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## Karnataka 2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.3

Part- A

**2nd PUC Basic Maths Application of Derivatives Ex 19.3 Two or Three Marks Questions with Answers.**

Question 1.

Find the maximum and minimum value of the following function.

(i) f(x) = x^{3} – 3x

(ii) f(x) = x^{3} – 6x^{2} + 9x + 15(0 ≤ x ≤ 6)

(iii) f(x) = x^{4} – 62x^{2} + 120x + 9

(iv) f(x) = 2x^{3} – 3x^{2} – 12x + 12

(v) f(x) = 2x^{3} – 3x^{2} – 36x + 10

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

(vii) f(x) = 2x^{3} – 15x^{2} + 36x + 10

(viii) f(x) = 2x^{3} – 21x^{2} + 36x – 20

(ix) f(x) = 2x^{3} – 15x^{2} + 36x + 10

(x) f(x) = 12x^{5} – 45x^{4} + 40x^{3} + 6

Answer:

(i) Given f(x) = x^{3} – 3x …..(1)

f'(x) = 3x^{2} – 3 = 3(x^{2} – 1) = 3(x – 1) (x + 1) …..(2)

f'(x) = 0 ⇒ x = ±1

f”(x) = 6x – (3)

At x = 1, f”(1) = 6 > 0, f(x) is minimum at x = 1

& minimum value is f(1) = 1 – 3 = -2

At x = -1, f”(-1) = -6 < 0 f(x) is maximum at x = -1

Maximum value is f(-1) = -1 + 3 = 2

(ii) f(x) = x^{3} – 6x^{2 }+ 9x + 15 (0 ≤ x ≤ 6) – (1)

f'(x) = 3x^{2} – 12x + 9

= 3(x^{2} – 4x + 3) = 3 (x – 3) (x + 1] = 0

f “(x) = 6x – 12

f'(x) = 0 ⇒ x = 1 or 3

At x = 1 f”(1) = 6 – 12 = -6 < 0 the function is maximum at x = 1.

And maximum value is f (1) = 1 – 6 + 9 + 15 = 19

At x = 3, f”(3) = 6 × 3 – 12 = 18 -12 = 6 > 0, f is minimum at x = 3

And minimum value is f(3) = 3^{3} – 6.3^{2} + 9.3 + 15 = 27 – 54 + 27 + 15 = 15.

(iii) Given f(x) = x^{4} – 62x^{2} + 120x + 9 …..(1)

f”(x) = 4x^{3} – 124x + 120 ….(2)

f”(x) = 12x^{2} – 124 …..(3)

for a function to be maximum or minimum of f'(x) = 0.

⇒ x^{3} – 31x + 30 = 0 here x = 1 is a root

⇒ x^{2} + x – 30 = 0 ⇒ (x + 6)(x – 5) = 0 ⇒ x = 5 – 6

Put x = 1 in (3] we get f”(x) = (12 – 124) < 0

f(x) att-ains maximum at = 1 & maximum at x = 1 & max value is

f(1) = 1 – 62 + 120 + 9 = 68.

At x = 5, f”(5) = 12(5)^{2} 124 = 300 – 124 > 0 f(x) attains

minimum at x = 5, & minimum value is f(5) = 625 – 1550 + 600 + 9 = – 316

At x = – 6, f “(-6) = 12 (-6)^{2} – 124 = 432 – 124 > 0

f(x) attains minimum at x = -6 & minimum value is

f (-6) = (-6)^{4} – 62(-6)^{2} + 120 (-6) + 9

= 1296 – 2232 – 720 + 9 = -1647.

