2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.1

Students can Download Basic Maths Exercise 18.1 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 18 Differential Calculus Ex 18.1

Part-A

2nd PUC Basic Maths Differential Calculus Ex 18.1 One Mark Questions and Answers

Question 1.
5ex – log x – 3\(\sqrt{x}\)
Answer:
Let y = 5ex – log x – 3\(\sqrt{x}\)
\(\frac{d y}{d x}=5 e^{x}-\frac{1}{x}-3 \frac{1}{2 \sqrt{x}}\)

Question 2.
log ex
Answer:
Let y = logee (constant)
\(\frac{d y}{d x}\) = 0

Question 3.
\(\frac{1}{x^{4 / 3}}-\frac{3}{x^{3 / 2}}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.1 - 1

KSEEB Solutions

Question 4.
\(\frac{4 x^{2}-3 x}{x}\)
Answer:
Let y = \(\frac{4 x^{2}}{x}-\frac{3 x}{x}\) = 4x – 3
∴ \(\frac{d y}{d x}\) = 4(1) – 0 = 4

Question 5.
\(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.1 - 2

Question 6.
\(\sqrt[3]{x^{2}}+\frac{4}{\sqrt[4]{x^{5}}}+\frac{1}{x^{7}}+x \sqrt{x}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.1 - 3

Part-B

2nd PUC Basic Maths Differential Calculus Ex 18.1 Two Marks Questions and Answers

Question 1.
(x – a)(x – b).
Answer:
Let (x – a) (x – b)
\(\frac{d y}{d x}\) = (x – a) . 1 + (x – b) . 1 = x – a + x – b = 2x – a – b

Question 2.
\(\frac{x-a}{x-b}\)
Answer:
Let y = \(\frac{x-a}{x-b}\)
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.1 - 4

KSEEB Solutions

Question 3.
(5x2 + 3x – 1) ( x – 1)
Answer:
Let y = (5x2 + 3x – 1) ( x – 1)
\(\frac{d y}{d x}\) = (5x2 + 3x – 1) \(\frac{d }{d x}\) (x – 1) + (x – 1) \(\frac{d }{d x}\) (5x2 + 3x – 1)
= (5x2 + 3x – 1) + (x – 1) (10x + 3) = 5x2 + 3x – 1 + 10x2 – 10x + 3x – 3 = 15x2 – 4x – 4

Question 4.
x-3(5 + 3x)
Answer:
let y = x-3(5 + 3x)
\(\frac{d y}{d x}\) = x-3(3) + (5 + 3x) (-3 . x-3-1)
= 3x-3 – 15x-4 -9x-3
= -15x-4 – 6x-3
= \(\frac{-3}{x^{4}}(2 x+5)\)

Question 5.
x5(3 – 6x-9).
Answer:
Let y = x5(3 – 6x-9)
\(\frac{d y}{d x}\) = x5(+54x-10) + (3 – 6x-9)(5x4)
= 54x-5 + 15x4 – 30x-5
= 24x-5 + 15x4

Question 6.
sin2x.
Answer:
Let y = (sin x)2
\(\frac{d y}{d x}\) = 2 sin x .(cos x) = sin 2x

Question 7.
\(\frac{x+1}{x}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.1 - 13

Question 8.
\(\frac{1}{a x^{2}+b x+c}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.1 - 5

KSEEB Solutions

Question 9.
\(\frac{\cos x}{1+\sin x}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.1 - 6

Question 10.
(x + cos x) ( x – tan x).
Answer:
Let y = (x + cos x)( x – tan x)
\(\frac{d y}{d x}\) = ( x + cosx)(1 – sec2x) + ( x tanx)(1 – sinx).

Question 11.
If f(x) = x2 – 3x + 10. Find f'(50) and f'(11).
Answer:
Let f(x) = x2 – 3x + 10
then f'(x) = 2x – 3
f'(50) = 100 – 3 = 97
f'(11) = 22 – 3 = 19

Question 12.
If f(x) = xn and if f ‘(1) = 10. Find the value of n.
Answer:
f(x) = xn and f1(1) = 10
f ‘(x) = n. xn-1; f ‘(1) = 10 = n.(1)n-1 ⇒ n = 10

Question 13.
If y = \(x+\frac{1}{x}\) show that x2 \(\frac{d y}{d x}\) – xy + 2 = 0
Answer:
Let y = \(x+\frac{1}{x}\)
\(\frac{d y}{d x}=1-\frac{1}{x^{2}}=\frac{x^{2}-1}{x^{2}} \Rightarrow x^{2} \frac{d y}{d x}=x^{2}-1\)
[∵ x2 + 1 = xy]
[ x2 = xy – 1]
⇒ x2 \(\frac{d y}{d x}\) = xy – 1 – 1
⇒ x2 \(\frac{d y}{d x}\) = xy + 2 = 0

KSEEB Solutions

Part-C

2nd PUC Basic Maths Differential Calculus Ex 18.1 Three Marks Questions and Answers

Question 1.
\(\frac{x^{5}-\cos x}{\sin x}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.1 - 7

Question 2.
\(\frac{x^{n}-a^{n}}{x-a}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.1 - 8

Question 3.
\(\frac{2}{x+1}-\frac{x^{2}}{3 x-1}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.1 - 9

Question 4.
\(\frac{\sqrt{a}+\sqrt{x}}{\sqrt{a}-\sqrt{x}}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.1 - 10

Question 5.
\(\frac{2^{x} \log x}{\sqrt{x}}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.1 - 11

KSEEB Solutions

Question 6.
Find the derivative of f(x) = \(\frac { 1 }{ x }\) wit respect to x from the first principle.
Answer:
2nd PUC Basic Maths Question Bank Chapter 18 Differential Calculus Ex 18.1 - 12

Question 7.
If y = x + tan x. Show that cos2x . \(\frac{d y}{d x}\) = 2 – sin2x
Answer:
Given y = x + tan x
\(\frac{d y}{d x}\) = 1 + sec2 x = 1 + \(\frac{1}{\cos ^{2} x}=\frac{\cos ^{2} x+1}{\cos ^{2} x}\)
∴ cos2x \(\frac{d y}{d x}\) = cos2x + 1 = 1 – sin2x + 1 = 2 – sin2x
∴ cos2x \(\frac{d y}{d x}\) = 2 – sin2x

2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.2

Students can Download Basic Maths Exercise 7.2 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 7 Ratios and Proportions Ex 7.2

Part – A

2nd PUC Basic Maths Ratios and Proportions Ex 7.2 One Mark Questions and Answers ( 1 × 4 = 4)

Question 1.
State which of the following are proportions.
(1) 2 : 3 :: 6 : 9
(ii) 1 : 3 :: 4 : 15
(iii) 3 : 5 :: 6 : 15
(iv) 7 : 9 :: 14 : 18
Answers:
(i) 2 × 9 = 3 × 6
18 = 18 It is a proportion

(ii) 1 × 15 ≠ 3 × 4
It is not a proportion

(iii) 3 × 15 ≠ 5 × 6
∴not a proportion

(iv) 7 × 18 = 9 × 14 ii
126 = 126 It is a proportion

KSEEB Solutions

Question 2.
Find the fourth proportional to each of the following.
(1) 4, 5, 24
(ii) 6, 12, 15
(ii) 1.5, 4.5, 3.5
(iv) \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\)
Answer:
Let x be the fourth proportional
(i) 4 : 5 :: 24 : x
4x = 5 x 24 ⇒ x = \(\frac{5 \times 24}{4}\) ⇒ X=30

