Application Of Derivatives Class 12th Mathematics Part I CBSE Solution

Class 12^th Mathematics Part I CBSE Solution

Exercise 6.1

1. Find the rate of change of the area of a circle with respect to its radius r…
2. The volume of a cube is increasing at the rate of 8 cm^3 /s. How fast is the…
3. The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the…
4. An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the…
5. A stone is dropped into a quiet lake and waves move in circles at the speed of 5…
6. The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate…
7. The length x of a rectangle is decreasing at the rate of 5 cm/minute and the…
8. A balloon, which always remains spherical on inflation, is being inflated by…
9. A balloon, which always remains spherical has a variable radius. Find the rate…
10. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled…
11. A particle moves along the curve 6y = x^3 +2. Find the points on the curve at…
12. The radius of an air bubble is increasing at the rate of 1/2 cm/s . At what…
13. A balloon, which always remains spherical, has a variable diameter 3/2 (2x+1) .…
14. Sand is pouring from a pipe at the rate of 12 cm^3 /s. The falling sand forms a…
15. The total cost C(x) in Rupees associated with the production of x units of an…
16. R(x) = 13x^2 + 26x + 15. Find the marginal revenue when x = 7. The total…
17. The rate of change of the area of a circle with respect to its radius r at r =…
18. R(x) = 3x^2 + 36x + 5. The marginal revenue, when x = 15 is The total revenue…

Exercise 6.2

1. Show that the function given by f (x) = 3x + 17 is strictly increasing on R.…
2. Show that the function given by f (x) = e2x is strictly increasing on R.…
3. Show that the function given by f (x) = sin x is (a) strictly increasing in (0 ,…
4. Find the intervals in which the function f given by f (x) = 2x^2 - 3x is (a)…
5. Find the intervals in which the function f given by f (x) = 2x^3 - 3x^2 - 36x +…
6. x^2 + 2x - 5 Find the intervals in which the following functions are strictly…
7. 10 - 6x - 2x^2 Find the intervals in which the following functions are strictly…
8. -2x^3 - 9x^2 - 12x + 1 Find the intervals in which the following functions are…
9. 6 - 9x - x^2 Find the intervals in which the following functions are strictly…
10. (x + 1)^3 (x - 3)^3 Find the intervals in which the following functions are…
11. Show that y = log (1+x) - 2x/2+x , x-1 , is an increasing function of x…
12. Find the values of x for which y = [x(x - 2)]^2 is an increasing function.…
13. Prove that y = 4sintegrate heta /(2+costheta) - theta is an increasing function…
14. Prove that the logarithmic function is strictly increasing on (0, ∞).…
15. Prove that the function f given by f (x) = x^2 - x + 1 is neither strictly…
16. Which of the following functions are strictly decreasing on (0 , pi /2) ? A.…
17. On which of the following intervals is the function f given by f (x) = x^100 +…
18. Find the least value of a that the function f given by f (x) = x^2 + ax + 1 is…
19. Let I be any interval disjoint from [-1, 1]. Prove that the function f given by…
20. Prove that the function f given by f (x) = log sin x is strictly increasing on…
21. Prove that the function f given by f (x) = log |cos x| is strictly decreasing…
22. Prove that the function given by f (x) = x^3 - 3x^2 + 3x - 100 is increasing in…
23. The interval in which y = x^2 e-x is increasing isA. (- ∞, ∞) B. (- 2, 0) C.…

Exercise 6.3

1. Find the slope of the tangent to the curve y = 3x^4 - 4x at x = 4.…
2. Find the slope of the tangent to the curve y = x-1/x-2 , x not equal 2 at x =…
3. Find the slope of the tangent to curve y = x^3 - x + 1 at the point whose…
4. Find the slope of the tangent to the curve y = x^3 -3x + 2 at the point whose…
5. Find the slope of the normal to the curve x = acos^3 θ, y = asin^3 θ at theta =…
6. Find the slope of the normal to the curve x = 1− a sinθ, y = bcos^2 θ at theta =…
7. Find points at which the tangent to the curve y = x^3 - 3x^2 - 9x + 7 is…
8. Find a point on the curve y = (x - 2)^2 at which the tangent is parallel to the…
9. Find the point on the curve y = x^3 - 11x + 5 at which the tangent is y = x -11.…
10. Find the equation of all lines having slope -1 that are tangents to the curve y…
11. Find the equation of all lines having slope 2 which are tangents to the curve y…
12. Find the equations of all lines having slope 0 which are tangent to the curve y…
13. Find points on the curve x^2/9 + y^2/16 = 1 at which the tangents are (i)…
14. Find the equations of the tangent and normal to the given curves at the…
15. y = x^4 - 6x^3 + 13x^2 - 10x + 5 at (0, 5) Find the equations of the tangent…
16. y = x^4 - 6x^3 + 13x^2 - 10x + 5 at (1, 3) Find the equations of the tangent…
17. y = x^3 at (1, 1) Find the equations of the tangent and normal to the given…
18. y = x^2 at (0, 0) Find the equations of the tangent and normal to the given…
19. Find the equation of the tangent line to the curve y = x^2 - 2x +7 which is (a)…
20. Show that the tangents to the curve y = 7x^3 + 11 at the points where x = 2 and…
21. Find the points on the curve y = x^3 at which the slope of the tangent is equal…
22. For the curve y = 4x^3 - 2x^5 , find all the points at which the tangent passes…
23. Find the points on the curve x^2 + y^2 - 2x - 3 = 0 at which the tangents are…
24. Find the equation of the normal at the point (am^2 , am^3) for the curve ay^2 =…
25. Find the equation of the normals to the curve y = x^3 + 2x + 6 which are…
26. Find the equations of the tangent and normal to the parabola y^2 = 4ax at the…
27. Prove that the curves x = y^2 and xy = k cut at right angles* if 8k^2 = 1.…
28. Find the equations of the tangent and normal to the hyperbola x^2/a^2 - y^2/b^2…
29. Find the equation of the tangent to the curve y = root 3x-2 which is parallel…
30. The slope of the normal to the curve y = 2x^2 + 3 sin x at x = 0 isA. 3 B. 1/3…
31. The line y = x + 1 is a tangent to the curve y^2 = 4x at the pointA. (1, 2) B.…

Exercise 6.4

1. root 25.3 Using differentials, find the approximate value of each of the…
2. root 49.5 Using differentials, find the approximate value of each of the…
3. root 0.6 Using differentials, find the approximate value of each of the…
4. (0.009)^1/3 Using differentials, find the approximate value of each of the…
5. (0.999)^1/10 Using differentials, find the approximate value of each of the…
6. (15)^1/4 Using differentials, find the approximate value of each of the…
7. (26)^1/3 Using differentials, find the approximate value of each of the…
8. (255)^1/4 Using differentials, find the approximate value of each of the…
9. (82)^1/4 Using differentials, find the approximate value of each of the…
10. (401)^1/2 Using differentials, find the approximate value of each of the…
11. (0.0037)^1/2 Using differentials, find the approximate value of each of the…
12. (26.57)^1/3 Using differentials, find the approximate value of each of the…
13. (81.5)^1/4 Using differentials, find the approximate value of each of the…
14. (3.968)^3/2 Using differentials, find the approximate value of each of the…
15. (32.15)^1/5 Using differentials, find the approximate value of each of the…
16. Find the approximate value of f (2.01), where f (x) = 4x^2 + 5x + 2.…
17. Find the approximate value of f (5.001), where f (x) = x^3 - 7x2 + 15.…
18. Find the approximate change in the volume V of a cube of side x metres caused by…
19. Find the approximate change in the surface area of a cube of side x metres…
20. If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find…
21. If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find…
22. If f(x) = 3x^2 + 15x + 5, then the approximate value of f (3.02) isA. 47.66 B.…
23. The approximate change in the volume of a cube of side x metres caused by…

