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Class 12 RD Sharma Solutions – Chapter 18 Maxima and Minima – Exercise 18.3
  • Last Updated : 30 Apr, 2021

Question 1. Find the points of local maxima or local minima and corresponding local maximum and local minimum values of the following function. Also, find the points of inflection, if any:

(i) f(x) = x4 – 62x2 + 120x + 9

Solution:

Given function: f(x) = x4 – 62x2 + 120x + 9

Differentiating with respect to x

f'(x) = 4x3 – 124x + 120 = 4(x3 – 31x + 30)

Again differentiating with respect to x



f”(x) = 12x2 – 124 = 4(3x2 – 31)

Now, for maxima and minima  

Put f'(x) = 0

⇒ 4(x3 – 31x + 30) = 0

⇒ x3 – 31x + 30 = 0

⇒ x = 5, 1, -6

Now,

f”(5) = 176 > 0

⇒ x = 5 is point of local minima  

f”(1) = -112 < 0

⇒ x = 1 is point of local maxima  

f”(-6) = 308 > 0

⇒ x = -6 is point of local minima  

Now, we find the local maximum value = f(1) = 68

and local minimum value = f(5) = -316

and = f(-6) = -1647

(ii) f(x) = x3 – 6x2 + 9x + 15

Solution:

Given function: f(x) = x3 – 6x2 + 9x +15



Differentiating with respect to x

f'(x) = 3x2 -12x + 9 = 3(x2 – 4x + 3)

Again differentiating with respect to x

f”(x) = 6x – 12

Now, for maxima and minima  

Put f'(x) = 0

⇒ 3(x2 – 4x + 3) = 0

⇒ x2 – 4x + 3 = 0

⇒ (x – 3)(x – 1) = 0

⇒ x = 3, 1  

Now,

f”(3) = 6 > 0

⇒ x = 3 is point of local minima  

f”(1) = -6 < 0

⇒ x = 1 is point of local maxima  

⇒ x = -6 is point of local minima  

Now we find the local maximum value = f(1) =19

and local minimum value = f(3) = 15                                  

(iii) f(x) = (x – 1)(x + 2)2

Solution:

Given function: f(x) = (x – 1)(x + 2)2



Differentiating with respect to x

f ‘(x) = (x + 2)2 + 2(x – 1)(x + 2)

= (x + 2)(x + 2 + 2x – 2)

= (x + 2)(3x)

Again differentiating with respect to x

f ”(x) = 3(x + 2) + 3x

= 6x + 6

For maxima and minima  

Put f'(x) = 0

⇒ 3x(x + 2) = 0

⇒ x = 0, -2  

Now,

f”(0) = 6 > 0

⇒ x = 0 is point of local minima  

f”(-2) = -6 < 0

⇒ x = -2 is point of local maxima  

Now we find the local maximum value = f(-2) =0

Local minimum value = f(0) = -4                

(iv) f(x) = 2/x – 2/x2, x > 0

Solution:

Given function: f(x) = 2/x – 2/x2, x > 0

Differentiating with respect to x

f'(x) = -2/x2 + 4/x3

Again differentiating with respect to x

f”(x) = -4/x3 – 12/ x4

For maxima and minima  

Put f'(x) = 0

⇒ -2/x2 + 4/x3 = 0

⇒ -2(x – 2)/x3 = 0 

⇒ x = 2  

Now,

f”(2) = 4/8 – 12/6 = 1/2 – 3/4 = -1/4 < 0

⇒ x = 2 is point of local maxima   

Now, local maximum value = f(2) = 1/2

(v) f(x) = xex

Solution:

Given function: f(x) = xex

Differentiating with respect to x

f'(x) = ex + xex = ex(x + 1) 

Again differentiating with respect to x

f”(x) = ex(x + 1) + ex = ex(x + 2) 

For maxima and minima  

Put f'(x) = 0

⇒ ex(x + 1) = 0

⇒ x = -1

Now,

f”(-1) = e-1 = 1/e > 0

⇒ x = -1 is point of local minima

Now, local minimum value = f(-1) = -1/e

(vi) f(x) = x/2 + 2/x, x > 0

Solution:

Given function: f(x) = x/2 + 2/x, x > 0

Differentiating with respect to x

f'(x) = 1/2 – 2/x2

Again differentiating with respect to x

f”(x) = 4/x3

For maxima and minima  

Put f'(x) = 0

⇒ 1/2 – 2/x2 = 0

⇒ (x2 – 4)/2x2 = 0

⇒ x = -2, +2  

Now,

f”(2) = 4/8 = 1/2 > 0

Thus, x = 2

x=-2 not taken because x > 0 is given. 

