NCERT Solutions for Class 12 Maths Exercise Miscellaneous Chapter 6 Application of Derivatives – FREE PDF Download

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 6 Exercise Miscellaneous (Ex Misc) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 6 Application of Derivatives Exercise Miscellaneous Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex Misc) Exercise Miscellaneous

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

(a)

(b)

Ans. (a)

Let ……….(i)

= ……….(ii)

Changing to and to in eq. (i), we have

……….(iii)

Here and

From eq. (iii),

0.677

(b)

Let ……….(i)

= ……….(ii)

Changing to and to in eq. (i), we have

…….(iii)

Here and

From eq. (iii),

$(33)^{- 1 / 5} \sim \frac{1}{2} - \frac{1}{5 {(2)}^{6}}$ 0.497

2. Show that the function given by has maximum value at

Ans. Here

……….(i)

……..(ii)

And

……….(iii)

Now

From eq. (iii),

= = < 0

is a point of local maxima and maximum value of is at .

3. The two equal sides of an isosceles triangle with fixed base 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?

Ans. Let BC =

be the fixed base and AB = AC =

be
the two equal sides of given isosceles triangle.

Since  cm/s ……….(i)
Area of  x BC x AM
=
=

=  [By chain rule]

=  cm²/s
Now when
  cm²/s
Therefore, the area is decreasing at the rate of  cm²/s.

4. Find the equation of the normal to the curve at the point (1, 2).

Ans. Equation of the curve is

……….(i)

Slope of the tangent to the curve at the point (1, 2) to curve (i) is

Slope of the normal to the curve at (1, 2) is

Equation of the normal to the curve (i) at (1, 2) is

5. Show that the normal at any point to the curve is at a constant distance from the origin.

Ans. The parametric equations of the curve are

And

Slope of tangent at point

Slope of normal at any point

And Equation of normal at any point

i.e., at

Distance of normal from origin (0, 0)
=

which is a constant.

6. Find the intervals in which the function given by is (i) increasing (ii) decreasing.

Ans. Given:

= ……….(i)

Now for all real as . Also > 0

(i) is increasing if , i.e., from eq. (i),

lies in I and IV quadrants, i.e., is increasing for

and

and (ii) is decreasing if , i.e., from eq. (i),

lies in II and III quadrants, i.e., is decreasing for

7. Find the intervals in which the function given by is (i) increasing (ii) decreasing.

Ans. (i) Given:

= =

= ……….(i)

Now

= 0

Here, is positive for all real

or [Turning points]

Therefore, or divide the real line into three disjoint sub intervals and

For , from eq. (i) at (say),

Therefore, is increasing at

For from eq. (i) at (say)

Therefore, is decreasing at

For from eq. (i) at (say),

Therefore, is increasing at

Therefore, is (i) an increasing function for and for and (ii) decreasing function for

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

Ans. Equation of the ellipse is

……….(i)

Comparing eq. (i) with we have and

and

Any point on ellipse is P

Draw PM perpendicular to axis and produce it to meet the ellipse in the point Q.

OM = and PM =

We know that the ellipse (i) is symmetrical about axis, therefore, PM = QM and hence triangle APQ is isosceles.

Area of APQ x Base x Height

= PQ.AM = . 2PM.AM = PM (OA – OM)

Now

= 0

i.e., or

is impossible

At ,

= [Negative]

is maximum at

From eq. (i), Maximum area

= =

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 70 per sq. meter for the base and 45 per square meter for sides. What is the cost of least expensive tank?

Ans. Given: Depth of tank = 2 m

Let m be the length and m be the breadth of the base of the tank.

Volume of tank = 8 cubic meters

Cost of building the base of the tank at the rate of ` 70 per sq. meter is

And cost of building the four walls of the tank at the rate of ` 45 per sq. meter is

= `

Let be the total cost of building the tank.

= $70 x (4 / x) + 180 x + 180 (4 / x)$

= $280 + 180 x + \frac{720}{x}$

and

Now

[length cannot be negative]

At [Positive]

is minimum at .

Minimum cost =

= 280 + 360 + 360 = Rs 1000

10. The sum of the perimeter of a circle and square is where is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Ans. Let

be the radius of the circle and

be the side of square.

According to question, Perimeter of circle + Perimeter of square =

……….(i)

Let be the sum of areas of circle and square.

[From eq. (i)]

and

Now

At [Positive]

is minimum when

From eq. (i),

= =

Therefore, sum of areas is minimum when side of the square is double the radius of the circle.

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.

Ans. Let

m be the radius of the semi-circular opening of the window. Then one side of rectangle part of window is

and

m be the other side of rectangle.