Exercise 6.5

1. f (x) = (2x - 1)^2 + 3 Find the maximum and minimum values, if any, of the…
2. f (x) = 9x^2 + 12x + 2 Find the maximum and minimum values, if any, of the…
3. f(x) = - (x - 1)^2 + 10 Find the maximum and minimum values, if any, of the…
4. g(x) = x^3 + 1 Find the maximum and minimum values, if any, of the following…
5. f(x) = |x + 2| - 1 Find the maximum and minimum values, if any, of the…
6. g(x) = -|x + 1| + 3 Find the maximum and minimum values, if any, of the…
7. h(x) = sin(2x) + 5 Find the maximum and minimum values, if any, of the…
8. f (x) = |sin 4x + 3| Find the maximum and minimum values, if any, of the…
9. h(x) = x + 1, x ∈ (-1, 1) Find the maximum and minimum values, if any, of the…
10. f (x) = x^2 Find the local maxima and local minima, if any, of the following…
11. g(x) = x^3 - 3x Find the local maxima and local minima, if any, of the…
12. h (x) = sinx+cosx, 0x pi /2 Find the local maxima and local minima, if any, of…
13. f (x) = sin x - cos x, 0 x 2π Find the local maxima and local minima, if any,…
14. f (x) = x^3 - 6x^2 + 9x + 15 Find the local maxima and local minima, if any, of…
15. g (x) = x/2 + 2/x , x0 Find the local maxima and local minima, if any, of the…
16. g (x) = 1/x^2 + 2 Find the local maxima and local minima, if any, of the…
17. f (x) = x root 1-x , 0x1 Find the local maxima and local minima, if any, of the…
18. f (x) = ex Prove that the following functions do not have maxima or minima:…
19. g(x) = log x Prove that the following functions do not have maxima or minima:…
20. h(x) = x^3 + x^2 + x +1 Prove that the following functions do not have maxima…
21. f (x) = x^3 , x ∈ [-2, 2] Find the absolute maximum value and the absolute…
22. f (x) = sin x + cos x, x ∈ [0, π] Find the absolute maximum value and the…
23. f (x) = 4x - 1/2 x^2 , x in[-2 , 9/2] Find the absolute maximum value and the…
24. f (x) = (x − 1)^2 + 3, x ∈ [−3, 1] Find the absolute maximum value and the…
25. Find the maximum profit that a company can make, if the profit function is given…
26. Find both the maximum value and the minimum value of 3x^4 - 8x^3 + 12x^2 - 48x +…
27. At what points in the interval [0, 2π], does the function sin 2x attain its…
28. What is the maximum value of the function sin x + cos x?
29. Find the maximum value of 2x^3 - 24x + 107 in the interval [1, 3]. Find the…
30. It is given that at x = 1, the function x^4 - 62x^2 + ax + 9 attains its…
31. Find the maximum and minimum values of x + sin 2x on [0, 2π].
32. Find two numbers whose sum is 24 and whose product is as large as possible.…
33. Find two positive numbers x and y such that x + y = 60 and xy3 is maximum.…
34. Find two positive numbers x and y such that their sum is 35 and the product x^2…
35. Find two positive numbers whose sum is 16 and the sum of whose cubes is…
36. A square piece of tin of side 18 cm is to be made into a box without top, by…
37. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top,…
38. Show that of all the rectangles inscribed in a given fixed circle, the square…
39. Show that the right circular cylinder of given surface and maximum volume is…
40. Of all the closed cylindrical cans (right circular), of a given volume of 100…
41. A wire of length 28 m is to be cut into two pieces. One of the pieces is to be…
42. Prove that the volume of the largest cone that can be inscribed in a sphere of…
43. Show that the right circular cone of least curved surface and given volume has…
44. Show that the semi-vertical angle of the cone of the maximum volume and of…
45. Show that semi-vertical angle of right circular cone of given surface area and…
46. The point on the curve x^2 = 2y which is nearest to the point (0, 5) isA.…
47. For all real values of x, the minimum value of 1-x+x^2/1+x+x^2 isA. 0 B. 1 C. 3…
48. The maximum value of [x (x-1) + 1]^1/3 , 0 less than equal to x less than equal…

Miscellaneous Exercise

1. (17/81)^1/4 Using differentials, find the approximate value of each of the…
2. (33)^1/5 Using differentials, find the approximate value of each of the…
3. Show that the function given by f (x) = logx/x has maximum at x = e.…
4. The two equal sides of an isosceles triangle with fixed base b are decreasing at…
5. Find the equation of the normal to curve x2 = 4y which passes through the point…
6. Show that the normal at any point θ to the curve x = a cosθ + a θ sinθ, y = a…
7. Find the intervals in which the function f given by f (x) =
8. Find the intervals in which the function f given by f (x) = x^3 + 1/x^3 , x not…
9. Find the maximum area of an isosceles triangle inscribed in the ellipse x^2/a^2…
10. A tank with rectangular base and rectangular sides, open at the top is to be…
11. The sum of the perimeter of a circle and square is k, where k is some constant.…
12. A window is in the form of a rectangle surmounted by a semi - circular opening.…
13. A point on the hypotenuse of a triangle is at distance a and b from the sides…
14. Find the points at which the function f given by f (x) = (x - 2)^4 (x + 1)^3…
15. Find the absolute maximum and minimum values of the function f given by f (x) =…
16. Show that the altitude of the right circular cone of maximum volume that can be…
17. Let f be a function defined on [a, b] such that f ′(x) 0, for all x ∈ (a, b).…
18. Show that the height of the cylinder of maximum volume that can be inscribed in…
19. Show that height of the cylinder of greatest volume which can be inscribed in a…
20. A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314…
21. The slope of the tangent to the curve x = t^2 + 3t - 8, y = 2t^2 - 2t - 5 at…
22. The line y = mx + 1 is a tangent to the curve y^2 = 4x if the value of m isA. 1…
23. The normal at the point (1,1) on the curve 2y + x^2 = 3 isA. x + y = 0 B. x - y…
24. The normal to the curve x^2 = 4y passing (1,2) isA. x + y = 3 B. x - y = 3 C. x…
25. The points on the curve 9y^2 = x^3 , where the normal to the curve makes equal…

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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:

(a) We know that area of a circle (A) is A = πr²

Then, Rate of change of the area with respect to its radius is given by,

When r = 3cm

Then 

Therefore, the area of the circle is changing at the rate of 6πcm²/s when its radius is 3cm.

(b) We know that area of a circle (A) is A = πr²

Then, Rate of change of the area with respect to its radius is given by,

When r = 4cm

Then 

Therefore, the area of the circle is changing at the rate of 8πcm²/s when its radius is 4cm.

Question 2.

The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?

Answer:

Let a be the length of a side, V be the volume and S be the surface area of the cube.

Then, V = a³ and S = 6a² where a is the function of time t.

Now, It is given that 

Then, by using the chain rule, we get,

………………(1)

So, when a = 12cm, then, 

Therefore, if the length of the edge of the cube is 12cm, then the surface area is increasing at the rate of .

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:

We know that area of a circle (A) is A = πr²

Then, Rate of change of the area with respect to time is given by,

It is given that

Then, 

Thus, when r = 10cm

Then 

Therefore, the rate at which the area of the circle is increasing when the radius is 10 cm is 60π cm²/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 = x³

…….. by chain rule

It is given that

Then, 

Thus, when x = 10cm

Then 

Therefore, the volume of the cube increasing when the edge is 10 cm long is 900 cm³/s.

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:

We know that area of a circle (A) is A = πr²

Then, Rate of change of the area with respect to time (t) is given by,

 ……………….. by chain rule

It is given that

Then, 

Thus, when r = 8cm

Then 

Therefore, when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing is 80π cm²/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:

We know that circumference of a circle (C) is C = 2πr

Then, Rate of change of circumference with respect to time (t) is given by,

 ……………….. by chain rule

It is given that

Then, 

Therefore, the rate of increase of its circumference is 1.4π cm/s.

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

Answer:

(a) It is given that 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, then we have,

The perimeter (P) of a rectangle is given by:

P = 2( x + y )

= 2(-5 +4) = -2 cm/min

Therefore, the rate of change of the perimeter is -2cm/min.

(b) It is given that 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, then we have,

Now, the area (A) of the rectangle is given by

A = x × y

= -5y + 4x

So, when x = 8cm and y = 6cm, then,

Therefore, the rate of change of the area of the rectangle is 2cm²/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:

Let V be the volume of the sphere, then

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

…….. by chain rule

It is given that

Then, 

Thus, when r = 15cm

Then, 

Therefore, the rate at which the radius of the balloon increases when the radius is 15 cm is  cm/s.

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:

Let V be the volume of the sphere, then

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

…….. by chain rule

Thus, when r = 10cm

Then, 

Therefore, the rate at which its volume is increasing with the radius when the later is 10 cm is 400π cm³/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 2cm/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 be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x m away from the wall.

we know that, by Pythagoras theorem,

x² + y² = 25

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

Now, it is given that 

Therefore,

And when x= 4 m, then

=

Therefore, the height of the ladder on the wall is decreasing at the rate of  m/s.

Question 11.

A particle moves along the curve 6y = x³ +2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Answer:

It is given that equation of the curve is 6y = x³ + 2.

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

 ………….. (1)

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

then, we have,

So, putting the value in (1), we get,

So, when x = +4, then, 

And

So, when x = -4, then, y = -62/6 = -31/2

Therefore, the points required on the curve are (4,11) are (-4, -31/2).

Question 12.

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

Answer:

As, the air bubble is in the shape of a sphere.

Therefore, V be the volume of the sphere, then

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

…….. by chain rule

It is given that

when, r = 1cm

Then, 

Therefore, the rate is the volume of the bubble increasing when the radius is 1 cm is 2πcm³/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:

Let V be the volume of the sphere, then

It is given that, Diameter = 

Now, the rate of change of its volume with respect to x is as

Therefore, the rate of change of its volume with respect to x is cm³/s.

Question 14.

Sand is pouring from a pipe at the rate of 12 cm³/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:

Let V be the volume of the cone, then

It is given that, h= 

⇒ r = 6h

Therefore, 

Now, the rate of change of its volume with respect to time (t) is given by:

Now, the rate of change of its volume with respect to x is as

……… by chain rule

 …………….(1)

Now, it is also given that  and h = 4cm

So, putting the value in equation (1), we get

Therefore, the height of the sand cone increasing when the height is 4 cm is  cm/s.

Question 15.

The total cost C(x) in Rupees associated with the production of x units of an item is given by

C(x) = 0.007x³ – 0.003x² + 15x + 4000.

Find the marginal cost when 17 units are produced.

Answer:

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

Then, 

= 0.021x² – 0.006x +15

So, when x = 17 then,

MC = 0.021(17²) – 0.006(17) +15

= 0.021(289) – 0.102 + 15

= 20.967

Therefore, the marginal cost when 17 units are produced is Rs. 20.967.

Question 16.

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

R(x) = 13x² + 26x + 15.

Find the marginal revenue when x = 7.