Now, local minimum value = f (2) = 2

(vii) f(x) = (x + 1)(x + 2)1/3, x ≥ -2

Solution:

Given function: f(x) = (x + 1)(x + 2)1/3, x ≥ -2

Differentiating with respect to x

f ‘(x) = (x + 2)1/3 + 1/3(x + 1)(x + 2)-2/3

= (x + 2)-2/3(x + 2 + 1/3(x + 1))

= 1/3(x + 2)-2/3(4x + 7)

Again differentiating with respect to x

f”(x) = -2/9(x + 2)-5/3(4x + 7) + 1/3(x + 2)-2/3 × 4 

For maxima and minima  

Put f'(x) = 0

⇒ 1/3(x + 2)-2/3(4x + 7) = 0

⇒ (4x + 7) = 0

⇒ x = -7/4

Now,

f “(-7/4) = 4/3(-7/4 + 2)-2/3 

Thus, x = -7/4 is point of local minima

Now, local minimum value = f(-7/4) = \frac{-3}{4^\frac{4}{3}}

(viii) f(x) = x\sqrt{32-x^2}, -5 ≤ x ≤ 5

Solution:

Given function: f(x)=x\sqrt{32-x^2}, -5 ≤ x ≤ 5

Differentiating with respect to x

f'(x) = \sqrt{32-x^2}+\frac{x}{2\sqrt{32-x^2}}      × (-2x)

\frac{2(32-x^2)-2x^2}{2\sqrt{32-x^2}}

\frac{64-4x^2}{2\sqrt{32-x^2}}

Again differentiating with respect to x

f”(x) = \frac{2\sqrt{32-x^2}×(-8x)\frac{-2(64-4x^2)}{2\sqrt{32-x^2}}×(-2x)}{2\sqrt{32-x^2}}

\frac{-4(32-x^2)8x+4x(64-x^2)}{8(32-x^2)^\frac{3}{2}}        

For maxima and minima  

Put f'(x) = 0

⇒ \frac{64-4x^2}{2\sqrt{32-x^2}}      = 0

⇒ 64 – 4x2 = 0

⇒ x = ±4

Now,

f”(4) = \frac{-4(32-16)32+16(64-16)}{8(32-16)^{3/2}}      <0

Thus, x = 4 is point of maxima

Now, local maximum value = f(4) = 16

x = -4 is point of minima

Now, local minimum value = f(-4) = -16

(ix) f(x) = x3 – 2ax2 + a2x, a > 0, x ∈ R

Solution:

Given function: f(x) = x3 – 2ax2 + a2x

Differentiating with respect to x

f ‘(x) = 3x2 – 4ax + a2

Again differentiating with respect to x

f”(x) = 6x – 4a

For maxima and minima  

Put f'(x) = 0

⇒ 3x2 – 4ax + a2 = 0

⇒ x = \frac{4a±\sqrt{16a^2-12a^2}}{6}

= (4a ± 2a)/6 = a, a/3

Now, 

f”(a) = 2a > 0 as a > 0

x = a is point of local minima

f”(a/3) = -2a < 0 as a < 0

x = a/3 is point of local maxima

Hence,

The local maximum value = f(a/3) = 4a3/27

and local minimum value = f(a) = 0

(x) f(x) = x + a2/x, a > 0, x ≠ 0

Solution:

Given function: f(x) = x + a2/x

Differentiating with respect to x

f’ (x) = 1 – a2/x2

Again differentiating with respect to x

f”(x) = 2a2/x3

For maxima and minima  

Put f'(x) = 0

⇒ 1 – a2/x2 = 0

⇒ x2 – a2 = 0

x = ± a

Now,

f”(a) = 2/a > 0 as a > 0

x = a is point of minima

f”(-a) = -2/a < 0 as a > 0

x = -a is point of maxima

Hence, the local maximum value = f(-a) = -2a

and local minimum value = f(a) = 2a

(xi) f(x) = x\sqrt{2-x^2} -\sqrt{2}, −√2 ​≤ x ≤ √2​

Solution:

Given function: f(x) = x\sqrt{2-x^2}

Differentiating with respect to x

f'(x) = \sqrt{2-x^2} -\frac{2x^2}{x\sqrt{2-x^2} }

\frac{2(2-x^2)-2x^2}{2\sqrt{2-x^2}}

 = \frac{2-2x^2}{\sqrt{2-x^2}}

Again differentiating with respect to x

f”(x) = \frac{\sqrt{2-x^2}(-4x)+\frac{(2-2x^2)2x}{\sqrt{2-x^2}}}{(\sqrt{2-x^2})^2}

\frac{-(2-x^2)4x+4x-4x^3}{(2-x^2)^\frac{3}{2}}

For maxima and minima  

Put f'(x) = 0

⇒ \frac{2(1-x^2)}{\sqrt(2-x^2)}=0

⇒x = ±1

Now,  

f”(1) < 0

⇒ x = 1 is point of local maxima

f”(-1) > 0

⇒ x = -1 is point of local maxima

Hence, the local maximum value = f (1) = 1

and local minimum value = f (-1) = -1

(xii) f(x) = x+ \sqrt{1-x}, x ≤ 1

Solution:

Given function: f(x) = x + \sqrt{1-x}

Differentiating with respect to x

f'(x) = 1-\frac{1}{2\sqrt{1-x}}=\frac{2\sqrt{1-x}-1}{2\sqrt{1-x}}

f'(x) = \frac{2\sqrt{1-x}(\frac{-1}{\sqrt{1-x}})+\frac{2\sqrt{1-x}-1}{\sqrt{1-x}}}{4(1-x)}

For maxima and minima  

Put f'(x) = 0

⇒ \frac{2\sqrt{1-x}-1}{2\sqrt{1-x}}=0

⇒ \sqrt{1-x}   =1/2

⇒ x= 1 – 1/4 = 3/4

Now,

f”(3/4) < 0

⇒ x = 3/4 is point of local maxima

Hence, the local maximum value = f (3/4) = 5/4

Question 2. Find the local extremum values of the following functions:

(i) f (x) = (x – 1)(x – 2)2

Solution:

Given function: f(x) = (x – 1)(x – 2)2

Differentiating with respect to x

f'(x) = (x – 2)2 + 2(x – 1)(x – 2)

= (x – 2)(x – 2 + 2x – 2)

= (x – 2)(3x – 4)

Again differentiating with respect to x

f”(x) =(3x – 4) + 3(x – 2)

For maxima and minima  

Put f'(x) = 0

⇒ (x – 2)(3x – 4) = 0

⇒ x = 2, 4/3

Now,

f”(2) > 0

x = 2 is local minima

f “(4/3) = -2 < 0

x = 4/3 is point of local maxima

Hence, the local maximum value = f(4/3) = 4/27

and the local minimum value = f(2) =0    

(ii) f (x) = x\sqrt{1-x}, x ≤ 1

Solution:

Given function: f (x) = x\sqrt{1-x}

Differentiating with respect to x

f ‘(x) = \sqrt{1-x}+\frac{x}{2\sqrt{1-x}}(-1)

\frac{2(1-x)-x}{2\sqrt{1-x}}

= (2 – 3x)/2\sqrt{1-x}

Again differentiating with respect to x

f”(x) = \frac{2\sqrt{1-x}(-3)+\frac{2-3x}{\sqrt{1-x}}}{4(1-x)}

For maxima and minima  

Put f'(x) = 0

⇒ (2 – 3x)/2\sqrt{1-x}   =0

⇒ x = 2/3

Now,

f “(2/3) < 0

x = 2/3 is point of maxima

Hence, the local maximum value = f (2/3) = 2/3√3

(iii) f(x) = -(x – 1)3(x + 1)2

Solution:

Given function: f(x) = -(x – 1)3(x + 1)2  

Differentiating with respect to x

  f ‘(x) = -3(x – 1)2 (x + 1)2 – 2(x – 1)3 (x + 1)  

= -(x – 1)2 (x + 1) (3x + 3 + 2x – 2)

= -(x – 1)2(x + 1) (5x + 1)

Again differentiating with respect to x

f”(x) = -2(x – 1) (x + 1) (5x + 1) – (x – 1)2 (5x + 1) – 5 (x – 1)2 (x + 1)

