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, c omposite 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. There are five exercises and a miscellaneous exercise.
If you are good in differentiation, it won't take much effort to learn its applications. T his 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 s olutions of NCERT class 12 maths chapter 6 a pplication of d erivatives 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 Chapter6 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 DerivativesExercise: 6.1
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
Solutions of NCERT for class 12 maths chapter 6 Applications of DerivativesExercise: 6.2
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) = logcos x
value of cos x is always +ve in both these cases
So, we can write logcos x = log(cos x)
Now,
We know that in interval
,
Hence, f(x) = logcos x is decreasing in interval
We know that in interval
,
Hence, f(x) = logcos 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
CBSE NCERT solutions for class 12 maths chapter 6 Application of DerivativesExercise: 6.3
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 xcoordinate 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 xcoordinate 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 xaxis.
Answer:
We are given :
Differentiating the equation with respect to x, we get :
or
or
It is given that tangent is parallel to the xaxis, 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=x11
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 xaxis
Answer:
Parallel to xaxis 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 yaxis
Answer:
Parallel to yaxis 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 ycoordinate 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 ycoordinate 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 xaxis.
Answer:
parellel to xaxis 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 curve
is
(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
NCERT solutions for class 12 maths chapter 6 Application of DerivativesExercise: 6.4
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
Solutions of NCERT for class 12 maths chapter 6 Application of DerivativesExercise: 6.5
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 there is no minimum value of the given function
Question:2(iii)
Find the maximum and minimum values, if any, of the following functions
given by
Answer:
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
Question:2(iv)
Find the maximum and minimum values, if any, of the following functions
given by
Answer:
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
Question:2(v)
Find the maximum and minimum values, if any, of the following functions
given by
Answer:
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
Answer:
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
Answer:
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
Answer:
Given function is
Now, we use the second derivative test
Hence,
is the point of maxima and the maximum value is
which is
Answer:
Given function is
Now, we use second derivative test
Hence,
is the point of maxima and maximum value is
which is
Answer:
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
Answer:
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
Answer:
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
Question:3(viii)
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:
Answer:
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
Question:4(i)
Prove that the following functions do not have maxima or minima:
Answer:
Given function is
But exponential can never be 0
Hence, the function
does not have either maxima or minima
Question:4(ii) Prove that the following functions do not have maxima or minima:
Answer:
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
Question:4(iii) Prove that the following functions do not have maxima or minima:
Answer:
Given function is
But, it is clear that there is no
such that
Hence, the function
does not have either maxima or minima
Question:5(i)
Find the absolute maximum value and the absolute minimum value of the following
functions in the given intervals:
Answer:
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
Question:5(ii)
Find the absolute maximum value and the absolute minimum value of the following
functions in the given intervals:
Answer:
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
Question:5(iii)
Find the absolute maximum value and the absolute minimum value of the following
functions in the given intervals:
Answer:
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
Question:5(iv) Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
Answer:
Given function is
Hence, x = 1 is the critical point of function
Now, we need to check the value of function
at x = 1 and at the end points of given range i.e. at x = 3 and x = 1
Hence, absolute maximum value of function
occurs at x = 3 and value is 19
and absolute minimum value of function
occurs at x = 1 and value is 3
Question:6
.
Find the maximum profit that a company can make, if the profit function is
given by
Answer:
Profit of the company is given by the function
x = 2 is the only critical point of the function
Now, by second derivative test
At x = 2
Hence, maxima of function
occurs at x = 2 and maximum value is
Hence, the maximum profit the company can make is 113 units
Question:7
.
Find both the maximum value and the minimum value of
on the interval [0, 3].
Answer:
Given function is
Now, by hit and trial let first assume x = 2
Hence, x = 2 is one value
Now,
which is not possible
Hence, x = 2 is the only critical value of function
Now, we need to check the value at x = 2 and at the end points of given range i.e. x = 0 and x = 3
Hence, maximum value of function
occurs at x = 0 and vale is 25
and minimum value of function
occurs at x = 2 and value is 39
Question:8 . At what points in the interval does the function attain its maximum value?
Answer:
Given function is
So, values of x are
These are the critical points of the function
Now, we need to find the value of the function
at
and at the end points of given range i.e. at x = 0 and
Hence, at
function
attains its maximum value i.e. in 1 in the given range of
Question:9 What is the maximum value of the function ?