(ii) 6 : 12 = 15 : x
⇒ x =\( \frac{12 \times 15}{6}\) = 30 ⇒ x = 30

(iii) 1.5 : 4.5 = 3.5 : x
x = \(\frac{4.5 \times 3.5}{1.5}\) = 10.5

(iv)
2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.2 - 1

Question 3.
Find the mean proportional to each of the following.
(i) 1 and 9
(ii) 2 and 8
(iii) \(\frac{1}{16}\) and \(\frac{1}{25}\)
(iv) 0.8 and 1.8
Answers:
Let x be the mean proportional

(i) 1&9
1 : x= x : 9
x2 = 9
x = 3

(ii) 2 & 8
2 : x = x : 8
x2 = 16
X = 4

(iii)
2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.2 - 2

(iv) 0.8 and 1.8
x2 = 0.8 x 1.8
x = 1.2

KSEEB Solutions

Question 4.
Find the third proportional to each of the following,
(i) 4, 6
(ii) 3,12
(iii) 2.4, 3.6
(iv) 2\(\frac{2}{3}\) , 4
Answers:
Let x be the third proportional
(i) 4,6
4:6 = 6 : x
x = \(\frac{36}{4}\) = 9

(ii) 3, 12
3 : 12 = 12 : x
x = \(\frac{12 \times 12}{3}\) = 48

(iii) 2.4, 3.6
2.4:3.6 = 3.6 : x
x = \(\frac{3.6 \times 3.6}{2.4}\) = 5.4

(iv) 2\(\frac{2}{3}\) , 4
\(\frac{8}{3}\):4 = 4 : x
2.4 : 3.6 = 3.6 : x
x = \(\frac{4 \times 4}{\frac{8}{3}}\) = 6

Part – B and C

2nd PUC Basic Maths Ratios and Proportions Ex 7.2 Three Marks Questions and Answers ( 3 × 6 = 18)

Question 1.
If \(\frac{a}{b}=\frac{c}{d}\) then prove that \(\frac{3 a+5 b}{3 c+5 d}=\frac{3 a-5 b}{3 c-5 d}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.2 - 3

Question 2.
If \(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{a}{b}\)find x
Answer:
2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.2 - 4

KSEEB Solutions

Question 3.
Find the numbers which added to the numerator and denominator of 25 : 37 make it 5 : 6.
Answer:
Let x be added to numerator and denominator of 25 : 37.
∴ \(\frac{25+x}{37+x}=\frac{5}{6}\)
150 + 6x = 185 + 5x
x = 185 – 150 = 35
∴ x = 35.

Question 4.
The ages of a father and his for son in the ratio 6:1. After 14 years their age will be in the ratio 8:3 what are their present ages?
Answer:
Let the father’s age be 6x and Son’s age be x
After 14 years father’s age is 6x + 14 and Son’s is x + 14
Given \(\frac{6 x+14}{x+14}=\frac{8}{3}\)
18x + 42 = 8x + 112
10x = 70 ⇒ x = 7
∴ father’s age is 6x = 6 x 7 = 42 years and son’s age = 7 years.

Question 5.
Four numbers formed by adding 1, 5, 10 and 15 to a certain number are in proportion. Find the number?
Answer:
Let the number be x
∴ 1 + x, 5 + x, 10 + x, 15 + x are in proportion
∴ (5 + x) (10 + x) = (1 + x) (15 + x)
50 + 10x + 5x + x2 = 15 + 15x + x + x2
⇒ x= 35

KSEEB Solutions

Question 6.
A concrete mixture was made of cement, chips and sand. The ratio of the cement and chip is the same as that of chips and the sand. If 144 bags of cement and 225 bags of sand were used is the mixture. Find the
number of bags of chips used in the mixture.
Answer:
Cement: Chip = Chip : Sand
∴ (Chip)2 = Cement × Sand
∴ (Chip)2 = 144 × 225
Chip = 12× 15.= 180
∴ Number of bags of chips is 180.

2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.1

Students can Download Basic Maths Exercise 7.1 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 7 Ratios and Proportions Ex 7.1

Part – A

2nd PUC Basic Maths Ratios and Proportions Ex 7.1 One Mark Questions and Answers ( 1 × 10 = 10)

Question 1.
If 3:5 is a ratio, find the antecedent and consequent.
Answer:
Antecedent = 3, Consequent = 5.

Question 2.
x gets a salary of7 20,000, y gets a salary of 5,000. Find the ratio of their salaries.
Answer:
Ratio of their salaries is 20000:50000 = 2:5.

Question 3.
Find the inverse ratio of 4:5.
Answer:
Inverse ratio of 4 : 5 is 5 : 4

KSEEB Solutions

Question 4.
A house consumes 30 kgs of wheat and 4kg of sugar compare the consumption of wheat and sugar in the form of ratio.
Answer:
Wheat: Sugar = 30 kg: 4 kg = 15:2. 6.

Question 5.
Mr. X completes a job is 3 hours and Mr. Y completes the same job is 45 minutes compare their performances.
Answer:
3 hrs.: 45 minutes = 3 × 60 : 45 = 180 : 45
= 4:1.

Question 6.
Find the compound ratio of 3 : 4 and 4: 7.
Answer:
Compound ratio = \(\frac{3}{4} \cdot \frac{4}{7}=\frac{3}{7}=3: 7\)

Question 7.
Find the compound ratio of 1:2, 2:3 and 3:5.
Answer:
Compound ratio = \(\frac{1}{2} \times \frac{2}{3} \times \frac{3}{5}=\frac{1}{5}=1: 5\)

Question 8.
Find the duplicate ratio of 5 : 4.
Answer:
Duplicate ratio of 5 : 4 is 52: 42 = 25: 16.

Question 9.
Find the triplicate ratio of 3 : 5.
Answer:
Triplicate ratio of 3 : 5 is 33: 53 = 27 : 125.

Question 10.
Find the subduplicate ratio of 9:49.
Answer:
Subduplicate ratio of 9 : 49 is √9:√49 = 3:7

KSEEB Solutions

Question 11.
Find the subtriplicate ratio of 125: 64.
Answer:
Subtriplicate ratio of 125 : 64
= 3√125:3√64 = 5:4

Question 12.
Find the value of x of 5 : 20 = 3 :x.
Answer:
5x = 20 x 3⇒ x = \(\frac{20 \times 3}{5}=\) = 12 ⇒ X = 12

Part – B

2nd PUC Basic Maths Ratios and Proportions Ex 7.1 Two Marks Questions and Answers (2 × 10 = 20)

Question 1.
Find the ratio between two numbers such that their sum is 40 and their difference is 8.
Answer:
Let the two numbers be x & y
Given x + y = 40 ……. (1)
& x – y = 8 … (2)
Solving we get
Adding 2x = 48 ⇒ x = 24 & y = 40 – 24 = 16
∴Ratio of the numbers is 24 : 16 ⇒ 3:2.

Question 2.
A ratio is the lowest term is 3:8. If the difference between the quantities is 25. Find the quantities.
Answer:
Let the numbers are 3x and 8x
Given 8x – 3x = 25
5x = 25 ⇒ X = 5
∴ Two numbers are 3(5) & 8(5)
i.e., 15 and 40.

KSEEB Solutions

Question 3.
Two numbers are is the ratio 3:5. If 5 is added to each , they are is the ratio. 2:3 find the numbers.
Answer:
Let two numbers be 3x & 5x
If 5 is added to each then \(\frac{3 x+5}{5 x+5}=\frac{2}{3}\)
9x + 15 = 10x + 10
x = 5
∴ The two numbers are 3(5) & 5(5) = 15 & 25.

Question 4.
What must be added to each term is the ratio 2 : 3 so that is becomes 5:6.
Answer:
Let x must be added to each term of 2 : 3 then
\(\frac{2+x}{3+x}=\frac{5}{6}\)
⇒ 12 + 6x = 15 + 5x
x = 3.

