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

#### Page No 242:

#### 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 Δ*y*and is given by,
Hence, the approximate value of = 0.667 + 0.010

= 0.677.

(b) Consider. Let

*x*= 32 and Δ*x*= 1.
Then,

Now,

*dy*is approximately equal to Δ*y*and is given by,
Hence, the approximate value of

= 0.5 − 0.003 = 0.497.

#### Question 2:

Show that the function given byhas maximum at

*x*=*e*.#### Answer:

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 ΔABC be isosceles where BC is the base of fixed length

*b*.
Let the length of the two equal sides of ΔABC be

*a*.
Draw AD⊥BC.

Now, in ΔADC, 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

*x*, we have:
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

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 byis
(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 (−*x*_{1},*y*_{1}) and the coordinates of B to be (−*x*_{1}, −*y*_{1}).
Now, we have.

∴Coordinates of A are and the coordinates of B are

As the point (

*x*_{1},*y*_{1}) lies on the ellipse, the area of triangle ABC (*A)*is given by,
But,

*x*_{1}cannot be equal to*a*.
Also, when, then

Thus, the area is the maximum when

∴ Maximum area of the triangle is given by,

#### Question 9:

A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m

^{3}. If building of tank costs Rs 70 per sq meters for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?#### Answer:

Let

*l*,*b*, and*h*represent the length, breadth, and height of the tank respectively.
Then, we have height (

*h)*= 2 m
Volume of the tank = 8m

^{3}
Volume of the tank =

*l*×*b*×*h*
∴ 8 =

*l*×*b*× 2
Now, area of the base =

*lb*= 4
Area of the 4 walls (

*A)*= 2*h*(*l*+*b*)
However, the length cannot be negative.

Therefore, we have

*l*= 4.
Thus, by second derivative test, the area is the minimum when

*l*= 2.
We have

*l*=*b*=*h*= 2.
∴Cost of building the base = Rs 70 × (

*lb*) = Rs 70 (4) = Rs 280
Cost of building the walls = Rs 2

*h*(*l*+*b*) × 45 = Rs 90 (2) (2 + 2)
= Rs 8 (90) = Rs 720

Required total cost = Rs (280 + 720) = Rs 1000

Hence, the total cost of the tank will be Rs 1000.

#### Question 10:

The sum of the perimeter of a circle and square is

*k*, where*k*is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.#### Answer:

Let

*r*be the radius of the circle and*a*be the side of the square.
Then, we have:

The sum of the areas of the circle and the square (

*A)*is given by,
Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle.

#### Page No 243:

#### Question 11:

A window is in the form of rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.

#### Answer:

Let

*x*and*y*be the length and breadth of the rectangular window.
Radius of the semicircular opening

It is given that the perimeter of the window is 10 m.

∴Area of the window (

*A)*is given by,
Thus, when

Therefore, by second derivative test, the area is the maximum when length.

Hence, the required dimensions of the window to admit maximum light is given by

#### 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:

Let ΔABC be right-angled at B. Let AB =

*x*and BC =*y*.
Let P be a point on the hypotenuse of the triangle such that P is at a distance of

*a*and*b*from the sides AB and BC respectively.
Let ∠C =

*θ*.
We have,

Now,

PC =

*b*cosec*θ*
And, AP =

*a*sec*θ*
∴AC = AP + PC

⇒ AC =

*b*cosec*θ*+*a*sec*θ*… (1)
Therefore, by second derivative test, the length of the hypotenuse is the maximum when

Now, when, we have:

Hence, the maximum length of the hypotenuses is.

#### Question 13:

Find the points at which the function

*f*given byhas
(i) local maxima (ii) local minima

(ii) point of inflexion

#### Answer:

The given function is

Now, for values of

*x*close toand to the left of Also, for values of*x*close to and to the right of
Thus, is the point of local maxima.

Now, for values of

*x*close to 2 and to the left of Also, for values of*x*close to 2 and to the right of 2,
Thus,

*x*= 2 is the point of local minima.
Now, as the value of

*x*varies through −1,does not changes its sign.
Thus,

*x*= −1 is the point of inflexion.#### Question 14:

Find the absolute maximum and minimum values of the function

*f*given by#### Answer:

Now, evaluating the value of

*f*at critical pointsand at the end points of the interval (i.e., at*x*= 0 and*x*= π), we have:
Hence, the absolute maximum value of

*f*is occurring at and the absolute minimum value of*f*is 1 occurring at#### Question 15:

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius

*r*is.#### Answer:

A sphere of fixed radius (

*r)*is given.
Let

*R*and*h*be the radius and the height of the cone respectively.
The volume (

*V)*of the cone is given by,
Now, from the right triangle BCD, we have:

∴

*h*
∴ The volume is the maximum when

Hence, it can be seen that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius

*r*is.#### Question 16:

Let

*f*be a function defined on [*a*,*b*] such that*f*‘(*x*) > 0, for all*x*∈ (*a*,*b*). Then prove that*f*is an increasing function on (*a*,*b*).#### Answer:

Let

x1, x2∈(a,b)such that

x1

x1, x2]. Since

*f*(*x*) is differentiable on (a, b) and
[x1, x2]⊂(a,b). Therefore, f(x) is continous on [

x1, x2] and differentiable on

(x1, x2). By the Lagrange’s mean value theorm, there exists

c∈(x1, x2)such that

f'(c)=f(x2)-f(x1)x2-x1 …(1)Since

*f*‘(*x*) > 0 for all
x∈(a,b), so in particular,

*f*‘(*c*) > 0
f'(c)>0⇒f(x2)-f(x1)x2-x1>0 [Using (1)]

⇒f(x2)-f(x1)>0 [âˆµ

x2-x1>0 when x1

⇒f(x2)>f(x1)⇒f(x1)

x1, x2are arbitrary points in

(a,b). Therefore,

x1f

*x*) is increasing 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:

A sphere of fixed radius (

*R)*is given.
Let

*r*and*h*be the radius and the height of the cylinder respectively.
From the given figure, we have

The volume (

*V)*of the cylinder is given by,
Now, it can be observed that at.

∴The volume is the maximum when

When, the height of the cylinder is

Hence, the volume of the cylinder is the maximum when the height of the cylinder is.

#### Question 18:

Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height

*h*and semi vertical angle*α*is one-third that of the cone and the greatest volume of cylinder istan^{2}*α*.#### Answer:

The given right circular cone of fixed height (

*h)*and semi-vertical angle (*α**)*can be drawn as:
Here, a cylinder of radius

*R*and height*H*is inscribed in the cone.
Then, ∠GAO =

*α*, OG =*r*, OA =*h*, OE =*R*, and CE =*H*.
We have,

*r*=

*h*tan

*α*

Now, since ΔAOG is similar to ΔCEG, we have:

Now, the volume (

*V)*of the cylinder is given by,
And, for, we have:

∴By second derivative test, the volume of the cylinder is the greatest when

Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest.

Now, the maximum volume of the cylinder can be obtained as:

Hence, the given result is proved.

#### Question 19:

A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic mere per hour. Then the depth of the wheat is increasing at the rate of

(A) 1 m/h (B) 0.1 m/h

(C) 1.1 m/h (D) 0.5 m/h

#### Answer:

Let

*r*be the radius of the cylinder.
Then, volume (

*V)*of the cylinder is given by,
Differentiating with respect to time

*t*, we have:
The tank is being filled with wheat at the rate of 314 cubic metres per hour.

∴

Thus, we have:

Hence, the depth of wheat is increasing at the rate of 1 m/h.

The correct answer is A.

#### Question 20:

The slope of the tangent to the curveat the point (2, −1) is

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

#### Answer:

The given curve is

The given point is (2, −1).

At

*x*= 2, we have:
The common value of

*t*is 2.
Hence, the slope of the tangent to the given curve at point (2, −1) is

The correct answer is B.

#### Page No 244:

#### Question 21:

The line

*y*=*mx*+ 1 is a tangent to the curve*y*^{2}= 4*x*if the value of*m*is
(A) 1 (B) 2 (C) 3 (D)

#### Answer:

The equation of the tangent to the given curve is

*y*=*mx*+ 1.
Now, substituting

*y*=*mx*+ 1 in*y*^{2}= 4*x*, we get:
Since a tangent touches the curve at one point, the roots of equation (i) must be equal.

Therefore, we have:

Hence, the required value of

*m*is 1.
The correct answer is A.

#### Question 22:

The normal at the point (1, 1) on the curve 2

*y*+*x*^{2}= 3 is
(A)

*x*+*y*= 0 (B)*x*−*y*= 0
(C)

*x*+*y*+ 1 = 0 (D)*x*−*y*= 1#### Answer:

The equation of the given curve is 2

*y*+*x*^{2}= 3.
Differentiating with respect to

*x*, we have:
The slope of the normal to the given curve at point (1, 1) is

Hence, the equation of the normal to the given curve at (1, 1) is given as:

The correct answer is B.

#### Question 23:

The normal to the curve

*x*^{2}= 4*y*passing (1, 2) is
(A)

*x*+*y*= 3 (B)*x*−*y*= 3
(C)

*x*+*y*= 1 (D)*x*−*y*= 1#### Answer:

The equation of the given curve is

*x*^{2}= 4*y*.
Differentiating with respect to

*x*, we have:
The slope of the normal to the given curve at point (

*h*,*k*) is given by,
∴Equation of the normal at point (

*h*,*k*) is given as:
Now, it is given that the normal passes through the point (1, 2).

Therefore, we have:

Since (

*h*,*k*) lies on the curve*x*^{2}= 4*y*, we have*h*^{2}= 4*k.*
From equation (i), we have:

Hence, the equation of the normal is given as:

The correct answer is A.

#### Question 24:

The points on the curve 9

*y*^{2}=*x*^{3}, where the normal to the curve makes equal intercepts with the axes are
(A) (B)

(C) (D)

#### Answer:

The equation of the given curve is 9

*y*^{2}=*x*^{3}.
Differentiating with respect to

*x*, we have:
The slope of the normal to the given curve at point is

∴ The equation of the normal to the curve at is

It is given that the normal makes equal intercepts with the axes.

Therefore, We have:

Also, the pointlies on the curve, so we have

From (i) and (ii), we have:

From (ii), we have:

Hence, the required points are

The correct answer is A.

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