Plus Two Maths Chapter Wise Questions and Answers Chapter 6 Application of Derivatives are part of Plus Two Maths Chapter Wise Questions and Answers. Here we have given Plus Two Maths Chapter Wise Questions and Answers Chapter 6 Application of Derivatives.

Board | SCERT, Kerala |

Text Book | NCERT Based |

Class | Plus Two |

Subject | Maths Chapter Wise Questions |

Chapter | Chapter 6 |

Chapter Name | Application of Derivatives |

Number of Questions Solved | 62 |

Category | Plus Two Kerala |

## Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 6 Application of Derivatives

**Short Answer Type Questions**

**(Score 3)**

Question 1.

The radius of a balloon is ¡n creasing at the rate of 10cm/sec. At what rate is the surface area of the balloon increasing when the radius is 15 cm.

Answer.

= 10 cm/sec ; S = 4πr²

Question 2.

Find the equation of the tangent to the curve (1+x²) y = 2-x where it crosses the X-axis.

Answer.

(1+x²) y = 2-x

Putting y = 0, we get x = 2.

∴ curve crosses x-axis at (2,0) ,

Question 3.

What is the distance between the origin and the normal to the curve y = e^{2x }+ x² at x = 0.

Answer.

Question 4.

Find the slope of the normal to the curve y = 2x² + 3sinx at x = 0.

Answer.

y = 2x² + 3sinx

D.w.r.to x

Question 5.

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

b. A manufacture can sell x Items at a price of Rs. () each. The cost price of x Items is c(x) = () . Write the selling price S(x) of x Items.

Answer.

a. 8, 8

b. S(x) = Selling price of x items

Question 6.

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

Answer.

Let x be one positive number, then 16 – x is the other positive number.

Let S be the sum of their cubes

∴ S = x^{3} + (16-x)^{3}.

∴ S is minimum when x = 8.

When x = 8. the other number is 16 – 8 = 8.

∴ The two numbers are 8,8.

Question 7.

a. A ball is thrown vertically upwards which satisfies the equation S = 80t – 16t². Find the time required to reach the maximum height.

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

Answer.

a. At max. height, v = = 0

i.e. 80-32t = 0

t = = 2.5

t = 2.5

b. f(x) = 3x+17

Since f(x) is a polynomial function, it is continuous and differentiable in R.

∴ f ‘(x) = 3 > 0 for all x ∈ R.

Hence f(x) is strictly increasing on R.

Question 8.

Find the slope of the tangent to the curve y = x^{3} – x + 1 at the point whose x coordinates is 2.

Answer.

y = x^{3} – x + 1

Differentiating w.r.t.x

= 3x² – 1

(at x = 2) = (3)(2²)-1 = 11

Slope of the tangent to the curve (at x = 2) = 11

Question 9.

Find the equation of tangent and normal to the given curve y = x^{3} at (1, 1)

Answer.

y = x^{3}

Differentiating w.r.t.x, = 3x²

At (1 ,1), = 3 x 1² = 3

At(1, 1), the slope of the tangent = 3

Equation of the tangent at (1,’ 1) is

y – 1 = 3(x – 1)

i.e., y = 3x – 2

At (1, 1), the slope of the normal =

Equation of the normal at (1, 1) is

y – 1 = (x – 1)

i.e.. x + 3y – 4 = 0

Question 10.

Find the slope of the normal to the curve y = 2x² + 3sinx at x = 0.

Answer.

y = 2x² + 3sinx

Differentiating w.r.to. x

= 4x+3cosx

**Long Answer Type Questions**

**(Score 4)**

Question 1.

Using differentials find the approximate value of (0.009)^{1/3} upto 3 places of decimal.

Answer.

Let y =

Question 2.

Consider the function

i. Find f ‘(x)

ii. Find points of local maxima & minima and corresponding maximum and minimum values.