Answer:
Given function is
Hence,
is the critical point of the function
Now, we need to check the value of the function
at
Value is same for all cases so let assume that n = 0
Now
Hence, the maximum value of the function
is
Question:10.
Find the maximum value of
in the interval [1, 3]. Find the
the maximum value of the same function in [–3, –1].
Answer:
Given function is
we neglect the value x = 2 because
Hence, x = 2 is the only critical value of function
Now, we need to check the value at x = 2 and at the end points of given range i.e. x = 1 and x = 3
Hence, maximum value of function
occurs at x = 3 and vale is 89 when
Now, when
we neglect the value x = 2
Hence, x = 2 is the only critical value of function
Now, we need to check the value at x = 2 and at the end points of given range i.e. x = 1 and x = 3
Hence, the maximum value of function
occurs at x = 2 and vale is 139 when
Question:11. It is given that at x = 1, the function attains its maximum value, on the interval [0, 2]. Find the value of a.
Answer:
Given function is
Function
attains maximum value at x = 1 then x must one of the critical point of the given function that means
Now,
Hence, the value of a is 120
Question:12 . Find the maximum and minimum values of
Answer:
Given function is
So, values of x are
These are the critical points of the function
Now, we need to find the value of the function
at
and at the end points of given range i.e. at x = 0 and
Hence, at
function
attains its maximum value and value is
in the given range of
and at x= 0 function
attains its minimum value and value is 0
Question:13 . Find two numbers whose sum is 24 and whose product is as large as possible.
Answer:
Let x and y are two numbers
It is given that
x + y = 24 , y = 24  x
and product of xy is maximum
let
Hence, x = 12 is the only critical value
Now,
at x= 12
Hence, x = 12 is the point of maxima
Noe, y = 24  x
= 24  12 = 12
Hence, the value of x and y are 12 and 12 respectively
Question:14 Find two positive numbers x and y such that x + y = 60 and is maximum.
Answer:
It is given that
x + y = 60 , x = 60 y
and
is maximum
let
Now,
Now,
hence, 0 is niether point of minima or maxima
Hence, y = 45 is point of maxima
x = 60  y
= 60  45 = 15
Hence, values of x and y are 15 and 45 respectively
Question:15 Find two positive numbers x and y such that their sum is 35 and the product is a maximum.
Answer:
It is given that
x + y = 35 , x = 35  y
and
is maximum
Therefore,
Now,
Now,
Hence, y = 35 is the point of minima
Hence, y= 0 is neither point of maxima or minima
Hence, y = 25 is the point of maxima
x = 35  y
= 35  25 = 10
Hence, the value of x and y are 10 and 25 respectively
Question:16 . Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Answer:
let x an d y are positive two numbers
It is given that
x + y = 16 , y = 16  x
and
is minimum
Now,
Hence, x = 8 is the only critical point
Now,
Hence, x = 8 is the point of minima
y = 16  x
= 16  8 = 8
Hence, values of x and y are 8 and 8 respectively
Answer:
It is given that the side of the square is 18 cm
Let assume that the length of the side of the square to be cut off is x cm
So, by this, we can say that the breath of cube is (182x) cm and height is x cm
Then,
Volume of cube
=
But the value of x can not be 9 because then the value of breath become 0 so we neglect value x = 9
Hence, x = 3 is the critical point
Now,
Hence, x = 3 is the point of maxima
Hence, the length of the side of the square to be cut off is 3 cm so that the volume of the box is the maximum possible
Answer:
It is given that the sides of the rectangle are 45 cm and 24 cm
Let assume the side of the square to be cut off is x cm
Then,
Volume of cube
But x cannot be equal to 18 because then side (24  2x) become negative which is not possible so we neglect value x= 18
Hence, x = 5 is the critical value
Now,
Hence, x = 5 is the point of maxima
Hence, the side of the square to be cut off is 5 cm so that the volume of the box is maximum
Question:19 Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Answer:
Let assume that length and breadth of rectangle inscribed in a circle is l and b respectively
and the radius of the circle is r
Now, by Pythagoras theorem
a = 2r
Now, area of reactangle(A) = l
b
Now,
Hence,
is the point of maxima
Since, l = b we can say that the given rectangle is a square
Hence, of all the rectangles inscribed in a given fixed circle, the square has the maximum area
Answer:
Let r be the radius of the base of cylinder and h be the height of the cylinder
we know that the surface area of the cylinder
