ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.7

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.7

Question 1.
Find the turning points of the following functions and distinguish between them. Also find the local maximum and minimum values of the functions :
(i) f(x) = 2x³ – 21x² + 36x – 20
(ii) f(x) = x³ – 3x + 3x. (NCERT)
Solution:
(i) Given f(x) = 2x³ – 21x² + 36x – 20 ………….(1)
Diff. both sides of eqn. (1) w.r.t. x ; we have
f'(x) = 6x² – 42x + 36
= 6 (x² – 7x + 6)
For maxima / minima, we put f'(x) = 0
⇒ 6 (x² – 7x + 6) = 0
⇒ 6 (x – 1) (x – 6) = 0
⇒ x = 1, 6
∴ f”(x) = 12x – 42.

Case – I :
When x = 1
f”(1)= 12 – 42 = – 30 < 0
∴ x = 1 is a point of local maxima. and local maximum value
= 2 – 21 + 36 – 20 = – 3

Case – II :
When x = 6
∴ f”(6) = 72 – 42
= 30 > 0
Thus, x = 6 be a point of local minima, and local minimum value = f(6)
= 432 – 756 + 216 – 20
= 648 – 776 = – 128.

(ii) Given f(x) = x³ – 3x² + 3x
∴ f (x) = 3x² – 6x + 3
∴ f”(x) = 6x – 6
For turning points, we have f'(x) = 0
⇒ 3x² – 6x + 3 = 0
⇒3(x²2 – 2x + 1) = 0
⇒ (x – 1)² = 0
⇒ x = 1
Thus, x = 1 be the only critical point.
f”(1) = 6 – 6 = 0
Now f”‘ (x) = 6 i.e. f” (1) = 6 ≠ 0
∴ x = 1 be the point of inflexion.
i.e. x = 1 be a point of neither maxima nor minima.

Question 1(old).
(ii) f(x) = 4x³ + 19x² – 14x + 3
Solution:
Given f(x) = 4x³ + 19x² – 14x + 3 ……………(1)
Diff. both sides eqn. (1) w.r.t. x ; we have
f'(x)= 12x² + 38x – 14
For maxima/minima, we put f'(x) = 0
⇒ 2 (6x² + 19x – 7) = 0
⇒ 6x² + 21x – 2x – 7 = 0
⇒ 3x (2x + 7) – 1 (2x + 7) = 0
⇒ (2x + 7) (3x – 1) = 0
⇒ x = – \(\frac{7}{2}\), \(\frac{1}{3}\)
∴ f”(x) = 24x + 38

Case – I :
at x = – \(\frac{7}{2}\)
∴ f”(- \(\frac{7}{2}\)) = 24 (- \(\frac{7}{2}\)) + 38
= – 84 + 38
= – 46 < 0
∴ x = – \(\frac{7}{2}\) be a point of local maxima. and max. value
= f(- \(\frac{7}{2}\))
= \(-\frac{343}{2}+\frac{931}{4}\) + 49 + 3
= \(\frac{453}{4}\)

Case – II :
When x = \(\frac{1}{3}\)
∴ f”(\(\frac{1}{3}\)) = 8 + 38
= 46 > 0
Thus, x = \(\frac{1}{3}\) be a point of local minima.
and local minimum value = f(\(\frac{1}{3}\))
= \(\frac{4}{27}+\frac{19}{9}-\frac{14}{3}\) + 3
= \(\frac{4+57-126+81}{27}\)
= \(+\frac{16}{27}\)

Question 2.
Find the coordinates of stationary points on the curve y = x³ – 3x² – 9x + 7, and distinguish between points of local maxima and minima.
Solution:
Given f(x) = x³ – 3x² – 9x + 7
Diff. both sides w.r.t. x, we have
f'(x) = 3x² – 6x – 9
∴ f” (x) = 6x – 6
For critical points, we put f’ (x) = 0
⇒ 3 (x² – 2x – 3) = 0
⇒ (x + 1) (x – 3) = 0
⇒ x = – 1, 3.

Case – I :
When x = – 1
∴ f”(- 1) = – 6 – 6 = – 12 < 0
∴ x = – 1 be a point of local maxima and local maximum value
= f (- 1) = – 1 – 3 + 9 + 7 = 12.

Case-II :
When x = 3
∴ f” (3) = 18 – 6 = 12 > 0
Thus, x = 3 be a point of local minima, and local minimum value = f(3)
= 27 – 27 – 27 + 7 = – 20.

Question 3.
Find the points of local maxima and minima (if any) of each of the following functions. Find also the local maximum and minimum values :
(i) f{x) = x² (NCERT)
(ii) f(x) = x³ – 3x (NCERT)
(iii) f(x) = x³ – 12x
(iv) f(x) = x³ – 6x² + 9x + 15 (NCERT)
(v) f(x) = x³ – 3x + 3 (NCERT)
(vi) f(x) = 2x³ – 6x² + 6x + 5 (NCERT)
(vii) f(x) = \(\frac{x}{2}+\frac{2}{x}\), x > 0 (NCERT)
(viii) f(x) = \(\frac{1}{x^2+2}\) (NCERT)
(ix) f(x) = 2 sin x – x,
(x) f(x) = sin x + cos x, 0 < x < \(\frac{\pi}{2}\) (NCERT)
(xi) f(x) = 3x⁴ + 4x³ – 12x² + 12. (NCERT)
Solution:
(i) f (x) = x²
∴ f'(x) = 2x
For local max. or min. f'(x) = 0
⇒ 2x = 0
⇒ x = 0
Now f” (x) = 2
∴ f” (0) = 2 > 0
∴ x = 0 is a pt. of minima and Min. value = f( 0) = 0.