Answer.

i. f ‘(x) = -3x^{3} – 24x^{2} – 45x

= -3x(x²+8x+15)

= 3x(x+5)(x+3)

ii. f ‘(x) = 0 ⇒x = 0,-5,-3

f ”(x) = -9x² – 48x – 45

When x = 0, f “(0) = -45 <0

So x = 0 is a point of local maxima.

Local maximum value at x = 0 is

f(0) = 105

When x = -5, f “(-5) = -30 < 0

So x = -5 is a point of local maxima.

Local maximum value at x = -5 is

f(-5) = When x = -3, f “(-3) = 18>0

So x = -3 ¡s a point of local minima.

Local minimum value at x = -3 is

f(-3) =

Question 3.

Consider the curve 2y = 3 – x²

i. Find the slope of the tangent to this curve at (x_{1}, y_{1}).

ii. Find the points on the curve at which tangent ¡s parallel to the line x+y = 0.

Answer.

Slope of tangent at (x_{1}y_{1}) = -x_{1}

ii. Slope of the line x + y = 0 is -1 since tangent is parallel to x + y = 0

we have -x_{1} = -1 or x_{1} = 1

(x_{1}, y_{1}) lies on the curve 2y = 3 – x²

when x_{1} = 1 , y_{1} = 1

∴ Required point is (1, 1).

Question 4.

Consider the curves x = y² and xy = k

i. Differentiate both the equations with respect to x.

ii. Prove that the curves cut at right angles if 8k² = 1.

Answer.

x = y² ———–(A)

Question 5.

The total profit y (in rupee) of a drug company from the manufacture and sale of x bottles of drug is given by

i. How many bottles of drug must the company sell to obtain maximum profit.

ii. What is the maximum profit?

Answer.

Question 6.

Show that the height of a closed cylinder of given surface and maximum volume is equal to the diameter of the base.

Answer.

Question 7.

A man 160 cm tall, walks away from a source of light situated at the top of a pole 6 m high, at the rate of 1.1 ml sec. How fast is the length of his shadow increasing when he is 1m away from the pole?

Answer.

Let AB be the lamp post Then,AB = 6m.

Let CD be the position of the man such that AC = x. Let CE be the shadow of the man (CE = s) from the similar triangles BAE and DCE.

Question 8.

It is given that at x = 1, the function x^{4} – 62x² + ax + 9 attains its maximum value, on the Interval (0,2). Find the value of a.

Answer.

Let f(x) = x^{4} – 62x² + ax + 9, x ∈ [0, 2]

f ‘(x) = 4x^{3} – 124x + a

Since f attains the maximum value at x = 1 and 1 ∈ [0,2]. we get f ‘(1) = 0.

f ‘(1) = 0 = 4 – 124 + a = 0

⇒ a = 120

f “(x) = 12x² – 124

f “(1) = 12 – 124 = -112<0

∴ At x = 1, f(x) attains its maximum value.

Question 9.

The total profit y (in rupees) of a drug company from the manufacture and sale of x bottles of drug is given by

i. How many bottles of drug must the company sell to obtain the maximum profit?

ii. What is the maximum profit?

Answer.

Question 10.

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 In creasing when the height ¡s 4cm?

Answer.

Let V be the volume,r be the base radius and h be the height of the sand cone at time ‘t’.

**Very Long Answer Type Questions**

**(Score 6)**

Question 1.

A cylinder is inscribed in a sphere.

i. Find an expression for volume of the cylinder.

ii. Show that height of the cylinder of maximum volume

Answer.

Hence the volume of the inscribed cylinder is maximum when its height

Question 2.

Water is running into a conical vessel, 15cm deep and 5cm in radius, at the rate of 0.1 cm³/sec. When the water is 6cm deep, find at what rate is

i. The water level rising.

ii. The water surface area increasing.

iii. The wetted surface of the vessel increasing.

Answer.

i. Let α be the semi vertical angle

Question 3.