Answer:

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

Then, 

So, when x = 7 then,

MR = 26(7) + 26 = 208

Therefore, the marginal revenue when x = 7 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:

We know that area of a circle (A) is A = πr²

Then, Rate of change of the area with respect to its radius is given by,

When r = 6cm

Then 

Therefore, the area of the circle is changing at the rate of 12πcm²/s when its radius is 6cm.

Question 18.

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

R(x) = 3x² + 36x + 5. The marginal revenue, when x = 15 is
A. 116

B. 96

C. 90

D. 126

Answer:

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

Then, 

So, when x = 15 then,

MR = 6(15) + 36 = 126

Therefore, the marginal revenue when x = 15 is Rs. 126.

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Exercise 6.2

Question 1.

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

Answer:

Let x₁ and x₂ be any two numbers in R.

Then, we have,

x₁ < x₂

⇒ 3x₁ < 3x₂

⇒ 3x₁ +17 < 3x₂ +17

⇒ f(x₁) < f(x₂)

Therefore, f is strictly increasing on R.

Question 2.

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

Answer:

Let x₁ and x₂ be any two numbers in R.

The, we have,

x₁ < x₂

⇒ 2x₁ < 2x₂

⇒ e^2x₁ < e^2x₂

⇒ f(x₁) < f(x₂)

Therefore, 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:

(a) The function is f (x) = sin x

Then, f’(x) = cos x

Since for each x ϵ , cos x > 0, we have f’(x) > 0

Therefore, f’ is strictly increasing in.

(b) The function is f (x) = sin x

Then, f’(x) = cos x

Since for each , cos x < 0, we have f’(x) < 0

Therefore, f’ is strictly decreasing in.

(c) The function is f (x) = sin x

Then, f’(x) = cos x

Since for each x ϵ , cos x > 0, we have f’(x) >0

Therefore, f’ is strictly increasing in……………….(1)

Now, The function is f (x) = sin x

Then, f’(x) = cos x

Since, for each x ϵ, cos x < 0, we have f’(x) < 0

Therefore, f’ is strictly decreasing in …………(2)

From (1) and (2),

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) = 2x² – 3x is

(a) strictly increasing (b) strictly decreasing

Answer:

(a) It is given that function f(x) = 2x² – 3x

⇒ f’(x) = 4x – 3

If f’(x) = 0, then we get,

So, the points  divides the real line into two disjoint intervals,  and 

So, in interval, f’(x) = 4x -3 >0

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

(b) It is given that function f(x) = 2x² – 3x

⇒ f’(x) = 4x – 3

If f’(x) = 0, then we get,

So, the points  divides the real line into two disjoint intervals,  and 

So, in interval  f’(x) = 4x -3 < 0

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

Question 5.

Find the intervals in which the function f given by f (x) = 2x³ – 3x² – 36x + 7 is

(a) strictly increasing (b) strictly decreasing

Answer:

(a) It is given that function f(x) = 2x³ – 3x² – 36x + 7

⇒ f’(x) = 6x² – 6x + 36

⇒ f’(x) = 6(x² – x + 6)

⇒ f’(x) = 6(x + 2)(x – 3)

If f’(x) = 0, then we get,

⇒ x = -2, 3

So, the points x = -2 and x = 3 divides the real line into two disjoint intervals, (-∞,2), (-2,3) and (3,∞)

So, in interval 

f’(x) = 6(x + 2)(x – 3) >0

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

(b) It is given that function f(x) = 2x³ – 3x² – 36x + 7

⇒ f’(x) = 6x² – 6x + 36

⇒ f’(x) = 6(x² – x + 6)

⇒ f’(x) = 6(x + 2)(x – 3)

If f’(x) = 0, then we get,

⇒ x = -2, 3

So, the points x = -2 and x = 3 divides the real line into two disjoint intervals, (-∞,2), (-2,3) and (3,∞)

So, in interval

f’(x) = 6(x + 2)(x – 3) < 0

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

Question 6.

Find the intervals in which the following functions are strictly increasing or decreasing:

x² + 2x – 5

Answer:

It is given that function f(x) = x² + 2x – 5

f’(x) = 2x + 2

If f’(x) = 0, then we get,

⇒ x = -1

So, the point x = -1 divides the real line into two disjoint intervals, (-∞,-1) and (1,∞)

So, in interval (-∞,-1)

f’(x) = 2x + 2 < 0

Therefore, the given function (f) is strictly decreasing in interval (-∞,-1).

And in interval (1,∞)

f’(x) = 2x + 2 > 0

Therefore, the given function (f) is strictly increasing in interval (1,∞).

Thus, f is strictly increasing for x > -1.

Question 7.

Find the intervals in which the following functions are strictly increasing or decreasing:

10 – 6x – 2x²

Answer:

It is given that function f(x) = 10 – 6x – 2x²

f’(x) = -6 – 4x

If f’(x) = 0, then we get,

⇒ x = 

So, the point x =  divides the real line into two disjoint intervals,

So, in interval

x < -3/2
-4x > 6
-6 - 4x > 0

f’(x) = -6 – 4x > 0

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

And in interval 

f’(x) = -6 – 4x < 0

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

Thus, f is strictly decreasing for x > .

Question 8.

Find the intervals in which the following functions are strictly increasing or decreasing:

–2x³ – 9x² – 12x + 1

Answer:

It is given that function f(x) = –2x³ – 9x² – 12x + 1

⇒ f’(x) = -6x² – 18x + 12

⇒ f’(x) = -6(x² +3x + 6)

⇒ f’(x) = -6(x + 1)(x + 2)

If f’(x) = 0, then we get,

⇒ x = -1 and -2

So, the points x = -1 and x = -2 divides the real line into two disjoint intervals,

(-∞,-2), (-2,-1) and (-1,∞)

So, in interval (-∞,-2),(-1,∞)

f’(x) = -6(x + 1) (x +2) < 0

Therefore, the given function (f) is strictly decreasing for x < -2 and x>-1.

So, in interval (-2.-1)

f’(x) = -6(x + 1)(x+2) > 0

Therefore, the given function (f) is strictly increasing for -2 < x < -1.

Question 9.

Find the intervals in which the following functions are strictly increasing or decreasing:

6 – 9x – x²

Answer:

It is given that function f(x) = 6 – 9x – x²

f’(x) = -9 – 2x

If f’(x) = 0, then we get,

⇒ x = 

So, the point x =  divides the real line into two disjoint intervals,

So, in interval

f’(x) = -9 – 2x > 0

Therefore, the given function (f) is strictly increasing for x < .

And in interval

f’(x) = -9 – 2x < 0

Therefore, the given function (f) is strictly decreasing for x>.

Thus, f is strictly decreasing for x>.

Question 10.

Find the intervals in which the following functions are strictly increasing or decreasing:

(x + 1)³ (x – 3)³

Answer:

It is given that function f(x) = –(x + 1)³ (x – 3)³

⇒ f’(x) = 3(x + 1)² (x – 3)³ +3(x + 1)³ (x – 3)²

⇒ f’(x) = 3(x + 1)² (x – 3)²[x – 3 + x + 1]

⇒ f’(x) = 6(x + 1)² (x – 3)²(x-1)

If f’(x) = 0, then we get,

⇒ x = -1,3 and 1

So, the points x = -1, x =1 and x = 3 divides the real line into four disjoint intervals,

So, in interval

f’(x) = 6(x + 1)² (x – 3)²(x-1) < 0

Therefore, the given function (f) is strictly decreasing in intervals .

So, in interval 

f’(x) = 6(x + 1)² (x – 3)²(x-1) > 0

Therefore, the given function (f) is strictly increasing in intervals.

Question 11.

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

Answer:

Let y = 

Now, x > -1

⇒ 1 + x >0

Also, for all x> -1, 

 for x > -1

Therefore, f is an increasing function throughout its domain.

Question 12.

Find the values of x for which y = [x(x – 2)]² is an increasing function.

Answer:

It is given that y = [x(x – 2)]², then,

= 4x(x - 2)(x - 1)

Now if 

⇒ x = 0, 1,2

So, the points x = 0, x =1 and x = 2 divides the real line into four disjoint intervals,

(-∞,0), (0,1), (1, 2) and (2,∞).

So, in interval(-∞,0),(1,2)

 < 0

Therefore, the given function (f) is strictly decreasing in intervals .

So, in interval (0,1) and (2,∞)

Therefore, the given function (f) is strictly increasing for 0 < x < 1 and x>2

Question 13.

Prove that  is an increasing function of θ in 

Answer:

We have, y = 

Now, 

⟹ 8cosθ + 4 = 4 + cos²θ + 4cosθ

⟹ cos²θ - 4cosθ = 0

⟹ cosθ(cosθ-4) = 0

⟹ cosθ = 0 or cosθ = 4

Since, cosθ≠4, cosθ = 0

⟹ cosθ = 0 ⟹ θ = π/2

Now, 

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

⇒ 4 – cosθ > 0

Therefore, cosθ(4 – cosθ) > 0 and also (2 + cosθ)² > 0

Therefore, y is strictly increasing in interval.

Also, the given function is continuous at x = 0 and x = .

Therefore, y is increasing in interval.

Question 14.

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

Answer:

The given function is f(x) = logx

It is clear that for x>0, 

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

Question 15.

Prove that the function f given by f (x) = x² – x + 1 is neither strictly increasing nor strictly decreasing on (– 1, 1).

Answer:

It is given that function f(x) = x² – x + 1

f’(x) = 2x – 1

If f’(x) = 0, then we get,

⇒ x = 

So, the point x =  divides the interval (-1,1) into two disjoint intervals,

So, in interval

f’(x) = 2x – 1 < 0

Therefore, the given function (f) is strictly decreasing in interval 

So, in interval

f’(x) = 2x -1 > 0

Therefore, the given function (f) is strictly increasing in interval for.

