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Class XII Chapter 6 – Application of Derivatives Maths

Exercise 6.1

Question 1:

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

(a) *r *= 3 cm (b) *r *= 4 cm Answer

The area of a circle (*A) *with radius (*r*) is given by,

Now, the rate of change of the area with respect to its radius is given by,

1.
When *r *= 3 cm,

Hence, the area of the circle is changing
at the rate of 6π
cm^{2}/s when its radius is 3 cm.

2.
When *r *= 4 cm,

Hence, the area of the circle is changing
at the rate of 8π
cm^{2}/s when its radius is 4 cm.

Question 2:

The volume
of a cube is increasing at the rate of 8 cm^{3}/s. How fast is the surface
area increasing when the length
of an edge is 12 cm?

Answer

Let *x *be the length of a
side, *V *be the volume,
and *s *be the surface
area of the cube. Then,
*V *= *x*^{3} and *S *= 6*x*^{2} where *x *is a function
of time *t*.

It is given that .

Then, by using the chain rule, we have:

∴

⇒

Thus, when *x *= 12 cm,

Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing

at the rate of cm^{2}/s.

Question 3:

The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

Answer

The area of a circle (*A)
*with radius *(r*) is given by,

Now, the rate of change of area (*A*) with respect to time (*t)
*is given by,

It is given that,

∴

Thus, when *r *= 10 cm,

Hence, the rate at which the area
of the circle is increasing when the radius
is 10 cm is 60π cm^{2}/s.

Question 4:

An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?

Answer

Let *x *be the length of a side and *V *be the volume of the cube.
Then,

*V *= *x*^{3.}

∴ (By chain rule)

It is given that,

∴

Thus, when *x *= 10 cm,

Hence, the volume of the cube is increasing at the rate of 900 cm^{3}/s when the edge is 10 cm
long.

Question 5:

A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

Answer

The area of a circle (*A*) with radius (*r*) is given by .

Therefore, the rate of change of area (*A)
*with respect to time (*t)
*is given by,

[By chain rule]

It is given that . Thus, when

Hence, when the radius of
the circular wave is 8 cm, the enclosed area is increasing at the rate of
80π cm^{2}/s.

Question 6:

The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?

Answer

The circumference of a circle
(*C*) with radius (*r) *is
given by

*C *= 2π*r.*

Therefore, the rate of change of circumference (C) with respect
to time (*t)
*is given by,

(By chain rule)

It is given that .

Hence, the rate of increase of the circumference

Question 7:

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 *= 8 cm and *y *=
6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

Answer

Since the length (*x)
*is decreasing at the rate of 5 cm/minute and the width (*y) *is
increasing at the rate of 4 cm/minute, we have:

and

(a)
The perimeter (*P) *of a rectangle is given by,

*P *= 2(*x
+ y*)

Hence, the perimeter is decreasing at the rate of 2 cm/min.

(b)
The area (*A) *of
a rectangle is given by,

*A *= *x × y*

* *

* *

∴

When *x *= 8 cm and *y *= 6 cm,

Hence, the area of the rectangle
is increasing at the rate of 2 cm^{2}/min.

Question 8:

A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Answer

The volume of a sphere (*V)
*with radius (*r*) is given by,

∴Rate of change of volume (*V)
*with respect to time (*t)
*is given by,

[By chain rule]

It is given that .

Therefore, when radius = 15 cm,

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

is

Question 9:

A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.

Answer

The volume of a sphere (*V*) with radius (*r*) is given by .

Rate of change of volume (*V)
*with respect to its radius
(*r) *is given by,

Therefore, when radius = 10 cm,

Hence, the volume of the balloon
is increasing at the rate of 400π cm^{3}/s.

Question 10:

A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Answer

Let *y *m be the height
of the wall at which the ladder
touches. Also, let the foot of the ladder be *x
*maway from the wall.

Then, by Pythagoras theorem, we have:

*x*^{2} + *y*^{2} = 25 [Length
of the ladder = 5 m]

Then, the rate of change of height (*y) *with respect
to time (*t) *is given by,

It is given that .

Now, when *x *= 4 m, we have:

Hence, the height of the ladder on the wall is decreasing at the rate of .

Question 11:

A particle
moves along the curve . Find the
points on the curve at which the *y*- coordinate is changing 8 times as fast
as the *x*-coordinate.

Answer

The equation of the curve is given as:

The rate of change of the position of the particle
with respect to time (*t) *is given by,

When the *y*-coordinate of the particle
changes 8 times
as fast as the

*x*-coordinate i.e., , we have:

When .

When .

Hence, the points required on the curve are (4, 11) and

Question 12:

The radius of an air bubble is increasing at the rate of cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Answer

The air bubble is in the shape of a sphere.

Now, the volume of an air bubble (*V*) with radius (*r*) is given by,

The rate of change
of volume (*V)
*with respect to time (*t)
*is given by,

It is given that . Therefore, when

Hence, the rate at which the volume of the bubble increases is 2π cm^{3}/s.

Question 13:

A balloon,
which always remains
spherical, has a variable diameter Find the rate of change
of its volume with respect to *x*.

Answer

The volume of a sphere (*V*) with radius (*r*) is given by,

It is given that:

Diameter

Hence, the rate of change of volume with respect to *x *is
as

Question 14:

Sand is pouring from a pipe at the rate of 12 cm^{3}/s. The falling sand forms a cone on the ground in such a way that the height of
the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height
is 4 cm?

Answer

The volume of a cone (*V)
*with radius (*r*) and height (*h)
*is given by,

It is given that,

The rate of change of volume
with respect to time (*t) *is given by,

[By chain rule]

It is also given that . Therefore, when

Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate

of .

Question 15:

The total cost

Find the marginal cost when 17 units are produced. Answer

Marginal cost is the rate of change of total cost with respect to output.

∴Marginal cost (MC)

When *x *= 17, MC = 0.021 (17^{2}) − 0.006 (17) + 15

= 0.021(289) − 0.006(17) + 15

= 6.069 − 0.102 + 15

= 20.967

Hence, when 17 units are produced, the marginal cost is Rs. 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 is the rate of change of total revenue with respect to the number of units sold.

∴Marginal Revenue
(MR) = 13(2*x*) + 26
= 26*x *+ 26

When *x *= 7,

MR = 26(7) + 26 = 182 + 26 = 208

Hence, the required marginal revenue is Rs 208.

Question 17:

The rate of change
of the area of a circle
with respect to its radius
*r *at *r *= 6 cm is (A) 10π
(B) 12π (C) 8π
(D) 11π

Answer

The area of a circle (*A) *with radius (*r*) is given by,

Therefore, the rate of change of the area with respect
to its radius *r *is

.

∴When *r *= 6 cm,

Hence, the required rate of change of the area of a circle is 12π
cm^{2}/s. The correct
answer is B.

Question 18:

The total revenue in Rupees received
from the sale of *x *units of a product
is given by

. The marginal revenue, when is (A) 116 (B) 96 (C) 90 (D) 126

Answer

Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

∴Marginal Revenue
(MR) = 3(2*x*) + 36 = 6*x
*+ 36

∴When *x *= 15,

MR = 6(15) + 36 = 90 + 36 = 126

Hence, the required marginal revenue is Rs 126. The correct answer is D.

Exercise 6.2

Question 1:

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

Let be any two numbers in R. Then, we have:

Hence, *f *is strictly increasing on R.

Alternate method:

*f*'(*x*) = 3 > 0, in every interval
of R.

Thus, the function is strictly increasing on R.

Question 2:

Show that the function
given by *f*(*x*) = *e*^{2x}* *is
strictly increasing on R. Answer

Let be any two numbers in R. Then, we have:

Hence, *f *is strictly increasing on R.

Question 3:

Show that the function
given by *f*(*x*) = sin *x *is

(a) strictly increasing in (b) strictly decreasing in

(c) neither increasing nor decreasing in (0, π) Answer

The given function is *f*(*x*) = sin *x*.