A jet of an enemy is flying along the curve y = x² + 2. A soldier is placed at the point (3, 2). Let (x, y) be a point on the curve nearest to (3, 2). Let ‘S’ be the distance between these two points:

i. Find S².

ii. If S² is minimum, find x.

iii. What is the minimum distance between the soldier and jet.

Answer.

Question 4.

Prove the following:

i. Of all the rectangles with given perimeter, square has the largest area.

ii. Of all the rectangles with given area, square has the smallest perimeter.

Answer.

i. Perimeter = C

2(x+y) = C; x+y = C/2 ;

y = (C/2)-x ;

Area, A = xy = x (C/2 – x)

Question 5.

a. Find the slope of the curve x² + 3y = 3 at the point (1, 2).

b. Find the equation of the tangent to the curve x² + 3y = 3 which is parallel to the line y – 4x + 5 = 0. Find also the equation of the normal to the curve at the point of contact.

Answer.

Question 6.

Show that none of the following functions has a local maximum or a local minimum.

i. x^{3}+x^{2}+x+1

ii. e^{x}

iii. logx

Answer.

i. Let f(x) = x^{3 }+ x^{2} + x + 1.

Then f ‘(x) = 3x^{2} + 2x + 1

At points of local maximum or minimum,

we have f ‘(x) = 0 ⇒ 3x^{2} + 2x + 1 = 0

But, this equation gives imaginary values of x. So , f ‘(x) ≠ 0 for any values of x .

Hence, f(x) does not have a maximum or minimum.

ii. Let f(x) = e^{x}. Then, f ‘(x) = e^{x}. Clearly,

f ‘(x) ≠ 0 for any value of x.

So, f(x) = e^{x} does not have a maximum or minimum.

iii. Let f(x) = logx. Then, f ‘(x) =

Clearly, f ‘(x) ≠ 0 for any value of x ∈ Domain (f)

So, f(x) = logx does not have a maximum or a minimum.

Question 7.

The combined resistance R of two resistors R1 and R2 (R1, R2 > 0) is given by

If R1 + R2 = C (a constant), show that the maximum resistance R is obtained by choosing R1 = R2.

Answer.

Hence, R is maximim when R1 = R2 = C/2

Question 8.

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

Answer.

Let ABCD be a rectangle inscribed in a given circle with centre at O and radius

a. Let AB = 2x and BC = 2y. Applying Pythagoras theorem in right triangle OAM, we obtain

Question 9.

Show that the height of the cylinder of maximum volume that can he inscribed in a sphere of radius R is Also find the maximum volume.

Answer.

Let y be the radius and x be the height of the cylinder.

Question 10.

a. x 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?

b. Find the local maxima and local minima of the following function g(x)= x^{3} – 3x . Also, find the local maximum and the local minimum values.

Answer.

a. Let x be the length of an edge of cube and V be the volume of the cube.

Then, V = x^{3}

∴ Rate of change of volume w.r.t. time

**Edumate Questions & Answers**

Question 1.

i. Choose the correct answer from the bracket. The rate of change of the area of a circle with respect to its radius r at r = 10cm is

{a.10π b. 20π c. 30π d. 40π}

ii. Find the intervals in which the function f given by f(x) = x² – 6x + 5 is

a. Strictly increasing

b. Strictly decreasing

iii. Find the local minimum and local maximum value, if any, of the function f(x) = x^{3} – 6x^{2} + 9x + 8.

Answer.

i. b. A = πr²

f is minimum, the local minimum value of f = 8

At x = 1,f “(x) = 6 × 1 – 12 = -6<0,

f is maximum, the local maximum value of f = 12.

Question 2.

i. Choose the correct answer from the bracket. The slope of the tangent to the curve y = x^{3} – 2x + 3 at x = 1 is …….

{a.0 b.1 c.2 d.3}

ii. Find points on the curve at which the tangents are

a. Parallel to x-axis

b. Parallel to y -axis

iii. Use differential to approximate √25.6

Answer.

