# Class 12 NCERT Solutions – Mathematics Part I – Chapter 6 Application of Derivatives – Exercise 6.5 | Set 1

### Question 1. Find the maximum and minimum values, if any, of the following function given by

### (i) f(x) = (2x – 1)^{2} + 3 ** **

**Solution:**

Given that, f(x) = (2x – 1)

^{2 }+ 3From the given function we observe that

(2x – 1)

^{2 }≥ 0 ∀ x∈ R,So,

(2x – 1)

^{2 }+ 3 ≥ 3 ∀ x∈ R,Now we find the minimum value of function f when 2x-1 = 0

So, x = 1/2

f = f(1/2) = (2(1/2) – 1)

^{2 }+ 3 = 3Hence, the minimum value of the function is 3 and this function does not contain maximum value.

** **(ii) f(x) = 9x^{2} + 12x + 2

**Solution:**

Given that, f(x) = 9x

^{2 }+ 12x + 2we can also write as f(x) = (3x + 2)

^{2 }– 2From the given function we observe that

(3x + 2)

^{2 }≥ 0 ∀ x∈ R,So,

(3x + 2)

^{2 }– 2 ≥ 2 ∀ x∈ R,Now we find the minimum value of function f when 3x + 2 = 0

So, x = -2/3

f = f(-2/3) = (3(-2/3) + 2)

^{2 }– 2 = -2Hence, the minimum value of the function is -2 and this function does not contain maximum value.

### (iii) f(x) = -(x – 1)^{2} + 10

**Solution:**

Given that, f(x) = -(x – 1)

^{2}+ 10From the given function we observe that

(x – 1)

^{2 }≥ 0 ∀ x∈ R,So,

-(x – 1)

^{2}+ 10 ≤ 10 ∀ x∈ R,Now we find the maximum value of function f when x – 1 = 0

So, x = 1

f = f(1) = -(1 – 1)

^{2 }+10 = 10Hence, the maximum value of the function is 10 and this function does not contain minimum value.

### (iv) g(x) = x^{3} + 1

**Solution:**

Given that, g(x) = x

^{3}+ 1When x —> ∞, then g(x) —> ∞

When x —> -∞, then g(x) —> -∞

So, this function has neither minimum nor maximum value

### Question 2. Find the maximum and minimum values, if any, of the following functions given by

### (i) f(x) = |x + 2| – 1

**Solution:**

Given that, f(x) = |x + 2| – 1

At x = -2 f'(x) change sign from negative to positive, hence by first derivative test, x = -2

is a point of local minima.

So, the minimum value f = f(-1)= |(-1)+ 2| – 1 = -1

So this function doesn’t contain maximum value

### (ii) g(x) = -|x + 1| + 3

**Solution:**

Given that, g(x) = -|x + 1| + 3

At x = -1 f'(x) change sign from positive to negative, hence by first derivative test, x = -1

is a point of local minima.

So, the maximum value of f = f(-1) = -|(-1) + 1| + 3 = 3

So, this function doesn’t contain minimum value

### (iii) h(x) = sin 2x + 5

**Solution:**

Given that, h(x) = sin(2x) + 5

On differentiate both side w.r.t x, we get

h'(x) = 2cos2x

Now put h'(x) = 0

2cos 2x = 0

2x = (2x – 1)π/2

x = (2x – 1)π/4

Let’s perform second derivative test,

h”(x) = -4sin2x

h”(π/4) < 0

& so on.

So,

of local maxima.

& are points of local minima.

So, the minimum value of the given function is 4 and the maximum value of the given function is 6

### (iv) f(x) = |sin(4x + 3)|

**Solution:**

Given that, f(x) = |sin(4x + 3)|

Now for any value of x, sin4x has the least value as -1. i.e., sin 4x + 3 ≥ 2

So f(x) = |sin(4x + 3)| = sin 4x + 3

On differentiate both side w.r.t x, we get

f'(x) = 4cos4x

Now put f'(x) = 0

4cos 4x = 0

4x = (2x – 1)π/2

x = (2x – 1)π/8

& son on

Let’s perform second derivative test,

f”(x) = -16 sin4x

f”(π/8) < 0; f”(3π/8) > 0; f”(5π/8) < 0

So, …. are points of local maxima. Minimum value = 4.

& are points of local minima. Minimum value = 2.

### (v) h(x) = x + 1, x ∈ (-1, 1)

**Solution:**

Given that, h(x) = x + 1, x ∈ (-1, 1)

As we can clearly see from the function that h(x) is a strictly increasing function.

So, the minimum value of x will give minimum value of h(x).

Now, x ∈ (-1, 1)

So, this function has no minimum nor maximum value.

