# NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives

NCERT solutions for class 12 maths chapter 6 Application of Derivatives: In the previous chapter, you have already learnt the differentiation of inverse trigonometric functions, exponential functions, logarithmic functions, composite functions, implicit functions, etc. In this article, you will get NCERT solutions for class 12 maths chapter 6 application of derivatives. The derivative has many applications in various fields like social science, physics, optimization, science, engineering etc. CBSE NCERT solutions for class 12 maths chapter 6 applications of derivatives will cover questions on some specific applications like graph, functions and approximations. Questions based on the topics like finding the rate of change of quantities, equations of tangent and normal on a curve at a point, local minima and maxima of a function, intervals on which a function is increasing or decreasing, and approximate value of certain quantities are covered in the solutions of NCERT class 12 maths chapter 6 application of derivatives. Check all NCERT solutions at a single place which will help you to learn CBSE maths and science.

If you are good in differentiation, it won't take much effort to learn its applications. This chapter alone has 11% weightage in 12 board final examination, which means you can score very easily by your basic knowledge of maths and basic differentiation. This chapter seems to be very easy but there are chances of silly mistakes as it requires knowledge of other chapters also. So, practice all the NCERT questions on your own, you can take help of these CBSE NCERT solutions for class 12 maths chapter 6 application of derivatives. There are five exercises with 102 questions. All these questions are explained in this solutions of NCERT class 12 maths chapter 6 application of derivatives article.

## What is the derivative?

The derivative  is the rate of change of distance(S) with respect to the time(t). In a similar manner, whenever one quantity (y) varies with another quantity (x), and also satisfy ,then  or  represents the rate of change of y with respect to x  and or represents the rate of change of y with respect to x at .  Let's take an example of a derivative

 Example- Find the rate of change of the area of a circle per second with respect to its radius r when r = 5 cm. Solution-  The area A of a circle with radius r is given by . Therefore, the rate of change of the area(A) with respect to its radius(r) is given by -                                            When   Thus, the area of the circle is changing at the rate of

## Topics of NCERT Grade 12 Maths Chapter-6 Application of Derivatives

6.1 Introduction

6.2 Rate of Change of Quantities

6.3 Increasing and Decreasing Functions

6.4 Tangents and Normals

6.5 Approximations

6.6 Maxima and Minima

6.6.1 Maximum and Minimum Values of a Function in a Closed Interval

## NCERT solutions for class 12 maths chapter 6 Application of Derivatives-Exercise: 6.1

Area of the circle (A) =
Rate of change of the area of a circle with respect to its radius r =  =  =
So, when r = 3, Rate of change of the area of a circle =   =
Hence, Rate of change of the area of a circle with respect to its radius r when r = 3 is

Area of the circle (A) =
Rate of change of the area of a circle with respect to its radius r =  =  =
So, when r = 4, Rate of change of the area of a circle =   =
Hence, Rate of change of the area of a circle with respect to its radius r when r = 4 is

The volume of the cube(V) =      where x is the edge length of the cube
It is given that the volume of a cube is increasing at the rate of

we can write         ( By chain rule)

- (i)
Now, we know that the surface area of the cube(A) is

- (ii)

from equation (i) we know that

put this value in equation (i)
We get,

It is given in the question that the value of  edge length(x) = 12cm
So,

Radius of a circle is increasing uniformly at the rate  =  3 cm/s
Area of circle(A) =
(by chain rule)

It is given that the value of r = 10 cm
So,

Hence,  the rate at which the area of the circle is increasing when the radius is 10 cm  is

It is given that the rate at which edge of cube increase    = 3 cm/s
The volume of cube =
(By chain rule)

It is given that the value of x is 10 cm
So,

Hence, the rate at which the volume of the cube increasing when the edge is 10 cm long is

Given =

To find =      at  r = 8 cm

Area of the circle (A) =
(by chain rule)

Hence, the rate at which the area increases when the radius of the circular wave is 8 cm is

Given =
To find =               , where C is circumference
Solution :-

we know that the circumference of the circle (C) =
(by chain rule)

