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

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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

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.

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