### Question 3. Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

### (i) f(x) = x^{2}

**Solution:**

Given that f(x) = x

^{2}On differentiate both side w.r.t x, we get

f'(x) = 2x

Now put f'(x) = 0

2x = 0

x = 0

Let’s do second derivative test,

f”(x) = 2 > 0

At, x = 0, f'(x) = 0 and f”(x) > 0,

So x = 0 is a point of local minima. Local minimum value.

### (ii) g(x) = x^{3 }– 3x

**Solution:**

Given that g(x) = x

^{3 }– 3xOn differentiate both side w.r.t x, we get

g'(x) = 3x

^{2 }– 3Now put g'(x) = 0

3x

^{2 }– 3 = 0x

^{2 }= 1x = ±1

Let’s do the second derivative test,

g”(x) = 6x ….(i)

g”(1) = 6 > 0

g”(-1) = -6 > 0

So by second derivatives test, x = 1 is a point of local maxima and the maximum value is

g(1) = (1)

^{3 }– 3(1) = -2So by second derivatives test, x = -1 is a point of local minima and the minimum value is

g(-1) = (-1)

^{3 }– 3(-1) = 2Hence, the local minimum value is -2 and the local maximum value is 2

### (iii) h(x) = sin x + cos x, 0 < x < π/2

**Solution:**

h(x) = sin x + cos x, x∈(0,π/2)

On differentiate both side w.r.t x, we get

h'(x) = cos x – sin x

Now put h'(x) = 0

cos x – sin x = 0

cos x = sin x, x ∈ (0, π/2)

Clearly x = π/4 [both cos x and sin x attain 1/√2 at π/4]

Let’s do second derivative test,

h”(x) = -sin x – cos x

h”(π/4) =

At is a point of local maxima and the maximum value is

h(π/4) = sin π/4 + cos π/4

= 1/√2 + 1/√2 = √2

### (iv) f(x) = sin x – cos x, 0 < x < 2π

**Solution:**

Given that, f(x) = sin x – cos x, x ∈ (0, 2π)

On differentiate both side w.r.t x, we get

f'(x) = cos x + sin x

Now put f'(x) = 0

cos x + sin x = 0

x = in (0, 2π)

Now let’s do the second derivative test

f”(x) = -sin x + cos x

f”(3

π/4) = – √2 > 0f”(7

π/4) = √2 > 0So by second derivatives test, x = is a point of local maxima and the maximum value is

f() = -sin 3

π/4 + cos 3π

4 = 1/√2 + 1/√2 = √2 > 0So by second derivatives test, x = is a point of local minima and the minimum value is

f() = -sin 7

π/4 + cos 7π

4 = -1/√2 – 1/√2 = -√2 > 0Hence, the local minimum value is -√2 and the local maximum value is √2.

### (v) f(x) = x^{3 }– 6x^{2 }+ 9x + 15

**Solution:**

Given that, f(x) = x

^{3 }– 6x^{2 }+ 9x + 15On differentiate both side w.r.t x, we get

f'(x) = 3x

^{2 }– 12x + 9Now put f'(x) = 0

3x

^{2 }– 12x + 9 = 03(x

^{2 }– 4x + 3) = 0x = 1, 3

Let’s do the second derivative test,

f”(x) = 6x – 12

f”(1) = -6 < 0

f”(3) = 6 > 0

So by second derivatives test, x = 1 is a point of local maxima and the maximum value is

f'(1) = 3(1)

^{2 }– 12(1) + 9 = 19So by second derivatives test, x = 3 is a point of local minima and the minimum value is

f'(3) = 3(3)

^{2 }– 12(3) + 9 = 15Hence, the local minimum value is 15 and the local maximum value is 19.

### (vi) , x > 0

**Solution:**

Given that, , x > 0

On differentiate both side w.r.t x, we get

g'(x)=

Now put g'(x) = 0

but ‘x > 0’

So, x = 2

Now we will do the second derivative test,

g”(x)=

Hence, x = 2 is a point of local minima.

Local maximum value = g(2) = 2

### (vii)

**Solution:**

Given that,

On differentiate both side w.r.t x, we get

Now put g'(x) = 0

Now, let’s perform the second derivative test,

= -8/16 = -1/2 < 0

At x = 0, g'(x) = 0 and g”(x) < 0

Hence, ‘x = 0’ is a point of local maxima.

Now the domain of g(x) is (-∞, ∞).

Value of g(x) at the extreme values of x is 0

So the global maxima of g(x)=is at x = 0.

The maximum value is g(0) = 1/2

### (viii) , x > 0

**Solution:**

Given that,

Now put f'(x) = 0

2(1 – x) = x

2 – 2x = x

3x = 2

x = 2/3

Now let’s do the second derivative test,

x = 2/3 is a point of local maxima f(2/3) =

Now, f (x) = x

For domain, 1 – x ≥ 0 or x ≤ 1

So x ∈ [0, 1]

Local maxima is at x = 2/3 and the local maximum value is

### Question 4. Prove that the following function do not have maxima or minima:

### (i) f(x) = e^{x }

**Solution:**

Given that, f(x) = e

^{x}f'(x) = e

^{x}Now e

^{x }> 0, f'(x) > 0Hence, f(x) is a strictly increasing function with no maxima or minima.