Therefore, f is neither strictly increasing and decreasing in interval (-1,1).

Question 16.

Which of the following functions are strictly decreasing on ?

A. cos x

B. cos 2x

C. cos 3x

D. tan x

Answer:

(A) Let f₁(x) = cosx

In interval, 

Therefore, f₁(x) = cosx is strictly decreasing in interval.

(B) Let f₂(x) = cos2x

Now, 0 < x < 

⇒ 0 < 2x < π

⇒ sin2x > 0

⇒ -2sin2x < 0

Therefore, f₂(x) = cos2x is strictly decreasing in interval.

(C) Let f₃(x) = cos3x

Now, 

⇒ sin3x = 0

⇒ 3x = π, as xϵ

⇒ x = 

The point x = divides the interval  into two distinct intervals.

i.e.  and 

Now, in interval, ,

f₃'(x) = -3sin3x < 0 as (0 < x <  => 0 < 3x < π)

Therefore, f₃ is strictly decreasing in interval.

Now, in interval 

f₃'(x)=-3sin3x > 0 as 

Therefore, f₃ is strictly increasing in interval.

(D) Let f₄ = tanx

In interval,

Therefore, f4 is strictly increasing in interval .

Question 17.

On which of the following intervals is the function f given by f (x) = x¹⁰⁰ + sin x –1 strictly decreasing?
A. (0,1)

B. 

C. 

D. None of these

Answer:

It is given that f (x) = x¹⁰⁰ + sin x –1

Then, f’(x) = 100x⁹⁹ + cosx

In interval (0,1), cos x >0 and 100x⁹⁹ > 0

⇒ f’(x)>0

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

In interval, cos x < 0 and 100x⁹⁹ > 0.

Also, 100x⁹⁹ > cos x

⇒ f’(x) > 0 in 

Therefore, function f is strictly increasing in interval .

In interval , cos x < 0 and 100x⁹⁹ > 0.

Also, 100x⁹⁹ > cos x

⇒ f’(x) > 0 on 

Therefore, function f is strictly increasing in interval .

Hence, function f is strictly decreasing on none of the intervals.

Question 18.

Find the least value of a that the function f given by f (x) = x² + ax + 1 is strictly increasing on [1, 2].

Answer:

It is given that function f(x) = x² + ax + 1

f’(x) = 2x + a

Now, function f will be increasing in [1, 2],

if f’(x) >0 in [1, 2]

⇒ 2x +a > 0
⇒ 2x > -a
⇒ a < -2x

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

⇒ a < -2x when x ϵ [1, 2]

Now, 1 ≤ x ≤ 2
⇒ -4 ≤ -2x ≤ -2

Therefore, the least value of a for f to be increasing on [1, 2] is given by

⇒ a = -4

Therefore, the least value of a is -4

Question 19.

Let I be any interval disjoint from [–1, 1]. Prove that the function f given by  is strictly increasing on 1.

Answer:

It is given that 

Now, f’(x) =0

The points x =1 and x = -1 divide the real line in three disjoint intervals

(-∞,-1),(-1,1) and (1,∞)

Now in interval, (-1,1)

it is clear that -1 < x < 1

⇒ x² < 1

Therefore, f’(x) = 1- < 0 (-1,1) ~ {0}

Therefore, f is strictly decreasing on (-1,1) ~ {0}

x < -1 or 1 < x

⇒ x² > 1

Therefore, f’(x) = 1- > 0 (-∞, -1) and (1,∞)

Therefore, f is strictly increasing in interval I disjoint from (-1,1)

Hence Proved.

Question 20.

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

Answer:

It is given that f (x) = log sin x

In interval, f’(x) = cot x >0

Therefore, f is strictly increasing in.

In interval, f’(x) = cot x < 0

Therefore, f is strictly decreasing in.

Question 21.

Prove that the function f given by f (x) = log |cos x| is strictly decreasing onand strictly increasing on.

Answer:

It is given that f (x) = log |cos x|

In interval, f’(x) = -tanx < 0

Therefore, f is strictly decreasing on.

In interval, f’(x) = -tanx > 0

Therefore, f is strictly increasing in.

Question 22.

Prove that the function given by f (x) = x³ – 3x² + 3x – 100 is increasing in R.

Answer:

We have, f (x) = x³ – 3x² + 3x – 100

=> f’(x) = 3x² -6x + 3

= 3(x² -2x + 1)

= 3(x-1)2

For any x ϵ R, (x -1)2 > 0

Thus, f’(x) is always positive in R.

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

Question 23.

The interval in which y = x² e^–x is increasing is
A. (– ∞, ∞)

B. (– 2, 0)

C. (2, ∞)

D. (0, 2)

Answer:

it is given that y = x² e^–x

then 

Now if 

⇒ x = 0 and x =2

The points x = 0 and x= 2 divide the real line into three disjoint intervals ie, (-∞,0), (0,2) and (2,∞).

In interval (-∞,0) and (2,∞),

f’(x) < 0 as e^-x is always positive.

Therefore, f is decreasing on (-∞,0) and (2,∞).

In interval (0,2), f’(x)>0

Therefore, f is strictly increasing in interval (0.2).

---

Exercise 6.3

Question 1.

Find the slope of the tangent to the curve y = 3x⁴ – 4x at x = 4.

Answer:

The given curve y = 3x⁴ – 4x

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

= 12(64) – 4

= 764

Therefore, the slop of the tangent is 764.

Question 2.

Find the slope of the tangent to the curve

 at x = 10.

Answer:

The given curve is y = 

Then, the slope of the tangent

Therefore, the slope of the tangent is .

Question 3.

Find the slope of the tangent to curve y = x³ – x + 1 at the point whose x-coordinate is 2.

Answer:

The given curve is y = x³ – x + 1

Then, the slope of the tangent

= 12 -1 = 11

Therefore, the slope of the tangent 11.

Question 4.

Find the slope of the tangent to the curve y = x³ –3x + 2 at the point whose x-coordinate is 3.

Answer:

The given curve is y = x³ – 3x + 2

Then, the slope of the tangent

= 27 -3 = 24

Therefore, the slope of the tangent 24.

Question 5.

Find the slope of the normal to the curve x = acos³ θ, y = asin³ θ at 

Answer:

The given curve is y = acos³θ and y = asin³θ

Then, the slope of the tangent  is given by:

Then, the slope of the tangent  is given by:

Therefore, the slope of the tangent 1.

Question 6.

Find the slope of the normal to the curve x = 1− a sinθ, y = bcos² θ at 

Answer:

The given curve is x = 1− a sin θ and y = b cos² θ

Then, the slope of the tangent  is given by:

Then, the slope of the tangent  is given by:

Therefore, the slope of the tangent.

Question 7.

Find points at which the tangent to the curve y = x³ – 3x² – 9x + 7 is parallel to the x-axis.

Answer:

It is given that the curve y = x³ – 3x² – 9x + 7

We know that the tangent is parallel to the x-axis if the slope of the tangent is zero.

Therefore, 3x² -6x-9 = 0

⇒ x² -2x-3 = 0

⇒ (x-3)(x+1) =0

⇒ x =3 and x = -1

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

Therefore, 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:

We know that if a tangent is parallel to the chord joining the points (2,0) and (4,4), then

Slope of the tangent = slope of the curve………….(1)

And, the slope of the curve = 

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

Now, from (1) we have,

2(x -2) = 2

⇒ x-2 = 1

⇒ x =3

So, when x = 3 then y = (3-2)² = 1

Therefore, required points are (3,1).

Question 9.

Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x –11.

Answer:

It is given that equation of the curve y = x³ – 11x + 5

At which the tangent is y = x –11

⇒ Slope of the tangent = 1

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

⇒ 3x² -11 = 1

⇒ 3x² = 12

⇒ x² = 4

⇒ x = 2

So, when x = 2 then y = (2)³ -11(2) + 5 = -9

And when x = -2 then y = (-2)³ -11(-2) + 5 = 19

Therefore, 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:

It is given that equation of the curve y = 

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

Now, if the slope of the tangent is -1, then we get,

⇒ (x-1)² = 1

⇒ (x-1) = 1

⇒ x = 2, 0

So, when x = 2 then y = 1

And when x = 0 then y = 1

Therefore, required points are (0, -1) and (2, 1).

Now, the equation of the tangent (0,1) is given by:

y – (-1) = -1(x-0)

⇒ y + 1 = -x

⇒ y + x+ 1 = 0

And the equation of the tangent (2,1) is given by:

y – 1 = -1(x-2)

⇒ y - 1 = -x +2

⇒ y + x - 3 = 0

Therefore, 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:

It is given that equation of the curve y = 

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

Now, if the slope of the tangent is 2, then we get,

⇒ 2(x-3)² = -1

⇒ (x-3)² = 

This is not possible since the L.H.S. is positive while the R.H.S. is negative.

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

Question 12.

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

Answer:

It is given that equation of the curve y = ,

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

Now, if the slope of the tangent is 0, then we get,

⇒ -2(x-1)=0

⇒ x =1

So, when x = 1 then y = 

Now, the equation of the tangent (0,) is given by:

y – = 0(x-1)

⇒ y - = 0

⇒ y = 

Therefore, the equations of the required line is y = .

Question 13.

Find points on the curve  at which the tangents are

(i) parallel to x-axis (ii) parallel to y-axis.