(a) Since for each we have .

Hence, *f *is strictly
increasing in .

(b) Since for each , we have .

Hence, *f *is strictly
decreasing in .

(c) From
the results obtained
in (a) and (b), it is clear that *f *is
neither increasing nor decreasing in (0,
π).

Question 4:

Find the intervals in which the function
*f *given by *f*(*x*) = 2*x*^{2} −
3*x *is

(a) strictly increasing (b) strictly decreasing Answer

The given function is *f*(*x*) = 2*x*^{2} −
3*x*.

Now, the point divides the real line into two disjoint intervals i.e., and

In interval

Hence, the given function
(*f) *is strictly
decreasing in interval .

In interval

Hence, the given function
(*f) *is strictly
increasing in interval .

Question 5:

Find the intervals in which the function *f *given by *f*(*x*) = 2*x*^{3} −
3*x*^{2} − 36*x
*+ 7 is

(a) strictly increasing (b) strictly decreasing Answer

The given function is *f*(*x*) = 2*x*^{3} −
3*x*^{2} −
36*x *+ 7.

∴ *x *= −
2, 3

The points *x *= −2 and *x *= 3 divide
the real line into three
disjoint intervals i.e.,

In intervals is positive while in interval (−2, 3), is negative.

Hence, the given function
(*f) *is strictly
increasing in intervals

, while function (*f) *is strictly
decreasing in interval

(−2, 3).

Question 6:

Find the intervals in which the following functions
are strictly increasing or decreasing: (a) *x*^{2} + 2*x
*− 5 (b) 10 − 6*x
*− 2*x*^{2}

(c) −2*x*^{3} −
9*x*^{2} − 12*x
*+ 1 (d) 6 − 9*x *− *x*^{2}

(e) (*x *+ 1)^{3} (*x *− 3)^{3}

Answer

(a) We have,

Now,

*x *= −1

Point *x *= −1
divides the real line into two disjoint
intervals i.e., In interval

∴*f *is
strictly decreasing in interval

Thus, *f *is strictly
decreasing for *x *< −1. In interval

∴ *f *is strictly increasing in interval

Thus, *f *is strictly
increasing for *x *> −1.

(b) We have,

*f*(*x*) = 10 − 6*x
*− 2*x*^{2}

The point divides the real line into two disjoint intervals

i.e.,

In interval i.e., when ,

∴ *f *is strictly increasing for .

In interval i.e., when ,

∴ *f *is strictly decreasing for .

(c) We have,

*f*(*x*) = −2*x*^{3} −
9*x*^{2} − 12*x
*+ 1

Points *x *= −1
and *x *= −2
divide the real line into three disjoint
intervals i.e.,

In intervals i.e., when *x *<
−2 and *x *> −1,

.

∴ *f *is strictly decreasing for *x *< −2
and *x *>
−1.

Now,
in interval (−2, −1) i.e., when −2
< *x *<
−1, .

∴ *f *is strictly
increasing for .

(d) We have,

The point divides the real line into two disjoint intervals i.e.,

.

In interval i.e., for , .

∴ *f *is strictly increasing for .

In interval i.e., for ,

∴ *f *is strictly
decreasing for .

(e) We have,

*f*(*x*) = (*x
*+ 1)^{3} (*x *− 3)^{3}

The points
*x *= −1, *x *= 1, and *x *= 3 divide the real line into four disjoint intervals i.e., , (−1, 1), (1, 3), and .

In intervals and (−1, 1), .

∴ *f *is strictly
decreasing in intervals and (−1, 1).

In intervals (1, 3) and , .

∴ *f *is strictly
increasing in intervals
(1, 3) and .

Question 7:

Show that , is an increasing function of *x *throughout its domain.

Answer We have,

Since *x *> −1, point *x *= 0 divides
the domain (−1, ∞) in two disjoint
intervals i.e., −1
<

*x *< 0 and *x *> 0.

When −1 < *x *< 0, we have:

Also, when *x *>
0:

Hence, function *f *is
increasing throughout this domain.

Question 8:

Find the values of *x *for
which is an increasing function. Answer

We have,

The points *x *= 0, *x *= 1, and *x *= 2 divide the real line into four disjoint intervals i.e.,

In intervals , .

∴ *y *is strictly
decreasing in intervals .

However, in intervals (0, 1) and (2, ∞),

∴ *y *is strictly increasing in intervals (0, 1) and (2, ∞).

*y *is
strictly increasing for 0 < *x
*< 1 and *x *>
2.

Question 9:

Prove that is an increasing function
of *θ *in
. Answer

We have,

Since cos *θ *≠ 4, cos *θ
*= 0.

Now,

In interval , we have
cos *θ *>
0. Also, 4 > cos *θ *⇒ 4
− cos *θ *> 0.

Therefore, *y *is strictly
increasing in interval .

Also, the given function is continuous at

Hence, *y *is increasing in interval .

Question 10:

Prove that the logarithmic function is strictly increasing on (0, ∞). Answer

It is clear that for *x *> 0,

Hence, *f*(*x*) = log *x *is
strictly increasing in interval (0, ∞).

Question 11:

Prove that the function
*f *given by *f*(*x*) = *x*^{2} −
*x *+ 1 is neither
strictly increasing nor strictly decreasing on (−1, 1).

Answer

The given function is *f*(*x*) = *x*^{2} −
*x *+ 1.

The point divides the interval (−1, 1) into two disjoint intervals

i.e.,

Now, in interval

Therefore, *f *is strictly
decreasing in interval .

However, in interval

Therefore, *f *is strictly
increasing in interval .

Hence, *f *is neither
strictly increasing nor decreasing in interval (−1, 1).

Question 12:

Which of the following functions are strictly decreasing on ?

(A) cos *x *(B) cos 2*x
*(C) cos 3*x *(D) tan *x*

Answer

(A) Let

In interval

is strictly decreasing in interval .

(B) Let

is strictly decreasing in interval .

(C) Let

The point divides the interval into two disjoint intervals

i.e., 0

∴ *f*_{3} is strictly
decreasing in interval .

∴ *f*_{3} is strictly
increasing in interval .

Hence, *f*_{3} is neither increasing nor decreasing in interval .

(D) Let

In interval

∴ *f*_{4} is strictly increasing in interval

Therefore, functions cos *x *and
cos 2*x *are
strictly decreasing in Hence, the correct answers
are A and B.

Question 13:

On which of the following intervals is the function
*f *given by strictly
decreasing?

(A) (B)

(C) (D) None of these Answer

We have,

In interval

Thus, function *f *is
strictly increasing in interval (0, 1).

In interval

Thus, function
*f *is strictly
increasing in interval .

∴ *f *is strictly
increasing in interval .

Hence, function
*f *is strictly
decreasing in none of the intervals. The correct answer
is D.

Question 14:

Find the least value of *a *such that the function
*f *given is strictly increasing on (1, 2).

Answer We have,

Now, function
*f *will be increasing
in (1, 2), if in
(1, 2).

⇒ 2*x *+ *a *>
0

⇒ 2*x *> −*a*

* *

* *

* *

* *

* *

⇒

Therefore, we have to find the least value of *a *such that

Thus, the least value of *a *for *f *to be increasing on (1, 2) is given by,

Hence, the required value of *a *is −2.

Question 15:

Let I be any interval disjoint
from (−1, 1). Prove that the function
*f *given by

is strictly increasing on I.

Answer We have,

The points *x *= 1 and *x *= −1 divide
the real line in three disjoint intervals i.e.,

.

In interval (−1, 1), it is observed that:

∴ *f *is strictly
decreasing on .

In intervals , it is observed that:

∴ *f *is strictly
increasing on .

Hence, function
*f *is strictly
increasing in interval
I disjoint from (−1, 1). Hence, the given
result is proved.