### (ii) g(x) = log x

**Solution:**

Given that, g(x) = log x

g'(x) = 1/x

Now the domain of log x is x > 0

So, 1/x > 0, i.e., g'(x) > 0

Hence, g(x) is a strictly increasing function with no maxima or minima.

### (iii) h(x) = x^{3 }+ x^{2 }+ x + 1

**Solution:**

Given that, h(x) = x

^{3 }+ x^{2 }+ x + 1h'(x) = 3x

^{2 }+ 2x + 1Now for this quadratic expression 3x

^{2 }+ 2x + 1,Its discriminant 0 = 2

^{2 }– 4(3)(1) = -8 < 0So, 3x

^{2 }+ 2x + 1 > 0Hence, h(x) is a strictly increasing function with no maxima or minima.

### Question 5. Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

### (i) f(x) = x^{3}, x ∈ [-2, 2]

**Solution:**

Given that, f(x) = x

^{3}, x ∈ [-2, 2]f'(x) = 3x

^{2}f'(x) = 0 at x = 0

f”(x) = 6x

f”(0) = 0, second derivative failure

Now f'(3

^{+}) > 0 and f'(3^{–}) > 0f'(x) does not change sign at x = 0.

x = 0 is neither maxima nor minima

f(x) = x

^{3}is a strictly increasing function.

### (ii) f(x) = sin x + cos x, x ∈ [0, π]

**Solution:**

Given that, f(x) = sin x + cos x, x ∈ [0, π]

First derivative

f'(x) = cos x – sin x

Now put f'(x) = 0

cos x = sin x

x = π/4

On applying second derivative test,

f”(x) = -sin x – cos x

f”(π/4) =

Hence, x = π/4 is pof local maxima . f(π/4)=

Now, for global maxima = max{f(0), f(π/4), f(π)}

= max{1, √2, -1}

For global maxima is at x = π/4 and the global maximum value is √2.

Now, for global minima = max{f(0), f(π/4), f(π)}

= max{1, √2, -1}

Global minima is at x = π and the global minimum value is -1.

### (iii)

**Solution:**

Given that,

f'(x) = 4 – x

Now put f'(x) = 0

4 – x = 0

x = 4

Now applying second derivative test f”(x) = -1 < 0

Hence, x = 4 is a pt. of local maxima.

f(4) =

Global maxima = max{f(-2), f(4), f(9/2)}

= max{-10, 8, 7.8}

= 8

Global maxima occur at x = 9/2 and global maximum value is f(9/2) = 8

Global minima = min{f(-2), f(4), f(9/2)}

= max{-10, 8, 16.9}

= -10

Global minima occur at x = -2 and the global minimum value is f(-2) = -10.

### (iv) f(x) = (x – 1)^{2 }+ 3, x ∈ [-3, 1]

**Solution:**

Given that, f(x) = (x – 1)

^{2 }+ 3, x ∈ [-3, 1]f'(x) = 2(x – 1)

Now put f'(x) = 0

2(x – 1) = 0

x = 1

Now applying second order derivative test,

f”(x) = 2 > 0

Hence, x = 1 is a point of local minima. f(1) = 3

Global maxima = max{f(-3), f(1)}

= max{19, 3}

= 19

The global or absolute maxima occurs at x = -3 and the absolute maximum value is f(-3) = 19

Global minima = min{f(-3), f(1)}

= min{19, 3]

= 3

The global or absolute minima occurs at x = 1 and the absolute value is f(1) = 3

### Question 6. Find the maximum profit that a company can make, if the profit function is given by p(x) = 41 – 24x – 18x^{2}

**Solution:**

Given that p(x) = 41 – 24x – 18x

^{2}p'(x) = -24 – 36x

Now put p'(x) = 0

-24 – 36x = 0

x = -24/36

x = -2/3

Now, doing the second order derivative test,

p”(x) = -36 < 0

Hence, x = -2/3 is point of local maxima.

Now in quadratic function with domain R, if there is a local maxima, it is the global maxima also. BC

_{3}p(-∞)⇢ -∞ and p(+∞)⇢ -∞The maximum profit is p(-2/3) = 49

If negative units (x) do not exist, then maximum profit is p(0) = 41.

### Question 7. Find both the maximum value and the minimum value of 3x^{4} – 8x^{3} + 12x^{2} – 48x + 25 on the interval [0, 3].