(iv) f(x) = 2x^{3} – 3x^{2} – 12x + 12 ….(1)

f'(x) = 6x^{2} – 6x – 12 = 6 (x^{2} – x – 2) …(2)

f'(x) = 12x – 6 ….(3)

for a function to be maximum or minimum f ‘(x) = 0 ⇒ (x – 2) (x + 1) = 0

⇒ x = 2 or – 1

Put x = (-1) in (3) we get f “(-1) = -12 – 6 = -18 < 0

f(x) att-ains maximum at x = -1 & maximum value is

f(-1) = 2 (-1)3 – 3 (-1)2 – 12(-1) + 12 = 19

Put x = 2 in(3) f”(2) = 24 – 6 = 18 > 0

f(x) attains minimum at x = 2 & minimum value is

f(2) = 2 (2)^{3} – 3(2)^{2} – 12(2) + 12 = 6- ^-24+ >^=8

(v) Given f(x) = 2x^{3} – 3x^{2} – 36x + 10 ….(1)

f ‘(x) = 6x^{2} – 6x – 36 = 6 (x^{2} – x – 6) …..(2)

f “(x) = 12x – 6 ……(3)

For a function to be maximum of minimum f ‘(x) = 0

⇒ (x^{2} – x – 6) = 0 => (x – 3) (x + 2) = 0

⇒ x = 3 or – 2

Put x = 3 in equation (3) we get

f “(3) = 36 – 6 = 30 >0

⇒ f(x) attains minimum at x = 3

Minimum value is f(3) = 2(3)^{3} – 3 (3^{2}) – 36 (3) + 10

f(3) = 54 – 27 – 108 + 10 = -71

Put x = -2 in equation (3) we get

f”(-2) = -24 – 6 = – 30 < 0 ⇒ f (x) attains maximum at x = -2

Maximum value is f(-2) = 2(-2)^{3} -3(-2)^{2} – 36 (-2] + 10

f(-2) -16-12 + 72 + 10 = 54

(vi) Given f(x) = 9x^{2} + 12x + 2 ….(1)

f'(x) = 18x + 12 …(2)

f”(x) = 18 > 0 ……(3)

⇒ f(x) attains minimum

f'(x) = 0 ⇒ 18x+ 12 = 0 ⇒ x = \(-\frac{2}{3}\)

& f” \(\left(-\frac{2}{3}\right)\) 18 > 0 ⇒ f(x) is minimum & the minimum value is

f\(\left(-\frac{2}{3}\right)\) = 9\(\left(\frac{4}{9}\right)\) + 12 \(\left(-\frac{2}{3}\right)\) + 2

= 4 – 8 + 2 = -2

(vii) f(x) = 2x^{3} – 15x^{2} + 36x + 10 …….(1)

f ‘(x) = 6x^{2} – 30x + 36 = 6 (x^{2} – 5x + 6) ……..(2)

f”(x) = 12x – 30 …..(3)

f'(x) = 0 ⇒ x^{2} -5x + 6 = 0 ⇒ (x – 3)(x – 2) = 0 ⇒ x = 3 or 2

when x = 3 f “(x) = 12x – 30

f “(3) = 36 – 30 = 6 > 0 ⇒ f(x) has minimum

Minimum value is f(3) = 2(3)^{3} – 15(3)^{2} + 36(3) + 10

f(3) = 54 – 135 + 108 + 10 = 37

when x = 2, f”(2) = 24 – 30 = -6 < 0 ⇒ f(x) has maximum

maximum value is f(2) = 2(2)^{3} – 15(2) + 36(2) + 10

f(2) = 16 – 60 + 72 + 10 = 38.

(viii) Given f(x) = 2x^{3} – 21x^{2} + 36x – 20 ….. (1)

f'(x) = 6x^{2} – 42x + 36 …… (2)

= 6(x^{2} – 7x + 6)

f'(x) = 6 (x – 1) (x – 6) = 0 ⇒ x = 1 or 6

f”(x) = 12x – 42 … (3)

when x = l,f “(1) = 12 – 42 = -30 < 0 ⇒ f(x) is maximum

maximum value is f(1) = 2 – 21 + 36 – 20 = -3

when x = 6, f “(6) = 72 – 42 = 30 > 0 ⇒ f(x) is minimum

minimum value is f(6) = 2(6)3^{3} – 21 (6)^{2} + 36(6) – 20

f(6) = 432 – 756 + 216 – 20 = -128

(ix) Given f(x) = 12x^{5} – 45x^{4} + 40 x^{3} + 6 ….(1)

f'(x) = 60x^{4} – 180x^{3} + 120x^{2} ….(2)