(ii) Given f(x) = x³ – 3x
∴ f (x) = 3x² – 3
= 3 (x – 1) (x + 1)
For maxima/minima, f’ (x) = 0
⇒ 3 (x – 1) (x + 1) = 0
⇒ x = 1, – 1
∴ f”(x) = 6x
f”(1) = 6 > 0
∴ x = 1 be a point of minima and min. value of f(x)
= f(1) = 1 – 3 = – 2
and f”(- 1) = – 6 < 0 x = – 1 be a point maxima
and max. value of f(x) = f(- 1)
= – 1 + 3 = 2

Aliter :
Given f(x) = x³ – 3x
∴ f'(x) = 3x² – 3 = 3(x² – 1)
= 3 (x – 1) (x + 1)
For local ‘maxima/minima, f'(x) = 0
⇒ x² – 1 = 0
⇒ x = ± 1

Case – I:
At x = 1 when x slightly < 1

Case – II:
At x = -1 when x slightly < -1 ⇒ \(\frac{d y}{d x}\) = (- ve) (+ ve) = (- ve) when x slightly > -1
∴ \(\frac{d y}{d x}\) = (+ ve) (+ ve) = (+ ve)
∴ f'(x) changes its sign from – ve to + ve
∴ x = 1 is a point of maxima and min. value = f (1) = 1 – 3 = – 2.

Case II:
At x = – 1
When x slightly < – 1 ∴ \(\frac{d y}{d x}\) = (- ve) (- ve) = (+ ve) when x slightly > – 1
∴ \(\frac{d y}{d x}\) = (- ve) (+ ve) = (- ve)
Thus \(\frac{d y}{d x}\) changes its sign from + ve to – ve
∴ x = – 1 is a point of maxima.
∴ max. value = f (- 1) = – 1 + 3 = 2.

(iii) Given f(x) = x³ – 12x ;
f’ (x) = 3×2 – 12 ;
f” (x) = 6x
For local max/minima, f’ (x) = 0
⇒ 3x² – 12 = 0
⇒ x = ± 2
Now, f” (2) = 12 > 0
∴ x = 2 is a pt. of minima
and min value = 8 – 24 = – 16
also, f” (- 2) = – 12 < 0
∴ x = – 2 is a pt. of maxima
and max. value = – 8 + 24 = 16.

(iv) Given f (x) = x³ – 6x² + 9x + 15 ;
Diff. both sides w.r.t. x, we have
∴ f’ (x) = 3x² – 12x + 9 ;
f” (x) = 6x – 12
For maxima/minima, f’ (x) = 0
⇒ 3 (x² – 4x + 3) = 0
⇒ (x – 1) (x – 3) = 0
⇒ x = 1, 3

Case – I :
When x = 1,
f” (1) = 6 – 12 = – 6 < 0
∴ x = 1 be a point of maxima (local) and local maximum value
f(1) = 1 – 6 + 9 + 15 = 19

Case – II :
When x = 3 f” (3) = 18 – 12 = 6 > 0
∴ x = 3 be a point of local minima
and local min value = f(3)
= 27 – 54 + 27 + 15 = 15

(v) Given f(x) = x³ – 3x + 3
∴ f(x) = 3x² – 3;
f”(x) = 6x
For local maxima / minima,
∴ f(x) = 0
⇒ 3 (x² – 1) = 0
⇒ x = ± 1
Now f” (1) = 6 > 0
∴ x = 1 is a point, of minima
and min value = 1 – 3 + 3 = 1
and f” (- 1) = – 6 < 0
∴ x = – 1 is a point of maxima
and max. value = – 1 + 3 + 3 = 5.

(vi) Given f(x) = 2x³ — 6x² + 6x + 5
Diff, both sides w.r.t. x; we have
∴ f(x) = 6x² – 12x + 6
∴ f”(x) = 12x – 12
For maxima/minima, we put f'(x) = 0
⇒ 6(x² – 2x + 1) = 0
⇒ 6(x – 1)²
⇒ x = 1
Now, f” (1) = 12 – 12 = 0
Now, f”‘ (x) = 12
⇒ f”‘(1) = 12 ≠ 0
∴ x = 1 be a point of inflexion.
Thus, x = 1 be a point of neither maxima nor minima.

