**Class 12 Maths NCERT solutions of exercise 6.5-Application of Derivations**

**Class 12 maths NCERT solutions of chapter 6.5- Application of Derivation** are explained here beautifully,we are sure every student of** 12 class** will under the **Maths NCERT solutions of** **exercise 6.5-Application of Derivations** completely. You can also download pdf of **NCERT solutions of exercise 6.5-Application of Derivation** from the link is given here.

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**Exercise 6.1 – Application of Derivatives**

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**Exercise 6.3 – Application of Derivatives**

**Exercise 6.4- Application of Derivatives**

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**Class 12 Maths NCERT solutions of exercise 6.5-Application of Derivations**

**Q1.Find the maximum and minimum values, if any, of the following functions given by:**

**(i) f(x) = (2x – 1)² + 3**

**(ii) f(x) = 9x² + 12x + 2**

**(iii) f(x) = -(x – 1)² + 10**

**(iv) g(x) =x³ + 1**

Ans. We are given the function f(x) = (2x – 1)² + 3

Let y = f(x)

y = (2x – 1)² + 3 = 4x² +1 -4x + 3

y = 4x² -4x + 4

We can get the slope of the tangent of the curve by differentiating y with respect to x.

The slope of the tangent drawn on the point of maxima or minima is =0

So, 8x – 4 = 0

x = 4/8 = 1/2

In order to get x = 1/2 is located at maxima or minima,double differenting y

The second derivative is positive indicates that the direction of the slope is – to +, means the slope is increasing so the point x= 1/2 is minima.

f(x) has no maximum value

Therefore the minimum value of f(x) at x = 1/2 is = 4x² -4x + 4 = 4(1/2)² – 4×1/2 +4 =3

The point of minima = (1/2,3)

On the same way, you can do other sections of the question number 1.

**Hint: If the second derivative is < 0 (i.e a negative value), then the slope is supposed to be decreasing, means that is tending from + to -. then f(x) is maxima at the value of x which comes after writing the second derivative equal to 0.**

**Q2.Find the maximum and minimum values, if any, of the following functions given by:**

Ans. We are given the function

Since a modulus function is not differentiable,therefore finding its extreme points(maxima or minima) by other methods.

As we know

It is clear that minimum value of x +2 = 0⇒ x = -2

At x = -2 the minimum value of f(x) is = -1

The maxima of f(x) →∞, so f(x) has no maxima.

(ii) You can do second part of the Q2 by the same method.

We know the value of sin θ

-1 ≤ sin θ ≤ 1

-1 ≤ sin 2x ≤ 1

Adding 5 in each side of an inequality

-1 +5≤ sin 2x+5 ≤ 1+5

4 ≤ h(x) ≤ 6

Therefore the minimum value of the f(x) is =4 and the maximum value is 6

Solve (iv) and (v) part of Q2. by the same method.

**Q3**.**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) = x²**

**(ii) g(x) = x³ – 3x**

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

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

**(v) f(x) = x³ – 6x² +9x + 15**

Ans.(i) f(x) = x²

Let y = f(x)

y = x²

The slope of the curve

Putting the slope equal to 0

2x = 0

x = 0

hence x = 0 is the extreme point of the f(x)

Second derivation (i.e the rate of change of slope)

The value of second derivative at x = 0 is = 2,which is positive,it shows that the slope is increasing from – to + therefore f(x) has minima at x = 0.

The value of minima of f(x) is f(0) = 0² = 0

(ii) g(x) = x³ – 3x

let y = x³ – 3x

3x² – 3 = 0

3(x² – 1) = 0

3(x + 1) 3(x – 1)

x = -1, x = 1

Second derivative of y

The value of second derivative at x = -1 is 6 × -1 = -6,it shows that the slope is decreasing from + to –

Hence, g(x) has maxima at x = -1, and its value is g(-1) = (-1)³ – 3× – 1 = -1 + 3 = 2

The value of second derivative at x = 1, is 6 × 1 = 6, it shows that slope is increasing from – to +

Hence g(x) has minima at x = 1 and its value is

g(1) = 1³ -3 × 1 =1 – 3 = -2

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

Let y = h(x)

y = sin x + cos x

The slope of the curve is as follows

Putting the slope equal to 0

⇒cos x – sin x = 0

⇒sinx = cos x

tan x = 1

tan x is positive in I st and III rd quadrant

Since, we are given 0 < x < π/2, so tan x lies in I st quadrant

Therefore, tan x = tan π/4⇒ x = π/4

x = π/4 is the terminal point of f(x)

Determining second derivative of y

The value of second derivative at x = π/4 is as follows

⇒-sin π/4 – cos π/4 = -1/√2 -1/√2 = -√2

The value of second derivative is -means h(x) has local maxima at x =π/4

So,value of local maxima of h(x)

h(π/4) = sin(π/4) + cos (π/4) =1/√2 +1/√2= √2

You can do other parts of question 3 by following the same method

**Q4**.**Prove that the following functions do not have maxima or minima:**

Ans.

Let y = f(x)

Putting the slope equal to 0

x ≠ real value, so f(x) can not have maxima or minima

Let y = g(x)

y = log x

Putting the slope equal to 0

It is impossible,so g(x) can not have maxima or minima

Let y = h(x)

y = x³ + x² + x +1

Putting the slope equal to 0

3x² + 2x + 1 = 0, calculating its diopter

Roots of h(x) are imaginary,hence h(x) can not have maxima or minima

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**Q5.Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:**

Ans.(i) f(x) = x³

Let f(x) = y

y = x³

3x² = 0

x = 0

x =0 , x = -2 and x = 2 are three terminal points of f(x)

Value of f(x) at x =0,x = -2 and x =2 are following

f(0) =0, f(-2) = -2³ =-8, f(2) =2³ =8

Therefore absolute maximum and minimum value of f(x) are 8 and -8 respectively.

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