Question 5.
What must be added to each term is the ratio 4:5 so that is becomes 7:8.
Answer:
Let x must be added to each of 4:5 then
\(\frac{7-x}{4-x}=\frac{5}{2}\)
x = 3

KSEEB Solutions

Question 6.
What must be subtracted from each term is the ratio 7:4 so that it becomes 5:2.
Answer:
Let x must be subtracted from each of 7 : 4 then \(\frac{7-x}{4-x}=\frac{5}{2}\)
⇒ 14 – 2x = 20 – 5x 5x – 2x = 20 – 14
3x = 6 ⇒ x= 2.

Question 7.
What must be subtracted from each term is the ratio 8:7 so that it becomes 4:3.
Answer:
Let x must be subtracted from each of 8:7 then
\(\frac{8-x}{7-x}=\frac{4}{3}\)
24 – 3x = 28 – 4x
4x – 3x = 28 – 24
x = 4

KSEEB Solutions

Question 8.
If a: b = 2:3, b:c=3:5 and c:d= 5:7 find a:d.
Answer:
2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.1 - 12nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.1 - 1

Question 9.
If a: b = 2:3 and b:c= 6:13 find a:b:c.
Answer:
Given
\(\frac{a}{b}=\frac{2}{3}, \frac{b}{c}=\frac{6}{13} \)
\(\frac{a}{b}=\frac{2 \times 2}{2 \times 2}=\frac{4}{6} \)
∴ a:b:c = 4 : 6 : 13

Question 10.
If a : b = 3:4, b:c=8 = 15 find a:b:c.
Answer:
\(\frac{a}{b}=\frac{3}{4}, \frac{b}{c}=\frac{8}{15}\)
\(\frac{a}{b}=\frac{3 \times 2}{4 \times 2}=\frac{6}{8} \frac{b}{c}=\frac{8}{15}\)
∴ a : b : c = 6 : 8 : 15

Part-C

2nd PUC Basic Maths Ratios and Proportions Ex 7.1 Three Marks Questions and Answers (3 × 16 = 48)

Question 1.
Divide 1800 is the ratio 3 : 4 : 5.
Answer:
Sum of the ratios 3 + 4 + 5 = 12
∴  \(\frac{3}{12}\) × 1800 = 450
\(\frac{4}{12}\)1800 = 600
\(\frac{4}{12}\)1800 = 750

KSEEB Solutions

Question 2.
If x : y = 3; 4 find \(\frac{2 x^{2}+3 y^{2}}{x^{2}+y^{2}}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.1 - 2

Question 3.
If a:b= 2:3 and x:y = 4:7.
Find \(\frac{5 a x+4 b y}{8 a x+3 b y}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.1 - 3

Question 4.
The angles of a triangle are in the ratio 3:4:5. Find the angles.
Answer:
Sum of the ratios 3 + 4 + 5 = 12
W. K. T sum of the angles of a triangle = 75°
∴  \(\frac{3}{12}\) × 180 = 45°, 12
\(\frac{3}{12}\) × 180 = 60°,
\(\frac{3}{12}\) × 180 = 75°
∴ The three angles of a triangle are 45°, 60°, 75°

KSEEB Solutions

Question 5.
An article is sold at 40% gain an the cost price. Find the ratio of the selling price and cost price.
Answer:
An article is sold at 40% gain
100 + 40 = 40
If the cost price is 100 then the selling price 100 + expenditure is 4x – 1000 40 = 140
∴ \(\frac{S P}{C P}=\frac{140}{100}=\frac{7}{5}\)
∴ SP : CP = 7 : 5

Question 6.
If the monthly incomes of A and B are is the ratio 3:4 and their expenditures are is the ratio 1 :2. If each saves 1000 find the monthly incomes.
Answer:
Let monthly income of A is 3x and expenditure
is 3x – 1000 and monthly income of B is 4x and
Given \(\frac{3 x-1000}{4 x-1000}=\frac{1}{2}\)
6x – 4000 = 4x – 2000
⇒ 2x = 2000 ⇒ x = 1000
∴ The monthly income of A = 3x = 3 × 1000 = 3000
∴ The monthly income of B is 3(500) = 2000.

Question 7.
If the monthly incomes of A and Bare in the ratio 3:4 and their expenditure are in the ratio 1:2.If each saves 2000. Find their monthly incomes.
Answer:
Let their monthly incomes are 3x and 4x and their expenditure are 3x – 2000 and 4x – 2000
Given difference of their squares is \(\frac{3 x-2000}{4 x-2000}=\frac{1}{2}\)
6x – 4000 = 4x – 2000
⇒ 2x = 2000 ⇒ x = 1000
The monthly income of A = 3x = 3 × 1000 = 3000
The monthly income of B = 4x = 4 x 1000 = 4000

Question 8.
Two numbers are is the ratio 6:7. If the difference of their squares is 117 find the two numbers.
Answer:
Let the two numbers are 6x and 7x
i.e., (7x)2 – (6x)2= 117
49x2 – 36x2= 117
13x2 = 117
⇒ x = 6
∴ The numbers are 3(6) and 46) = 18 and 21.

KSEEB Solutions

Question 9.
Two numbers are is the ratio 3:4. If the sum of their squares is 900 find the two numbers.
Answer:
Given \(\frac{x}{y}=\frac{3}{4}, \frac{y}{z}=\frac{7}{9}\)
Given sum of their squares = 900
(3x)2 + (4x)2= 900
9x2 + 16x2 = 900
25x2 = 900
⇒ x2 = \(\frac{900}{25}\) = 36 ⇒ x = 6
∴ The numbers are 3(6) and 4 (6) = 18 and 24

Question 10.
If x:y= 3:4 and y:z= 7:9. Find x:y:z.
Answer:
2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.1 - 4

Part – D

2nd PUC Basic Maths Ratios and Proportions Ex 7.1 Five Marks Questions and Answers (5 × 16 = 80)

Question 1.
Divide ₹ 800 is three parts such that three times of the first, five times of the second and six times the third are equal.
Answer:
Let the 3 parts be x, y and z
Given 3x = 5y = 62
Let 3x = 6z and 5y = =\(\frac{6}{5} \)z
x = 2z, y = 2
∴ x : y : z = 2z : \(\frac{6}{5}\)z : z= 10:6:5
Sum of the ratios = 10 + 6 + 5 = 21
2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.1 - 5

Question 2.
Divide ₹ 3262 among x, y and z such that if  ₹35, ₹15 and ₹12 are deducted from their respective shares, the remainders are in the ratio 3:5:8.
Answer:
x + y + z = 3262 and x – 35:y – 15: Z – 12 = 3:5:8 x-35 3 y-155 y -15 5
2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.1 - 10
5x – 175 = 3y – 45,
5x – 3y = 130,
5x = 130 + 3y,
x= 26+\(\frac{3}{5} \)y
8y – 120 = 52 – 60
8y – 5z = 60
5z = 8y – 60
z = \(\frac{8}{5} \)y – 12
Substitute for x and z
Buty x + y + z = 3262
2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.1 - 6
= 26 + 609
x = 635
x = 635, y = 1015 and z = 1612
z = 1624 – 12
z = 1612

KSEEB Solutions

Question 3.
If x:y= 2:3. Find the value of \(\frac{2 x^{3}+3 y^{3}}{x^{3}+y^{3}}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.1 - 7

Question 4.
Divide 5880 is to three parts, such that ‘B’ receive thrice as ‘A’ and C receives \(\frac{5^{t h}}{6}\) of what B receives.
Answer:
Let A receive ₹x and B receives ₹ 3x and C receive \(\frac{5}{6}\)(3x) = \(\frac{5}{2}\)x
∴ A:B:C= 1:3:\(\frac{5}{2}\)= 2:6:5
∴ A’s share = \(\frac{2}{13}\) × 5880 = 904.6
B’s share = \(\frac{6}{13}\) x 5880 = 2713.8 :
C’s share = \(\frac{5}{13}\)x 5880 = 2261.5