Perimeter of window

= Semi-circular arc AB + Length (AD + DC + BC)

……….(i)

Area of window

= Area of semi-circle + Area of rectangle

and =

Now

At [Negative]

is maximum at

From eq. (i),

$y = \frac{10 π + 40 - 10 π - 20}{2 (π + 4)}$

= = m

Therefore, Length of rectangle = m and Width of rectangle = m

And Radius of semi-circle = m

12. A point on the hypotenuse of a triangle is at distances and from the sides of the triangle. Show that the maximum length of the hypotenuse is

Ans. Let P be a point on the hypotenuse AC of a right triangle ABC such that PL

AB =

and PM

BC =

and let

BAC =

MPC =

, then in right angled

AP = PL =

And in right angled PMC,

PM = PM

Let AC = then

= AP + PC = ……….(i)

Now

……….(ii)

And

[ and is +ve as )

is minimum when

Also

Putting these values in eq. (i),

Minimum length of hypotenuse =

= =

13. Find the points at which the function given by has:

(i) local maxima

(ii) local minima

(iii) point of inflexion.

Ans. Given:

……….(i)

$f^{'} (x) = (x - 2)^{4} \frac{d}{d x} (x + 1)^{3} + (x + 1)^{3} \frac{d}{d x} (x - 2)^{4}$

= $3 (x - 2)^{4} (x + 1)^{2} + 4 (x + 1)^{3} (x - 2)^{3}$

= $(x - 2)^{3} (x + 1)^{2} [3 (x - 2) + 4 (x + 1)]$

Now

= 0

or or

Now, for values of close to and to the left of . Also for values of close to and to the right of .

Therefore, is the point of local maxima.

Now, for values of close to 2 and to the left of . Also for values of close to 2 and to the right of .

Therefore, is the point of local minima.

Now as the values of varies through does not change its sign. Therefore, is the point of inflexion.

14. Find the absolute maximum and minimum values of the function given by

Ans. Given:

……….(i)

= =

Now

= 0

[Turning points]

thus two turning points are $π / 2 a n d π / 6$

Now

= 0 + 1 = 1

1 + 0 = 1

= 1

Therefore, absolute maximum is and absolute minimum is 1.

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

Ans. Let

be the radius of base of cone and

be the height of the cone inscribed in a sphere of radius

OD = AD – AO =

In right angled triangle OBD,

OD² + BD² = OB²

……….(i)

Volume of cone (V) = = [From eq. (i)]

V =

and

Now

and $y \neq 0$

= [Negative]

Volume is maximum at

16. Let be a function defined on such that for all Then prove that is an increasing function on

Ans. Let

I

be the interval

Given: for all in an interval $I$ . Let $I$ with

By Lagrange’s Mean Value Theorem, we have,

where

Now

……….(i)

Also, for all in an interval I

From eq. (i),

Thus, for every pair of points I with

Therefore, is strictly increasing in $I$ .

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.

Ans. Let

be the radius and

be the height of the cylinder inscribed in a sphere having centre O and radius R.

In right triangle OAM, AM² + OM² = OA²

……….(i)

Volume of cylinder (V) = ……….(ii)

V =

= ……….(iii)

and

Now

= [Negative]

V is maximum at

From eq. (iii),

Maximum value of cylinder =

18. Show that the height of the cylinder of greatest volume which can be inscribed in a right circular cone of height and having semi-vertical angle is one-third that of the cone and the greatest volume of the cylinder is

Ans. Let

be the radius of the right circular cone of height

Let the radius of the inscribed cylinder be

and height

In similar triangles APQ and ARC,

Volume of cylinder (V) = ……….(ii)

V =

= ……….(iii)

$\frac{d V}{d x} = \frac{π h}{r} (2 r x - 3 x^{2})$ and $\frac{d^{2} V}{d x^{2}} = \frac{π h}{r} (2 r - 6 x)$

Now $\frac{d V}{d x} = 0$

and $x \neq 0$

At , $\frac{d^{2} V}{d x^{2}} = \frac{π h}{r} (2 r - \frac{12 r}{3})$

= [Negative]

V is maximum at

From eq. (iii),

Maximum value of cylinder = $\frac{π h}{r} [r \frac{4 r^{2}}{9} - \frac{8 r^{3}}{27}]$

Choose the correct answer in the Exercises 19 to 24:

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

(A) 1 cu m/h

(B) 0.1 cu m/h

(D) 0.5 cu m/h

Ans. Let

be the depth of the wheat in the cylindrical tank of radius 10 m at time