Question 16:

Prove that the function
*f *given by *f*(*x*) = log sin *x *is
strictly increasing on and

strictly decreasing on Answer

We have,

In interval

∴ *f *is strictly
increasing in .

In interval

∴*f *is
strictly decreasing in

Question 17:

Prove that the function
*f *given by *f*(*x*) = log cos *x
*is strictly decreasing on and

strictly increasing on Answer

We have,

In interval

∴*f *is
strictly decreasing on .

In interval

∴*f *is
strictly increasing on .

Question 18:

Prove that the function given by is increasing in R. Answer

We have,

For any *x*∈R, (*x *− 1)^{2} > 0.

Thus, is always positive in R.

Hence, the given function
(*f) *is increasing in R.

Question 19:

The interval in which is increasing is (A) (B) (−2, 0) (C) (D) (0, 2)

Answer We have,

The points
*x *= 0 and *x *= 2 divide the real line into three
disjoint intervals i.e.,

In intervals is always positive.

∴*f *is
decreasing on

In interval (0, 2),

∴ *f *is strictly increasing on (0, 2).

Hence, *f *is strictly
increasing in interval
(0, 2). The correct answer is
D.

Exercise 6.3

Question 1:

Find the slope of the tangent
to the curve *y *= 3*x*^{4} −
4*x *at *x *= 4. Answer

The given curve is *y *= 3*x*^{4} −
4*x*.

Then, the slope of the tangent
to the given curve at *x *= 4 is given by,

Question 2:

Find the slope of the tangent
to the curve , *x *≠ 2 at *x *= 10. Answer

The given curve is .

Thus, the slope of the tangent
at *x *=
10 is given by,

Hence, the slope of the tangent
at *x *=
10 is

Question 3:

Find the slope of the tangent
to curve *y *= *x*^{3} −
*x *+ 1 at the point whose *x*-coordinate is 2.

Answer

The given curve is .

The slope of the tangent to a curve
at (*x*_{0}, *y*_{0}) is . It is given
that *x*_{0} = 2.

Hence, the slope of the tangent
at the point where the *x*-coordinate is 2 is given by,

Question 4:

Find the slope of the tangent
to the curve *y *= *x*^{3} −
3*x *+ 2 at the point whose *x*- coordinate is 3.

Answer

The given curve is .

The slope of the tangent to a curve
at (*x*_{0}, *y*_{0}) is .

Hence, the slope of the tangent
at the point where the *x*-coordinate is 3 is given by,

Question 5:

Find the slope of the normal
to the curve *x *=
*a*cos^{3}*θ*,
*y *= *a*sin^{3}*θ *at . Answer

It is given that *x *= *a*cos^{3}*θ *and
*y *= *a*sin^{3}*θ*.

Therefore, the slope of the tangent at is given by,

Hence, the slope of the normal at

Question 6:

Find the slope of the
normal to the curve *x *= 1 −
*a *sin *θ*, *y *= *b *cos^{2}*θ *at . Answer

It is given that *x *=
1 − *a *sin *θ
*and *y *= *b *cos^{2}*θ*.

Therefore, the slope of the tangent at is given by,

Hence, the slope of the normal at

Question 7:

Find points
at which the tangent to the curve *y *= *x*^{3} −
3*x*^{2} − 9*x *+ 7 is parallel
to the *x*- axis.

Answer

The equation of the given curve is

Now, the tangent is parallel to the *x*-axis if the slope of the tangent is zero.

When *x *= 3, *y
*= (3)^{3} −
3 (3)^{2} −
9 (3) + 7 = 27 − 27 − 27 + 7 = −20.

When *x *= −1, *y *= (−1)^{3} −
3 (−1)^{2} − 9 (−1) + 7 = −1 −
3 + 9 + 7 = 12.

Hence, the points at which the tangent is parallel to the *x*-axis are (3, −20) and (−1, 12).

Question 8:

Find a point
on the curve *y *=
(*x *− 2)^{2} at which the tangent
is parallel to the chord
joining the points (2, 0) and (4, 4).

Answer

If a tangent is parallel to the chord joining the points (2, 0) and (4, 4), then the slope of the tangent = the slope of the chord.

The slope of the chord is

Now, the slope of the tangent
to the given curve at a point (*x*, *y*) is given by,

Since the slope of the tangent = slope of the chord, we have:

Hence, the required point is (3, 1).

Question 9:

Find the point on the curve
*y *= *x*^{3} −
11*x *+ 5 at which
the tangent is *y
*= *x *− 11. Answer

The equation
of the given curve is *y *= *x*^{3} −
11*x *+ 5.

The equation
of the tangent to the given curve
is given as *y *= *x *− 11 (which is of the form
*y *= *mx *+ *c)*.

∴Slope of the tangent = 1

Now, the slope of the tangent
to the given curve at the point (*x*, *y*)
is given by,

Then, we have:

When *x *= 2, *y *= (2)^{3} −
11 (2) + 5 = 8 −
22 + 5 = −9.

When *x *= −2, *y *=
(−2)^{3} − 11 (−2) + 5 = −8 + 22 + 5 = 19.

Hence, the required points are (2, −9) and (−2, 19).

Question 10:

Find the equation of all lines having slope −1 that are tangents to the curve

.

Answer

The equation of the given curve is .

The slope of the tangents to the given
curve at any point (*x*, *y*) is given by,

If the slope of the tangent is −1, then we have:

When *x *= 0, *y
*= −1 and when *x *= 2, *y *= 1.

Thus, there are two tangents to the given curve having slope −1. These are passing through the points (0, −1) and (2, 1).

∴The equation of the tangent through (0, −1) is given by,

∴The equation of the tangent through (2, 1) is given by,

*y *− 1 = −1 (*x *− 2)

⇒ *y *− 1 =
− *x *+ 2

⇒ *y *+ *x *− 3 = 0

Hence, the equations of the required
lines are *y *+ *x *+ 1 = 0 and *y *+ *x *− 3 = 0.

Question 11:

Find the equation of all lines having slope 2 which are tangents to the

curve .

Answer

The equation of the given curve is .

The slope of the tangent to the given curve at any point (*x*, *y*) is given by,

If the slope of the tangent is 2, then we have:

Hence, there is no tangent to the given curve having slope 2.

Question 12:

Find the equations of all lines having slope 0 which are tangent to the curve

.

Answer

The equation of the given curve is .

The slope of the tangent to the given curve at any point (*x*, *y*) is given by,

If the slope of the tangent is 0, then we have:

When *x *= 1,

∴The equation of the tangent through is given by,

Hence, the equation of the required line is

Question 13:

Find points on the curve at which the tangents are

(i) parallel
to *x*-axis (ii) parallel to *y*-axis Answer

The equation of the given curve is .

On differentiating both sides with respect
to *x*, we have:

(i) The
tangent is parallel
to the *x*-axis if the slope of the tangent is i.e.,
0 which is possible if *x
*= 0.

Then, for *x *= 0

Hence, the points at which the tangents are parallel to the *x*-axis are (0, 4) and (0,
− 4).

(ii)
The tangent is parallel to the *y*-axis if the slope of the normal is 0, which

gives ⇒ *y *= 0.

Then, for *y *= 0.

Hence, the points at which the tangents are parallel to the *y*-axis are (3, 0) and (− 3, 0).

Question 14:

Find the equations of the tangent
and normal to the given curves at the indicated points: (i) *y *= *x*^{4} −
6*x*^{3} + 13*x*^{2} −
10*x *+ 5 at (0, 5)

(ii) *y *= *x*^{4} −
6*x*^{3} + 13*x*^{2} −
10*x *+ 5 at (1, 3)

(iii)
*y *= *x*^{3} at (1, 1)

(iv)
*y *= *x*^{2} at (0, 0)

(v) *x *= cos *t*, *y *= sin *t
*at Answer

(i)

The equation of the curve is

Thus, the slope of the tangent at (0, 5) is −10. The equation of the tangent is given as:

*y *− 5 = − 10(*x
*− 0)

⇒ *y *− 5 = − 10*x*

* *

* *

* *

* *

* *

⇒ 10*x *+ *y *= 5

The slope of the normal at (0, 5) is Therefore, the equation of the normal at (0, 5) is given as:

(ii)
The equation
of the curve is *y *= *x*^{4} −
6*x*^{3} + 13*x*^{2} −
10*x *+ 5. On differentiating with respect to *x*,
we get:

Thus, the slope of the tangent at (1, 3) is 2. The equation of the tangent is given as:

The slope of the normal at (1, 3) is Therefore, the equation of the normal at (1, 3) is given as:

(iii) The equation of the curve is *y *= *x*^{3}.