Question 5.
If \(\frac{2 x^{2}+3 y^{2}}{x^{3}+y^{2}}=\frac{2}{41}\) Find x:y.
Answer:
2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.1 - 8

Question 6.
Divide 6000 into three parts is the ratio \(\frac{1}{2}: \frac{1}{3}: \frac{1}{6}\)
Answer:
Given \(\frac{1}{2}: \frac{1}{3}: \frac{1}{6}\) multiply by 6, we get 3:2:1
1st part = \(\frac{3}{6}\) × 6000 = 3000
2nd part = \(\frac{3}{6}\) × 6000 = 2000
3rd part = \(\frac{3}{6}\) × 6000= 1000

KSEEB Solutions

Question 7.
Divide 17,640 a many P, Q, R and S such that O gets \(7^{\text {th }}5\) of P, R gets \(5^{\text {th }}8\) of Q and S gets \(\frac{2^{t h}}{13}\) of the sum of Q and R.
Answer:
2nd PUC Basic Maths Question Bank Chapter 7 Ratios and Proportions Ex 7.1 - 9

KSEEB Solutions

Question 8.
The Rajdhani express takes 18 hours to reach Delhi from Bhubaneshwar while Nilachala Express takes
24 hours for the same of Delhi is 2880 kms from Bhubaneshwar find the ratio between the average speeds of the two trains.
Answer:
Distance Speed = \frac{\text { Distance }}{\text { Time taken }}
Speed of Ragdhari express =\(=\frac{2880}{18}\)
Speed of Nilachala express = \(\frac{2880}{24}\)
Ratio of average speed = \(\frac{2880}{18}: \frac{2880}{24}\) = 24:18 = 4:3

2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4

Students can Download Maths Chapter 6 Application of Derivatives Ex 6.4 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 6.4

2nd PUC Maths Application of Derivatives NCERT Text Book Questions and Answers Ex 6.4

Using differentials, find the approximate value of each of the following up to 3 places of decimal.

(i) \(\sqrt{25.3}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.1

(ii) \(\sqrt{49.5}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.2

KSEEB Solutions

(iii) \(\sqrt{0.6}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.3

(iv)\((0.009)^{1 / 3}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.4

(v) \((0.009)^{1 / 3}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.5

(vi)\((15)^{1 / 4}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.6

(vii)\((26)^{1 / 3}\)
Answer:

2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.7
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.8

KSEEB Solutions

(viii) \((255)^{1 / 4}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.9

(ix)\((82)^{1 / 4}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.10

(x) \((401)^{1 / 2}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.11
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.12

(xi) \((0.0037)^{1 / 2}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.13

(xii) \((26.57)^{1 / 3}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.14

(xiii) \((81.5)^{1 / 4}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.15
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.16

(xiv)\((3.968)^{3 / 2}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.17

(xv) \((32.15)^{1 / 5}\)
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.18

KSEEB Solutions

Question 2.
Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2.
Answer:
f (x) = 4x2 + 5x + 2         .
f’ (x) = 8x + 5
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.19

Question 3.
Find the approximate value of f (5.001), where f (x) = x3 – 7x2 + 15.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.20

Question 4.
Find the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1%.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.21

Question 5.
Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.22
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.23

Question 6.
If the radius of a sphere is measured as 7 m with an error of 02 m, then find the approximate error in calculating its volume.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.24

Question 7.
If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.
Answer:
2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.4.25

Question 8.
If f(x) = 3x2 + 15x + 5, then the approximate value of f (3.02) is
(A) 47.66
(B) 57.66
(C) 67.66
(D) 77.66
Answer:
The correct answer is (D)

KSEEB Solutions

Question 9.
The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is
(A) 0.06 x3 m3
(B) 0.6 x3 m3
(C) 0.09 x3 m3
(D) 0.9 x3 m3
Answer:
The correct answer is (C)

 

2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.4

Students can Download Basic Maths Exercise 19.4 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 19.4

Part-B

Application to Economics

2nd PUC Basic Maths Application of Derivatives Ex 19.4 Two or Three Marks Questions and Answers

Question 1.
Find the Average cost and Marginal cost if the total cost function of an article is given by C = 5x2 + 2x + 3 (x = quantity).
Answers:
Given T.C = C(x) = 5x2 + 2x + 3; M.C = \(\frac{d}{d x}\) (T.C) = 10x + 2
A.C = \(\frac{C(x)}{x}=\frac{T \cdot C}{x}=\frac{5 x^{2}+2 x+3}{x}\) = 5x + 2 + \(\frac{3}{x}\)

Question 2.
The total cost of a commodity is given by C = -x2 + 5x + 7 (x = number of unit) and the price per unit is ₹12. Find the profit function.
Answer:
Given T.C = -x2 + 5x + 7; T.R = 12x
profit function = p(x) = T.R. – T.C = 12x -(-x2 + 5x + 7) = x2 + 7x + 7

Question 3.
For the demand function 2x – 5y = 7 (x = number of unit and y is the price / unit). Find the Total Revenue, Marginal Revenue and Average Revenue.
Answer:
Given 2x – 5y = 7 ⇒ y = \(\frac{2 x-7}{5}\) = price
T.R. = price y quantity = \(\frac{2 x-7}{5} \cdot x=\frac{2 x^{2}-7 x}{5}\)
M.R. = \(\frac{d}{d x}\left(\frac{2 x^{2}-7 x}{5}\right)=\frac{1}{5}(4 x-7)\)
A.R. = \(\frac{T \cdot R}{x}=\frac{2 x-7}{5}\)

KSEEB Solutions

Question 4.
The total cost of L ot output Q is given by C = 300Q – 10Q2 + \(\frac{\mathrm{Q}^{3}}{3}\) Find the output level at which the marginal costand the Average cost attain their respective minimum
Answer:
Given T.C = c(x) = 300Q – 10Q2 + \(\frac{\mathrm{Q}^{3}}{3}\)
2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.4 - 1
\(\frac{d}{d q}\) (M.C) = 0 ⇒ 2Q – 20 = 0
⇒ Q = 10
\(\frac{\mathrm{d}^{2}}{\mathrm{d} \mathrm{q}^{2}}\) (M.C) = 2 > 0 ⇒ MC is Maximum
\(\mathrm{A.C}=\frac{\mathrm{T.C}}{\mathrm{x}}=\frac{300 \mathrm{Q}-10 \mathrm{Q}^{2}+\frac{\mathrm{Q}^{3}}{3}}{\mathrm{Q}}=300-10 \mathrm{Q}+\frac{\mathrm{Q}^{2}}{3}\)
\(\frac{\mathrm{d}}{\mathrm{d} q}(\mathrm{A.C})=-10+\frac{2 \mathrm{Q}}{3}=0 \Rightarrow 2 \mathrm{Q}=30 \Rightarrow \mathrm{Q}=15\)
\(\frac{d^{2}(A C)}{d Q^{2}}=\frac{2}{3}>0\) ⇒ AC is Maximum.

Question 5.
The total cost of the production of a firm in given by the following function C = 0.7x +18 (x = output) Find
(i) the Total cost for an output 10 unit
(ii) the Average cost for an output 9 unit
(iii) the Marginal cost for an output of 6 unit
Answer:
Given C(x) = 0.7x + 18
(a) Total cost when x = 10
C(x) = (0.7 × 10) + 18 = 25

(b) Ave cost when x = 9
\(\mathrm{A.C}=\frac{\mathrm{T.C}}{\mathrm{x}}=\frac{0.7 \mathrm{x}+18}{\mathrm{x}}=\frac{0.7(9)+18}{9}=\frac{24.3}{9}=2.7\)

(C) Marginal cost when x = 6
MC = \(\frac{d}{d x}\) (TC) = \(\frac{d}{d x}\) (0.7x + 18) = 0.7.

