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Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 18 |

Exercise | 18.3 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Class 12 Ex 18.3 Solutions Chapter 18 Maxima and Minima**

### 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) = x^{4} – 62x^{2} + 120x + 9

**Solution:**

Given function: f(x) = x

^{4}– 62x^{2}+ 120x + 9Differentiating with respect to x

f'(x) = 4x

^{3}– 124x + 120 = 4(x^{3 }– 31x + 30)Again differentiating with respect to x

f”(x) = 12x

^{2}– 124 = 4(3x^{2 }– 31)Now, for maxima and minima

Put f'(x) = 0

⇒ 4(x

^{3 }– 31x + 30) = 0⇒ x

^{3 }– 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) = x^{3} – 6x^{2} + 9x + 15

**Solution:**

Given function: f(x) = x

^{3}– 6x^{2}+ 9x +15Differentiating with respect to x

f'(x) = 3x

^{2}-12x + 9 = 3(x^{2 }– 4x + 3)Again differentiating with respect to x

f”(x) = 6x – 12

Now, for maxima and minima

Put f'(x) = 0

⇒ 3(x

^{2 }– 4x + 3) = 0⇒ x

^{2 }– 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/x^{2}, x > 0

**Solution:**

Given function: f(x) = 2/x – 2/x

^{2}, x > 0Differentiating with respect to x

f'(x) = -2/x

^{2}+ 4/x^{3}Again differentiating with respect to x

f”(x) = -4/x

^{3}– 12/ x^{4}For maxima and minima

Put f'(x) = 0

⇒ -2/x

^{2}+ 4/x^{3}= 0⇒ -2(x – 2)/x

^{3}= 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) = xe^{x}

**Solution:**

Given function: f(x) = xe

^{x}Differentiating with respect to x

f'(x) = e

^{x}+ xe^{x}= e^{x}(x + 1)Again differentiating with respect to x

f”(x) = e

^{x}(x + 1) + e^{x}= e^{x}(x + 2)For maxima and minima

Put f'(x) = 0

⇒ e

^{x}(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/x

^{2}Again differentiating with respect to x

f”(x) = 4/x

^{3}For maxima and minima

Put f'(x) = 0

⇒ 1/2 – 2/x

^{2}= 0⇒ (x

^{2 }– 4)/2x^{2}= 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 ≥ -2Differentiating 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 }× 4For 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 – 4x

^{2 }= 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) = x^{3} – 2ax^{2} + a^{2}x, a > 0, x ∈ R

**Solution:**

Given function: f(x) = x

^{3}– 2ax^{2}+ a^{2}xDifferentiating with respect to x

f ‘(x) = 3x

^{2}– 4ax + a^{2}Again differentiating with respect to x

f”(x) = 6x – 4a

For maxima and minima

Put f'(x) = 0

⇒ 3x

^{2}– 4ax + a^{2}= 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) = 4a

^{3}/27and local minimum value = f(a) = 0

### (x) f(x) = x + a^{2}/x, a > 0, x ≠ 0

**Solution:**

Given function: f(x) = x + a

^{2}/xDifferentiating with respect to x

f’ (x) = 1 – a

^{2}/x^{2}Again differentiating with respect to x

f”(x) = 2a

^{2}/x^{3}For maxima and minima

Put f'(x) = 0

⇒ 1 – a

^{2}/x^{2 }= 0⇒ x

^{2}– a^{2}= 0x = ± 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

### 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, 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 + bx^{2} + x has extreme values at x = 1 and x = 2. Find a and b.

**Solution:**

We have,

y = alogx + bx

^{2}+ xDifferentiating with respect to x

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

Again differentiating with respect to x

d

^{2}y/dx^{2}= -a/x^{2}+2bFor 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)/e^{3} = -1/e^{3} < 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 }+ 1Again 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⇒ x

^{2}+ 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) = sec

^{2}x – 2Again differentiating with respect to x

f”(x) = 2sec

^{2}x tanxFor maxima and minima

Put f'(x) = 0

⇒ sex

^{2}x = 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) = x^{3} + ax^{2} + b x + c has a maximum at x = -1 and minimum at x = 3. Determine a, b and c.

**Solution:**

Given function: f(x) = x

^{3}+ ax^{2}+ b x + cDifferentiating with respect to x

f ‘(x) = 3x

^{2}+ 2ax + bIt is given that f (x) is maximum at x = -1

f ‘(-1) = 3(-1)

^{2 }+ 2a(-1) + b = 0f ‘(-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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