On differentiating with respect
to *x*, we get:

Thus, the slope of the tangent at (1, 1) is 3 and the equation of the tangent is given as:

The slope of the normal at (1, 1) is

Therefore, the equation of the normal at (1, 1) is given as:

(iv) The equation of the curve is *y *= *x*^{2}.

On differentiating with respect
to *x*, we get:

Thus, the slope of the tangent at (0, 0) is 0 and the equation of the tangent is given as:

*y *− 0 = 0
(*x *− 0)

⇒ *y *= 0

The slope of the normal at (0, 0) is , which is not defined.

Therefore, the equation of the normal at (*x*_{0,} *y*_{0}) = (0, 0) is given by

(v) The equation of the curve
is *x *=
cos *t*, *y *= sin *t*.

∴The slope of the tangent at is −1.

When

Thus, the equation of the tangent to the given curve at is

The slope of the normal at is

Therefore, the equation of the normal to the given curve at is

Question 15:

Find the equation of the tangent
line to the curve *y *= *x*^{2} −
2*x *+ 7 which is

(a)
parallel to the line 2*x *− *y *+ 9 = 0

(b) perpendicular to the line 5*y *− 15*x *= 13. Answer

The equation of the given curve is . On differentiating with respect to

(a) The
equation of the line is 2*x
*− *y *+
9 = 0. 2*x *− *y *+ 9 = 0 ∴ *y *= 2*x *+ 9

This is of the form *y *= *mx *+ *c*.

∴Slope of the line = 2

If a tangent is parallel to the line 2*x *− *y *+ 9 =
0, then the slope of the tangent
is equal to the slope of the line.

Therefore, we have:

2
= 2*x *− 2

Now, *x *= 2

*y *= 4
− 4 + 7 = 7

Thus, the equation of the tangent passing through (2, 7) is given by,

Hence, the equation of the tangent
line to the given curve (which is parallel to line 2*x *−

*y *+ 9
= 0)
is .

(b)
The equation of the line is 5*y *− 15*x
*= 13.

5*y *− 15*x *= 13 ∴

This is of the form *y *= *mx *+ *c*.

∴Slope of the line = 3

If a tangent is perpendicular to the line 5*y *− 15*x *= 13, then the slope of the tangent
is

Thus, the equation of the tangent passing through is given by,

Hence, the equation of the tangent
line to the given curve (which is perpendicular to line 5*y
*− 15*x *= 13)
is .

Question 16:

Show that the tangents
to the curve *y *= 7*x*^{3} + 11 at the points
where *x *= 2 and *x *= −2
are parallel.

Answer

The equation of the given curve is *y *= 7*x*^{3} + 11.

The slope of the tangent to a curve
at (*x*_{0}, *y*_{0}) is .

Therefore, the slope of the tangent
at the point where *x *= 2 is given by,

It is observed that the slopes of the tangents at the points
where *x *=
2 and *x *= −2
are equal.

Hence, the two tangents are parallel.

Question 17:

Find the points
on the curve *y *= *x*^{3} at which the slope of the tangent
is equal to the *y*-
coordinate of the point.

Answer

The equation of the given curve is *y *= *x*^{3}.

The slope of the tangent at the point (*x*, *y*) is given by,

When the slope of the tangent
is equal to the *y*-coordinate of the point, then *y *= 3*x*^{2}. Also,
we have *y *= *x*^{3}.

∴3*x*^{2} = *x*^{3}

□ *x*^{2}
(*x** *− 3) = 0

□ *x *= 0, *x *= 3

When *x *= 0, then *y *= 0 and when *x *= 3, then *y
*= 3(3)^{2} = 27. Hence,
the required points
are (0, 0) and
(3, 27).

Question 18: For the curve
*y *= 4*x*^{3} − 2*x*^{5}, find all the points at which
the tangents passes
through the origin.

Answer

The equation of the given curve is *y *= 4*x*^{3} −
2*x*^{5}.

Therefore, the slope of the tangent
at a point (*x*, *y*) is 12*x*^{2} −
10*x*^{4}. The equation
of the tangent at (*x*, *y*) is given by,

When the tangent passes through the origin (0, 0), then *X *= *Y *= 0. Therefore, equation (1) reduces
to:

Also, we have

When *x *= 0, *y *=

When *x *= 1, *y *= 4 (1)^{3} − 2 (1)^{5} = 2.

When *x *= −1,
*y *= 4 (−1)^{3} −
2 (−1)^{5} = −2.

Hence, the required points are (0, 0), (1, 2), and (−1, −2).

Question 19:

Find the points
on the curve *x*^{2} + *y*^{2} −
2*x *− 3 = 0 at which the tangents
are parallel to the *x*-axis.

Answer

The equation of the given curve is

Now, the tangents are parallel to the *x*-axis if the slope of the tangent is 0.

But, *x*^{2} + *y*^{2} −
2*x *− 3 = 0 for *x *= 1.

*y*^{2} = 4 ∴

Hence, the points at which the tangents are parallel to the *x*-axis are (1, 2) and (1, −2).

Question 20:

Find the equation of the normal
at the point (*am*^{2}, *am*^{3}) for the curve *ay*^{2} = *x*^{3}. Answer

The equation of the given curve is

The slope of a tangent to the curve at (*x*_{0}, *y*_{0}) is .

The slope of the tangent to the given
curve at (*am*^{2}, *am*^{3}) is

□ Slope
of normal at (*am*^{2}, *am*^{3})

=

Hence, the equation of the normal
at (*am*^{2}, *am*^{3}) is given by,

*y *− *am*^{3} =

Question 21:

Find the equation of the normals
to the curve *y *= *x*^{3} + 2*x *+ 6 which are parallel
to the line *x *+ 14*y
*+ 4 = 0.

Answer

The equation of the given curve
is *y *= *x*^{3} + 2*x *+ 6.

The slope of the tangent to the given curve at any point (*x*, *y*) is given by,

□ Slope of the normal to the
given curve at any point (*x*, *y*)

The equation of the given line is *x *+ 14*y *+ 4 = 0.

*x *+ 14*y *+ 4 = 0 ∴ (which is of the form *y *= *mx *+ *c)*

* *

* *

∴Slope of the given line =

If the normal is parallel to the line, then we must have the slope of the normal being equal to the slope of the line.

When *x *= 2, *y
*= 8 + 4 + 6 = 18.

When *x *= −2, *y *= − 8 − 4 + 6 = −6.

Therefore, there are two normals to the given curve with slope and passing through the points (2, 18) and (−2, −6).

Thus, the equation of the normal through (2, 18) is given by,

And, the equation of the normal through (−2, −6) is given by,

Hence, the equations of the normals to the given curve (which are parallel to the given line) are

Question 22:

Find the equations of the tangent
and normal to the parabola
*y*^{2} = 4*ax
*at the point (*at*^{2}, 2*at*).

Answer

The equation of the given parabola
is *y*^{2} = 4*ax*.

On differentiating *y*^{2} = 4*ax *with
respect to *x*, we have:

∴The slope of the tangent at is

Then, the equation of the tangent at is given by,

*y *− 2*at *=

Now, the slope of the normal at is given by,

Thus, the equation of the normal
at (*at*^{2}, 2*at*) is given as:

Question 23:

Prove that the
curves *x *= *y*^{2} and *xy = k *cut at right angles if
8*k*^{2} = 1. [Hint: Two curves
intersect at right angle if the tangents to the curves at the point of
intersection are perpendicular to each
other.]