Question 6.
If R = 250x + 45x2 – x3, (R = Total Revenue, x = no. of unit) what will be the Marginal Revenue if x = 25 unit and the Average Revenue if x = 10 unit.
Answer:
\(\text { A.R }=\frac{T R}{x}=\frac{250 x+45 x^{2}-x^{3}}{x}=250+45 x-x^{2}\)
When x = 10, A.R = 250 + 450 – 100 = 600
MR = \(\frac{d}{d x}\) (T.R) = \(\frac{d}{d x}\) (250x + 45x2 – x3)
= 250 – 90x – 3x2
At x = 25, M.R = 250 + 25(90) – 75 = 2425

KSEEB Solutions

Question 7.
The total cost function of a manufacturer is C = 5x2 + 500x + 50000. Find the output (x) when AC = MC.
Answer:
Given T.C. = 5x2 + 500x + 50000
A.C. = 5x + 500 + \(\frac{50000}{x}\)
M.C = \(\frac{d}{d x}\) (T.C) = \(\frac{d}{d x}\) (5x2 + 500x + 50,000)
= 10x + 500
2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.4 - 2
⇒ \(\frac{50000}{x}\) = 5x ⇒ x2 = 10,000
⇒ x = 100

Question 8.
If R = x \(\left(15-\frac{x}{30}\right)\) What is the Marginal Revenue function and what will be the Marginal Revenue if 100 units are produced?
Answer:
Given T.R = 15x – \(\frac{x^{2}}{30}\)
∴ M.R = 15 – \(\frac{2 x}{30}\)
At x = 100 M.R = \(15-\frac{200}{30}=\frac{450-200}{30}=\frac{25}{3}\)

Question 9.
The Total Revenue (R) and the total cost (C) function of a company are given by R(Q) = 300Q – Q2 and C(Q) = 20 + 4Q (Q = output). Find the equilibrium output. (Hint: equilibriums MR = MC).
Answer:
Given T.R = R(Q) = 300Q – Q2
M.R = \(\frac{d}{d q}\) (T.R) = 300 – 2Q
T.C = 20 + 4Q
M.C = \(\frac{d}{d q}\) (T.C) = 4
M.R = M.C ⇒ 300 – 2Q = 4
⇒ 296 = 2Q
⇒ Q = 148.

KSEEB Solutions

2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2

Students can Download Basic Maths Exercise 6.2 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 6 Mathematical Logic Ex 6.2

Part – A

2nd PUC Basic Maths Mathematical Logic Ex 6.2 One Mark Questions and Answers

Question 1.
Negate the following propositions.
(1) p ∨ ~ q
(ii) ~p → q
(iii) ~p ∧ ~q
(iv) p ∧ ~q
(v) ~p → ~q
Answers:
(i) ~ (p ∨~q)  ≡ ~p ∧~(~9) = ~p ∧ q ( ∵ ~(~q) = q )
(ii) ~(~p → q) ≡ ~p ∧ ~q
(iii) ~(~p ∧~q) ≡ ~(~p) ∨~(~q) = p ∨ q   (∵ ~(~p) = p)
(iv) ~ (p ∧ ~q) ≡ ~p ∨~(~q) = ~p ∨ q
(v) ~(~p →~q) ≡~p ∧~(~q) =~p ∧ q.

KSEEB Solutions

Question 2.
Negate the following.
(i) 4 is an even integer or 7 is a prime number.
(ii) He likes to run and he does not like to sit.
(iii) He likes Mathematics and he does not like Logic.
(iv) If 6 is a divisor of 120 then 486 is not dividiable by 6.
(v) If 2 triangles are similar then their ares are equal.
(vi) It is cold or it is raining,
Answers:
(i) Let p: 4 is even integer, q: 7 is a prime number given (p v q)
~(p ∨ q) = ~ p ∧ ~q
4 is not an even integer & 7 is not a prime number

(ii) Let p: He likes to run, q: He likes to sit
Given is ~ p ∧ ~q
∴ ~ (p ∧~ ) = ~p ∨ ~(~q) ≡ ~p ∨ q
He does not like to run or he likes to sit

(iii) Let p: He likes mathematices, q: He like logic
Given (p ∧ ~q)
~ (p ∧ ~q) ≡ ~p ∨ q
He does not like mathematics or he likes logic

(iv) Let p: 6 is a divisor or 120
q: 486 is divisible by 6
Given (p → ~ q)
~(p → ~ q) = p ∧ ~(~q)

(v) Let p: 2 triangles are similar.
q: Areas are equal
Given (p → q)
∴ ~(p → q) = (p ∧ ~q)
2 triangles are similar & areas are not equal.

(vi) p: It is cold
q: It is raining Given p ∨ a
~(p ∨ q) =~p∧~q
It is not cold & it is not raining

KSEEB Solutions

Part B

2nd PUC Basic Maths Mathematical Logic Ex 6.2 Two Marks Questions and Answers

Question 1.
Negate
(i) p → (q ∧ r)
(ii) q ∨ [~(p ∧ r)] |
(iii) (p → q)∧( q → p/q)
(iv) p→ (q ∧ ~r)
Answers:
(i) ~ [p → (p ∧ r)] = p ∧~(q ∧ r) ≡ p ∧ (~q ∨ ~r)
(ii) ~ [q ∨ (~ (p ∧ r)] ≡ ~q ∧ ~[~ (p ∧ r) ≡ ~q ∧ (p ∧ r)
(iii) ~[(p →q) ∧ (q → p)] ≡ ~ (p →q) ∨ ~(q → p)
≡ (p ∧ ~q) ∨ (q ∧ ~p)
(iv) ~[p → (q^~r)] = p ∧ ~(q ∧ ~r)
= p ∧ [(-q ∨ ~(~r)]
= p ∧(~q ∨ r)

KSEEB Solutions

Question 2.
Negate the following:
(1) If an integer is greater than 3 and less than 5 then it is a multiple of 5.
(ii) If ‘x’ is divisible by ‘y’ then it is divisible by ‘a’ and ‘b.
(iii) Weather is fine and my friends are not coming or we do not go to a movie.
(iv) If a triangle is equilateral then it’s sides are equal and angles are equal.
(v) 14 is a divisor of 48 and 28 is not divisible by 82.
Answers:
(i) Let p: An integer is greater than 3
q: An integer is less than 5
r: An integer is multiple of 5
Given[(p ∧ q) → r]
∴ ~ ((p ∧ q) → r] = (p ∧ q) ∧~r
An integer greater than 3 and less than 5 and not divisible by b

(ii) Let P:x is divisible by y
q: x is divisible by a
r: x is divisible by b
Given [p → (q ∧ r)]
∴ ~P (q ∧ r)] = p ∧~(q ∧ ~r)
= p ∧( ~ ( q ∨ ~r)
x is divisible by y and x is not divisible by a or not a multiple of 5

(iii) Let p: Weather is fine
q: Friends are coming
r: we go to a movie
Given[p∧~q ∨~r)]
~(p ∧ (~q ∨ ~r)) ≡ ~p ∨ ~(~ q ∨ ~r) :
≡ ~p ∨ (q ∧ r)

(iv) Let p = a triangle is equilateral
q: Sides are equal
r: angles are equal
Given p → (q ∧ r)
∴ ~(p → (~q ∨ ~r)) ≡ ~p ∧ ~q ∨ ~r)
A triangle is equilateral and sides are not.
equal or angles are not equal.
Let p: 14 is a divisor of 48
q: 28 is divisible by 82
Given p ∧ ~q
∴  ~(p ∧ ~q) ≡ ~p ∨ q
14 is not a divisor of 48 or 28 is divisible by 82