Volume of cylinder
Hence,
is the critical point
Now,
Hence,
is the point of maxima
Hence, the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter(D = 2r) of the base
Answer:
Let r be the radius of base and h be the height of the cylinder
The volume of the cube (V) =
It is given that the volume of cylinder = 100
Surface area of cube(A) =
Hence,
is the critical point
Hence,
is the point of minima
Hence,
and
are the dimensions of the can which has the minimum surface area
Answer:
Area of the square (A) =
Area of the circle(S) =
Given the length of wire = 28 m
Let the length of one of the piece is x m
Then the length of the other piece is (28  x) m
Now,
and
Area of the combined circle and square
= A + S
Now,
Hence,
is the point of minima
Other length is = 28  x
=
Hence, two lengths are
and
Answer:
Volume of cone (V) =
Volume of sphere with radius r =
By pythagoras theorem in
we ca say that
V =
Now,
Hence, point
is the point of maxima
Hence, the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is
Volume =
Hence proved
Answer:
Volume of cone(V)
curved surface area(A) =
Now , we can clearly varify that
when
Hence,
is the point of minima
Hence proved that the right circular cone of least curved surface and given volume has an altitude equal to
time the radius of the base
Question:25 Show that the semivertical angle of the cone of the maximum volume and of given slant height is
Answer:
Let a be the semivertical angle of cone
Let r , h , l are the radius , height , slent height of cone
Now,
we know that
Volume of cone (V) =
Now,
Now,
Now, at
Therefore,
is the point of maxima
Hence proved
Question:26 Show that semivertical angle of the right circular cone of given surface area and maximum volume is
Answer:
Let r, l, and h are the radius, slant height and height of cone respectively
Now,
Now,
we know that
The surface area of the cone (A) =
Now,
Volume of cone(V) =
On differentiate it w.r.t to a and after that
we will get
Now, at
Hence, we can say that
is the point if maxima
Hence proved
Question:27 The point on the curve which is nearest to the point (0, 5) is
Answer:
Given curve is
Let the points on curve be
Distance between two points is given by
Hence, x = 0 is the point of maxima
Hence, the point
is the point of minima
Hence, the point
is the point on the curve
which is nearest to the point (0, 5)
Hence, the correct answer is (A)
Question:28
For all real values of x, the minimum value of
is
(A) 0 (B) 1 (C) 3 (D) 1/3
Answer:
Given function is
Hence, x = 1 and x = 1 are the critical points
Now,
Hence, x = 1 is the point of minima and the minimum value is
Hence, x = 1 is the point of maxima
Hence, the minimum value of
is
Hence, (D) is the correct answer
Question:29
The maximum value of
Answer:
Given function is
Hence, x = 1/2 is the critical point s0 we need to check the value at x = 1/2 and at the end points of given range i.e. at x = 1 and x = 0
Hence, by this we can say that maximum value of given function is 1 at x = 0 and x = 1
option c is correct
NCERT solutions for class 12 maths chapter 6 Application of DerivativesMiscellaneous Exercise
Question:1(a) Using differentials, find the approximate value of each of the following:
Answer:
Let
and
Now, we know that
is approximate equals to dy
So,
Now,
Hence,
is approximately equal to 0.677
Question:1(b)
Using differentials, find the approximate value of each of the following:
Answer:
Let
and
Now, we know that
is approximately equals to dy
So,
Now,
Hence,
is approximately equals to 0.497
Question:2. Show that the function given by has maximum at x = e.
Answer:
Given function is
Hence, x =e is the critical point
Now,
Hence, x = e is the point of maxima
Answer:
It is given that the base of the triangle is b
and let the side of the triangle be x cm ,
We know that the area of the triangle(A) =
now,
Now at x = b
Hence, the area decreasing when the two equal sides are equal to the base is
Question:4 Find the equation of the normal to curve which passes through the point (1, 2).
Answer:
Given the equation of the curve
We know that the slope of the tangent at a point on the given curve is given by
We know that
At point (a,b)
Now, the equation of normal with point (a,b) and
It is given that it also passes through the point (1,2)
Therefore,
(i)
It also satisfies equation
(ii)
By comparing equation (i) and (ii)
Now, equation of normal with point (2,1) and slope = 1
Hence, equation of normal is x + y  3 = 0
Question:5 . Show that the normal at any point to the curve is at a constant distance from the origin.