Answer

The equations of the given curves are given as Putting

Thus, the point of intersection of the given curves is . Differentiating

Therefore, the slope of the tangent to the curve

□ Slope of
the tangent to the curve *xy = k *at is given by,

We know that two curves intersect at right angles if the tangents to the curves at the

point of intersection i.e., at are perpendicular to each other. This implies that we should have the product of the tangents as − 1.

Thus, the given two curves cut at right angles if the product of the slopes of their

respective tangents at is −1.

Hence, the given two curves cut at right angels if 8*k*^{2} = 1.

Question 24:

Find the equations of the tangent and normal to the hyperbola at the point .

Answer

Differentiating with respect
to *x*, we have:

Therefore, the slope of the tangent at is . Then, the equation of the tangent at is given by,

Now, the slope of the normal at is given by,

Hence, the equation of the normal at is given by,

Question 25:

Find the equation of the tangent
to the curve which
is parallel to the line 4*x*

— 2*y *+ 5 = 0.

Answer

The equation of the given curve is

The slope of the tangent to the given curve at any point (*x*, *y*) is given by,

The equation
of the given line is 4*x *− 2*y *+ 5 = 0.

4*x *− 2*y *+ 5 = 0 ∴ (which is of the form

∴Slope of the line = 2

Now, the tangent to the given curve is parallel to the line 4

∴Equation of the tangent passing through the point is given by,

Hence, the equation of the required tangent is .

Question 26:

The slope of the normal to the curve *y *= 2*x*^{2} + 3 sin *x *at *x *= 0 is

(A) 3 (B) (C) −3 (D)

Answer

The equation of the given curve is .

Slope of the tangent
to the given curve at *x *= 0 is given by,

Hence, the slope of the normal to the given curve at *x *= 0 is

The correct answer is D.

Question 27:

The line *y *= *x *+
1 is a tangent to the curve *y*^{2} = 4*x *at the point
(A) (1, 2) (B) (2, 1) (C) (1, −2)
(D) (−1, 2)

Answer

The equation of the given curve is . Differentiating with respect to

Therefore, the slope of the tangent
to the given curve at any point (*x*, *y*)
is given by,

The given line is *y
*= *x *+ 1 (which is of the form *y *=
*mx *+ *c)*

□ Slope of the line = 1

The line *y *= *x *+ 1 is a tangent
to the given curve if the slope of the line is equal to the slope of the tangent. Also, the line must
intersect the curve.

Thus, we must have:

Hence, the line *y *= *x *+
1 is
a tangent to the given
curve at the point (1, 2). The correct answer
is A.

Exercise 6.4

Question 1:

1. Using differentials, find the approximate value of each of the following up to 3 places of decimal

(i) (ii) (iii)

(iv) (v) (vi)

(vii) (viii) (ix)

(x) (xi) (xii)

(xiii) (xiv) (xv)

Answer (i)

Consider . Let

Now, *dy *is approximately equal to ∆*y *and is given by,

Hence, the approximate value of is 0.03 + 5 = 5.03. (ii)

Consider . Let *x *= 49 and ∆*x *= 0.5. Then,

Now, *dy *is approximately equal to ∆*y *and is given by,

Hence, the approximate value of is 7 + 0.035 = 7.035. (iii)

Consider . Let

Now, *dy *is approximately equal to ∆*y *and is given by,

Hence, the approximate value of is 1 + (−0.2) = 1 − 0.2 = 0.8.

(iv)

Consider . Let

Now, *dy *is approximately equal to ∆*y *and is given by,

Hence, the approximate value of is 0.2 + 0.008 = 0.208.

(v)

Consider . Let

Now, *dy *is approximately equal to ∆*y *and is given by,

Hence, the approximate value of is 1 + (−0.0001) = 0.9999.

(vi)

Consider . Let

Now, *dy *is approximately equal to ∆*y *and is given by,

Hence, the approximate value of is 2 + (−0.03125) = 1.96875.

(vii)

Consider . Let

Now, *dy *is approximately equal to ∆*y *and is given by,

Hence, the approximate value of is 3 + (−0.0370) = 2.9629.

(viii)

Consider . Let

Now, *dy *is approximately equal to ∆*y *and is given by,

Hence, the approximate value of is 4 + (−0.0039) = 3.9961.

(ix)

Consider . Let

Now, *dy *is approximately equal to ∆*y *and is given by,

Hence, the approximate value of is 3 + 0.009 = 3.009.

(x)

Consider . Let

Now, *dy *is approximately equal to ∆y
and is given by,

Hence, the approximate value of is 20 + 0.025 = 20.025.

(xi)

Consider . Let

Now, *dy *is approximately equal to ∆*y *and is given by,

Thus, the approximate value of is 0.06 + 0.00083 = 0.06083.

(xii)

Consider . Let

Now, *dy *is approximately equal to ∆*y *and is given by,

Hence, the approximate value of is 3 + (−0.015) = 2.984.

(xiii)

Consider . Let

Now, *dy *is approximately equal to ∆*y *and is given by,

Hence, the approximate value of is 3 + 0.0046 = 3.0046. (xiv)

Consider . Let

Now, *dy *is approximately equal to ∆*y *and is given by,

Hence, the approximate value of is 8 + (−0.096) = 7.904. (xv)

Consider . Let

Now, *dy *is approximately equal to ∆*y *and is given by,

Hence, the approximate value of is 2 + 0.00187 = 2.00187.

Question 2:

Find the approximate value of *f *(2.01), where
*f *(*x*) = 4*x*^{2} + 5*x *+ 2
Answer

Let *x *= 2 and ∆*x *= 0.01. Then, we have:

*f*(2.01) = *f*(*x *+ ∆*x*) = 4(*x
*+ ∆*x*)^{2} + 5(*x *+
∆*x*) + 2 Now,
∆*y *= *f*(*x *+ ∆*x*) − *f*(*x*)

□ *f*(*x *+ ∆*x*) = *f*(*x*) + ∆*y*

* *

≈

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

Question 3:

Find the approximate value of *f *(5.001), where *f *(*x*) = *x*^{3} −
7*x*^{2} + 15. Answer

Let *x *= 5 and ∆*x *= 0.001. Then, we have:

Hence, the approximate value of *f *(5.001) is −34.995.

Question 4:

Find the approximate change in the volume *V *of a cube of side *x *metres
caused by increasing side by 1%.

Answer

The volume of a cube (*V)
*of side *x *is given by *V *= *x*^{3}.

Hence, the approximate change in the volume of the cube is 0.03*x*^{3} m^{3}.

Question 5:

Find the approximate change in the surface area of a cube of side *x *metres caused
by decreasing the side by 1%

Answer

The surface area of a cube (*S) *of
side *x *is given
by *S *= 6*x*^{2}.

Hence, the approximate change in the surface area of the cube is 0.12*x*^{2} m^{2}.

Question 6:

If the radius of a sphere is measured as 7 m with an error of 0.02m, then find the approximate error in calculating its volume.

Answer

Let *r *be the radius of the sphere
and ∆*r *be
the error in measuring the radius. Then,

*r *= 7 m and ∆*r *= 0.02 m

Now, the volume *V *of the sphere is given by,

Hence, the approximate error
in calculating the volume is 3.92 π
m^{3}.

Question 7:

If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating in surface area.

Answer

Let *r *be the radius of the sphere
and ∆*r *be
the error in measuring the radius. Then,

*r *= 9 m and ∆*r *= 0.03 m

Now, the surface area of the sphere (*S)
*is given by,

*S *= 4πr^{2}

Hence, the approximate error
in calculating the surface area is 2.16π m^{2}.