Hence,  the rate of increase of its circumference is

the perimeter of rectangle

Given =   Length x of a rectangle is decreasing at the rate   =   -5  cm/minute                  (-ve sign indicates decrease in rate)
the width y is increasing at the rate    = 4 cm/minute
To find =      and      at  x = 8 cm and y = 6 cm                     , where P is perimeter
Solution:-

Perimeter of rectangle(P) = 2(x+y)

Hence, Perimeter decreases at the rate of

Given same as previous question
Solution:-
Area of rectangle = xy

Hence, the rate of change of area is

Given =
To find =      at r = 15 cm
Solution:-

Volume of sphere(V) =

Hence,  the rate at which the radius of the balloon increases when the radius is 15 cm is

We need to find the value of   at r =10 cm
The volume of the sphere (V) =

Hence,  the rate at which its volume is increasing with the radius when the later is 10 cm is

Let h be the height of the ladder and x be the distance of the foot of the ladder from the wall
It is given  that
We need to find  the rate at  which the height of the ladder decreases
length of ladder(L) = 5m and x = 4m    (given)
By Pythagoras theorem, we can say that

Differentiate on both sides w.r.t.  t

at x = 4

Hence, the rate at which the height of ladder decreases is

We need to find the point at which
Given the equation of curve =
Differentiate both sides w.r.t. t

(required condition)

when x = 4 ,
and
when x = -4 ,
So , the coordinates are

It is given that
We know that the shape of the air bubble is spherical
So, volume(V) =

Hence, the rate of change in volume is

Volume of sphere(V) =
Diameter =

Given =        and
To find =     at h = 4 cm
Solution:-

Volume of cone(V) =

Find the marginal cost when 17 units are produced.

Marginal cost (MC) =

Now, at x = 17
MC

Hence, marginal cost when 17 units are produced is 20.967

Find the marginal revenue when x = 7

Marginal revenue =

at x = 7

Hence, marginal revenue when x = 7 is 208

Area of circle(A) =

Now, at r = 6cm

Hence, the rate of change of the area of a circle with respect to its radius r at r = 6 cm is
Hence, the correct answer is B

Marginal revenue =

at x = 15

Hence, marginal revenue when x = 15 is 126
Hence, the correct answer is D

## Solutions of NCERT for class 12 maths chapter 6 Applications of Derivatives-Exercise: 6.2

Let     are two numbers in R

Hence, f is strictly increasing on R

Let  are two numbers in R

Hence, the function is  strictly increasing in R

Given f(x) = sinx

Since,

Hence, f(x) = sinx is strictly increasing in

Question:3 b) Show that the function given by f (x) = is

decreasing in

f(x) = sin x

Since,    for each
So, we have
Hence, f(x) = sin x is strictly decreasing in

We know that sin x is strictly increasing in    and strictly decreasing in
So, by this, we can say that f(x) = sinx is neither increasing or decreasing in range

Now,

4x - 3 = 0 So, the range is
So,
when          Hence, f(x) is strictly decreasing in this range
and
when               Hence, f(x) is strictly increasing in this range
Hence,    is strictly increasing in

Now,
&nnbsp;
4x - 3 = 0 So, the range is
So,
when          Hence, f(x) is strictly decreasing in this range
and
when               Hence, f(x) is strictly increasing in this range
Hence,    is strictly decreasing  in

It is given that

So,

x = -2 ,  x = 3 So, three ranges are there
Function      is positive in interval     and negative in the interval  (-2,3)
Hence,   is strictly increasing in
and  strictly decreasing in the interval  (-2,3)

We have

Differentiating the function with respect to x,  we get  :

or

When          ,   we have  :

or So, three ranges are there
Function      is positive in the interval     and negative in the interval  (-2,3)

So,               f(x) is decreasing  in  (-2, 3)

f(x)  =

Now, The range is from
In interval     is -ve
Hence, function f(x)  =  is strictly decreasing in interval
In interval    is +ve
Hence, function f(x)  =  is strictly increasing in interval

