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

### (i) f(x) = (2x – 1)2 + 3

Solution:

Given that, f(x) = (2x – 1)2 + 3

From 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 = 3

Hence, the minimum value of the function is 3 and this function does not contain maximum value.

### (ii) f(x) = 9x2 + 12x + 2

Solution:

Given that, f(x) = 9x2 + 12x + 2

we can also write as f(x) = (3x + 2)2 – 2

From 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 = -2

Hence, 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 + 10

From 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 = 10

Hence, the maximum value of the function is 10 and this function does not contain minimum value.

### (iv) g(x) = x3 + 1

Solution:

Given that, g(x) = x3 + 1

When x —> ∞, then g(x) —> ∞

When x —> -∞, then g(x) —> -∞

So, this function has neither minimum nor maximum value

### (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.

### (i) f(x) = x2

Solution:

Given that f(x) = x2

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) = x3 – 3x

Solution:

Given that g(x) = x3 – 3x

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

g'(x) = 3x2 – 3

Now put g'(x) = 0

3x2 – 3 = 0

x2 = 1

x = ±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) = -2

So by second derivatives test, x = -1 is a point of local minima and the minimum value is

g(-1) = (-1)3 – 3(-1) = 2

Hence, 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”(3π​/4) = – √2 > 0

f”(7π​/4) = √2 > 0

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

f( ) = -sin 3π​/4 + cos 3π
4 = 1/√2 + 1/√2 = √2 > 0

So by second derivatives test, x = is a point of local minima and the minimum value is

f( ) = -sin 7π​/4 + cos 7π
4 = -1/√2 – 1/√2 = -√2 > 0

Hence, the local minimum value is -√2 and the local maximum value is √2.

### (v) f(x) = x3 – 6x2 + 9x + 15

Solution:

Given that, f(x) = x3 – 6x2 + 9x + 15

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

f'(x) = 3x2 – 12x + 9

Now put f'(x) = 0

3x2 – 12x + 9 = 0

3(x2 – 4x + 3) = 0

x = 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 = 19

So 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 = 15

Hence, 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 ### (i) f(x) = ex

Solution:

Given that, f(x) = ex

f'(x) = ex

Now ex > 0, f'(x) > 0

Hence, 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) = x3 + x2 + x + 1

Solution:

Given that, h(x) = x3 + x2 + x + 1

h'(x) = 3x2 + 2x + 1

Now for this quadratic expression 3x2 + 2x + 1,

Its discriminant 0 = 22 – 4(3)(1) = -8 < 0

So, 3x2 + 2x + 1 > 0

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

### (i) f(x) = x3, x ∈ [-2, 2]

Solution:

Given that, f(x) = x3, x ∈ [-2, 2]

f'(x) = 3x2

f'(x) = 0 at x = 0

f”(x) = 6x

f”(0) = 0, second derivative failure

Now f'(3+) > 0 and f'(3) > 0

f'(x) does not change sign at x = 0.

x = 0 is neither maxima nor minima

f(x) = x3 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 – 18x2

Solution:

Given that p(x) = 41 – 24x – 18x2

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. BC3 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 3x4 – 8x3 + 12x2 – 48x + 25 on the interval [0, 3].

Solution:

Given that f(x) = 3x4 – 8x3 + 12x2 – 48x + 25, x ∈ [0, 3]

f'(x) = 12x2 – 24x2 + 24x – 48

Now put f'(x) = 0

12x3 – 24x2 + 24x – 48 = 0

12(x2 – 2x2 + 2x – 4) = 0

12(x2(x – 2) + 2(x – 2)) = 0

12(x2 + 2)(x – 2) = 0

x = 2 because x2 + 2 ≠ 0

Now applying second derivative test,

f”(x) = 12(3x2 – 4x + 2)

f”(2) = 12(3.22 – 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 2x3 – 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [-3, -1].

Solution:

Given that f(x) = 2x3 – 24x + 107

On differentiating w.r.t. x we get

f'(x) = 6x2 – 24

Now, put f'(x) = 0

6x2 = 24

x2 = 4

x = ±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 x4 – 62x2 + ax + 9 attains its maximum value on the interval [0, 2]. Find the value of a.

Solution:

Give that, f(x) = x4 – 62x2 + ax + 9

On differentiating w.r.t. x we get

f'(x) = 4x3 – 124x + a

The 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.