Answer:

(i) It is given that 

Now, differentiating both sides with respect to x, we get

We know that the tangent is parallel to the x –axis if the slope is 0 ie,

, which is possible if x =0

Then,  for x =0

⇒ y² = 16

⇒ 

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

(ii) It is given that 

Now, differentiating both sides with respect to x, we get

We know that the tangent is parallel to the y–axis if the slope of the normal is 0 ie,

,

⇒ y = 0

Then,  for y =0

⇒ 

Therefore, 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:
x = cos t, y = sin t at t=π/4

Answer:

It is given that equation of curve is x = cost, y = sint

On differentiating with respect to x, we get

Therefore, the slope of the tangent at ) is -1.

When 

Then, the equation of the tangent is  is

y –  = -1(x – )

⇒ x + y -- = 0

⇒ x + y - = 0

Then, slope of normal at 

=

Now, equation of the normal at 

y –  = 1(x –)

⇒ x = y

Question 15.

Find the equations of the tangent and normal to the given curves at the indicated points:

y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5)

Answer:

It is given that equation of curve is y = x⁴ – 6x³ + 13x² – 10x + 5

On differentiating with respect to x, we get

 = 4x³ - 18x² +26x -10

Now, Slope of tangent will be the value of  at x = 0 i.e. 
m₁ = 4(0)³ - 18(0)² + 26(0) - 10 = -10

Therefore, the slope of the tangent at (0, 5) is -10.

Then, the equation of the tangent is

y – 5 = -10(x – 0)

⇒ y – 5 = 10x

⇒ 10x +y =5

Also, slope of normal at (0,5)

Now, equation of the normal at (0,5)

⇒ 10y - 50 = x

⇒ x -10y +50 = 0

Question 16.

Find the equations of the tangent and normal to the given curves at the indicated points:

y = x⁴ – 6x³ + 13x²– 10x + 5 at (1, 3)

Answer:

It is given that equation of curve is y = y = x⁴ – 6x³ + 13x²– 10x + 5

On differentiating with respect to x, we get

= 4x³ - 18x² +26x - 10

Therefore, the slope of the tangent at (1, 3) is 2.

Then, the equation of the tangent is

y – 3 = 2(x – 1)

⇒ y – 3 = 2x - 2

⇒ y = 2x +1

Then, slope of normal at (1,3)

=

Now, equation of the normal at (1,3)

y – 3 = (x – 1)

⇒ 2y -6 =- x + 1

⇒ x + 2y - 7 = 0

Question 17.

Find the equations of the tangent and normal to the given curves at the indicated points:

y = x³ at (1, 1)

Answer:

It is given that equation of curve is y = x³

On differentiating with respect to x, we get

= 3x²

Therefore, the slope of the tangent at (1, 1) is 3.

Then, the equation of the tangent is

y – 1 = 3(x – 1)

⇒ y = 3x - 2

Then, slope of normal at (1,1)

=

Now, equation of the normal at (1,1)

y – 1 = (x – 1)

⇒ 3y -3 =- x + 1

⇒ x + 3y - 4 = 0

Question 18.

Find the equations of the tangent and normal to the given curves at the indicated points:

y = x² at (0, 0)

Answer:

It is given that equation of curve is y = x²

On differentiating with respect to x, we get

= 2x

Therefore, the slope of the tangent at (0, 0) is 0.

Then, the equation of the tangent is

y – 0 = 0(x – 0)

⇒ y = 0

Then, slope of normal at (0,0)

=

Now, equation of the normal at (0,0)

x = 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:

(a) It is given that equation of the curve is y = x² – 2x +7

On differentiating with respect to x, we get

The equation of the line is 2x – y + 9 = 0

⇒ y = 2x + 9

⇒ Slope of the line = 2

Now we know that if a tangent is parallel to the line 2x – y + 9 = 0, then

Slope of the tangent = Slope of the line

⇒ 2 = 2x – 2

⇒ 2x = 4

⇒ x = 2

Now, putting x = 2, we get

y =4 -4 + 7 = 7

Then, the equation of the tangent passing through (2,7)

⇒ y – 7 = 2(x – 2)

⇒ y – 2x – 3 = 0

Therefore, the equation of the tangent line to the given curve which is parallel to line 2x – y + 9 = 0 is y – 2x – 3 = 0.

(b) It is given that equation of the curve is y = x² – 2x +7

On differentiating with respect to x, we get

The equation of the line is 5y – 15x = 13

⇒ y = 

⇒ Slope of the line = 3

Now we know that if a tangent is perpendicular to the line 5y – 15x = 13, then

⇒ 2x – 2=

⇒ 2x = 

⇒ x = 

Now, putting x = , we get

y =

Then, the equation of the tangent passing through 

⇒ y – = (x – )

⇒ 

⇒ 36y – 217 = -2(6x -5)

⇒ 36y+12x – 227 = 0

Therefore, the equation of the tangent line to the given curve which is perpendicular to line 5y – 15x = 13 is 36y+12x – 227 = 0.

Question 20.

Show that the tangents to the curve y = 7x³ + 11 at the points where x = 2 and

x = – 2 are parallel.

Answer:

The given curve y = 7x³ + 11

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

It is cleared that the slopes of the tangents at the points where x = 2 and x = -2 are equal.

Therefore, the two tangents are parallel.

Question 21.

Find the points on the curve y = x³ at which the slope of the tangent is equal to the y-coordinate of the point.

Answer:

The given curve y = x³

Then, the slope of the tangent at the point (x, y) is given by:

We know that, when the slope of the tangent is equal to the y- coordinate of the point,

Then y = 3x²

Also, we have y = x³

⇒ 3x² = x3

⇒ x²(x-3) = 0

⇒ x = 0, x = 3

When x = 0 then y = 0

and when x = 3 then y = 27

Therefore, the required points are (0, 0) and (3, 27).

Question 22.

For the curve y = 4x³ – 2x⁵, find all the points at which the tangent passes through the origin.

Answer:

The given curve y = 4x³ – 2x⁵

Then, the slope of the tangent at the point (x, y) is 12x² – 10x⁴

The equation of the tangent at (x,y) is given by,

Y – y = (12x² – 10x⁴)(X – x) ………….(1)

When the tangent passes through the origin (0,0), then X =Y=0

Then equation (1) becomes,

-y = (12x² – 10x⁴)(– x)

⇒ y = (12x³ – 10x⁵)

Also, we have y = 4x³ – 2x⁵

⇒ (12x³ – 10x⁵) = 4x³ – 2x⁵

⇒ 8x⁵ – 8x³ = 0

⇒ x⁵ – 2x³ = 0

⇒ x³(x² – 1) = 0

⇒ x = 0 , 1

When x = 0 then y = 0

When x = 1 then y = 2

And when x = -1 then y = -2

Therefore, the required points are (0, 0), (1,2) and (-1, -2).

Question 23.

Find the points on the curve x² + y² – 2x – 3 = 0 at which the tangents are parallel to the x-axis.

Answer:

It is given that x² + y² – 2x – 3 = 0

Now, differentiating both sides with respect to x, we get

We know that the tangents are parallel to the x –axis if the slope of the tangent is 0 ie,

⇒ 1-x = 0

⇒ x = 1

But, x² + y² – 2x – 3 = 0 for x = 1

⇒ y² = 4

⇒ y = 2

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

Question 24.

Find the equation of the normal at the point (am², am³) for the curve ay² = x³.

Answer:

It is given that ay² = x³

Now, differentiating both sides with respect to x, we get

Then, the slope of the tangent to the given curve at (am², am³) is

Then, slope of normal at (am², am³)

=

Therefore, equation of the normal at (am², am³) is given by:

y - am³ =

⇒ 3my – 3am⁴ = -2x + 2am²

⇒ 2x + 3my – am²(2 + 3m²) = 0

Therefore, equation of the normal at (am², am³) is 2x + 3my – am²(2 + 3m²) = 0

Question 25.

Find the equation of the normals to the curve y = x³ + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

Answer:

It is given that the equation of the normal to the curve y = x³ + 2x + 6

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

Then, slope of normal to the given curve at any point (x,y)

=Mediumy = 

⇒ Slope of the given line = 

We know that 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.

⇒ 

⇒ 3x2 + 2 = 14

⇒ 3x² = 12

⇒ x² = 4

⇒ 

So, when x = 2 then y = 18

and x -2 then y = -6

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

Then, the equation of the normal through (2, 18) is:

y -18 = 

⇒ 14y -252 = -x +2

⇒ x +14y-254 = 0

And, the equation of the normal through (-2, -6) is:

y –(-6) = 

⇒ y + 6 = 

⇒ 14y +84 = -x -2

⇒ x +14y+86 = 0

Therefore, the equation of the normals to the curve y = x³ + 2x + 6 which are parallel to the line x + 14y + 4 = 0 are x +14y-254 = 0 and x +14y+86 = 0.

Question 26.

Find the equations of the tangent and normal to the parabola y² = 4ax at the point (at², 2at).

Answer:

The equation of a parabola is y² = 4ax, then,

On differentiating it with respect to x, we get

2y

Then, the slope of the tangent at (at², 2at) is 

Then, the equation of the tangent at (at², 2at) is given by,

y – 2at = 

⇒ ty -2at² = x –at²

⇒ ty = x + at2

Now, Then, slope of normal at (at², 2at)

=

Then, the equation of the normal at (at², 2at) is given by:

y – 2at = -t(x-at²)

⇒ y – 2at = -tx + at³

⇒ y = -tx + 2at + at³

Therefore, the equation of the normal at (at², 2at) is y = -tx + 2at + at³.

