∨ ~s = F is false
T∧ T = T
F ∨ F= F is false
∴ p = T, q = T, r = F, S = T

Students can Download Basic Maths Exercise 18.2 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 18 Differential Calculus Ex 18.2

Part-A

2nd PUC Basic Maths Differential Calculus Ex 18.2 One or Two Marks Questions and Answers

Question 1.
(a² – x²)¹⁰
Answer:
Let y(a² – x²)¹⁰
\(\frac{d y}{d x}\) = 10(a² – x²)^{10 – 1}. \(\frac{d y}{d x}\)(a² – x²)
= 10(a² – x²)⁹ (-2x) = -20x (a² – x²)⁹

Question 2.
log[log(log x)]
Answer:
Let y = log x (log log(x))
\(\frac{d y}{d x}=\frac{1}{\log (\log x)} \cdot \frac{1}{\log x} \cdot \frac{1}{x}\)

Question 3.
cos x³
Answer:
Let y = cosx³
\(\frac{d y}{d x}\) = -sinx³ .3x²

Question 4.
sin³\(\sqrt{x}\)
Answer:
Let y = sin³(\(\sqrt{x}\)) = (sin \(\sqrt{x}\)))³
\(\frac{d y}{d x}\) = 3.sin² \(\sqrt{x}\) . cos \(\sqrt{x}\) . \(\frac{1}{2 \sqrt{x}}\)

Question 5.
[log(cos x)]²
Answer:
Let y = [log(cos x)]²
\(\frac{d y}{d x}\) = 2log(cos x). \(\frac{1}{\cos x}\)(-sin x)
= -2 tan x log(cos x)

Question 6.
\(\sec \left(x+\frac{1}{x}\right)\)
Answer:

Question 7.
\(7^{\sin \sqrt{x}}\)
Answer:
Let y = \(7^{\sin \sqrt{x}}\)
\(\frac{d y}{d x}\) = \(7^{\sin \sqrt{x}}\) . log 7 . cos \(\sqrt{x}\) . \(\frac{1}{2 \sqrt{x}}\)

Question 8.
\(\sqrt{\cot \sqrt{x}}\)
Answer:

Question 9.
log(sin \(\sqrt{x}\))
Answer:

Question 10.
log[log (tan x)]
Answer:
Let y = log(log (tan x))
\(\frac{d y}{d x}\) = \(\frac{1}{\log (\tan x)} \cdot \frac{1}{\tan x} \cdot \sec ^{2} x\)

Question 11.
cos 3x . sin 5x
Answer:
Let y = cos 3x . sin 5x (Trans using formula)
sy = 2[sin 8x – sin(-2x)] = 2 [sin 8x] + 2sin 2x
\(\frac{d y}{d x}\) = 16 cos 8x + 4 cos 2x
OR
Let y = cos 3x . sin 5x
\(\frac{d y}{d x}\) = cos 3x(5 cos 5x) + sin 5x (-3 sin 3x) = 5 cos 3x sin5x – 3 sin 5x.sin 3x.

Question 12.
sin x . sin 2x
Answer:
Let y = sin x . sin 2x
\(\frac{d y}{d x}\) = sin x(2 cos2x) + sin 2x cos x = 2 sin x cos 2x + cos x sin 2x

Question 13.
elog_e(x + \(\sqrt{x^{2}+a^{2}}\)).
Answer:

Question 14.
e^2x . sin 3x.
Answer:
Let y = e^2x . sin 3x
\(\frac{d y}{d x}\) = e^2x(3 cos 3x) + sin 3x(2e^2x)

Question 15.
cos⁵x . cos(x⁵).
Answer:
Let y = cos⁵x . cos(x⁵)
\(\frac{d y}{d x}\) = cos⁵x(-sin(x⁵)5x⁴) + cos(x⁵).5cos⁴x.(-sin x)
= – cos⁵x sin(x⁵) 5x⁴ – 5 cos(x⁵) . cos⁴x . sin x

Question 16.
3^x2 .log x.
Answer:
Let y = 3^x2 .log x.
\(\frac{d y}{d x}\) = 3^x2 . \(\frac { 1 }{ x }\) + logx . 3^x2 . log_e 3 . 2x .

Question 17.
\(\frac{x}{\sqrt{x^{2}-1}}\)
Answer:

Question 18.
\(\frac{x}{\sqrt{2 x-1}}\)
Answer:

Question 19.
\(\frac{\mathrm{e}^{\sin \mathrm{x}}}{\sqrt{\log \mathrm{x}}}\)
Answer:

Question 20.
\(\log \left(\frac{1+\sin x}{1-\sin x}\right)\)
Answer:

Part-B

2nd PUC Basic Maths Differential Calculus Ex 18.2 Three Marks Questions and Answers

Question 1.
If y = \(\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\) , show that \(\frac{d y}{d x}\) = sec² \(\left(x+\frac{\pi}{4}\right)\)
Answer:

Question 2.
If y = log \(\left[\frac{1-\cos x}{1+\cos x}\right]\) , prove that \(\frac{d y}{d x}\) = 2 cosec x.
Answer:

Question 3.
Differentiate e^2x w.r.t x from first principles
Answer:

Question 4.
Differentiate sin 2x w.r.t x from first principles.
Answer:

Question 5.
Differentiate tan ax w.r.t x froom the principles.
Answers:

Students can Download Basic Maths Exercise 7.3 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 7 Ratios and Proportions Ex 7.3

Part – A

2nd PUC Basic Maths Ratios and Proportions Ex 7.3 Three Marks Questions and Answers ( 3 × 4 = 12)

Question 1.
If ₹ 150 maintains a family of 4 persons for 30 days. How long 7600 maintain a family of 6 persons?
Answer:

Question 2.
300 workers can finish a work in 8 days. How many workers will finish the same work in 5 days.
Answer:

Workers and days are in inverse proportion 300 : x = 5 : 8
x = \(\frac{300 \times 8}{5}\) = 480 workers

Question 3.
5 carpenters can earn ₹540 in 6 days working 9 hours a day. How much will 8 carpenters can earn in 12 days working 6 hours a day?
Answer:

Days and amount are in direct proportion, hours and amount are in direct proportion

Question 4.
A mixture contains milk and water in the ratio 6:1 on adding 5 litres of water, the ratio of milk and water becomes 7 : 2, find the quantity of milk in the original mixture.
Answer:
Quantity of milk is 6x and water is 1x, 5 liters of water is added, the new ratio is 7:2
\(\frac{6 x}{x+5}=\frac{7}{2}\)
12x = 7x + 35 ⇒ 5x = 35
x = 7
The quantity of milk is 6(x) = 6 (7) = 42

Part – B

2nd PUC Basic Maths Ratios and Proportions Ex 7.3 Five Mark Questions and Answers 

Question 1.
A jar contains two liquids X and Y in the ration 7:5. When 6 litres of the mixture is drawn and the jar in filled with the same quantity of Y, the ratio of Xand Y becomes 7:9. Find the quantity X in the jar initially,
Answer:
Let quantity of liquid X is 7x and Y is 5x 6 liters of the mixture is drawn.
i.e.,\(\frac{7 \times 6}{12}=\frac{7}{2}\) litres of X is removed
\(\frac{5 \times 6}{12}=\frac{5}{2}\) litres of Y is removed
∴ The remaining quantity of X and Y is 7x – \(\frac{7}{2}\) and 5x –\(\frac{5}{2}\) respectively.
6 litres of Y is added to get ratio 7:9.

Question 2.
Two taps fill a cistern separately in 20 minutes and 40 minutes respectively and a drain pipe can drain off 30 litres per minute. If all the three pipes are opened, the cistern fills in 72 minutes what is the capacity of the cistern?
Answer:
Time taken by tap A is 20 min
∴ \(\frac{1}{20}\)b of the Cistern is filled by tap A
Similarly \(\frac{1}{40}\) of the Cistern is filled by tap B
Both the taps can fill \(\frac{1}{20}+\frac{1}{40}=\frac{3}{40}\)
“204040 A drain tap can drain 30 liters per minute the cistern is filled in 60 minute.

The drain tap can drain in 17 minutes in 1minute it drains 30 litres
In 17 minutes it drains 30 × 17litres
∴ The capacity of the cistern = 510 litres

Question 3.
If ten persons can do a job in 60 days. In how many days can twenty persons do the same job?
Answer:

Persons & days are in inverse proportion
∴ 10:20 = x : 60
X= \(\frac{60 \times 10}{20}\) = 30days 20

Question 4.
A can do a piece of work in 20 days, B in 30 days and C in 60 days. All of them began to work together. However, A left the job after 6 days and B quit work 6 days before the completion of work. How many days
did the work last?
Answer:
In 1 day the work done by A, B & C is

In 6 days the work done is \(\frac{6}{10}=\frac{3}{5}\)
Remainin work is \(1-\frac{3}{5}=\frac{2}{5}\) of the work
Let the number of days to complete the work be x.C does \(\frac{x}{60}\) of the work & B does \(\frac{x-6}{30}\) of the work in x days

Question 5.
8 men and 16 women can finish a job in 6 days | but 12 men & 24 women can finish it in 8 days. How many days will 26 men and 20 women take to finish the job?
Answer:
8 men & 16 women can finish a Job in 6 days
∴ In 1 day the work done is of 48men and 96 women.
12 men & 24 women can finish a job in 8 days. In 1 day the work done is of 96 men & 192 women.
∴ 48 men + 96 women = 96 men + 192 women
26M + 20W = 52W + 20W = 72W
Let the required number of days be x.
192 : 72 = x : 5
X = \(\frac{192 \times 5}{72}=\frac{45}{3}\) = 15 days

Question 6.
4 men and 12 boys can do a piece of work in 5 … days by working 8 hours per day. In how many days 2 men & 4 boys can do the same piece of work working 12 hours a day.
Answer:
Given 4 m = 12 B
1 m = 3 B
2men & 4 boys = 6 boys + 4 boys = 10 boys

Boys & days are in inverse proposition days & hours are in inverse proportion.
∴ 10 : 12 :: 5 : x 12 : 8
x = \(\frac{12 \times 8 \times 5}{10 \times 12}=\) = 4days.

Question 7.
A railway train 100 metres long is running at the speed of 30 kmph. In what time will it pass (i) a man standing near the line (ii) a bridge 100 metres long?
Answer:
d = 100m Speed = 30km ph
Speed = \(\frac{\mathrm{d}}{\mathrm{t}}\)

Length of the bridge is 100m
d = 100 + 100 = 200
t = \(\frac{200 \times 18}{30 \times 5}\) = = 24 Sec.

Question 8.
The driver of car is traveling at a speed of 36 kmph and spots a bus 80 metres ahead of him. After 1 hour the bus is 120 metres behind the car. What is the speed of the bus?
Answer:
Speed of the car = 36 kmph
Speed of the bus = x kmph
Relative speed = (36 – x) kmph
Distance = 200m
Bus by 120m
∴ t = \(\frac{200}{36-x}\)
1 = \(\frac{200}{(36-x) 1000}\)
36 – x = 0.2
36 -0.2 = x
x = 35.8
∴ Speed of the bus is 35.8 kmph.