Question 3.

i. Choose the correct answer from the bracket. The function f(x) = cosx is strictly decreasing In the interval is ….

{a. () b. (0, 2π)

c. (0, π) d. (-π , π)}

ii. Find the equation of the tangent to the curve y = x² – 4x + 5 which is parallel to the line 2x + y + 5 = 0.

iii. Find the absolute maximum and minimum values of a function f given by f(x) = x^{3} + 3x² – 9x + 8 on [-4, 2].

Answer.

i. c. (0, π)

ii.

2x-4 = -2

x = 1, y = 2

Equation of tangent is y – 2 = -2 (x – 1)

iii. f ‘(x) = 3x² + 6x – 9 = 3(x² + 2x – 3)

f ‘(x) = 3(x + 3)(x – 1)

f “(x) = 6x + 6

For maxima, minima f ‘(x) = 0

3(x+3)(x-1 ) = 0

3(x+3)(x-1) = 0

x = -3, x = 1

f(1) = 21, f(-3) = 35. f(-4) = 28,

f(2) = 20

The absolute maximum value of

f[-4, 2] is 35, at x = -3.

Question 4.

Prove that f(x) = logsinx is strictly increasing in [o,] and strictly decreasing in [,π]

ii. Consider the parametric forms and of the curve then

a.Find

b. Find the equation of tangent at t = 2.

c. Find the equation of normal at t = 2.

Answer.

Question 5

i. Find the approximate change in volume of a cube of side x metre caused by an increase in the side by 3%.

ii. A wire of length 28m 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 using differentiation ?

Answer.

i. Let x be the side and V be the volume of the cube .

Change in side, ∆x = 3% of x =

Change in volume,

Question 6.

Consider the function , x>0

i. Find the extreme points of f(x).

ii. Find the maximum or minimum values if any.

Answer.

Question 7.

A rectangle sheet of tin with adjuscent sides 45cm and 24cm is to be made into a box without top, by cutting off equal squares from the corners and folding up the flaps.

i. Taking the side of the square cut off as x, express the volume of the box as the function of x.

ii. For what values of x, the volume of the box will be maximum.

Answer.

i. Length of the box = 45 – 2x

Breadth of the box = 24 – 2x

Height of the box = x

Volume V = (45 – 2x) (24 – 2x)x

(1080 – 138x + 4x²)x

=4x^{3} – 138x² + 1080x

Question 8.

i. The slope of the tangent to the curve y = x^{3} inclined at an angle 60° to x-axis is ……

ii. Consider the function y² = 4x + 5

a. Find a point on the curve at which the tangent is parallel to the line y = 2x + 7.

b. Find the equation of the tangent at the point of tangency.

c. Find the approximate value of √0.037.

Answer.

i. Slope = tan60 = √3

ii. a. y² = 4x + 5

Question 9.

Consider the function f(x) = x² in [-2,1].

i. Find the local maximum or minimum if any

ii. Find the absolute maximum and minimum.

Answer.

i. f ‘(x) = 2x

f ‘(x) = 2>0

So f(x) has local minimum

For maximum or minimum f ‘(x) = 0

2x = 0, so x = 0

Local minimum values of f at x = 0 is

f(0) = 0

At the end points we have

x = -2, f(-2) = 4 and

x = 1, f(1) = 1

ii. Absolute maximum value of f

= max {4,1,0} = 4

Absolute minimum value of f

= min {4,1,0} = 0

Question 10.

Of all the cylinders with given surface area, show that the volume is maximum when height is equal to the diameter of the base.

Answer.

Let r be the radius, h be the height,

V be the volume ans S be the surface area

**NCERT Questions & Answers**

Question 1.

Sand is pouring from a pipe at a rate of 12cm^{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 ¡s the height of the sand cone in creasing when the height is 4 cm ?

