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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) = x^{4} – 8x^{3} + 22x^{2} – 24x + 5 at x = 0, -2

(ii) f(x) = 4x^{3} – 15x^{2} + 12x – 2 at x = 1,-1

(iii) f(x) = (x – 1)(x – 2)^{2} at x = 1,3.

Answer:

(i) Given

f(x) = x^{4} – 8x^{3} + 22x^{2} – 24x + 5 at x = 0, -2

f'(x) = 4x^{3} – 24x^{2} + 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) = 4x^{3} – 15x^{2} + 12x – 2 at x = 1,-1

f'(x) = 12x^{2} – 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) = 2x^{3} – 15x^{2} – 84x + 7

(ii) f(x) = x^{4} – 2x^{3} + 1

(iii) f(x) = x^{3} – 3x^{2} + 3x – 100

(iv) f(x) = 2x^{2} – 96x + 5

(v) f(x) = 10 – 6x – 2x^{2}

(vi) f(x) = 2x^{3} + 9x^{2} + 12x + 20

Answer:

(i) Given

f(x) = 2x^{3} – 15x^{2} – 84x + 7

f'(x) = 6x^{2} – 30x – 84

= 6(x^{2} – 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) = x^{4} – 2x^{3} + 1

f'(x) = 4x^{3} – 6x^{2}

= 2x^{2} (2x – 3) Here 2x^{2} > 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) = x^{3} – 3x^{2} + 3x – 100

f'(x) = 3x^{2} – 6x + 3

= 3(x^{2} -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) = 2x^{2} – 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 – 2x^{2}

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) = 2x^{3} + 9x^{2} + 12x + 20

f'(x) = 6x^{2} + 18x + 12

= 6(x^{2} + 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.