(vii) Given f(x) = \(\frac{x}{2}+\frac{2}{x}\)
∴ f'(x) = \(+\frac{1}{2}-\frac{2}{x^2}\)
= [/latex]\frac{x^2-4}{2 x^2}[/latex]
The critical points of f(x) are given by
∴ f(x) = 0
⇒ x² – 4 = 0
⇒ x = ± 2
but x > 0
∴ x = – 2 is rejected.
Hence x = 2 be the only critical point.
∴ f”(x) = \(\frac{4}{x^3}\)
f”(2) = \(\frac{4}{8}=\frac{1}{2}\) > 0
Thus x = 2 is a point of minima and local min. value of f(x) at x = 2.
= f(2)
= \(\frac{2}{2}+\frac{2}{2}\) = 2.

(viii) Given f(x) = \(\frac{1}{x^2+2}\)
∴ f'(x) = – \(\frac{1}{\left(x^2+2\right)^2}\) × 2x
The critical points of f(x) are given by putting f'(x) = 0
i.e. \(\frac{2 x}{\left(x^2+2\right)^2}\) = 0
⇒ x = 0
[∵ (x² + 2)² > 0]

signs of f'(x) for different values of x
Clearly f'(x) changes its sign from negative to positive as x increases through 0.
Thus x = 0 is a point of local minima.
∴ local minimum value f(0) = \(\frac{1}{0+2}=\frac{1}{2}\).

(ix) Given f(x) = 2 sin x – x,
Diff both sides w.r.t. x, we have
∴ f(x) = 2 cos x – 1
For maxima/minima, we have f'(x) = 0
⇒ 2 cos x – 1 = 0
⇒ cos x = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\)
⇒ x = 2nπ ± \(\frac{\pi}{3}\) ∀ n ∈ Z
But – \(\frac{\pi}{2}\) < x < \(\frac{\pi}{2}\)
∴ x = ± \(\frac{\pi}{3}\)
∴ f”(x) = – 2 sin x
When x = \(\frac{\pi}{3}\);
f”(\(\frac{\pi}{3}\)) = – 2 sin \(\frac{\pi}{3}\)
= – 2 × \(\frac{\sqrt{3}}{2}\) = – √3 < 0
∴ x = \(\frac{\pi}{3}\) be a point of local maxima and local maximum value
= f(\(\frac{\pi}{3}\)) = 3 sin \(\frac{\pi}{3}\) – \(\frac{\pi}{3}\)
= √3 – \(\frac{\pi}{3}\)
When x = – \(\frac{\pi}{3}\); f”(- \(\frac{\pi}{3}\))
= – 2 sin (- \(\frac{\pi}{3}\))
= √3 > 0
∴ x = – \(\frac{\pi}{3}\) be a point of local minima.
and local minimum value = f(- \(\frac{\pi}{3}\))
= 2 sin (- \(\frac{\pi}{3}\)) + \(\frac{\pi}{3}\)
= – √3 + \(\frac{\pi}{3}\)

(x) Given f(x) = sin x + cos x, 0 < x < \(\frac{\pi}{2}\)
∴ f'(x) = cos x – sin x
For local maxima or minima, f’ (x) = 0
⇒ cos x – sin x = 0
⇒ tan x = 1 = tan \(\frac{\pi}{4}\)
∴ x = \(\frac{\pi}{4}\) ∈ (0, \(\frac{\pi}{2}\))
[∵ tan x = tan α
⇒ x = nπ + α, n ∈ I]
∴ f”(x) = – sin x – cos x
Now at x = \(\frac{\pi}{4}\),
f”(\(\frac{\pi}{4}\)) = \(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\)
= – √2 < 0
∴ x = \(\frac{\pi}{4}\) is a pt. of maxima
and Max. value = \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\) = √2

(xi) Given f(x) = 3x⁴ + 4x³ – 12x² + 12
∴ f’ (x) = 12x³ + 12x² – 24x ;
f” (x) = 36x² + 24x – 24
For local maxima/minima, f’ (x) = 0
⇒ 12x (x² + x – 2) = 0
∴ x = 0, 1, – 2
at x = 0 ;
f” (x) = – 24 < 0
x = 0 is a point of local maxima and max. value = 12
at x = 1 ;
f” (1) = 36 + 24 – 24 = 36 > 0
∴ x = 1 is a pt. of local minima and min value
= 3 + 4 – 12 + 12 = 7
at x = – 2 ;
f”(- 2) = 144 – 48 – 24
= 72 > 0
∴ x = – 2 is a pt. of local minima and local minimum value
= 48 – 32 – 48 + 12 = – 20.

Question 4.
For what values of a and b, the function f(x) = x³ + ax² + bx – 3 has local maximum value at x = 0 and local minimum value at x = 1 ?
Solution:
Given f(x) = x³ + ax² + bx – 3
Diff. both sides w.r.t. x, we have
f ‘(x) = 3x² + 2ax + b
Since the function f(x) has local maximum value at x = 0
and local minimum value at x = 1.
f’ (0) = 0 = f’ (1)
Now f’ (0) = 0
⇒ 0 + 2a × 0 + b = 0
⇒ b = 0
and f'(1) = 0
⇒ 3 + 2a + b = 0
⇒ 2a + 3 = 0
⇒ a = – \(\frac{3}{2}\)
Thus a = – \(\frac{3}{2}\) and b = 0.