KSEEB Solutions

Question 3.
Determine whether the following propositions is a Tautology or a contradiction or neither.
(i) (p ∧ q) ∧ ~p
(ii) [~p ∧ (p ∨ q)]
(iii) (p ∧ q) → (p ∨ q)
(iv) (p ∧ q) →p
(v) ~ p ∧ ~q
Answers:
(i) (p ∧ q) ∧~p
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 1
From last column we conculde that it is a contradiction

(ii) [~p ∧ (p ∨ q)]
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 2
From last column we conclude it is neither tautology nor a contradiction

(iii) (p ∧ q) → (p ∨ q)
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 3
From last column we conclude it is a tautology

(iv) (p ∧ q) → p
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 4
From last column we conclude that it is a tautology

KSEEB Solutions

(v) ~ p ∧ ~q
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 5
It is neither tautology nor contradiction

Part C

2nd PUC Basic Maths Mathematical Logic Ex 6.2 Five Marks Questions and Answers

Question 1.
Check whether the following propositions is a Tautology or a contradiction.
(i) (p ∧ ~q) → (p ∧ q)
(ii) [~p ∧ (p ∨ q)] → q
(iii) (p →q) ↔ (~p →~q)
(iv) [~(p →~q)] ∨ (~ p ↔ q)
(v) (~p ∨ q) ↔ (p ∨~9)
Answers:
(i) (p ∧~q) → (p∧q)
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 6
It is neither Tauthology nor contradiction

(ii) [~p ∧ (p ∨ q)] → q
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 7
F From last column we conclude it is a tautology

(iii) (p →q) ↔ (~p →~q)
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 8
F It is neither a tautology nor contradiction

(iv) [~(p →~q)] ∨ (~ p ↔ q)
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 9
It is neither tautology nor contradiction

(v) (~p ∨ q) ↔ (p ∨~9)
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 10
It is neither tautology nor contradiction

Question 2.
Show that (p → q) + (~q→~p) is a Tautology.
Answer:
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 11
It is a tautology

KSEEB Solutions

Question 3.
Show that ~(p ∨ q) →(~p∧~q) is a Tautology.
Answer:
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 12
It is a tautology

Question 4.
Prove : [(p →q) ∧ (q →r)] → (P→r) is a Tautology.
Answer:
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 13
From last column we conclude that it is a tautology

Question 5.
Prove that (p ∨ q) ∧ (~p ∧~q) is a contradiction.
Answer:
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 14
From last column we conculde that it is a contradiction

KSEEB Solutions

Question 6.
Show that [(~p ^ q) ^ (q ^ r) ^ (~q)] is a contradiction.
Answer:
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 15
From last column we conclude that it is a contradiction

Question 7.
Examine whether the following are logically equivalent:
(i) p ↔ q and (p → q) ∧ (q → p)
(ii) p → (q → r) and (p → q) → r
(iii) (p ∧ ~q) ∨ q and p ∨ q.
(iv) p ↔ q and (~ p ∨ q) ∧ (~q ∨ p)
(v) p ∧ q and ~(p →~q)
(vi) ~ (p ↔ q ) and (p ∧ ~q) ∨ (q ∧~p)
(vii) p∨ (q ∧ r) and (p ∨ q) ∧ (p ∨ r)

(i)
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 16
From 3rd & 6th column we conclude that
p ↔ q (p ↔ q) ∧ (q ↔ p)

(ii) p → (q → r) and (p → q) → r
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 17
From last two columns we conclude that p → (q → r) and (p → q) → r are not logically equivalent

(iii) (p ∧ ~q) ∨ q and p ∨ q.
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 18
From last two columns we conclude (p ∧ ~q) ∨ q p ∨ q.

(iv) p↔ q and (~ p ∨ q) ∧ ( ~q ∨ p)
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 19
From 3 rd and 8 th columns we conclude that p ↔ q ≡ (~ p ∨ q) ∧ (~q ∨ p)

(v) P ∧ q and (p → ~q)
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 20
Column 3 and column 6 are identical :
∴ They are logically equivalent

(vi) ~ (p ↔ q ) and (p ∧ ~q) ∨ (q ∧~p)
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 21

4th & th columns are identical ∴ they are logically equivalent

KSEEB Solutions

(vii) p ∨ (q ∧ r) and (p ∨ q)∧ (p ∨ r)
2nd PUC Basic Maths Question Bank Chapter 6 Mathematical Logic Ex 6.2 - 22

5th column & 8th columns are identical
∴ p∨ (q∧r) = (p ∨ q)∧ (p ∨ r)

2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.6

Students can Download Basic Maths Exercise 20.6 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 20 Indefinite Integrals Ex 20.6

Part-A

Integration by Parts

2nd PUC Basic Maths Indefinite Integrals Ex 20.6 Three Marks Questions and Answers

Question 1.
∫2log x dx
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.6 - 1

Question 2.
∫x3 log x
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.6 - 2

KSEEB Solutions

Question 3.
\(\int \frac{1}{x^{2}} \log x d x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.6 - 3

Question 4.
∫xe3x+5 dx.
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.6 - 4

Question 5.
∫x sin x dx.
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.6 - 5
= x . (-cosx) – ∫-cos x .dx
= -x cos x + sin x + c

Question 6.
∫x cosec2 dx.
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.6 - 6
= x(-cot x) – ∫-cot x . dx
= -x cot x + log sin x + C

KSEEB Solutions

Question 7.
∫x2cos x dx.
Answer:
∫x2cos x dx = x2 – ∫sinx(2x) dx
= x2 -2∫xsinx dx
= x2 -2[x(-cos x) – ∫-cos x dx]
= x2sin x + cos x – 2 sinx + c

Question 8.
∫x sin(5x + 7)dx.
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.6 - 7

2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise

Students can Download Maths Chapter 5 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 5 Continuity and Differentiability Miscellaneous Exercise

2nd PUC Maths Chapter 5 Continuity and Differentiability NCERT Text Book Questions and Answers Miscellaneous Exercise

Differentiate w.r.t. x the function in Exercise 1 to 11.

Question 1.
(3x2 – 9x + 5)9
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 1

Question 2.
sin3 x + cos6 x
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 2

KSEEB Solutions

Question 3.
(5x)3cos2x
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 3
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 4

Question 4.
\(\sin ^{-1}(x \sqrt{x}), 0 \leq x \leq 1\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 5

Question 5.
\(\frac{\cos ^{-1}\left(\frac{x}{2}\right)}{\sqrt{2 x+7}},-2<x<2\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 6
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 7

KSEEB Solutions

Question 6.
\(\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right], 0<x<\frac{\pi}{2}\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 8

Question 7.
(log x)log x ,x > 1
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 9

Question 8.
cos (a cos x + b sin x), for some constant a and b.
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 10

Question 9.
\((\sin x-\cos x)^{(\sin x \cdot \cos x)}, \frac{\pi}{4}<x<\frac{3 \pi}{4}\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 11

KSEEB Solutions

Question 10.
xx + xn + ax + an, for some fixed a > x and x > 0
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 12

Question 11.
\(x^{x^{2}-3}+(x-3)^{x^{2}}, \text { for } x>3\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 13
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 14

Question 12.
Find
\(\frac{d y}{d x}, \text { if } y=12(1-\cos t), x=10(t-\sin t),-\frac{\pi}{2}<t<\frac{\pi}{2}\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 15

KSEEB Solutions

Question 13.
Find
\(\frac{d y}{d x}, \text { if } y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}},-1 \leq x \leq 1\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 16

Question 14.
If \(x \sqrt{1+y}+y \sqrt{1+x}=0 \text { for },-1<x<1, \text { prove that } \frac{d y}{d x}=-\frac{1}{(1+x)^{2}}\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 17
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 18