Answer.

Let V be the volume of the cone,

= 12cm^{3}/s

The height of the cone

Question 2.

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

Answer.

r = 9cm; ∆r = dr = 0.03 cm

Surface area of the sphere, S = 4πr²

Differentiating wrt r,

= 8πr ⇒dS = (8πr)dr

dS = (72 π) x 0.03, as dr = 0.03

dS = 2.16 π cm², Thus approximate error in calculating the surface area of sphere = 2.16 π cm²

Question 3.

Two equal sides of an Isosceles triangle with fixed base ‘a’ are decreasing at the rate of 9cm/s. How fast is the area of the triangle decreasing when the two sides are equal to’a’.

Answer.

Let A be the area of ∆ ABC in which

AB = AC = x and BC = a.

Question 4.

Consider the function

f(x) = (x+1)^{3}(x-3)^{3}

(i) Find f ‘(x)

(ii) Find critical points of f(x)

(iii) Find the intervals in which f(x) is increasing or decreasing.

Answer.

(i) f(x) = (x+1)^{3}(x-3)^{3}

f ‘(x) = (x + 1)^{3}3(x – 3)^{2} + (x – 3)^{3}.3(x + 1)^{2}

= 3(x + 1)^{2}(x – 3)^{2} [x – 3 + x + 1]

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

= 6(x+1)^{2}(x-3)^{2}(x-1)

(ii) Critical points are given by

f ‘(x)=0

i.e., x = -1, 3, 1

(iii) These three points divide the real line into 4 intervals.

Thus f(x) is decreasing in (-∞, 1) and increasing in (1,∞).

Question 5.

Find the points on the curve y = x^{3} – 11x + 5 at which the tangent has the equation y = x – 11.

Answer.

y = x^{3} – 11x + 5

= 3x^{2} – 11; y = x – 11 .

⇒ slope of tangent = 1; i.e., = 1

3x^{2} – 11 = 1; 3x^{2} = 12; x = ±2

When x = 2,y = 2^{3} – 11 x 2 + 5 = -9

When x = -2,y = (-2)^{3} – 11 x – 2 + 5 = 19

∴ The points are (2 -9) and (-2, 19). Of these two points only (2, -9) satisfies the equation of tangent.

∴y = x – 11

∴ Required point is (2, -9).

Question 6.

Prove that the volume of the largest cone that can be inscribed In a sphere of radius R is 8/27 of the volume of the sphere.

Answer.

From figure,

(h-R)² + r² = R²;

r² = R² – (h – R)²

r² = 2Rh – h²;

Question 7.

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

Answer.

Question 8.

A cone as given in figure is given

i. Find an expression for volume of the cone in terms of α.

ii. Prove that volume is maximum when α = tan^{-1}√2

Answer.

i. Let l be the given slant height of the cone and α, its semi vertical angle.

Then height of the cone = h

= OB l cosα and radius

of the base of the cone = r = OA = l

sinα ‘V’ denote the volume of the cone.

Then

Question 9.

A right circular cylinder is inscribed in a cone as in figure .

i. Find an expression for volume of cylinder in terms of x, h.

ii. Prove that volume Is maximum when y = h / 3.

Answer.

i. From similar triangles OCB and QRB

Question 10.

a. An open box is made by removing squares of equal size from the corners of a tin sheet of size 16 cm x 10 cm and folding up the sides of

the box so obtained.

i. With the help of figure, obtain the relation V = x(16 – 2x) (10 – 2x).

ii.What ¡s the value of x for which V is maximum?

b. What is the slope of the tangent and normal at (1, 1) on the curve y = x^{3}.

Answer.

a.(i)V = x(16 – 2x) (10 – 2x)

V = 160x – 52x^{2} + 4x^{3}

Question 11.

a. A water tank is in the shape of a right circular cone with its axis vertical and vertex down. Its height and diameter are same. Water is powered into it at a constant rate of 2m^{3}/minute.

i. With the help of figure obtain the relation

ii. Find the rate at which water level ¡s increasing when depth of water in the tank is 6m.

b. Find the interval in which the function x^{3} – 6x^{2 }+ 9x + 15 is increasing.