Given function is,

Now,

So, the  range is
In interval   ,   is +ve
Hence,  is strictly increasing in the interval
In interval   ,  is -ve
Hence,  is strictly decreasing in interval

Given function is,

Now, So, the range is
In interval    ,  is -ve
Hence,  is strictly decreasing in interval
In interval (-2,-1)  ,  is +ve
Hence,  is strictly increasing in the interval (-2,-1)

Given function is,

Now, So, the range is
In interval  ,    is +ve
Hence,    is strictly increasing in interval
In interval  ,  is -ve
Hence,    is strictly decreasing in interval

Given function is,

Now,

So, the intervals are Our function    is +ve in the interval
Hence,    is strictly increasing in the interval
Our function    is -ve in the interval
Hence,    is strictly decreasing in interval

Given function is,

Now, for    , is is clear that
Hence,     strictly increasing when

Given function is,

Now,

So, the intervals are
In interval   ,
Hence,  is an increasing function in the interval

Given function is,

Now,  for

So,
Hence,  is increasing function in

Let logarithmic function is log x

Now, for all values of x in  ,
Hence,  the logarithmic function   is increasing in the interval

Given function is,

Now, for interval  ,          and for interval
Hence, by this, we can say that    is neither strictly increasing nor decreasing in the interval (-1,1)

(A)

for x in
Hence,  is decreasing function in

(B)

Now, as

for 2x in
Hence,  is decreasing function in

(C)

Now, as

for    and
Hence, it is clear that   is neither increasing nor decreasing in

(D)

for x in
Hence,  is strictly increasing function in the interval

So, only (A) and (B) are decreasing functions in

(A) Given function is,

Now, in interval (0,1)

Hence,   is increasing function in interval  (0,1)

(B)  Now, in interval

,
Hence,   is increasing function in interval

(C)  Now, in interval

,
Hence,   is increasing function in interval

So,   is increasing for all cases
Hence, correct answer is (D) None of these

Given function is,

Now,  we can clearly see that  for every  value of

Hence,    is increasing for every value of      in the interval [1,2]

Given function is,

Now, So, intervals are from
In interval   ,

Hence,  is increasing   in  interval
In interval (-1,1) ,

Hence,  is decreasing   in  interval  (-1,1)
Hence, the function f given by   is increasing on I  disjoint from [–1, 1]

Given function is,

Now, we know that  cot x is+ve in the interval     and -ve  in the interval

Hence,   is increasing in the interval  and   decreasing in interval

Given function is,
f(x) =  log|cos x|
value of cos x is always +ve in both these  cases
So, we can write   log|cos x| = log(cos x)
Now,

We know that  in interval   ,

Hence, f(x) =  log|cos x| is decreasing in interval

We know that  in interval    ,

Hence, f(x) =  log|cos x| is increasing in interval

Given function is,

We can clearly see that for any value of x in R
Hence,   is an increasing function in R

Question:19 The interval in which is increasing is

(A) (B) (C) (D)

Given function is,

Now, it is clear that   only in the interval (0,2)
So,   is  an increasing function for the interval  (0,2)

## CBSE NCERT solutions for class 12 maths chapter 6 Application of Derivatives-Exercise: 6.3

Question:1. Find the slope of the tangent to the curve

Given curve is,

Now, the slope of the tangent at point  x =4  is given by

Question:2Find the slope of the tangent to the curve

Given curve is,

The slope of the tangent at x = 10 is given by

at x = 10

hence, slope of tangent at x = 10 is

Given curve is,

The slope of the tangent at x = 2 is given by

Hence, the slope of the tangent at point x = 2 is 11

Given curve is,

The slope of the tangent at x = 3 is given by

Hence, the slope of tangent at point x = 3 is 24

The slope of the tangent at a point on a given curve is given by

Now,

Similarly,

Hence, the slope of the tangent at        is -1
Now,
Slope of normal =   =
Hence, the slope of normal at        is 1

The slope of the tangent at a point on given curves is given by

Now,

Similarly,

Hence, the slope of the tangent at   is
Now,
Slope of normal =   =
Hence, the slope of normal at    is

We are given  :

Differentiating the equation with respect to x,  we get :

or

or

It is given that tangent is parallel to the x-axis, so the slope of the tangent is equal to 0.