Question 15.
If (x – a)2 + (y – b)2 = c2, for some c > 0, prove that
\(\frac{\left[1+\left(\frac{\mathrm{dy}}{\mathrm{d} \mathbf{x}}\right)^{2}\right]^{\frac{3}{2}}}{\frac{\mathbf{d}^{2} \mathbf{y}}{\mathbf{d x}^{2}}}\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 19

KSEEB Solutions

Question 16.
If cos y = x cos (a + y), with cos a ≠ ± 1, prove that
\(\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 20
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 21

Question 17.
If x = a (cos t + t sin t) and y = a (sin t -1 cos t), find \(\frac{d^{2} y}{d x^{2}}\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 22

Question 18.
If f (x) = |x|3, show that fn (x) exists for all real x and find it.
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 23

KSEEB Solutions

Question 19.
Using mathematical induction prove that \(\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\) for all positive integers n.
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 24

Question 20.
Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.
Answer:
sin (A + B) = sin A cos B + cos A sin B
Diff : w:r.to B
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 25
cos(A +b) = cos A cos B – sin A sin B
hence cosine formulae

Question 21.
Does there exist a function which is continuous everywhere but not differentiable at exactly two points ? Justify your answer.
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 26
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 27

KSEEB Solutions

Question 22.
It
\(\left|\begin{array}{ccc}{\mathbf{f}(\mathbf{x})} & {\mathbf{g}(\mathbf{x})} & {\mathbf{h}(\mathbf{x})} \\{\mathbf{1}} & {\mathbf{m}} & {\mathbf{n}} \\{\mathbf{a}} & {\mathbf{b}} & {\mathbf{c}}\end{array}\right|, \text { prove that } \quad \frac{\mathbf{d y}}{\mathbf{d x}}=\left|\begin{array}{ccc}{\mathbf{f}^{\prime}(\mathbf{x})} & {\mathbf{g}^{\prime}(\mathbf{x})} & {\mathbf{h}^{\prime}(\mathbf{x})} \\{1} & {\mathbf{m}} & {\mathbf{n}} \\{\mathbf{a}} & {\mathbf{b}} & {\mathbf{c}}\end{array}\right|\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 28

Question 23.
\(y=e^{a \cos ^{-1} x},-1 \leq x \leq 1\)\(\text { show that }\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 29

2nd PUC Maths Chapter 5 Continuity and Differentiability Miscellaneous Exercise Additional Questions and Answer

Question 1.
\(\text { If } x=a\left(\cos \theta+\log \tan \frac{\theta}{2}\right), y=a \sin \theta, \text { find } \frac{d y}{d x} \text { at } \theta=\frac{\pi}{4}\) (DB 2011)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 30

Question 2.
\(f(x)=\left\{\begin{aligned}\frac{\sin (a+1) x+\sin x}{x}, & x<0 \\c &, x=0 \\\frac{\sqrt{x+b x^{2}}-\sqrt{x}}{b x^{3 / 2}}, & x>0\end{aligned}\right.\)
If f(x) is continuous at x = 0 ,find a,b,c. (CBSE 2011, 2008, 2005)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 31

2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 32

KSEEB Solutions

Question 3.
\(\text { If } y=\cos ^{-1}\left\{\frac{3 x+4 \sqrt{1-x^{2}}}{5}\right\} \text { find } \frac{d y}{d x}\) (CBSE 2010)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 33

Question 4.
\(\text { If } x^{y}=e^{x \cdot y}, \text { show that } \frac{d y}{d x}=\frac{\log x}{\{\log (x e)\}^{2}}\)(CBSE 2009)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 34

Question 5.
Differentiate
\(\sin ^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right) \text { w:r:to }\)
\(\cos ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 35
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 36

KSEEB Solutions

Question 6.
\(\text { If } y=\sin ^{-1}\left[\frac{1-\sqrt{x}}{1+\sqrt{x}}\right]+\sec ^{-1}\left[\cfrac{1+\sqrt{x}}{1-\sqrt{x}}\right], \text { find } \frac{d y}{d x}\)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 37

Question 7.
Find the derivative of |x|
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 38

Question 8.
Diff: log10 x w : r : to logx10
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 39

Question 9.
\(\text { Let } f(x)=\left\{\begin{array}{ll}{\frac{3|x|+4 \tan x}{x},} & {x \neq 0} \\{k} & {x=0}\end{array}\right.\) is continuos at x = o,find K
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 40

Question 10.
\(\text { If } y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+\ldots \ldots \infty}}} \text { prove }\) that \((2 y-1) \frac{d y}{d x}=\cos x\)(Kerala CET)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 41

Question 11.
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 42
\(\text { prove that, } \frac{d y}{d x}=\cfrac{(1+y) \cos x+y \sin x}{1+2 y+\cos x-\sin x}\) (Kerala CET)
Answer:
2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 43

KSEEB Solutions

Question 12.
\(y=(\sqrt{x})^{(\sqrt{x})^{(\sqrt{x}}} \text { show that } \frac{d y}{d x}=\frac{y^{2}}{x(2-y \log x)}\)  (Kerala CET)
Answer:

2nd PUC Maths Question Bank Chapter 5 Continuity and Differentiability Miscellaneous Exercise 44

2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5

Students can Download Basic Maths Exercise 20.5 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 20 Indefinite Integrals Ex 20.5

Part-A

Integration by partial fractions

2nd PUC Basic Maths Indefinite Integrals Ex 20.5 Five Marks Questions and Answers

Question 1.
\(\int \frac{4 x+5}{(x-1)(x+2)} d x\)
Answer:
Let
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5 - 8
4x + 5 = A(x + 2) + B(x – 1)
put x = 1, 4 + 5 = A(1 + 2) + B(0)
9 = 3A ⇒ A = 3

put x = -2 -8 + 5 = A(0) + B(-2 -1)
-3 = -3B ⇒ B = 1

∴\(\int \frac{4 x+5}{(x-1)(x+2)} d x=\int\left(\frac{3}{x-1}+\frac{1}{x+2}\right) d x\) =3log(x – 1) + log(x + 2) +c

KSEEB Solutions

Question 2.
\(\int \frac{3 x+2}{(2 x+3)(3 x-1)} d x=\int \frac{A}{2 x+3}+\frac{B}{3 x-1} \cdot d x\)
Answer:
\(\int \frac{3 x+2}{(2 x+3)(3 x-1)} d x=\int \frac{A}{2 x+3}+\frac{B}{3 x-1} \cdot d x\)
Let 3x + 2 = A(3x – 1) + B(2x + 3)
comparing coefficients of x both sides & Constant terms
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5 - 1

Question 3.
\(\int \frac{1}{x(x+1)(x+2)} d x\)
Answer:
Let
\(\frac{1}{x(x+1)(x+2)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}\) …..(1)
1 = A ( x + 1) (x + 2) +B(x) (x + 2) + c × x(x + 1)
put x = 0, 1 = A(1)(2) ⇒ 1 = 2A ⇒ A = \(\frac { 1 }{ 2 }\)
put x = -1, 1 = A(0) + B(-1) (-1 + 2) + c(0) ⇒ 1 = -B ⇒ B = -1
put x = -2 1 =A(0) + B(0) + c(-2), (-2 + 1) ⇒ 1 = 2c ⇒ c = \(\frac { 1 }{ 2 }\)
∴\(\int \frac{d x}{x(x+1)(x+2)}=\int \frac{\frac{1}{2}}{x}+\frac{-1}{x+1}+\frac{\frac{1}{2}}{x+2} \cdot d x\)
= \(\frac { 1 }{ 2 }\)log 2 – log(x + 1) + \(\frac { 1 }{ 2 }\)log(x + 2) +c