**Solution:**

Given that f(x) = 3x

^{4}– 8x^{3}+ 12x^{2}– 48x + 25, x ∈ [0, 3]f'(x) = 12x

^{2 }– 24x^{2 }+ 24x – 48Now put f'(x) = 0

12x

^{3 }– 24x^{2 }+ 24x – 48 = 012(x

^{2 }– 2x^{2 }+ 2x – 4) = 012(x

^{2}(x – 2) + 2(x – 2)) = 012(x

^{2 }+ 2)(x – 2) = 0x = 2 because x

^{2 }+ 2 ≠ 0Now applying second derivative test,

f”(x) = 12(3x

^{2 }– 4x + 2)f”(2) = 12(3.2

^{2 }– 4.2 + 2)f”(2) = 12.6 = 72 > 0

Hence, x = 2 is point of local minima.

f(2) = -39

Global maxima = max{f(0), f(2), f(3)}

= max{25, -39, 16}

= 25

Global maxima occur at x = 0 and the global maximum is 25.

Global minima = min{f(0), f(2), f(3)}

= min{25, -39, 16}

= -39

Global minima occur at x = 2 andthe global minimum value is -39.

### Question 8. At what points in the interval[0, 2π], does the function sin 2x attain its maximum value?

**Solution:**

Given that, f(x) = sin 2x , x ∈ [0, 2π]

f'(x) = 2 cos 2x

Now put f'(x) = 0

2cos2x = 0

2x = (2x – 1)π/2

x = (2x – 1)π/4

x = π/4, 3π/4, 5π/4, 7π/4

Now let’s do second order derivative test.

f”(x) = -4 sin2x

f”(π/4) =

x = π/4 and x = 5π/4 are point of local maxima.

x = 3π/4 and x = 7π/4 are point of local minima.

f(π/4) = f(5π/4) = 1 and f(3π/4) = f(7π/4) = -1

Now,

Global maxima = max{f(0), f(π/4), f(3π/4), f{5π/4}, f(7π/4), f(2π)}

= max{0, 1, -1, 1, -1, 0}

= 1

Global maxima occur at the points x = π/4 and x = 5π/4 and the absolute maximum value is 1.

### Question 9. What is the maximum value of the function sin x + cos x?

**Solution:**

Given that, f(x) = sin x + cos x

f'(x) = cos x – sin x

Now put f'(x) = 0

cos x = sin x

= {-√2, √2, +√2, -√2, -√2}

Now, second order derivative test,

f”(x) = -sin x – cos x

f”(π/4) = f”(9π/4) = f”(17π/4)……….. = -√2 < 0

A liter ⇢ f(x) = sin x + cos x =

=

### Question 10. Find the maximum value of 2x^{3 }– 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [-3, -1].

**Solution:**

Given that f(x) = 2x

^{3 }– 24x + 107On differentiating w.r.t. x we get

f'(x) = 6x

^{2 }– 24Now, put f'(x) = 0

6x

^{2 }= 24x

^{2 }= 4x = ±2

Now second order test

f”(x) = 12x

f”(2) = 12.2 = 24 > 0

x = 2 is a pt. of local minima

f(2) = 75

f”(-2) = 12(-2) = -24 < 0

x = -2 is point of local maxima. f(-2) = 139

Now, in the interval [1, 3]

Global maxima = max{f(1), f(2), f(3)}

= max{85, 75, 89}

= 89

Now, in the interval [-3,-1]

Global maxima = max{f(-3), f(-2), f(-1)}

= max{125, 139, 129}

= 139

### Question 11. It is given that at x = 1, the function x^{4 }– 62x^{2 }+ ax + 9 attains its maximum value on the interval [0, 2]. Find the value of a.

**Solution:**

Give that, f(x) = x

^{4 }– 62x^{2 }+ ax + 9On differentiating w.r.t. x we get

f'(x) = 4x

^{3 }– 124x + aThe maximum value is attained at x = 1, and 1 lies between 0 and 2.

So, at x = 1, there must be a local maxima

That means, f'(1) = 0

f'(1) = 4(1)

^{3 }– 124(1) + a = 0-120 + a = 0

a = 120

### Question 12. Find the maximum and minimum values of x + sin2x on [0, 2π].

**Solution:**

Give that f(x) = x + sin2x, x ∈ [0, 2π]

On differentiating w.r.t. x we get

f'(x) = 1 + 2cos2x

Now put f'(x) = 0, we get

1 + 2cos2x = 0

cos2x = -1/2

∈ [0, 2π]

Now,

For global maxima = max{f(0), f(π/3), f(4π/3), f(2π)}

= max{0, π/3, }

= 2π

Global maxima occur at x = 2π and the maximum value is f(2π) = 2π.

For global minima = min{f(0), f(2π/3), f(5π/3), f(2π)}

= min{0, }

= 0

Global minima occur at x = 0 and the minimum value is 0.