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• RD Sharma Class 12 Solutions for Maths

# Class 12 RD Sharma Solutions – Chapter 18 Maxima and Minima – Exercise 18.3

### (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) = ### (viii) f(x) = x , -5 ≤ x ≤ 5

Solution:

Given function: f(x)=x , -5 ≤ x ≤ 5

Differentiating with respect to x

f'(x) = × (-2x)  Again differentiating with respect to x

f”(x) =  For maxima and minima

Put f'(x) = 0

⇒ = 0

⇒ 64 – 4x2 = 0

⇒ x = ±4

Now,

f”(4) = <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 = = (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) = , −√2 ​≤ x ≤ √2​

Solution:

Given function: f(x) = Differentiating with respect to x

f'(x) =  = Again differentiating with respect to x

f”(x) =  For maxima and minima

Put f'(x) = 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) , x ≤ 1

Solution:

Given function: f(x) = x + Differentiating with respect to x

f'(x) = f'(x) = For maxima and minima

Put f'(x) = 0

⇒ ⇒ =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

### (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 , x ≤ 1

Solution:

Given function: f (x) = x Differentiating with respect to x

f ‘(x) =  = (2 – 3x)/2 Again differentiating with respect to x

f”(x) = For maxima and minima

Put f'(x) = 0

⇒ (2 – 3x)/2 =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) Now, f'(x) = 0

⇒ 1 – logx = 0

⇒ logx = 1

⇒ logx = loge

⇒ x = e

Again differentiating with respect to x   Now, f ”(e) = =(-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.