KSEEB Solutions

Question 4.
\(\int \frac{5 x+7}{(x-2)^{2}(x+3)} d x\)
Answer:
Let
\(\frac{5 x+7}{(x-2)^{2}(x+3)}=\frac{A}{x-2}+\frac{B}{(x-2)^{2}}+\frac{C}{x+3}\)
5x + 7 = A(x – 2) (x + 3) + B(x +3) + (x – 2)2
put x = 2 10 + 7 = A(0) + B(2 + 3) + c(0)2 = 17 = 5B ⇒ B = \(\frac { 17 }{ 5 }\)
put x = -3 -15 + 7 = A(0) + B(0) + c(-5)2 = -8 = 25c ⇒ c = \(\frac { 17 }{ 5 }\)
Comparing the coefficient of ×2 both sides we get 0 = A + c ⇒ A = -c = \(\frac { 8 }{ 25 }\)
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5 - 2

Question 5.
\(\int \frac{5}{\left(x^{2}+6 x+9\right)(x-3)} d x\)
Answer:
\(\int \frac{5}{\left(x^{2}+6 x+9\right)(x-3)} d x=\int \frac{5 d x}{(x+3)^{2}(x-3)}=\int \frac{A}{x+3}+\frac{B}{(x+3)^{2}}+\frac{C}{(x-3)} d x\)
Let
\(\frac{5}{(x+3)^{2}(x-3)}=\frac{A}{x+3}+\frac{B}{(x+3)^{2}}+\frac{C}{x-3}\) ….(1)
5 = A(x + 3) (x – 3) + B(x – 3) + c(x + 3)2
put x = 3, 5 = A(0) + B(0) + C(3 +3)2 = 5 = 36C ⇒ c = \(\frac { 5 }{ 35 }\)
put x = -3, 5 = (A) (0) + B(-3-3) + C(0)2
5 = -6B ⇒C = \(\frac { 5 }{ 36 }\)
put x = -3, 5 = (A) (0) + B(-3-3) + C(0)2
5 = -6B ⇒ B = \(\frac { 5 }{ -6 }\)
Comparing coefficients of ×2 both sides we get
0 = A + C ⇒ A = -C = \(\frac { -5 }{ 36 }\)
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5 - 3

KSEEB Solutions

Question 6.
\(\int \frac{2 x-1}{\left(x^{2}-4\right)(x+1)} d x\)
Answer:
Let
\(\int \frac{(2 x-1) d x}{(x-2)(x+2)(x+1)}=\int \frac{A}{x-2}+\frac{B}{x+2}+\frac{C}{x+1} \cdot d x\) …(1)
2x – 1 = A(x + 2) (x + 1) +B(x – 2) (x + 1) + c(x – 2)(x + 2)
put x = 2 4 – 1 = A(4) (3) + B(0) + C(0)
3 = 12A ⇒ A = \(\frac { 1 }{ 4 }\)
put x = 2 -4 -1 = A(0) + B(-2) (-2 +1) + C(0)
-5 = 4B ⇒ B = \(\frac { -5 }{ 36 }\)
put x = -1 -2 -1 = A(0) + B(0) + C(-3) 1
-3 = -3c ⇒ c = 1
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5 - 4

Question 7.
\(\int \frac{3 e^{x}}{e^{2 x}+5 e^{x}+6} d x\)
Answer:
\(\int \frac{3 e^{x}}{\left(e^{x}\right)^{2}+5 e^{x}+6} d x=3 \int \frac{t}{t^{2}+5 t+6} d t\) put ex = t
\(=1 \int \frac{3 t d t}{(t+2)(t+3)}=\int \frac{A}{t+2}+\frac{B}{t+3} d t\)
Let 36 = A(t + 3) + B(t + 2)
put t = -2 -6 = A(1) ⇒ A = -6
put t = -3, -9 = A(0) + B(-1) ⇒ B = 9
∴ \(\int \frac{3 e^{x}}{\left(e^{x}\right)^{2}+5 e^{x}+6}=\int \frac{-6}{t+2}+\frac{9}{t+3} \cdot d t\)
= -6 log (t + 2) + 9 log (t + 3) + C = -6 log(ex + 3) + 9 log (ex + 3) + c

KSEEB Solutions

Question 8.
\(\int \frac{6}{x(2 \log x)^{2}+7 \log x+5} d x\)
Answer:
\(\int \frac{6 / x}{(2 \log x)^{2}+7 \log x+5} d x=\int \frac{6 \cdot d t}{2 t^{2}+7 t+5}\)
put log x = t
\(\int \frac{6 d t}{2 t^{2}+7 t+5}=\int \frac{6 d t}{(2 t+5)(t+1)}=\int \frac{A}{2 t+5}+\frac{B}{t+1} \cdot d t\)
Let 6 = A(t + 1) + B(2t + 5)
put t = -1, 6 = B(3) ⇒ B = 2
Equate the coefficient of t both sides
0 = A + 2B
A = -2B = -4
∴ \(\int \frac{6 d x}{x(2 \log x)^{2}+7 \log x+5}=\int \frac{-4}{2 t+5}+\frac{2}{t+1} \cdot d t=\frac{-4 \log (2 t+5)}{2}+2 \log (t+1)\)
= – 2log(2 log x + 5) + 2log(logx + 1) + C

Question 9.
\(\int \frac{3 x^{2}+2 x+3}{(x+1)(3 x+2)} d x\)
Answer:
\(\int \frac{3 x^{2}+2 x+3}{(x+1)(3 x+2)} d x\) this is an improper fraction.
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5 - 5

KSEEB Solutions

Question 10.
\(\int \frac{x^{2}+3 x-2}{x^{2}-4 x-12} d x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5 - 6
Let 7x + 10 = A(x + 2) + B(x – 6)
put x = 6 42 + 10 = A(6 + 2) + B(0); 52 = 8A ⇒ A = \(\frac{B}{2}\)
put x = -2 -14 + 10 = A(0) + B(-2 -6)
-4 = -8B ⇒ B = \(\frac{-4}{-8}=\frac{1}{2}\)
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.5 - 7

2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.4

Students can Download Basic Maths Exercise 20.4 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 20 Indefinite Integrals Ex 20.4

Part – A

2nd PUC Basic Maths Indefinite Integrals Ex 20.4 Two Marks Questions and Answers

Question 1.
∫2x(x2 + 2)2/3 dx
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.4 - 1

Question 2.
\(\int \frac{2}{x}(\log x)^{2} d x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.4 - 2

KSEEB Solutions

Question 3.
\(\int e^{x} \sqrt{e^{x}+1} d x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.4 - 3

Question 4.
∫cos2x sin x dx.
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.4 - 4

Question 5.
∫cosec4x cot x dx.
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.4 - 5

Part-B

2nd PUC Basic Maths Indefinite Integrals Ex 20.4 Three Marks Questions and Answers

Question 1.
∫(6x + 5)(3x2 + 5x – 4)5/3dx
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.4 - 6

KSEEB Solutions

Question 2.
\(\int \frac{3}{x(2+3 \log x)^{2 / 3}} d x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.4 - 7

Question 3.
\(\int \frac{1+e^{x}}{\left(x+e^{x}\right)^{5}} d x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.4 - 8

Question 4.
\(\int \frac{1}{2 \sqrt{x}(\sqrt{x}-1)^{2 / 5}} d x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.4 - 9

Question 5.
\(\int \frac{5^{x} \log 5}{\left(5^{x}+3\right)^{7}} d x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.4 - 10

KSEEB Solutions

Question 6.
\(\int \csc ^{2} x \sqrt{1+\cot x} d x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.4 - 11

Question 7.
\(\int \frac{\sin 2 x}{\left(1-\cos ^{2} x\right)^{3}} d x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.4 - 12

Question 8.
\(\int \frac{\left(\csc ^{2} x-\csc x \cot x\right)}{(\csc x-\cot x)^{1 / 4}} d x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 20 Indefinite Integrals Ex 20.4 - 13

error: Content is protected !!