Answer:
We know that the slope of tangent at any point is given by
Given equations are
We know that
equation of normal with given points and slope
Hence, the equation of normal is
Now perpendicular distance of normal from the origin (0,0) is
Hence, by this, we can say that
the normal at any point to the curve
is at a constant distance from the origin
Question:6(i) Find the intervals in which the function f given by is
Answer:
Given function is
But
So,
Now three ranges are there
In interval
,
Hence, the given function
is increasing in the interval
in interval
so function is decreasing in this inter
Question:6(ii) Find the intervals in which the function f given by f x is equal to
Answer:
Given function is
But
So,
Now three ranges are there
In interval
,
Hence, given function
is increasing in interval
in interval
Hence, given function
is decreasing in interval
Question:7(i) Find the intervals in which the function f given by
Answer:
Given function is
Hence, three intervals are their
In interval
Hence, given function
is increasing in interval
In interval (1,1) ,
Hence, given function
is decreasing in interval (1,1)
Question:7(ii) Find the intervals in which the function f given by
Answer:
Given function is
Hence, three intervals are their
In interval
Hence, given function
is increasing in interval
In interval (1,1) ,
Hence, given function
is decreasing in interval (1,1)
Answer:
Given the equation of the ellipse
Now, we know that ellipse is symmetrical about x and yaxis. Therefore, let's assume coordinates of A = (n,m) then,
Now,
Put(n,m) in equation of ellipse
we will get
Therefore, Now
Coordinates of A =
Coordinates of B =
Now,
Lenghth AB(base) =
And height of triangle ABC = (a+n)
Now,
Area of triangle =
Now,
Now,
but n cannot be zero
therefore,
Now, at
Therefore,
is the point of maxima
Now,
Now,
Therefore, Area (A)
Answer:
Let l , b and h are length , breath and height of tank
Then, volume of tank = l X b X h = 8
h = 2m (given)
lb = 4 =
Now,
area of base of tank = l X b = 4
area of 4 side walls of tank = hl + hl + hb + hb = 2h(l + b)
Total area of tank (A) = 4 + 2h(l + b)
Now,
Hence, b = 2 is the point of minima
So, l = 2 , b = 2 and h = 2 m
Area of base = l X B = 2 X 2 =
building of tank costs Rs 70 per sq metres for the base
Therefore, for
Rs = 4 X 70 = 280 Rs
Area of 4 side walls = 2h(l + b)
= 2 X 2(2 + 2) =
building of tank costs Rs 45 per square metre for sides
Therefore, for
Rs = 16 X 45 = 720 Rs
Therefore, total cost for making the tank is = 720 + 280 = 1000 Rs
Answer:
It is given that
the sum of the perimeter of a circle and square is k =
Let the sum of the area of a circle and square(A) =
Now,
Hence,
is the point of minima
Hence proved that the sum of their areas is least when the side of the square is double the radius of the circle
Answer:
Let l and bare the length and breadth of rectangle respectively and r will be the radius of circle
The total perimeter of window = perimeter of rectangle + perimeter of the semicircle
=
Area of window id given by (A) =
Now,
Hence, b = 5/2 is the point of maxima
Hence, these are the dimensions of the window to admit maximum light through the whole opening
Question:12 A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is
Answer:
It is given that
A point on the hypotenuse of a triangle is at a distance a and b from the sides of the triangle
Let the angle between AC and BC is
So, the angle between AD and ED is also
Now,
CD =
And
AD =
AC = H = AD + CD
=
+
Now,
When
Hence,
is the point of minima
and
AC =
=
Hence proved
Question:13 Find the points at which the function f given by has (i) local maxima (ii) local minima (iii) point of inflexion
Answer:
Given function is
Now, for value x close to
and to the left of
,
,and for value close to
and to the right of
Thus, point x =
is the point of maxima
Now, for value x close to 2 and to the Right of 2 ,
,and for value close to 2 and to the left of 2
Thus, point x = 2 is the point of minima
There is no change in the sign when the value of x is 1
Thus x = 1 is the point of inflexion
Question:14
Find the absolute maximum and minimum values of the function f given by
Answer:
Given function is
Now,
Hence, the point
is the point of maxima and the maximum value is
And
Hence, the point
is the point of minima and the minimum value is
Answer:
The volume of a cone (V) =
The volume of the sphere with radius r =
By Pythagoras theorem in
we ca say that
V =
Now,
Hence, the point
is the point of maxima
Hence, the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is
Answer:
Let's do this question by taking an example
suppose
Now, also
Hence by this, we can say that f is an increasing function on (a, b)
Question:17 Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is . Also, find the maximum volume.