Question 8:

If *f *(*x*) = 3*x*^{2} + 15*x *+ 5, 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 ∆*x *= 0.02. Then, we have:

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

Question 9:

The approximate change in the volume of a cube of side *x *metres caused by increasing the side by 3% is

A. 0.06 *x*^{3}
m^{3} B. 0.6 *x*^{3} m^{3} C. 0.09 *x*^{3} m^{3} D. 0.9 *x*^{3} m^{3}

Answer

The volume of a cube (*V*) of side *x *is given by *V *= *x*^{3}.

Hence, the approximate change in the volume of the cube is 0.09*x*^{3} m^{3}. The correct answer
is C.

Exercise 6.5

Question 1:

Find the maximum and minimum values,
if any, of the following
functions given by (i) *f*(*x*) = (2*x *− 1)^{2} + 3
(ii) *f*(*x*) = 9*x*^{2} + 12*x *+
2

(iii) *f*(*x*) = −(*x *− 1)^{2} + 10 (iv) *g*(*x*) = *x*^{3} + 1 Answer

(i) The given function
is *f*(*x*) = (2*x *− 1)^{2} + 3.

It can be observed that (2*x *− 1)^{2} ≥ 0 for every *x *∴ R. Therefore,
*f*(*x*) = (2*x *− 1)^{2} + 3 ≥ 3 for
every *x *∴ R. The minimum
value of *f *is attained
when 2*x *− 1 = 0.

2*x *− 1 = 0 ∴

∴Minimum value of *f *= =
3 Hence, function *f *does
not have a maximum value.

(ii) The
given function is *f*(*x*) = 9*x*^{2} + 12*x *+ 2 = (3*x *+
2)^{2} − 2. It
can be observed that (3*x *+ 2)^{2} ≥ 0 for every *x *∴ R. Therefore, *f*(*x*) = (3*x
*+ 2)^{2} − 2 ≥ −2
for every *x *∴ R.

The minimum value of *f *is attained when 3*x *+ 2 = 0.

3*x *+ 2 = 0 ∴

∴Minimum value of *f *=

Hence, function *f *does
not have a maximum value.

(iii) The given function
is *f*(*x*) = − (*x
*− 1)^{2} + 10.

It can be
observed that (*x *− 1)^{2} ≥ 0 for every *x *∴
R. Therefore,
*f*(*x*) = − (*x *− 1)^{2} + 10 ≤ 10
for every *x *∴ R. The maximum value of *f *is attained
when (*x *− 1) = 0. (*x *− 1) = 0 ∴ *x *= 0

∴Maximum value of *f *= *f*(1) = − (1 − 1)^{2} + 10 = 10

Hence, function *f *does
not have a minimum value.

(iv) The given function
is *g*(*x*) = *x*^{3} + 1.

Hence, function *g *neither has a maximum
value nor a minimum value.

Question 2:

Find the maximum and minimum values,
if any, of the following functions given by (i) *f*(*x*) = |*x *+ 2| −
1 (ii) *g*(*x*) = −
|*x *+ 1| + 3

(iii) *h*(*x*) = sin(2*x*) + 5 (iv) *f*(*x*) = |sin 4*x
*+ 3| (v) *h*(*x*) = *x *+ 4, *x *(−1, 1)

Answer

(i) *f*(*x*) =

We know that for
every *x *∴ R. Therefore, *f*(*x*) = for every
*x *∴ R.

The minimum value of *f *is attained
when .

∴Minimum value of *f *= *f*(−2) =

Hence, function
*f *does not have a maximum value.

(ii) *g*(*x*) =

We know that for
every *x *∴ R. Therefore, *g*(*x*) = for every *x *∴ R.

The maximum value of *g *is
attained when .

∴Maximum value of *g *= *g*(−1) =

Hence, function *g *does
not have a minimum value.

(iii) *h*(*x*) = sin2*x *+ 5

We know that −
1 ≤ sin 2*x *≤ 1.

□ − 1 + 5 ≤
sin 2*x *+ 5 ≤ 1 + 5

□ 4 ≤ sin 2*x *+ 5 ≤
6

Hence, the maximum and minimum values
of *h *are 6 and 4 respectively.

(iv) *f*(*x*) =

We know that −1
≤ sin 4*x *≤ 1.

□ 2 ≤ sin 4*x *+ 3 ≤
4

□ 2 ≤ ≤ 4

Hence, the maximum and minimum values
of *f *are 4 and 2 respectively. (v) *h*(*x*) = *x *+ 1, *x *∴ (−1, 1)

Here, if a point
*x*_{0} is closest
to −1, then we find for all *x*_{0} ∴
(−1, 1).

Also, if *x*_{1} is closest to 1, then for all *x*_{1} ∴
(−1, 1).

Hence, function *h*(*x*) has neither maximum
nor minimum value
in (−1, 1).

Question 3:

Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

(i). *f*(*x*) = *x*^{2} (ii). *g*(*x*) = *x*^{3} −
3*x*

* *

* *

(iii). *h*(*x*) = sin*x *+
cos, 0 < (iv). *f*(*x*) = sin*x *− cos *x*, 0 < *x *< 2π
(v). *f*(*x) *= *x*^{3} − 6*x*^{2} + 9*x
*+ 15

(vi).

(vii).

(viii). Answer

(i) *f*(*x*) = *x*^{2}

Thus, *x *= 0 is the only critical
point which could possibly be the point of local maxima
or local minima of *f*.

We have , which is positive.

Therefore, by second derivative test, *x *= 0 is a point of local minima and local minimum
value of *f *at *x *= 0 is *f*(0) = 0.

*(ii) **g*(*x*) = *x*^{3} −
3*x*

By second derivative test,
*x *= 1 is a point of local minima and local minimum
value of *g*

at *x *= 1 is *g*(1) = 1^{3} −
3 = 1 − 3 = −2.
However,

*x *= −1 is a point of local maxima and local maximum
value of *g *at

*x *= −1 is *g*(1) = (−1)^{3} − 3 (− 1) = − 1 + 3 = 2.

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

Therefore, by second derivative test, is a point of local maxima and the local

maximum value of *h *at is

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

Therefore, by second derivative test, is a point of local maxima and the local

maximum value of *f *at is

However, is a point of local minima and

the local minimum
value of *f *at is .

(v) *f*(*x*) = *x*^{3} −
6*x*^{2} + 9*x *+ 15

Therefore,
by second derivative test, *x *= 1 is a point of local maxima and the
local maximum value of *f *at *x *= 1 is *f*(1) = 1 −
6 + 9 + 15 = 19. However, *x *= 3 is a point
of local minima and the local
minimum value of *f *at *x *= 3 is *f*(3) = 27 − 54 + 27 +
15 = 15.

(vi)

Since *x *> 0, we take *x *= 2.

Therefore, by second derivative test, *x *= 2 is a point
of local minima
and the local

minimum value of *g *at *x *= 2 is
*g*(2) =

(vii)

Now, for values close
to *x *= 0 and to the left of 0, Also, for values close
to *x *= 0 and to the right
of 0, .

Therefore, by first derivative test, *x *= 0 is a point of local maxima and the local

maximum value of . (viii)

Therefore, by second derivative test, is a point of local maxima and the local

maximum value of *f *at is

Question 4:

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

(i) *f*(*x*) = *e ^{x} *(ii)

(iii) *h*(*x*) = *x*^{3} + *x*^{2} + *x *+
1 Answer

i. We have,

*f*(*x*) = e^{x}

Now, if . But, the exponential function can never assume 0 for any value of *x*.

Therefore, there
does not exist *c*∴ R such that Hence, function
*f *does not have maxima or minima.

ii. We have,

*g*(*x*) = log *x*

* *

Therefore, there does not exist *c*∴ R such that . Hence, function
*g *does not have maxima or minima.

iii. We have,

*h*(*x*) = *x*^{3} + *x*^{2} + *x *+ 1

Now,

*h*(*x*) = 0 ∴
3*x*^{2} + 2*x *+ 1 = 0 ∴

Therefore, there does not exist *c*∴ R such that . Hence, function
*h *does not have
maxima or minima.

