2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.2

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

a

KSEEB Solutions

(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)

KSEEB Solutions

(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

KSEEB Solutions

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.

a