For maxima and minima  

Put f'(x) = 0

⇒ -(x – 1)2 (x + 1) (5x + 1) = 0

⇒ x = 1, -1, -1/5

Now,

f”(1) = 0

x = 1 is inflection point

f “(-1) = -4 × -4 = 16 > 0

x = -1 is point of minima

f”(-1/5) = -5(36/25) × (4/5) = -144/25 < 0

x = -1/5 is point of maxima

Hence, the local maximum value = f (-1/5) = 3456/3125

and the local minimum value = f (-1) = 0

Question 3. The function y = a log x + bx2 + x has extreme values at x = 1 and x = 2. Find a and b.

Solution:

We have,

y = alogx + bx2 + x

Differentiating with respect to x

dy/dx = a/x + 2bx + 1

Again differentiating with respect to x

d2y/dx2 = -a/x2 +2b

For maxima and minima  

Put dy/dx = 0

⇒ a/x + 2bx +1 = 0

Given that extreme value exist at x = 1, 2

⇒ a + 2b = -1        …..(i)

a/2 + 4b = -1

⇒ a + 8b = -2       …..(ii)

On solving eq(i) and (ii), we get

a = -2/3, b = -1/6

Question 4. Show that (log x / x) has a maximum value at x = e.

Solution:

Given function: f(x) = logx / x

Differentiating with respect to x

f'(x)=\frac{x(1/x)-logx}{x^2}=\frac{1-logx}{x^2}

Now, f'(x) = 0

⇒ 1 – logx = 0

⇒ logx = 1

⇒ logx = loge

⇒ x = e

Again differentiating with respect to x

f ''(x)=\frac{x^2(-1/x)-(1-logx)(2x)}{x^4}

=\frac{-x-2x(1-logx)}{x^4}

\frac{-3+2logx}{x^3}

Now, f ”(e) = \frac{-3+2loge}{e^3}

=(-3 + 2)/e3 = -1/e3 < 0

Therefore, by second derivative test, f is the maximum at x = e.

Question 5. Find the maximum and minimum values of the function f (x) = 4/(x + 2) + x.

Solution:

Given function: f(x) = 4/(x + 2) + x

Differentiating with respect to x

f'(x) = -4/(x + 2)2 + 1

Again differentiating with respect to x

f”(x) = 8/(x + 2)3

For maxima and minima  

Put f'(x) = 0

⇒ -4/(x + 2)2 + 1 = 0

⇒ (x + 2)2 = 4

⇒ x2 + 4x = 0

⇒ x (x + 4) = 0

 x = 0, -4

Now,

f ”(0) = 1 > 0

x = 0 is point of minima

f ”(-4) = -1 < 0

x = -4 is point of maxima

Hence, the local maximum value = f (-4) = -6

and the local minimum value = f (0) = 2

Question 6. Find the maximum and minimum values of f (x) = tan x – 2x.

Solution:

Given function: f(x) = tanx – 2x

Differentiating with respect to x

f'(x) = sec2x – 2

Again differentiating with respect to x

f”(x) = 2sec2x tanx

For maxima and minima  

Put f'(x) = 0

⇒ sex2x = 2

⇒ secx = ±√2

⇒ x = π/4, 3π/4

f”(π/4) = 4 > 0

x = π/4 is point of minima

f”(3π/4) = -4 < 0

x = (3π/4) is point of maxima

Hence, the maximum value = f (3π/4) = -1 – 3π/2

and the minimum value = f (π/4) = 1 – π/2

Question 7. If f(x) = x3 + ax2 + b x + c has a maximum at x = -1 and minimum at x = 3. Determine a, b and c.

Solution:

Given function: f(x) = x3 + ax2 + b x + c

Differentiating with respect to x

f ‘(x) = 3x2 + 2ax + b

It is given that f (x) is maximum at x = -1

f ‘(-1) = 3(-1)2 + 2a(-1) + b = 0

f ‘(-1) = 3 – 2a + b = 0     …….(i)

At x = 3, f(x) is minimum 

f ‘(3) = 3(3)2 + 2a(3) + b = 0

⇒ f ‘(3) = 27 + 6a +b = 0     …..(ii)

On solving eq(i) and (ii), we get

a = -3 and b = -9

Since f ‘(x) is independent of constant c, so, it can be any real number.

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