**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.

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

When Thus, the area of the circle is changing at the rate of |

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

**Question:1** **a)** Find the rate of change of the area of a circle with respect to its radius r when

r = 3 cm

**Answer:**

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

**Question:1** **b)** Find the rate of change of the area of a circle with respect to its radius r when

r = 4 cm

**Answer:**

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

**Answer:**

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,

**Answer:**

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

**Answer:**

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

**Answer:**

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

**Answer:**

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

**Answer:**

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

**Question:7(b)** The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8cm and y = 6cm, find the rates of change of the area of the rectangle.

**Answer:**

Given same as previous question

Solution:-

Area of rectangle = xy

Hence, the rate of change of area is

**Answer:**

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

**Answer:**

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

**Answer:**

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

**Answer:**

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

**Answer:**

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

**Answer:**

Volume of sphere(V) =

Diameter =

So, radius(r) =

**Answer:**

Given = and

To find = at h = 4 cm

Solution:-

Volume of cone(V) =

**Question:15** The total cost C(x) in Rupees associated with the production of x units of an

item is given by

Find the marginal cost when 17 units are produced.

**Answer:**

Marginal cost (MC) =

Now, at x = 17

MC

Hence, marginal cost when 17 units are produced is 20.967

**Question:16** The total revenue in Rupees received from the sale of x units of a product is

given by

Find the marginal revenue when x = 7

**Answer:**

Marginal revenue =

at x = 7

Hence, marginal revenue when x = 7 is 208

**Answer:**

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

**Answer:**

Marginal revenue =

at x = 15

Hence, marginal revenue when x = 15 is 126

Hence, the correct answer is D

**Question:1**. Show that the function given by f (x) = 3x + 17 is increasing on R.

**Answer:**

Let are two numbers in R

Hence, f is strictly increasing on R

**Question:2.** Show that the function given by is increasing on R.

**Answer:**

Let are two numbers in R

Hence, the function is strictly increasing in R

**Question:3 a)** Show that the function given by f (x) = is increasing in

**Answer:**

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

**Answer:**

f(x) = sin x

Since, for each

So, we have

Hence, f(x) = sin x is strictly decreasing in

**Question:3** **c)** Show that the function given by f (x) = is neither increasing nor decreasing in

**Answer:**

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

**Question:4(a).** Find the intervals in which the function f given by is increasing

**Answer:**

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

**Question:4(b) **Find the intervals in which the function f given by is

decreasing

**Answer:**

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

**Question:5(a)** Find the intervals in which the function f given by is

increasing

**Answer:**

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)

**Question:5(b) **Find the intervals in which the function f given by is

decreasing

**Answer:**

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)

**Question:6(a) **Find the intervals in which the following functions are strictly increasing or

decreasing:

**Answer:**

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

**Question:6(b)** Find the intervals in which the following functions are strictly increasing or

decreasing

**Answer:**

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

**Question:6(c)** Find the intervals in which the following functions are strictly increasing or

decreasing:

**Answer:**

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)

**Question:6(d)** Find the intervals in which the following functions are strictly increasing or

decreasing:

**Answer:**

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

**Question:6(e)** Find the intervals in which the following functions are strictly increasing or

decreasing:

**Answer:**

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

**Question:7** Show that is an increasing function of x throughout its domain.

**Answer:**

Given function is,

Now, for , is is clear that

Hence, strictly increasing when

**Question:8** Find the values of x for which is an increasing function.

**Answer:**

Given function is,

Now,

So, the intervals are

In interval ,

Hence, is an increasing function in the interval

**Question:9** Prove that is an increasing function of

**Answer:**

Given function is,

Now, for

So,

Hence, is increasing function in

**Question:10** Prove that the logarithmic function is increasing on

**Answer:**

Let logarithmic function is log x

Now, for all values of x in ,

Hence, the logarithmic function is increasing in the interval

**Question:11** Prove that the function f given by is neither strictly increasing nor decreasing on (– 1, 1).

**Answer:**

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)

**Question:12** Which of the following functions are decreasing on

**Answer:**

(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

**Answer:**

(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

**Question:14** For what values of a the function f given by is increasing on

[1, 2]?

**Answer:**

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]

**Question:15** Let I be any interval disjoint from [–1, 1]. Prove that the function f given by is increasing on I.

**Answer:**

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]

**Question:16** Prove that the function f given by is increasing on

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

**Question:17 **Prove that the function f given by f (x) = log |cos x| is decreasing on

and increasing on

**Answer:**

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

**Question:18** Prove that the function given by is increasing in R.

**Answer:**

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

**Answer:**

Given function is,

Now, it is clear that only in the interval (0,2)

So, is an increasing function for the interval (0,2)

Hence, (D) is the answer

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

**Answer:**

Given curve is,

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

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

**Answer:**

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

**Question:3** Find the slope of the tangent to curve at the point whose x-coordinate is 2.

**Answer:**

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

**Question:4** Find the slope of the tangent to the curve at the point whose x-coordinate is 3.

**Answer:**

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

**Question:5** Find the slope of the normal to the curve

**Answer:**

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

**Question:6** Find the slope of the normal to the curve

**Answer:**

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

**Question:7** Find points at which the tangent to the curve is parallel to the x-axis.

**Answer:**

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).