Question 27.

Prove that the curves x = y² and xy = k cut at right angles* if 8k² = 1.

Answer:

It is given that the curves x = y² and xy = k

Now, putting the value of x in y = k, we get

y³ = k

Then, the point of intersection of the given curves is 

On differentiating x = y² with respect to x, we get

Then, the slope of the tangent at xy = k at 

is 

As we know that two curves intersect at right angles if the tangents to the curve at the point of intersection are perpendicular to each other.

So, we should have the product of the tangent as -1.

Then, the given two curves cut at right angles if the product of the slopes of their respective tangent at is -1.

Cubing both side, we get

⇒ 8k² = 1

Therefore, the given two curves cut at right angles if 8k² = 1.

Question 28.

Find the equations of the tangent and normal to the hyperbola  at the point (x₀, y₀).

Answer:

It is given that the equations of the tangent and normal to the hyperbola , then,

On differentiating it with respect to x, we get,

Therefore, the slope of the tangent at (x₀, y₀) is

Then, the equation of the tangent at (x₀, y₀) is given by:

y - y₀ =

Then, slope of normal at (x₀, y₀)

=

Therefore, the equation of the normal at (x₀, y₀) is

y - y₀ =  (x - x₀)

Therefore, the equation of the normal at (x₀, y₀) is.

Question 29.

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

Answer:

It is given that 

Then, the equation of the tangent at any given point (x, y) is given by,

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

⇒ y = 2x +

⇒ slope of the line = 2

Now, the tangent to the given curve is parallel to the line 4x − 2y + 5 = 0

if the slope of the tangent = the slope of the line

When x =,

y = 

Then, Equation of the tangent passing through the point is given by:

⇒ 24y – 18 = 48x – 41

⇒ 48x -24y =23

Therefore, the equation of the required tangent is 48x -24y =23

Question 30.

The slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is
A. 3

B. 

C. –3

D. 

Answer:

It I given that the slope of the normal to the curve y = 2x² + 3 sin x,

Then, the slope of the tangent at x = 0 is

Therefore, the slope of the normal to the curve at x = 0 is

Therefore, the slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is.

Question 31.

The line y = x + 1 is a tangent to the curve y² = 4x at the point
A. (1, 2)

B. (2, 1)

C. (1, – 2)

D. (– 1, 2)

Answer:

It is given that tangent to the curve y² = 4x

Then differentiating with respect to x, we have,

Then, the equation of the tangent at any given point (x,y) is given by,

The given line is y = x + 1

⇒ 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.

Then, we have,

⇒ y =2

Now, y = x+1

⇒ x = y -1

⇒ x = 2-1 = 1

Therefore, the line y = x+1 is a tangent to the given curve at the point (1, 2).

---

Exercise 6.4

Question 1.

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

Answer:

Consider y = √x.

Let x = 25 and Δx = 0.3. Then, we get

Δy = √25.3 - √25

Δy = √25.3 – 5

Δy = √25.3 = Δy + 5

Now, dy is approximately equal to Δy and is given by:

= 0.03

Therefore, the approximate value of  is 5.03.

Question 2.

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

Answer:

Consider y = √x.

Let x = 49 and Δx = 0.5. Then, we get

Δy = 

= √49.5-√49

= √49.5 – 7

= √49.5 = Δy + 7

Now, dy is approximately equal to Δy and is given by:

dy = 

= 0.035

Therefore, the approximate value of  is 7.035.

Question 3.

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

Answer:

Consider y = .

Let x = 1 and Δx = -0.4. Then, we get

= √0.6-1

= √0.6 = Δy + 1

Now, dy is approximately equal to Δy and is given by:

= -0.2

Therefore, the approximate value of  is 0.8.

Question 4.

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

Answer:

Consider y = 

Let x = 0.008 and Δx = 0.001. Then, we get

Δy = (x+∆x)^1/3 - (x)^1/3

= (0.009)^1/3 - (0.008)^1/3 = (0.009)^1/3 - 0.2

= (0.009)^1/3 = Δy + 0.2

Now, dy is approximately equal to Δy and is given by:

= 0.208

Therefore, the approximate value of  is 0.208.

Question 5.

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

Answer:

Consider y = 

Let x = 1 and Δx = -0.001. Then, we get

Δy = (x+∆x)^1/10 - (x)^1/10

Δy = (0.999)^1/10 - 1

(0.999)^1/10 = Δy + 1

Now, dy is approximately equal to Δy and is given by:

dy = 

= -0.0001

Therefore, the approximate value of  is 0.9999.

Question 6.

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

Answer:

Consider y = 

Let x = 16 and Δx = -1. Then, we get

Δy = 

=

⇒  = Δy + 2

Now, dy is approximately equal to Δy and is given by:

dy = 

= -0.03125

Therefore, the approximate value of  is -0.03125.

Question 7.

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

Answer:

Consider y = 

Let x = 27 and Δx = -1. Then, we get

Δy = (x+∆x)^1/3 - (x)^1/3

= (26)^1/3 - 3

= (26)^1/3 = Δy + 3

Now, dy is approximately equal to Δy and is given by:

= 

Therefore, the approximate value of is 2.9629.

Question 8.

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

Answer:

Consider y = 

Let x = 256 and Δx = -1. Then, we get

Δy = (x+∆x)^1/4 - (x)^1/4

= (225)^1/4 - (256)^1/4

= (225)^1/4 - 4

= (225)^1/4 = Δy + 4

Now, dy is approximately equal to Δy and is given by:

= -0.0039

Therefore, the approximate value of (255)^1/4 is 3.9961.

Question 9.

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

Answer:

Consider y = 

Let x = 81 and Δx = 1. Then, we get

Δy = 

=

⇒  = Δy + 3

Now, dy is approximately equal to Δy and is given by:

dy = 

= 0.009

Therefore, the approximate value of is 3.009.

Question 10.

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

Answer:

Consider y = 

Let x = 400 and Δx = 1. Then, we get

Δy = 

=

⇒  = Δy + 20

Now, dy is approximately equal to Δy and is given by:

dy = 

= 0.025

Therefore, the approximate value of is 20.025.

Question 11.

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

Answer:

Consider y = 

Let x = 0.0036 and Δx = 0.0001. Then, we get

Δy = 

=

⇒  = Δy + 0.06

Now, dy is approximately equal to Δy and is given by:

dy = 

= 0.00083

Therefore, the approximate value of is 0.06083.

Question 12.

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

Answer:

Consider y = 

Let x = 27 and Δx = -0.43. Then, we get

Δy = 

=

⇒  = Δy + 3

Now, dy is approximately equal to Δy and is given by:

dy = 

= -0.015

Therefore, the approximate value of is 2.984.

Question 13.

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

Answer:

Consider y = 

Let x = 81 and Δx = 0.5. Then, we get

Δy = 

=

⇒  = Δy + 3

Now, dy is approximately equal to Δy and is given by:

dy = 

= 0.0046

Therefore, the approximate value of is 3.0046.

Question 14.

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

Answer:

Consider y = 

Let x = 4 and Δx = -0.032. Then, we get

Δy = 

=

⇒  = Δy + 8

Now, dy is approximately equal to Δy and is given by:

dy = 

= 3(-0.032)

= -0.096

Therefore, the approximate value of is 7.904.

Question 15.

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

Answer:

Consider y = 

Let x = 32 and Δx = 0.15. Then, we get

Δy = 

=

⇒  = Δy + 2

Now, dy is approximately equal to Δy and is given by:

dy = 

= 0.00187

Therefore, the approximate value of is 2.00187.

Question 16.

Find the approximate value of f (2.01), where f (x) = 4x² + 5x + 2.

Answer:

Let x = 2 and Δx = 0.01. Then, we get,

f(2.01) = f(x + Δx) = 4(x +Δx)² + 5 ( x + Δx) + 2

Now, Δy = f (x + Δx) – f(x)

⇒ f (x + Δx) = f(x) + Δy

≈ f(x) + f’(x).Δx (as dx = Δx)

⇒ f(2.01) ≈ (4x² + 5x + 2) + (8x + 5) Δx

= [4(2)² + 5(2) + 2] +[8(2) + 5] (0.01)

= (16 + 10 + 2) + (16 + 5)(0.01)

= 28 + 0.21

= 28.21

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

Question 17.

Find the approximate value of f (5.001), where f (x) = x³ – 7x2 + 15.

Answer:

Let x = 5 and Δx = 0.001. Then, we get,

f(5.001) = f(x + Δx) = (x +Δx)³ - 7 ( x + Δx)² + 15

Now, Δy = f (x + Δx) – f(x)

⇒ f (x + Δx) = f(x) + Δy

≈ f(x) + f’(x).Δx (as dx = Δx)

⇒ f(5.001) ≈ (x³ – 7x² + 15) + (3x² – 14x) Δx

= [(5)² – 7(5)² + 15] +[3(5)² -14(5)] (0.001)

= (125 – 175 + 15) + (75 - 70)(0.001)

= -35 + 0.005

= -34.995

Therefore, the approximate value of f (5.001) is -34.995.

Question 18.

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

Answer:

It is given that the volume of a cube (V) of a side x

⇒ V = x³

= (3x²)Δx

= (3x²)(0.01x)

= 0.03x³

Therefore, the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1% is 0.03x³.

Question 19.

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

Answer:

It is given that the surface area of cube (S) of a side x

⇒ S = 6x²

= (12x)Δx

= (12x)(0.01x)

= 0.12x²

Therefore, the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1% is 0.12x².

