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

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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) = x3 – 3x
(ii) f(x) = x3 – 6x2 + 9x + 15(0 ≤ x ≤ 6)
(iii) f(x) = x4 – 62x2 + 120x + 9
(iv) f(x) = 2x3 – 3x2 – 12x + 12
(v) f(x) = 2x3 – 3x2 – 36x + 10
(vi) f(x) = 9x2 + 12x + 2
(vii) f(x) = 2x3 – 15x2 + 36x + 10
(viii) f(x) = 2x3 – 21x2 + 36x – 20
(ix) f(x) = 2x3 – 15x2 + 36x + 10
(x) f(x) = 12x5 – 45x4 + 40x3 + 6
Answer:
(i) Given f(x) = x3 – 3x …..(1)
f'(x) = 3x2 – 3 = 3(x2 – 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

KSEEB Solutions

(ii) f(x) = x3 – 6x2 + 9x + 15 (0 ≤ x ≤ 6) – (1)
f'(x) = 3x2 – 12x + 9
= 3(x2 – 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) = 33 – 6.32 + 9.3 + 15 = 27 – 54 + 27 + 15 = 15.

(iii) Given f(x) = x4 – 62x2 + 120x + 9 …..(1)
f”(x) = 4x3 – 124x + 120 ….(2)
f”(x) = 12x2 – 124 …..(3)
for a function to be maximum or minimum of f'(x) = 0.
⇒ x3 – 31x + 30 = 0 here x = 1 is a root
2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.3 - 1
⇒ x2 + 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) = 2x3 – 3x2 – 12x + 12 ….(1)
f'(x) = 6x2 – 6x – 12 = 6 (x2 – 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
2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.3 - 2

KSEEB Solutions

(v) Given f(x) = 2x3 – 3x2 – 36x + 10 ….(1)
f ‘(x) = 6x2 – 6x – 36 = 6 (x2 – x – 6) …..(2)
f “(x) = 12x – 6 ……(3)
For a function to be maximum of minimum f ‘(x) = 0
⇒ (x2 – 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 (32) – 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) = 9x2 + 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) = 2x3 – 15x2 + 36x + 10 …….(1)
f ‘(x) = 6x2 – 30x + 36 = 6 (x2 – 5x + 6) ……..(2)
f”(x) = 12x – 30 …..(3)
f'(x) = 0 ⇒ x2 -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.

KSEEB Solutions

(viii) Given f(x) = 2x3 – 21x2 + 36x – 20 ….. (1)
f'(x) = 6x2 – 42x + 36 …… (2)
= 6(x2 – 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)33 – 21 (6)2 + 36(6) – 20
f(6) = 432 – 756 + 216 – 20 = -128

(ix) Given f(x) = 12x5 – 45x4 + 40 x3 + 6 ….(1)
f'(x) = 60x4 – 180x3 + 120x2 ….(2)
= 60x2 (x2 – 3x + 2)
= 60x2 (x – 1) (x – 2)
f ‘(x) = 0 ⇒ 60x2 (x – 1) (x – 2) = 0 ⇒ x = 0, 1, 2
f “(x) = 60 (4x3 – 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 – x2.
\(\frac{d p}{d x}\) = 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.

KSEEB Solutions

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 = x2 + y2 where y = 14 – x
∴ S = x2 + (14 – x)2 = x2 + 142 + x2 – 28x = 2x2 – 28x + 142
\(\frac{d s}{d x}\) = 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 = x3 + y3 where y = 30 – x
S = x3 + (30 – x)3 = x3 + (30)3 – x3 – 2700x + 90x2
\(\frac{d s}{d x}\) = – 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 ⇒ x2 = 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.

KSEEB Solutions

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