**Answer:**

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

**Question:9** Find the point on the curve at which the tangent is

**Answer:**

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

**Question:10** Find the equation of all lines having slope –1 that are tangents to the curve

**Answer:**

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**

**Question:11** Find the equation of all lines having slope 2 which are tangents to the curve

**Answer:**

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

**Question:12** Find the equations of all lines having slope 0 which are tangent to the curve

**Answer:**

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

**Question:13(i)** Find points on the curve at which the tangents are parallel to x-axis

**Answer:**

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)

**Question:13(ii)** Find points on the curve at which the tangents are parallel to y-axis

**Answer:**

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)

**Question:14(i)** Find the equations of the tangent and normal to the given curves at the indicated

points:

**Answer:**

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

**Question:14(ii)** Find the equations of the tangent and normal to the given curves at the indicated

points:

**Answer:**

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

**Question:14(iii)** Find the equations of the tangent and normal to the given curves at the indicated

points:

**Answer:**

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

**Question:14(iv)** Find the equations of the tangent and normal to the given curves at the indicated points

**Answer:**

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

**Question:14(v)** Find the equations of the tangent and normal to the given curves at the indicated points:

**Answer:**

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

**Question:15(a)** Find the equation of the tangent line to the curve which is parallel to the line

**Answer:**

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

**Question:15(b)** Find the equation of the tangent line to the curve which is perpendicular to the line

**Answer:**

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

**Question:16** Show that the tangents to the curve at the points where x = 2 and x = – 2 are parallel.

**Answer:**

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

**Question:17** Find the points on the curve at which the slope of the tangent is equal to the y-coordinate of the point.

**Answer:**

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)

**Question:18** For the curve , find all the points at which the tangent passes

through the origin.

**Answer:**

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)

**Question:19** Find the points on the curve at which the tangents are parallel

to the x-axis.

**Answer:**

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)

**Question:20 **Find the equation of the normal at the point for the curve

**Answer:**

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

**Question:21 **Find the equation of the normals to the curve which are parallel

to the line

**Answer:**

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**

**Question:22** Find the equations of the tangent and normal to the parabola at the point

**Answer:**

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

**Question:23** Prove that the curves and xy = k cut at right angles*

**Answer:**

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

**Question:24** Find the equations of the tangent and normal to the hyperbola

at the point

**Answer:**

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

**Question:25** Find the equation of the tangent to the curve which is parallel to the line

**Answer:**

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

**Question:26** The slope of the normal to the curveis

(A) 3 (B) 1/3 (C) –3 (D) -1/3

**Answer:**

Equation of the given curve is

Slope of tangent =

at x = 0

Now, we know that

Hence, (D) is the correct option

**Question:27** The line is a tangent to the curve at the point

(A) (1, 2) (B) (2, 1) (C) (1, – 2) (D) (– 1, 2)

**Answer:**

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

**Question:1(i) **Using differentials, find the approximate value of each of the following up to 3

places of decimal.

**Answer:**

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

**Question:1(ii)** Using differentials, find the approximate value of each of the following up to 3 places of decimal.

**Answer:**

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

**Question:1(iii)** Using differentials, find the approximate value of each of the following up to 3 places of decimal.

**Answer:**

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

**Question:1(iv)** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

**Answer:**

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

**Question:1(v)** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

**Answer:**

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)

**Question:1(vi)** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

**Answer:**

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

**Question:1(vii)** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

**Answer:**

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

**Question:1(viii)** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

**Answer:**

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

**Question:1(ix)** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

**Answer:**

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

**Question:1(x)** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

**Answer:**

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

**Question:1(xi)** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

**Answer:**

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)

**Question:1(xii)** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

**Answer:**

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)

**Question:1(xiii)** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

**Answer:**

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

**Question:1(xiv)** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

**Answer:**

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

**Question:1(xv)** Using differentials, find the approximate value of each of the following up to 3

places of decimal.

**Answer:**

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

**Question:2** Find the approximate value of f (2.01), where

**Answer:**

Let x = 2 and

We know that is approximately equal to dy

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

**Question:3** Find the approximate value of f (5.001), where

**Answer:**

Let x = 5 and

We know that is approximately equal to dy

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

**Answer:**

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

**Answer:**

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

**Answer:**

Error in radius of sphere = 0.02 m

Volume of sphere =

Error in volume

Hence, the approximate error in its volume is

**Answer:**

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

**Question:8** If , then the approximate value of f (3.02) is

(A) 47.66 (B) 57.66 (C) 67.66 (D) 77.66

**Answer:**

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

**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

**Question:1(i)** Find the maximum and minimum values, if any, of the following functions

given by

(

**Answer:**

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

**Question:1(ii)** Find the maximum and minimum values, if any, of the following functions

given by

**Answer:**

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

**Question:1(iii)** Find the maximum and minimum values, if any, of the following functions

given by

**Answer:**

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

**Question:1(iv)** Find the maximum and minimum values, if any, of the following functions

given by

**Answer:**

Given function is,

value of varies from

Hence, function neither has a maximum or minimum value

**Question:2(i)** Find the maximum and minimum values, if any, of the following functions

given by

**Answer:**

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

**Question:2(ii)** Find the maximum and minimum values, if any, of the following functions

given by

**Answer:**

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 t