= 60x^{2} (x^{2} – 3x + 2)

= 60x^{2} (x – 1) (x – 2)

f ‘(x) = 0 ⇒ 60x^{2} (x – 1) (x – 2) = 0 ⇒ x = 0, 1, 2

f “(x) = 60 (4x^{3} – 6x + 4x) ……. (3)

when x = 0, f”(x) = 0 ⇒ f(x) has neither maximum nor minimum

when x = 1, f”(x) = -1 < 0 ⇒ f(x) has a maximum & maximum value is

f(1) = 12 – 45 + 40 + 6 = 13

When x =2 f”(x) = 4 > 0 ∴ f(x) has a minimum

minimum value is f(2) = 12(32) – 45(16) + 40(8) + 6

f(2) = 384 – 720 + 320 + 6 = -10.

Question 2.

The sum of two natural numbers is 48. Find the numbers when their product is maximum.

Answer:

Let the two numbers be x & y.

Given x + y = 48 & product: = xy where y 48 – x.

Let p = xy = x (48 – x) = 48x – x^{2}.

\(\frac{d p}{d x}\) = 48 – 2x

\(\frac{d p}{d x}\) = 0 ⇒ 48 – 2x = 0 x = 24

\(\frac{\mathrm{d}^{2} \mathrm{p}}{\mathrm{d} \mathrm{x}^{2}}\) = -2 < 0 ⇒ product is maximum

x = 24 ⇒ y = 48 – 24 = 24

⇒ the two numbers are 24, 24.

Question 3.

Find two positive numbers whose sum is 14 and the sum of whose square is minimum.

Answer:

Let the two numbers be x and y

Given x + y = 14 & S = x^{2} + y^{2} where y = 14 – x

∴ S = x^{2} + (14 – x)^{2} = x^{2} + 14^{2} + x^{2} – 28x = 2x^{2} – 28x + 14^{2}

\(\frac{d s}{d x}\) = 4x – 28 → (1) \(\frac{d s}{d x}\) = 0 ⇒ 4x – 28 = 0 ⇒ x = 7

\(\frac{\mathrm{d}^{2} \mathrm{s}}{\mathrm{d} \mathrm{x}^{2}}\) = 4 > 0, sum is minimum.

∴ y = 14 – x = 14 – 7 = 7

∴ the two positive number are 7 & 7.

Question 4.

Find two positive numbers whose sum is 30 and the sum of their cubes is minimum.

Answer:

Let the two numbers be x & y.

Given x + y = 30 & S = x^{3} + y^{3} where y = 30 – x

S = x^{3} + (30 – x)^{3} = x^{3} + (30)^{3} – x^{3} – 2700x + 90x^{2}

\(\frac{d s}{d x}\) = – 2700 + 180x

\(\frac{d s}{d x}\) = 0 ⇒ x = \(\frac{2700}{180}\) = 15

\(\frac{\mathrm{d}^{2} \mathrm{s}}{\mathrm{d} \mathrm{x}^{2}}\) = 180 > 0 ⇒ sum of cubes is minimum & y = 30 – 15 = 15

∴ two positive number are 15 & 15.

Question 5.

The product of two natural numbers is 64. Find the numbers is their sum is minimum

Answer:

Let the two numbers be x & y

Given xy = 64 ⇒ y = \(\frac{64}{x}\)

Let s = x + y = x + \(\frac{64}{x}\).

\(\frac{d s}{d x}\) = 1 – \(\frac{64}{x^{2}}, \frac{d s}{d x}\) = 0 ⇒ x^{2} = 64 ⇒ x = ±8

\(\frac{d^{2} s}{d x^{2}}=+\frac{128}{x^{3}}\)

When x = 8, \(\frac{d^{2} s}{d x^{2}}=\frac{128}{8^{3}}\) > 0 ⇒ s is minimum

When x = -8, \(\frac{d^{2} s}{d x^{2}}=\frac{128}{8^{3}}\) < 0 ⇒ s is maximum

The two numbers are 8 & 8.