So,

or

Thus,                                       Either       x =  -1     or          x = 3

When x = -1 we get y = 12          and         if  x =3 we get y = -20

So the required points are   (-1, 12)  and  (3, -20).

(4, 4).

Points joining the chord is (2,0) and (4,4)
Now, we know that the slope of the curve with given two points is

As it is given that the tangent is parallel to the chord, so their slopes are  equal
i.e.  slope of the tangent = slope of the chord
Given the equation of the curve is

Now, when
Hence, the coordinates are

We know that the  equation of a line is y = mx + c
Know the given equation of tangent is
y = x - 11
So, by comparing with the standard equation we can say that the slope of the tangent (m) = 1 and value of c if -11
As we know that  slope of the tangent at a point on the given curve is given by
Given the equation of curve is

When x = 2 ,
and
When x = -2 ,
Hence, the coordinates are (2,-9) and (-2,19), here (-2,19) does not satisfy the equation y=x-11

Hence, the coordinate is (2,-9) at which the tangent is

We know that the slope of the tangent of at the point of the given curve is given by

Given the equation of curve is

It is given thta slope is -1
So,

Now, when x = 0 ,
and
when x = 2 ,
Hence, the coordinates are (0,-1) and (2,1)
Equation of line passing through (0,-1) and having slope = -1 is
y = mx + c
-1 = 0 X -1 + c
c = -1
Now equation of line is
y = -x -1
y + x + 1 = 0
Similarly, Equation of line passing through (2,1) and having slope = -1 is
y = mx  + c
1 = -1 X 2   + c
c = 3
Now equation of line is
y = -x + 3
y + x - 3 = 0

We know that the slope of the tangent of at the point of the given curve is given by

Given the equation of curve is

It is given that slope is 2
So,

So, this is not possible as our coordinates are imaginary numbers
Hence, no tangent is possible with slope 2  to the curve

We know that the slope of the tangent  at a point on the given curve is given by

Given the equation of the curve as

It is given thta slope is 0
So,

Now, when x = 1 ,

Hence, the coordinates are
Equation of line passing through  and having slope = 0 is
y = mx + c
1/2 = 0 X 1 + c
c = 1/2
Now equation of line is

Parallel to x-axis means slope of tangent is 0
We know that slope of tangent at a given point on the given curve is given by
Given the equation of the  curve is

From this, we can say that
Now. when   ,
Hence, the coordinates are (0,4) and (0,-4)

Parallel to y-axis means the slope of the tangent is   , means the slope of normal is 0
We know that slope of the tangent at a given point on the given curve is given by
Given the equation of the curve is

Slope of normal =
From this we can say that y = 0
Now. when y = 0,
Hence, the coordinates are (3,0) and (-3,0)

We know that Slope of the tangent at a point on the given curve is given  by
Given the equation of the curve

at point (0,5)

Hence slope of tangent is -10
Now we know that,

Now, equation of tangent at point (0,5) with slope = -10 is

equation of tangent is

Similarly, the equation of normal at point (0,5) with slope = 1/10 is

equation of normal is

We know that Slope of tangent at a point on given curve is given  by
Given equation of curve

at point (1,3)

Hence slope of tangent is 2
Now we know that,

Now, equation of tangent at point (1,3) with slope = 2 is
y = 2x + 1
y  -2x = 1
Similarly, equation of normal at point (1,3) with slope = -1/2 is
y = mx + c

equation of normal is

We know that Slope of the tangent at a point on the given curve is given  by
Given the equation of the curve

at point (1,1)

Hence slope of tangent is 3
Now we know that,

Now, equation of tangent at point (1,1) with slope = 3 is

equation of tangent is

Similarly, equation of normal at point (1,1) with slope = -1/3 is
y = mx + c

equation of normal is

We know that Slope of the tangent at a point on the given curve is given  by
Given the equation of the curve

at point (0,0)