Question 5:

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

(i) (ii)

(iii)

(iv)

Answer

(i) The given function
is *f*(*x*) = x^{3}.

Then, we evaluate the value of *f *at
critical point *x *= 0 and at end points of the interval
[−2, 2].

*f*(0) = 0

*f*(−2) = (−2)^{3} = −8

*f*(2) = (2)^{3} = 8

Hence, we can conclude
that the absolute
maximum value of *f *on
[−2, 2] is 8 occurring
at *x *= 2. Also, the absolute
minimum value of *f *on [−2, 2] is −8
occurring at *x *= −2.

(ii)
The given function is *f*(*x*) = sin *x *+ cos *x*.

Then, we evaluate the value of *f *at
critical point and at the end points of the interval
[0, π].

Hence, we can conclude
that the absolute
maximum value of *f *on
[0, π] is occurring

at and the absolute minimum
value of *f *on [0, π] is −1 occurring at *x *= π.

(iii) The given function is

Then, we evaluate the value of *f *at
critical point *x *= 4 and at the end points of the

interval .

Hence, we can conclude
that the absolute
maximum value of *f *on is 8 occurring

at *x *= 4 and the absolute minimum
value of *f *on is −10 occurring at *x *= −2.

(iv) The given function is

Now,

2(*x
*− 1) = 0 ∴
*x *= 1

Then, we evaluate the value of

Hence, we can conclude
that the absolute
maximum value of *f *on
[−3, 1] is 19 occurring at *x *= −3 and the minimum
value of *f *on [−3, 1] is 3 occurring at *x *= 1.

Question 6:

Find the maximum profit that a company can make, if the profit function is given by

*p*(*x*) = 41 − 24*x *− 18*x*^{2}

Answer

The profit function is given as *p*(*x*) = 41 −
24*x *− 18*x*^{2}.

By second derivative test, is the point of local maxima
of *p*.

Hence, the maximum profit that the company can make is 49 units.

Question 7:

Find the intervals in which the function
*f *given by is

(i) increasing (ii) decreasing Answer

Now, the points *x *= 1 and *x *= −1
divide the real line into three disjoint
intervals i.e.,

In intervals i.e., when *x *<
−1 and *x *> 1, Thus, when *x *<
−1 and *x *> 1, *f
*is increasing.

In interval
(−1, 1) i.e., when −1
< *x *<
1, Thus, when −1 < *x *<
1, *f *is
decreasing.

Question 8:

At what points in the interval
[0, 2π], does the function
sin 2*x *attain its maximum value?
Answer

Let *f*(*x*) = sin 2*x*.

Then, we evaluate the values of

Hence, we can conclude
that the absolute
maximum value of *f *on
[0, 2π] is occurring

at and .

Question 9:

What is the maximum
value of the function sin *x *+ cos *x*? Answer

Let *f*(*x*) = sin *x *+ cos *x*.

Now, will be negative when (sin *x *+ cos *x*) is positive i.e.,
when sin *x *and cos *x*

are both positive. Also,
we know that sin *x *and cos *x *both
are positive in the first

quadrant. Then, will be negative when .

Thus, we consider .

∴By second derivative test, *f *will
be the maximum at and the maximum value of *f*

* *

* *

is .

Question 10:

Find the maximum value of 2*x*^{3} −
24*x *+ 107 in the interval [1, 3]. Find the maximum
value of the same function
in [−3, −1].

Answer

Let *f*(*x*) = 2*x*^{3} −
24*x *+ 107.

We first consider the interval [1, 3].

Then, we evaluate
the value
of
*f *at the critical point *x *= 2 ∴
[1, 3] and at the end points of the interval
[1, 3].

*f*(2) = 2(8) −
24(2) + 107 = 16 − 48 + 107 = 75

*f*(1) = 2(1) −
24(1) + 107 = 2 − 24 + 107 = 85

*f*(3) = 2(27) −
24(3) + 107 = 54 − 72 + 107 = 89

Hence, the absolute
maximum value of *f*(*x*) in the interval
[1, 3] is 89 occurring at *x *=
3.

Next, we consider the interval [−3, −1].

Evaluate the value of *f *at the critical point *x *= −2 ∴
[−3, −1] and at the end points
of the interval [1, 3].

*f*(−3) = 2 (−27) − 24(−3) + 107 = −54 + 72 + 107 = 125

*f*(−1) = 2(−1) − 24 (−1)
+ 107 = −2 + 24 + 107 = 129

*f*(−2) = 2(−8) − 24 (−2) + 107 = −16 + 48 + 107 = 139

Hence, the absolute maximum
value of *f*(*x*) in the interval
[−3, −1] is 139 occurring
at *x*

= −2.

Question 11:

It is given that at *x *= 1, the function
*x*^{4}− 62*x*^{2} + *ax *+ 9 attains
its maximum value,
on the interval [0, 2]. Find the value
of *a*.

Answer

Let *f*(*x*) = *x*^{4} −
62*x*^{2} + *ax *+ 9.

It is given that function *f *attains
its maximum value on the interval [0, 2] at *x
*= 1.

Hence, the value of *a *is 120.

Question 12:

Find the maximum and minimum values of *x *+ sin 2*x *on [0, 2π]. Answer

Let *f*(*x*) = *x *+ sin 2*x.*

* *

* *

Then, we evaluate the value of *f *at critical points and at the end points of the interval
[0, 2π].

Hence, we can conclude that the absolute
maximum value of *f*(*x*) in the interval
[0, 2π] is 2π occurring at *x *= 2π and the absolute
minimum value of *f*(*x*) in the interval
[0, 2π] is 0 occurring at *x *= 0.

Question 13:

Find two numbers whose sum is 24 and whose product is as large as possible. Answer

Let one number be *x*. Then, the other number is (24 −
*x*).

Let *P*(*x*) denote the product of the two numbers. Thus,
we have:

∴By second
derivative test, *x *= 12 is the point of local maxima of *P*. Hence, the product
of the numbers is the maximum when the numbers
are 12 and 24 −
12 = 12.

Question 14:

Find two positive numbers
*x *and *y *such that *x *+
*y *= 60 and *xy*^{3} is maximum.

Answer

The two numbers are *x
*and *y *such that *x *+ *y *= 60.

□ *y *= 60 − *x*

Let *f*(*x*) = *xy*^{3.}

∴By second derivative test, *x *= 15 is a point of local maxima of *f*. Thus, function
*xy*^{3} is maximum when *x *= 15
and *y *= 60 −
15 = 45.

Hence, the required numbers are 15 and 45.

Question 15:

Find two positive numbers
*x *and *y *such that their sum is 35 and the product
*x*^{2}*y*^{5} is a maximum

Answer

Let one number be *x*. Then, the other number is *y *=
(35 − *x*). Let *P*(*x*) =
*x*^{2}*y*^{5}. Then, we have:

*x *= 0, *x *=
35, *x *=
10

When *x *= 35, and *y *= 35 − 35 = 0. This will make the product *x*^{2} *y*^{5}

equal to 0.

When *x *= 0, *y *= 35 −
0 = 35 and the product *x*^{2}*y*^{2} will be 0.

□

□ By second derivative test, *P*(*x*) will be the maximum when *x *= 10 and *y *= 35 −
10 = 25.

Hence, the required numbers are 10 and 25.

Question 16:

Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum. Answer

Let one number be *x*. Then, the other number is (16 −
*x*).

Let the sum of the cubes
of these numbers
be denoted by S(*x*). Then,

Now,

□ By second derivative test, *x *= 8
is the point of local
minima of *S*.

Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16 − 8 = 8.

Question 17:

A square piece of tin of side 18 cm is to made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?

Answer

Let the side of the square
to be cut off be *x *cm.
Then, the length
and the breadth of the box will be (18 − 2*x*) cm each and the
height of the box is *x *cm.