Question 20.

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

Answer:

Let r be the radius of the sphere and Δr be the error in measuring the radius.

Now, it is given that r = 7m and Δr = 0.02m

We know that volume of sphere (V) = 

Now, 

= (4πr²)Δr

= 4π (7²)(0.02)m³

= 3.92π m³

Therefore, the approximate error in calculating its volume is 3.92π m³.

Question 21.

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 its surface area.

Answer:

Let r be the radius of the sphere and Δr be the error in measuring the radius.

Now, it is given that r = 9m and Δr = 0.03m

We know that surface area of sphere (S) = 

Now, 

= (8πr)Δr

= 8π (9)(0.03)m²

= 2.16π m³

Therefore, the approximate error in calculating its surface area is 2.16π m³.

Question 22.

If f(x) = 3x² + 15x + 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 Δx = 0.02. Then, we get,

f(3.02) = f(x + Δx) = 3(x +Δx)² + 15( x + Δx) + 5

Now, Δy = f (x + Δx) – f(x)

⇒ f (x + Δx) = f(x) + Δy

≈ f(x) + f’(x).Δx (as dx = Δx)

⇒ f(3.02) ≈ (3x² + 15x + 5) + (6x + 15) Δx

= [3(3)² + 15(3) + 5] +[6(3) + 15] (0.02)

= (27 + 45 + 5) + (18 + 15)(0.02)

= 77 + 0.66

= 77.66

Therefore, the approximate value of f (3.02) is 77.66.

Question 23.

The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is
A. 0.06 x³ m³

B. 0.6 x³ m³

C. 0.09 x³ m³

D. 0.9 x³ m³

Answer:

It is given that the volume of a cube (V) of a side x

⇒ V = x³

⇒ dV = 

= (3x²)Δx

= (3x²)(0.03x)

= 0.09x³

Therefore, the approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is 0.09x³.

---

Exercise 6.5

Question 1.

Find the maximum and minimum values, if any, of the following functions given by

f (x) = (2x – 1)² + 3

Answer:

It is given that f (x) = (2x – 1)² + 3

Now, we can see that (2x – 1)² ≥ 0 for every x ϵ R

⇒ f (x) = (2x – 1)² + 3 ≥ 3 for every x ϵ R

The minimum value of f is attained when 2x – 1 = 0

2x -1 = 0

Then, Minimum value of 

Therefore, function f does not have a maximum value.

Question 2.

Find the maximum and minimum values, if any, of the following functions given by

f (x) = 9x² + 12x + 2

Answer:

It is given that f (x) = 9x² + 12x + 2 = (3x + 2)² - 2

Now, we can see that (3x + 2)² ≥ 0 for every x ϵ R

⇒ f (x) = (3x + 2)² - 2 ≥ -2 for every x ϵ R

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

3x +2 =0

Then, Minimum value of 

Therefore, function f does not have a maximum value.

Question 3.

Find the maximum and minimum values, if any, of the following functions given by

f(x) = – (x – 1)² + 10

Answer:

It is given that f (x) = –(x – 1)² + 10

Now, we can see that (x - 1)² ≥ 0 for every x ϵ R

⇒ f (x) = –(x – 1)² + 10 ≤ 10 for every x ϵ R

The minimum value of f is attained when x - 1 = 0

x - 1 = 0

⇒ x = 1

Then, Maximum value of f = f(1) = -(1-1)² + 10 = 10

Therefore, function f does not have a minimum value.

Question 4.

Find the maximum and minimum values, if any, of the following functions given by

g(x) = x³ + 1

Answer:

It is given that g(x) = x³ + 1

Now,

x ∈ ℝ

⇒ -∞ ≤ x ≤ ∞

⇒ -∞ ≤ x³ ≤ ∞

⇒ -∞ ≤ x³ + 1 ≤ ∞

The function g neither has a maximum value nor a minimum value.

Question 5.

Find the maximum and minimum values, if any, of the following functions given by

f(x) = |x + 2| – 1

Answer:

It is given that f (x) = |x + 2| – 1

Now, we can see that |x + 2| ≥ 0 for every x ϵ R

⇒ f (x) = |x + 2| – 1 ≥ -1 for every x ϵ R

The minimum value of f is attained when |x + 2| = 0

|x + 2| =0

⇒ x = -2

Then, Minimum value of f = f(-2) = |-2 + 2| - 1 = -1

Therefore, function f does not have a maximum value.

Question 6.

Find the maximum and minimum values, if any, of the following functions given by

g(x) = –|x + 1| + 3

Answer:

It is given that g(x) = –|x + 1| + 3

Now, we can see that –|x + 1| ≤ 0 for every x ϵ R

⇒ g(x) = –|x + 1| + 3 ≤ 3 for every x ϵ R

The maximum value of f is attained when |x + 1| = 0

|x + 1| = 0

⇒ x = -1

Then, Maximum value of g = g(-1) = -|-1 + 1| + 3 = 3

Therefore, function f does not have a minimum value.

Question 7.

Find the maximum and minimum values, if any, of the following functions given by

h(x) = sin(2x) + 5

Answer:

It is given that h(x) = sin(2x) + 5

Now, we can see that -1 ≤ sin2x ≤ 1

⇒ -1 + 5 ≤ sin2x + 5 ≤ 1 + 5

⇒ 4 ≤ sin2x + 5 ≤ 6

Therefore, the maximum and minimum value of function h are 6 and 4 respectively.

Question 8.

Find the maximum and minimum values, if any, of the following functions given by

f (x) = |sin 4x + 3|

Answer:

It is given that f(x) = |sin 4x + 3|

Now, we can see that -1 ≤ sin4x ≤ 1

⇒ 2 ≤ sin 4x + 3 ≤ 4

⇒ 2 ≤ |sin 4x + 3| ≤ 4

Therefore, the maximum and minimum value of function h are 4 and 2 respectively.

Question 9.

Find the maximum and minimum values, if any, of the following functions given by

h(x) = x + 1, x ∈ (–1, 1)

Answer:

It is given that h(x) = x + 1, x ∈ (–1, 1)

Now, if a point x₀ is closest to -1, then,

We find  for all x₀∈ (–1, 1)

Also, if a point x₁ is closest to 1, then,

We find  for all x₁∈ (–1, 1)

Therefore, the function h (x) has neither maximum nor minimum value in (-1, 1).

Question 10.

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:

f (x) = x²

Answer:

f(x) = x²

⇒ f’(x) = 2x

Now, f’(x) = 0

⇒ x = 0

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

⇒ f’’(0) = 2, which is positive.

Then, by second derivative test,

⇒ x = 0 is point of local maxima and local minima of f at x = 0 is f(0) = 0.

Question 11.

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:

g(x) = x³ – 3x

Answer:

g(x) = x³ – 3x

⇒ g’(x) = 3x² - 3

Now, g’(x) = 0

⇒ 3x² - 3 = 0

⇒ 3x² = 3

⇒ x = � 1

g’’(x) = 6x

Now, g’(1) = 6>0

and g’(-1) = -6 < 0

Then, by second derivative test,

⇒ x = 1 is point of local maxima and local minima of g at x = 1 is

g(1) = 1³ – 3 = 1-3 =-2

And,

x = -1 is point of local maxima and local maximum value of g at x = -1 is

g(-1) = (-1)³ – 3(-1) = -1+3 = 2

Question 12.

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:

Answer:

h(x) = sin x + cos x, 

h’(x) = cosx - sinx

Now, h’(x) = 0

⇒ cosx - sinx = 0

⇒ cosx = sinx

⇒ tanx = 1

h’’(x) = -sinx – cosx = -(sinx + cosx)

Then, by second derivative test,

 is point of local maxima and local minima of h at  is

Question 13.

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:

f (x) = sin x – cos x, 0 < x < 2π

Answer:

f (x) = sin x – cos x, 0 < x < 2π

f’(x) = cosx + sinx

Now, f’(x) = 0

⇒ cosx + sinx = 0

⇒ cosx = -sinx

⇒ tanx = -1

’(x) = -sinx + cosx

Then, by second derivative test,

 is point of local maxima and the local maximum value of f at  is

And,

 is point of local minima and the local minimum value of f at  is

Question 14.

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:

f (x) = x³ – 6x² + 9x + 15

Answer:

f (x) = x³ – 6x² + 9x + 15

⇒ f’(x) = 3x² – 12x + 9

Now, f’(x) = 0

⇒ 3x² – 12x + 9 = 0

⇒ 3(x-1)(x-3) = 0

⇒ x = 1,3

g’’(x) = 6x – 12 =6(x-2)

Now, f’(1) = 6(1-2)=-6 < 0

and f’(3) = 6(3-2) = 6 > 0

Then, by second derivative test,

⇒ x = 1 is point of local maxima and local maximum of f at x = 1 is

f(1) = 1³ – 6 +9 +15 = 19

And,

x = 3 is point of local minima and local minimum value of f at x = 3 is

f(3) = 27 – 54 + 27 +15 = 15

Question 15.

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:

Answer:

Now, g’(x) = 0

 = 

⇒ x² = 4

⇒ x = �2

Since x > 0, we take x = 2

Then, by second derivative test,

⇒ x = 2 is point of local minima and local minimum of g at x = 2 is

Question 16.

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:

Answer:

⇒ x = 0

Now, for values close to x = 0 and to the left of 0,

g’(x) > 0.

Also, for values close to x = 0 and to the right of 0,

g’(x) < 0

Then, by first derivative test,

⇒ x = 0 is point of local maxima and local maximum of g at x = 0 is

Question 17.

