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

Students can get the NCERT solutions for Class 12 Maths chapter 6 miscellaneous exercise on this page. The Class 12 Maths chapter 6 miscellaneous exercise solutions are designed to make the students understand the applications of concepts studies in the NCERT book. Class 12 Maths chapter 6 miscellaneous solutions covers all the topics discussed in the previous exercises. Miscellaneous exercise chapter 6 Class 12 presents questions to practice the complete chapter. One should visit the NCERT Solutions for Class 12 Maths chapter 6 miscellaneous exercise.

After completing all the topics of NCERT syllabus Class 12 Maths chapter 6 and solved and unsolved questions of all other exercises the miscellaneous exercise can be attempted. The miscellaneous exercise combines questions from the complete chapter and the level of questions compared to exercises will be a bit higher. The following exercises are also present in the chapter.

Application of Derivatives Exercise 6.1

Application of Derivatives Exercise 6.2

Application of Derivatives Exercise 6.3

Application of Derivatives Exercise 6.4

Application of Derivatives Exercise 6.5

## Application of Derivatives Miscellaneous 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

** 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: **

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

increasing

** 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

is

decreasing

** 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

Increasing

** 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

decreasing

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

** Answer: **

Given the equation of the ellipse

Now, we know that ellipse is symmetrical about x and y-axis. 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,

Length 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)

** 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 m3. If building of tank costs Rs 70 per sq metres 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 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

** 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: **

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

** Question:11 ** A window is in the form of a 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 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

** 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 4r/3

** 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

** Question:16 ** Let f be a function defined on [a, b] such that , for all . Then prove that f is an increasing function on (a, b).

** 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

** 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 a is one-third that of the cone and the greatest volume of cylinder 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

** Question:19 ** A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314

cubic metre 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: **

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

A ) 22/7

B ) 6/7

C ) 7/6

D ) -6 /7

** 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

(B) 2

(C) 3

(D)1/2

** 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 ** The normal at the point (1,1) on the curve is

(A) x + y = 0

(B) x – y = 0

(C) x + y +1 = 0

(D) x – y = 1

** 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

(A) x + y = 3

(B) x – y = 3

(C) x + y = 1

(D) x – y = 1

** 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)

**More About NCERT Solutions for Class 12 Maths Chapter 6 Miscellaneous Exercise**

Practice questions related to all the 5 main topics covered in the Class 12 NCERT Mathematics chapter application of derivatives are covered in Class 12 Maths chapter 6 miscellaneous exercise solutions. All these solutions of miscellaneous exercise are detailed in this page and are solved by Mathematics experts. The NCERT solutions for Class 12 Maths chapter 6 miscellaneous exercise are given in detail and step by step manner.

**Also Read| **Application of Derivatives Class 12 Notes

**Benefits of NCERT Solutions for Class 12 Maths Chapter 6 Miscellaneous Exercise**

Class 12 Maths chapter 6 miscellaneous solutions can be used to prepare for board exam as well as copetitive exams to the admission for various engineering colleges across India.

The NCERT solutions for Class 12 Maths chapter 6 miscellaneous exercise helps in revising the whole chapter and also some basic derivatives.

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