Therefore, the volume *V*(*x*) of the box is given by,

*V*(*x*) = *x*(18 − 2*x*)^{2}

*x *= 9 or *x *= 3

If *x *= 9, then the length and the breadth
will become 0.

*x *≠ 9.

*x *= 3.

Now,

By second
derivative test, *x *= 3 is the point of maxima of *V*.

Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible.

Question 18:

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?

Answer

Let the side of the square
to be cut off be *x *cm.
Then, the height
of the box is *x*, the length is 45 −
2*x, *and the breadth is 24 −
2*x*.

Therefore, the volume *V*(*x*) of the box is given by,

Now, *x *=
18 and *x *= 5

It is not possible
to cut off a square of side 18 cm from each corner of the rectangular sheet. Thus, *x *cannot be equal to 18.

∴*x *= 5

Now,

By second
derivative test, *x *= 5 is the point of maxima.

Hence, the side of the square to be cut off to make the volume of the box maximum possible is 5 cm.

Question 19:

Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Answer

Let a rectangle of length *l *and breadth
*b *be inscribed in the given circle
of radius *a*. Then, the diagonal
passes through the centre and is of length 2*a *cm.

Now, by applying the Pythagoras theorem, we have:

∴Area of the rectangle,

By the second derivative test, when , then the area of the rectangle is the maximum.

Since , the rectangle is a square.

Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square has the maximum area.

Question 20:

Show that the right circular cylinder of given surface and maximum volume is such that is heights is equal to the diameter of the base.

Answer

Let

Let *V *be the volume of the cylinder. Then,

□ By second derivative test, the volume is the maximum when .

Hence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the diameter.

Question 21:

Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area? Answer

Let *r *and *h *be the radius and height of the cylinder
respectively.

Then, volume (*V) *of
the cylinder is given by,

Surface area (*S) *of the cylinder
is given by,

∴By second derivative test, the surface area is the minimum when the radius of the

cylinder is .

Hence, the required dimensions of the can which has the minimum surface area is given

by radius = and height =

Question 22:

A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Answer

Let a piece of length *l *be cut from the given wire to make a square.

Then, the other piece
of wire to be made into a circle is of length
(28 − *l*)
m.

Now, side of square = .

Let *r *be the radius of the circle.
Then,

The combined areas of the square and the circle (*A*) is given by,

Thus, when

By second
derivative test, the area
(*A) *is the minimum when .

Hence, the combined area is the minimum when the length of the wire in making the

square is cm while the length of the wire in making the circle

is .

Question 23:

Prove that the volume of the largest cone that can be inscribed in a sphere of radius *R *is

of the volume of the sphere.

Answer

Let *r *and *h *be the radius and height of the cone respectively inscribed in a sphere of radius
*R*.

Let *V *be the volume of the cone.

Then,

Height of the cone is given by,

*h *= *R *+ AB

□ By second derivative test, the volume of the cone is the maximum when

Question 24:

Show that the right circular cone of least curved surface and given volume has an

altitude equal to time the radius of the base. Answer

Let

The surface area (*S)
*of the cone is given
by,

*S *= π*rl *(where *l *is the slant
height)

Thus, it can be easily verified that when

□ By second derivative test, the surface area of the cone is the least when

Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to times the radius of the base.

Question 25:

Show that the semi-vertical angle of the cone of the maximum volume and of given slant

height is . Answer

Let *θ
*be the semi-vertical angle of the cone.

It is clear that

Let *r*, *h*, and *l *be the radius, height,
and the slant height
of the cone respectively. The slant
height of the cone is given as constant.

Now, *r *= *l *sin *θ
*and *h *=
*l *cos *θ*

The volume (*V) *of the cone is given by,

∴By second derivative test, the volume (*V) *is the maximum when .

Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is .

Question 27:

The point on
the curve *x*^{2} = 2*y *which is nearest to the point (0, 5) is

(A) (B)

(C) (0, 0) (D) (2, 2)

Answer

The given curve is *x*^{2} = 2*y*.

For each value of *x*, the position of the point will be

The distance *d*(*x*) between
the points and (0, 5) is given by,

When,

When,

∴By second derivative test, *d*(*x*) is the minimum at .

When

Hence, the point
on the curve *x*^{2} = 2*y *which is nearest to the point (0, 5) is . 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) Answer

Let

∴By second derivative test,
*f *is the minimum at *x *= 1 and the minimum value is given

by .

The correct answer is D.

Question 29:

The maximum value of is

(A) (B)

(C) 1 (D) 0 Answer

Let

Then, we evaluate the value of

Hence, we can conclude
that the maximum
value of *f *in the interval [0, 1] is 1. The correct answer
is C.

Miscellaneous Solutions

Question 1:

Using differentials, find the approximate value of each of the following.

(a) (b)

Answer

(a) Consider

Then,

Now, *dy *is approximately equal to ∆*y *and is given by,

Hence, the approximate value of = 0.667 + 0.010

= 0.677.

(b)
Consider . Let *x *= 32 and ∆*x
*= 1.

Then,

Now, *dy *is approximately equal to ∆*y *and is given by,

Hence, the approximate value of

= 0.5 − 0.003 = 0.497.

Question 2:

Show that the function given by has maximum at

Now,

1 −
log *x *=
0

Question 3:

The two equal sides of an isosceles triangle with fixed base *b *are
decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

Answer

Let ∆ABC be isosceles where BC is the base of fixed length *b*. Let the length of the two equal sides
of ∆ABC be *a*.

Draw AD∴BC.

Now, in ∆ADC, by applying the Pythagoras theorem, we have:

□ Area of triangle

The rate of change of the area
with respect to time (*t*) is given by,

It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second.

∴

Then, when *a *= *b*, we have:

Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of .

Question 4:

Find the equation of the normal
to curve *y*^{2} = 4*x *at
the point (1, 2). Answer

The equation of the given curve is . Differentiating with respect to

Now, the slope of the normal at point (1, 2) is

∴Equation of the
normal at (1, 2) is *y *− 2 = −1(*x *− 1).

□ *y *− 2 = − *x *+ 1

□ *x *+ *y *− 3 = 0

Question 5:

Show that the normal at any point *θ
*to the curve

is at a constant distance from the origin.

Answer

We have *x *= *a *cos
*θ *+ *a θ *sin
*θ.*

* *

* *

□ Slope of the normal at any point
*θ *is .

The equation of the normal at a given point (*x*, *y*) is given by,

Now, the perpendicular distance of the normal from the origin is

, which is independent of θ.

Hence, the perpendicular distance of the normal from the origin is constant.

Question 6:

Find the intervals in which the function
*f *given by

is (i) increasing (ii) decreasing Answer

Now,

cos *x *= 0 or cos *x *= 4 But, cos
*x *≠ 4

∴cos *x *= 0

divides (0, 2π) into three disjoint intervals i.e.,

In intervals ,

Thus, *f*(*x*) is increasing for

In the interval

Thus, *f*(*x*) is decreasing for .

Question 7:

Find the intervals in which the function
*f *given by is

(i) increasing (ii) decreasing Answer

Now, the points *x *= 1 and *x *= −1
divide the real line into three disjoint
intervals i.e.,

In intervals i.e., when *x *<
−1 and *x *> 1, Thus, when *x *<
−1 and *x *> 1, *f
*is increasing.

In interval
(−1, 1) i.e., when −1
< *x *<
1, Thus, when −1 < *x *<
1, *f *is
decreasing.

Question 8:

Find the maximum area of an isosceles triangle inscribed in the ellipse with its vertex at one end of the major axis.

Answer

The given ellipse is .

Let the major axis be along the *x *−axis.

Let ABC be the triangle inscribed in the ellipse
where vertex C is at (*a*, 0).

Since the ellipse is symmetrical with respect to the *x*−axis and *y *−axis, we can assume the coordinates of A to be