Hence slope of tangent is 0
Now we know that,

Now, equation of tangent at point (0,0) with slope = 0 is
y = 0
Similarly, equation of normal at point (0,0) with slope =  is

We know that Slope of the tangent at a point on the given curve is given  by
Given the equation of the curve

Now,
and
Now,

Hence slope of the tangent is -1
Now we know that,

Now, the equation of the tangent at the point  with slope = -1 is
and

equation of the tangent at

i.e.  is

Similarly, the equation of normal at   with slope = 1 is
and

equation of the tangent at

i.e.  is

Parellel to line  means slope of tangent and slope of line is equal
We know that the equation of line is
y = mx + c
on comparing with the given equation we get slope of line m = 2 and c = 9
Now, we know that the  slope of tangent at a given point to given curve is given by
Given equation of curve is

Now, when x = 2 ,
Hence, the coordinates are (2,7)
Now, equation of tangent paasing through (2,7) and with slope m = 2 is
y = mx + c
7 = 2 X 2 + c
c = 7 - 4 = 3
So,
y = 2 X x+ 3
y = 2x + 3
So, the equation of tangent is y - 2x = 3

Perpendicular to line    means
We know that the equation of the line is
y = mx + c
on comparing with the given equation we get the slope of line m = 3 and c = 13/5

Now, we know that the  slope of the tangent at a given point to given curve is given by
Given the equation of curve is

Now, when  ,
Hence, the coordinates are
Now, the equation of tangent passing through (2,7) and with slope   is

So,

Hence, equation of tangent is 36y + 12x = 227

Slope of tangent =
When x = 2

When  x = -2

Slope is equal when x= 2 and x = - 2
Hence, we can say that both the tangents to curve  is parallel

Given equation of curve is
Slope of tangent =
it is given that  the slope of the tangent is equal to the y-coordinate of the point

We have

So, when x = 0 , y = 0
and when x = 3 ,

Hence, the coordinates are (3,27) and (0,0)

Tangent passes through origin so, (x,y) = (0,0)
Given equtaion of curve is
Slope of tangent =

Now, equation of tangent is

at (0,0)  Y =  0 and X = 0

and we have

Now, when x = 0,

when x = 1 ,

when x= -1 ,

Hence, the coordinates are (0,0) , (1,2) and (-1,-2)

parellel to x-axis means slope is 0
Given equation of curve is

Slope of  tangent =

When x = 1 ,

Hence, the coordinates are (1,2) and (1,-2)

Given equation of curve is

Slope of tangent

at point

Now, we know that

equation of normal at point  and with slope

Hence, the equation of normal is

Equation of given curve is

Parellel to line  means slope of normal and line is equal
We know that, equation of line
y= mx + c
on comparing it with  our given equation. we get,

Slope of tangent =
We know that

Now, when x = 2,
and
When x = -2 ,
Hence, the coordinates are (2,18) and (-2,-6)
Now, the equation of at point (2,18) with slope

Similarly,  the equation of at point (-2,-6) with slope

Hence, the equation of the normals to the curve which are parallel
to the line

are x +14y - 254 = 0  and  x + 14y +86 = 0

Equation of the given curve is

Slope of tangent =
at point

Now, the equation of tangent with point  and slope  is

We know that

Now, the equation of at point    with slope -t

Hence, the equations of the tangent and normal to the parabola

at the point  are

Let suppose, Curve  and xy = k cut at the right  angle
then the slope of their tangent also cut  at the right angle
means,
-(i)

Now these values in equation (i)

Hence proved

Given equation is

Now ,we know that
slope of tangent =
at point

equation of tangent at point  with slope

Now, divide  both sides by

Hence, the equation of tangent is

We know that

equation of normal  at the point  with slope

Parellel to line  means the slope of tangent and slope of line is equal
We know that the equation of line is
y = mx + c
on comparing with the given equation we get the slope of line m = 2 and c = 5/2
Now, we know that the  slope of the tangent at a given point to given curve is given by
Given the equation of curve is