Answer:
The volume of the cylinder (V) =
By Pythagoras theorem in
h = 2OA
Now,
Hence, the point
is the point of maxima
Hence, the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is
and maximum volume is
Answer:
Let's take radius and height of cylinder = r and h ' respectively
Let's take radius and height of cone = R and h respectively
Volume of cylinder =
Volume of cone =
Now, we have
Now, since
are similar
Now,
Now,
Now,
at
Hence,
is the point of maxima
Hence proved
Now, Volume (V) at
and
is
hence proved
Answer:
It is given that
Volume of cylinder (V) =
Hence, (A) is correct answer
Question:20
The slope of the tangent to the curve
at the point
(2,– 1) is
Answer:
Given curves are
At point (2,1)
Similarly,
The common value between two is t = 2
Hence, we find the slope of the tangent at t = 2
We know that the slope of the tangent at a given point is given by
Hence, (B) is the correct answer
Question:21
The line y is equal to
is a tangent to the curve
if the value of m is
(A) 1
Answer:
Standard equation of the straight line
y = mx + c
Where m is lope and c is constant
By comparing it with equation , y = mx + 1
We find that m is the slope
Now,
we know that the slope of the tangent at a given point on the curve is given by
Given the equation of the curve is
Put this value of m in the given equation
Hence, value of m is 1
Hence, (A) is correct answer
Question:22
T
he normal at the point (1,1) on the curve
is
(A) x + y = 0
Answer:
Given the equation of the curve
We know that the slope of the tangent at a point on the given curve is given by
We know that
At point (1,1)
Now, the equation of normal with point (1,1) and slope = 1
Hence, the correct answer is (B)
Question:23 The normal to the curve passing (1,2) is
Answer:
Given the equation of the curve
We know that the slope of the tangent at a point on the given curve is given by
We know that
At point (a,b)
Now, the equation of normal with point (a,b) and
?
It is given that it also passes through the point (1,2)
Therefore,
(i)
It also satisfies equation
(ii)
By comparing equation (i) and (ii)
Now, equation of normal with point (2,1) and slope = 1
Hence, correct answer is (A)
Question:24 The points on the curve , where the normal to the curve makes equal intercepts with the axes are
Answer:
Given the equation of the curve
We know that the slope of the tangent at a point on a given curve is given by
We know that
At point (a,b)
Now, the equation of normal with point (a,b) and
It is given that normal to the curve makes equal intercepts with the axes
Therefore,
point(a,b) also satisfy the given equation of the curve
Hence, The points on the curve
, where the normal to the curve makes equal intercepts with the axes are
Hence, the correct answer is (A)
NCERT solutions for class 12 maths chapter wise
chapter 1 
Solutions of NCERT for class 12 maths chapter 1 Relations and Functions 
chapter 2 
CBSE NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions 
chapter 3 

chapter 4 
Solutions of NCERT for class 12 maths chapter 4 Determinants 
chapter 5 
CBSE NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability 
chapter 6 
NCERT solutions for class 12 maths chapter 6 Application of Derivatives 
chapter 7 

chapter 8 
CBSE NCERT solutions for class 12 maths chapter 8 Application of Integrals 
chapter 9 
NCERT solutions for class 12 maths chapter 9 Differential Equations 
chapter 10 
Solutions of NCERT for class 12 maths chapter 10 Vector Algebra 
chapter 11 
CBSE NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry 
chapter 12 
NCERT solutions for class 12 maths chapter 12 Linear Programming 
chapter 13 
Solutions of NCERT for class 12 maths chapter 13 Probability 
NCERT solutions for class 12 subject wise
NCERT solutions for class 12 maths chapter 6 application of derivatives are very helpful for the preparation of this chapter. Here are some tips to get command on this chapter
 You should cover the differentiation completely and then go for its applications
 You should solve every NCERT problem on your own including examples, if you find difficulties in solving the problems, CBSE NCERT solutions for class 12 maths chapter 6 application of derivatives will help you for the same
 Draw the figure first and label it, when it required. It makes easy to solve the problems as you are able to visualize it
Happy learning!!!