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# 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

Exercise 6.2 – Application of Derivatives

Exercise 6.3 – Application of Derivatives

Exercise 6.4- Application of Derivatives

## 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.

$\boldsymbol{\frac{dy}{dx}=8x-4}$

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

$\boldsymbol{\frac{d^{2}y}{dx^{2}}=8}$

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:

$\boldsymbol{\left ( i \right )\: f\left ( x \right )=\left | x +2 \right |-1}$

$\boldsymbol{\left ( ii \right )\: g\left ( x \right )=\left | x +1 \right |+3}$

$\boldsymbol{\left ( iii \right )\: h\left ( x \right )=sin\left ( 2x \right )+5}$

$\boldsymbol{\left ( iv \right )\: f\left ( x \right )=\left | sin\: 4x+3 \right |}$

$\boldsymbol{\left ( v \right )\: h\left ( x \right )=x+1,\: x\epsilon \left ( -1,1 \right )}$

Ans. We are given the function

$\boldsymbol{\left ( i \right )\: f\left ( x \right )=\left | x +2 \right |-1}$

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

As we know

$\boldsymbol{x+2\geq 0,for\: all \: x\epsilon R}$

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.

$\boldsymbol{\left ( iii \right )\: h\left ( x \right )=sin\left ( 2x \right )+5}$

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

$\boldsymbol{\left ( vi \right )\: g\left ( x \right )=\frac{x}{2}+\frac{2}{x},x>0}$

$\boldsymbol{\left ( vii \right )g\left ( x \right )=\frac{1}{x^{2}+2}}$

Ans.(i) f(x) = x²

Let y = f(x)

y = x²

The slope of the curve

$\boldsymbol{\frac{dy}{dx}=2x}$

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)

$\boldsymbol{\frac{d^{2}y}{dx^{2}}=2}$

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

$\boldsymbol{\frac{dy}{dx}=3x^{2}-3}$

3x² – 3 = 0

3(x² – 1) = 0

3(x + 1) 3(x – 1)

x = -1, x = 1

Second derivative of y

$\boldsymbol{\frac{d^{2}y}{dx^{2}}=6x}$

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

$\boldsymbol{\frac{dy}{dx}=cos\: x -sin\: x}$

Putting the slope equal to 0

⇒cos x – sin x = 0

⇒sinx = cos x

$\boldsymbol{\frac{sin\: x}{cos\: x}=1}$

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

$\boldsymbol{\frac{d^{2}y}{dx^{2}}=-sin \: x-cos\: x}$

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:

$\boldsymbol{\left ( i \right )f\left ( x \right )=e^{x}}$

$\boldsymbol{\left ( ii \right )g\left ( x \right )=log\: x}$

$\boldsymbol{\left ( ii i\right )h\left ( x \right )=x^{3}+x^{2}+x+1}$

Ans.

$\boldsymbol{\left ( i \right )f\left ( x \right )=e^{x}}$

Let y = f(x)

$\boldsymbol{y=e^{x}}$

$\boldsymbol{\frac{dy}{dx}=\frac{d}{dx}\left ( e^{x} \right )=e^{x}}$

Putting the slope equal to 0

$\boldsymbol{e^{x}=0}$

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

$\boldsymbol{\left ( ii \right )g\left ( x \right )=log\: x}$

Let y = g(x)

y = log x

$\boldsymbol{\frac{dy}{dx}=\frac{1}{x}}$

Putting the slope equal to 0

$\boldsymbol{\frac{1}{x}=0\Rightarrow 1=0}$

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

$\boldsymbol{\left ( ii i\right )h\left ( x \right )=x^{3}+x^{2}+x+1}$

Let y = h(x)

y = x³ + x² + x +1

$\boldsymbol{\frac{dy}{dx}=3x^{2}+2x +1}$

Putting the slope equal to 0

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

$\boldsymbol{D=\sqrt{b^{2}-4ac}=\sqrt{2^{2}-4\times 3\times 1}=\sqrt{-8}=2\sqrt{2}i}$

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

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

$\boldsymbol{\left ( i \right )f\left ( x \right )=x^{3},x\: \epsilon\: [-2,2]}$

$\boldsymbol{\left ( ii \right )f\left ( x \right )=sin\: x+cos\: x: [0,\pi ]}$

$\boldsymbol{\left ( iii \right )f\left ( x \right )\: =4x-\frac{1}{2}x^{2},x\epsilon \left [ -2,\frac{9}{2} \right ]}$

$\boldsymbol{\left ( iv \right )f\left ( x \right )=\left ( x-1 \right )^{2}+3,x\epsilon \left [ -3,\frac{9}{2} \right ]}$

Ans.(i) f(x) = x³

Let f(x) = y

y = x³

$\boldsymbol{\frac{dy}{dx}=3x^{2}}$

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.

Class 12 maths NCERT Solutions of exercise 6.5

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