Now, when

,

but y cannot be -ve so we take only positive value
Hence, the coordinates are

Now, equation of tangent paasing through

and with slope m = 2 is

Hence, equation of tangent paasing through  and with slope m = 2 is   48x - 24y = 23

Equation of the given curve is

Slope of tangent =
at x = 0

Now, we know that

Hence, (D) is the correct option

The slope of the given line  is 1
given curve equation is

If the line is tangent to the given curve than the slope of the tangent is equal to the slope of the curve
The slope of tangent =

Now, when y = 2,
Hence, the coordinates are (1,2)

Hence, (A) is the correct answer

## NCERT solutions for class 12 maths chapter 6 Application of Derivatives-Exercise: 6.4

Lets suppose  and let x = 25 and
Then,

Now, we can say that   is approximate equals to dy

Now,

Hence,  is approximately equals to 5.03

Lets suppose  and let x = 49 and
Then,

Now, we can say that   is approximately equal to dy

Now,

Hence,  is approximately equal to 7.035

Lets suppose  and let x = 1 and
Then,

Now, we cam say that   is approximately equals to dy

Now,

Hence,  is approximately equal to 0.8

Lets suppose  and let x = 0.008 and
Then,

Now, we cam say that   is approximately equals to dy

Now,

Hence,  is approximately equal to 0.208

Lets suppose  and let x = 1 and
Then,

Now, we cam say that   is approximately equals to dy

Now,

Hence,  is approximately equal to 0.999  (because we need to answer up to three decimal place)

Let's suppose  and let x = 16 and
Then,

Now, we can say that   is approximately equal to dy

Now,

Hence,  is approximately equal to 1.969

Lets suppose  and let x = 27 and
Then,

Now, we can say that   is approximately equal to dy

Now,

Hence,  is approximately equal to 2.963

Let's suppose  and let x = 256 and
Then,

Now, we can say that   is approximately equal to dy

Now,

Hence,  is approximately equal to 3.997

Let's suppose  and let x = 81 and
Then,

Now, we can say that   is approximately equal to dy

Now,

Hence,  is approximately equal to 3.009

Let's suppose  and let x = 400 and
Then,

Now, we can say that   is approximately equal to dy

Now,

Hence,  is approximately equal to 20.025

Lets suppose  and let x = 0.0036 and
Then,

Now, we can say that   is approximately equal to dy

Now,

Hence,  is approximately equal to 0.060  (because we need to take up to three decimal places)

Lets suppose  and let x = 27 and
Then,

Now, we cam say that   is approximately equals to dy

Now,

Hence,  is approximately equal to 0.060  (because we need to take up to three decimal places)

Lets suppose  and let x = 81 and 0.5
Then,

Now, we can say that   is approximately equal to dy

Now,

Hence,  is approximately equal to 3.004

Let's suppose  and let x = 4 and
Then,

Now, we can say that   is approximately equal to dy

Now,

Hence,  is approximately equal to 7.904

Lets suppose  and let x = 32 and
Then,

Now, we can say that   is approximately equal to dy

Now,

Hence,  is approximately equal to 2.001

Let x = 2 and

We know that  is approximately equal to dy

Hence, the approximate value of f (2.01), where is 28.21

Let x = 5 and

We know that  is approximately equal to dy

Hence, the approximate value of f (5.001), where

Side of cube increased by 1% = 0.01x m
Volume of cube =
we know that  is approximately equal to dy
So,

Hence, the approximate change in volume V of a cube of side x metres caused by increasing the side by 1% is

Side of cube decreased by 1% = -0.01x m
The surface area of cube =
We know that,  is approximately equal to dy

Hence,  the approximate change in the surface area of a cube of side x metres
caused by decreasing the side by 1%. is

Error in radius of sphere  = 0.02 m
Volume of sphere =
Error in volume

Hence,  the approximate error in its volume is

Error in radius of sphere  = 0.03 m
The surface area of sphere =
Error in surface area

Hence,  the approximate error in its surface area is

Let x = 3 and

We know that  is approximately equal to dy

Hence,  the approximate value of f (3.02) is 77.66
Hence, (D) is the correct answer