Answer.

Question 12.

A window is in the form of a rectangle surmounted by a semicircle as shown in the figure. The perimeter of the window is 5 metres.

a. If r is the radius of the semicircle and x is the length of the larger side of the rectangle, find a relation between r,x.

b. Find the area of the window in terms of r.

c. Find the dimensions of the window so that the greatest possible light may be admitted.

Answer.

a. Perimeter 5m

2x + 2r + πr = 5 ⇒2x + r(π+2) = 5

Question 13.

a. A rectangular sheet of tin with adjacent sides 45cm and 24cm is to be made into a box without top, by cutting off equal squares from the corners and folding up the flaps.

i. Taking the side of the square cut off as x, express the volume of the box as the function of x.

ii. For what value of x, the volume of the box will be maximum.

b. What Is the slope of the tangent and normal at (1, 1) on the curve y = x^{3}

Answer.

a.(i) .

Length of the box = 45-2x

Breadth of the box = 24 – 2x

Height of the box = x

Volume V = (45 – 2x)(24 – 2x) x

= (1080 – 90x – 48x + 4x²)x

V = 4x^{3} – 138x² + 1080x .

Question 14.

A wire of length 28m 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 using differentiation.

Answer.

Let ‘x’ be the length of the square piece & 28 – x be the length of the circular piece.

perimeter of square = x

4a = x; a = x/4

Question 15.

A car starts from a point P at time t = 0 seconds and stops at point Q. The distance x, in metres, covered by it, in t seconds ¡s given by

Find the time taken by it to reach Q and also find distance between P and Q.

Answer.

Differentiating w.r.t. t

Question 16.

Show that the function given by

f(x) = sinx is

a. strictly increasing in

b. strictly decreasing in

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

Answer.

f(x) = sinx

∴ f ‘(x) = cosx

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.

y = x^{3}

= 3x²

The slope of the tangent = 3x²

slope of the tangent = y coordinate,

3x^{3} = y or 3x² = x^{3}

⇒ 3x² – x^{3} = 0 ⇒x²(3-x) = 0

x = 3 or x = 0

When x = 0, y = (0)^{3} = 0

When x = 3, y = (3)^{3} = 27

∴(0,0) and (3,27) are the points at which the slope of the tangent is equal to the y coordinate of the point. .

Question 18.

Consider parametric forms given by

x = a sin^{3}t, y = b cos^{3}t

i. Find

ii. Find the equation of tangent at

Answer.

x = a sin^{3}t, y = b cos^{3}t

Hence, the equation of tangent to the given curve at (a ,0) is y – 0 = 0 (x – a), i.e., y = 0.

Question 19.

Find the equation of the tangent line to the curve. y = x² – 2x + 7 which is

a. parallel to the line 2x – y + 9 = 0

b. perpendicular to the line 5y – 15x = 13

Answer.

y = x² – 2x + 7

Question 20.

Using differentials, find the approximate value of (15)¼ upto 3 places of decimal.

Answer.

Question 21.

Prove that the function g(x) = logx does not have maxima or minima.

Answer.

g (x) = logx. x>0

g ‘(x) =

g ‘(x) = 0 ⇒ = 0 is not possible

∴ g does not have maxima or minima

Question 22.

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

Answer.

Consider the fig.

x² + y² = (2r)²

y² = 4r² – x²

We hope the given Plus Two Maths Chapter Wise Questions and Answers Chapter 6 Application of Derivatives will help you. If you have any query regarding Plus Two Maths Chapter Wise Questions and Answers Chapter 6 Application of Derivatives, drop a comment below and we will get back to you at the earliest.

## Leave a Reply