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

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