Side of cube increased by 3% = 0.03x m
The volume of cube =
we know that  is approximately equal to dy
So,

Hence, the approximate change in volume V of a cube of side x metres caused by increasing the side by 3% is
Hence, (C) is the correct answer

## Solutions of NCERT for class 12 maths chapter 6 Application of Derivatives-Exercise: 6.5

Given function is,

Hence, minimum value occurs when

Hence, the minimum value of function  occurs at
and the minimum value is

and it is clear that there is no maximum value of

Given function is,

add and subtract 2 in given equation

Now,
for every
Hence, minimum value occurs when

Hence, the minimum value of function  occurs at
and the minimum value is

and it is clear that there is no maximum value of

Given function is,

for every
Hence, maximum value occurs when

Hence, maximum value of function   occurs at  x = 1
and the maximum value is

and it is clear that there is no minimum value of

Given function is,

value of  varies from
Hence, function   neither has a maximum or minimum value

Given function is

Hence, minimum value occurs when |x + 2| = 0
x = -2
Hence, minimum value occurs at x = -2
and minimum value is

It is clear that there is no maximum value  of the given function

Given function is

Hence, maximum value occurs when -|x + 1| = 0
x = -1
Hence, maximum value occurs at x = -1
and maximum value is

It is clear that there is no minimum value  of the given function

Given function is

We know that value of sin 2x varies from

Hence, the maximum value of our function    is  6  and the minimum value is 4

Given function is

We know that value of sin 4x varies from

Hence, the maximum value of our function    is  4  and the minimum value is 2

Given function is

It is given that the value of
So, we can not comment about either maximum or minimum value
Hence, function    has neither has a maximum or minimum value

Given function is

So, x = 0 is the only critical point of the given function
So we find it through the 2nd derivative test

Hence, by this, we can say that 0 is a point of minima
and the minimum value is

Given function is

Hence, the critical points are 1 and - 1
Now, by second derivative test

Hence, 1 is the point of minima and the minimum value is

Hence, -1 is the point of maxima and the maximum value is

Given function is

Now, we use the second derivative test

Hence,    is the point of maxima and the maximum value is  which is

Given function is

Now, we use second derivative test

Hence,    is the point of maxima and maximum value is  which is

Givrn function is

Hence 1 and 3 are critical points
Now, we use the second derivative test

Hence, x = 1 is a point of maxima and the maximum value is

Hence, x = 1 is a point of minima and the minimum value is

Given function is
( but as  we only take the positive value of x i.e. x = 2)
Hence, 2 is the only  critical point
Now, we use the second derivative test

Hence, 2 is the point of minima and the minimum value is

Gien function is

Hence., x = 0 is only critical point
Now, we use the second derivative test

Hence, 0 is the point of local maxima and the maximum value is

Given function is

Hence,  is the only critical point
Now,  we use the second derivative test

Hence, it is the point of minima and the minimum value is

Given function is

But exponential can never be 0
Hence, the function  does not have either  maxima or minima

Given function is

Since log x deifne for positive x i.e.
Hence,   by this, we can say that  for any value of x
Therefore, there is no  such that
Hence, the function  does not have either maxima or minima

Given function is

But, it is clear that there is no  such that
Hence, the function  does not have  either maxima or minima

Given function is

Hence, 0 is the critical point of the function
Now, we need to see the value of the function  at x = 0 and as  we also need to check the value at end points of given range i.e. x = 2 and x = -2

Hence, maximum value of function  occurs at x = 2 and value is 8
and minimum value of function  occurs at x = -2 and value is -8

Given function is

as
Hence,  is the critical point  of the function
Now, we need to check the value of function   at   and at the end points of given range  i.e.

Hence, the absolute maximum value of function  occures at  and value is
and absolute minimum value of function  occurs at  and value is -1

Given function is

Hence, x = 4 is the critical point  of function
Now, we need to check the value of function   at  x = 4  and at the end points of given range  i.e. at x = -2 and x =  9/2

Hence,  absolute maximum value of function   occures at x = 4 and value is 8
and absolute minimum value of function   occures at x = -2 and value is -10