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:

Answer:

f’(x) = 0

⇒ 2 – 3x = 0

Question 18.

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

f (x) = e^x

Answer:

f (x) = e^x

⇒ f’(x) = e^x

Now, if f’(x) = 0, then e^x = 0.

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

Therefore, there does not exist c ϵ R such that f’(c) = 0

Hence, function f does not have maxima or minima.

Question 19.

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

g(x) = log x

Answer:

g(x) = logx

Since, log x is defined for a positive number x,

g’(x) > 0 for any x.

Therefore, there does not exist c ϵ R such that f’(c) = 0

Hence, function f does not have maxima or minima.

Question 20.

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

h(x) = x³ + x² + x +1

Answer:

h(x) = x³ + x² + x +1

⇒ h’(x) = 3x² + 2x +1

h(x) = 0

⇒ 3x² + 2x +1 = 0

Therefore, there does not exist c ϵ R such that h’(c) = 0

Hence, function h does not have maxima or minima.

Question 21.

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

f (x) = x³, x ∈ [–2, 2]

Answer:

It is given that f (x) = x³, x ∈ [–2, 2]

⇒ f’(x) = 3x²

Now, f’(x) = 0

⇒ x = 0

Now, 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

Therefore, we have the absolute maximum value of f on [-2, 2] is 8 occurring at x = 2.

And, the absolute minimum value of f on [-2, 2] is -8 occurring at x =-2.

Question 22.

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

f (x) = sin x + cos x, x ∈ [0, π]

Answer:

It is given that f (x) = sin x + cos x, x ∈ [0, π]

f’(x) = cosx - sinx

Now, f’(x) = 0

⇒ cosx - sinx = 0

⇒ cosx = sinx

⇒ tanx = 1

⇒ x = 

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

f(0) = sin0 +cos0 = 0+1 = 1

f(π) = sin π + cos π = 0 -1 = -1

Therefore, we have the absolute maximum value of f on [0, π] is √2 occurring at 

And, the absolute minimum value of f on [0, π] is -1 occurring at x = π.

Question 23.

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

Answer:

It is given that 

Now, f’(x) = 0

⇒ x = 4

Now, we evaluate the value of f at critical point x = 0 and at end points of the interval 

Therefore, we have 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.

Question 24.

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

f (x) = (x − 1)² + 3, x ∈ [−3, 1]

Answer:

It is given that f (x) = (x − 1)² + 3, x ∈ [−3,1]

⇒ f’(x) = 2(x – 1)

Now, f’(x) = 0

⇒ 2(x-1)

⇒ x = 1

Now, we evaluate the value of f at critical point x = 1 and at end points of the interval [-3, 1].

f(1) = (1 - 1)² + 3 = 0 + 3 = 3

f(-3) = (-3 - 1)² + 3 = 16 + 3 = 19

Therefore, we have the absolute maximum value of f on [-3, 1] is 19 occurring at x =-3.

And, the absolute minimum value of f on [-3,1] is 3 occurring at x = 1.

Question 25.

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

p(x) = 41 – 72x – 18x²

Answer:

It is given that the profit function p(x) = 41 – 72x – 18x²

⇒ p’(x) = -72 – 36 x

and p’’(x) = -36

Now, g’(x) = 0

Then, by second derivative test,

x = -2 is point of local maxima of p.

Therefore, Maximum Profit =


= 113

Therefore, the maximum profit that the company can make is 113 units.

Question 26.

Find both the maximum value and the minimum value of 3x⁴ – 8x³ + 12x² – 48x + 25 on the interval [0, 3].

Answer:

Let f (x) = 3x⁴ – 8x³ + 12x² – 48x + 25, x ∈[0,3]

⇒ f’(x) = 12x³ - 24x² + 24x – 48

=12(x³ - 2x² + 2x – 4)

=12[x² (x – 2)+ 2(x – 2)]

=12(x -2)( x²+ 2)

Now, f’(x) = 0

⇒ x =2 or (x²+ 2) = 0 for which there are no real roots.

Therefore, we will only consider x = 2

Now, we evaluate the value of f at critical point x = 2 and at end points of the interval [0,3].

f(2) = 3(2)⁴ – 8(2)³ + 12(2)² – 48(2) + 25

= 3(16) – 8(8) + 12(4) +25

=48 – 64 +48 – 96 + 25

= -39

f(0) = 3(0)⁴ – 8(0)³ + 12(0)² – 48(0) + 25

= 0+ 0 + 0 +25

= 25

f(3) = 3(3)⁴ – 8(3)³ + 12(3)² – 48(3) + 25

= 3(81) – 8(27) + 12(9) +25

=243 – 216 +108 – 144 + 25

= 16

Therefore, we have the absolute maximum value of f on [0,3] is 25 occurring at x =0.

And, the absolute minimum value of f on [0,3] is -39 occurring at x = 2.

Question 27.

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

Answer:

It is given that f (x) = sin2x, x ∈ [0, 2π]

f’(x) = 2cos2x

Now, f’(x) = 0

⇒ cos2x = 0

⇒ 2x = 0

⇒ x = 

Now, we evaluate the value of f at critical point x =  and at end points of the interval [0, 2π]

f’ = 

f’ = 

f’ = 

f’ = 

f(0) = sin0, f(2π) = sin2π = 0

Therefore, we have the absolute maximum value of f on [0, 2π] is 1 occurring at

x = and x =.

Question 28.

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

Answer:

Let f(x) = sin x + cos x,

⇒ f’(x) = cosx - sinx

Now, f’(x) = 0

⇒ cosx - sinx = 0

⇒ cosx = sinx

⇒ tanx = 1

Now,

If f’’(x) will be negative when (sinx + cosx) > 0, means both sinx and cosx are positive.

And, we know that sinx and cosx both are positive in the first quadrant.

Then, f’’(x) will be negative when 

f’’(x) = -sinx – cosx = -(sinx + cosx)

Now, let us take x =

Then, by second derivative test,

f will be maximum at x = 

And, the maximum value of f is

Question 29.

Find the maximum value of 2x³ – 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [–3, –1].

Answer:

Let f (x) = 2x³ – 24x + 107, x ∈ [1, 3]

⇒ f’(x) = 6x² – 24

=6(x² - 4)

Now, f’(x) = 0

⇒ 6(x² - 4) = 0

⇒ x² = 4

⇒ x = �2

Therefore, we will only consider the interval [1,3]

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

f(2) = 2(2)³ – 24(2) + 107

= 2(8) – 24(2) + 107

= 75

f(1) = 2(1)³ – 24(1)+ 107

= 2 – 24 +107

= 85

f(3) = 2(3)³ – 24(3) + 107

= 2(27) -24(3) +107

= 89

Therefore, we have the absolute maximum value of f on [1,3] is 89 occurring at x =3.

Now, we will only consider the interval [-3, -1]

Now, we evaluate the value of f at critical point x = -2 ϵ [-3, -1] and at end points of the interval [1,3].

f(-3) = 2(-3)³ – 24(-3) + 107

= 2(-27) – 24(-3) + 107

= 125

f(-1) = 2(-1)³ – 24(-1)+ 107

= -2 + 24 +107

= 129

f(-2) = 2(-2)³ – 24(-2) + 107

= 2(-8) -24(-2) +107

= 139

Therefore, we have the absolute maximum value of f on [-3, -1] is 89 occurring at x = -2.

Question 30.

It is given that at x = 1, the function x⁴ – 62x² + ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a.

Answer:

It is given that f(x) = x⁴ – 62x² + ax + 9

Then, f’(x) = 4x³ – 124x+ a

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

⇒ f’(1) = 0

⇒ 4 – 124 + a = 0

⇒ a = 120

Therefore, the value of a is 120.

Question 31.

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

Answer:

It is given that f (x) = x + sin 2x, x ∈ [0, 2π]

f’(x) = 1+ 2cos2x

Now, f’(x) = 0

2x = 2π � , n ϵ Z

⇒ x = nπ � , n ϵ Z

Now, we evaluate the value of f at critical point  and at end points of the interval [0, 2π]

f’(0) = 0 + sin 0 = 0

f’(2π) = 2π + sin 4π = 2π + 0 = 2π

Therefore, we have the absolute maximum value of f on [0, 2π] is 2π occurring at

x = 2π and absolute minimum value of f(x) in the interval [0, 2π] is 0 occuring at x = 0.

Question 32.

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.

Then, we get,

P(x) = x(24 –x) = 24x – x²

⇒ P’(x) = 24 – 2x

⇒ P’’(x) = -2

Now, P’(x) = 0

⇒ x = 12

And

P’’(12) = -2 <0

Then, by second derivative test,

x = 12 is the point of local maxima of P.

Therefore, the product of the numbers is the maximum when the numbers are 12 and 24 – 12 = 12.

Question 33.

Find two positive numbers x and y such that x + y = 60 and xy3 is maximum.

Answer:

Let the two numbers are x and y such that x + y = 60

⇒ y = 60 –x

Let f(x) = xy³

⇒ f(x) = x(60-x)³

⇒ f’(x) = (60 –x)³ -3x(60 –x)²

= (60-x)²[60 – x – 3x]

= (60-x)²[60 – 4x]

And f’’(x) = -2(60 – x)(60 -4x) -4(60-x)²

= -2(60 – x)[60 -4x + 2(60-x)]

= -2(60 – x)(180 – 6x)

= -12(60 – x)(30 -x)

Now, f’(x) =0

⇒ x = 60 or x =15

When x = 60, f’’(x) = 0.

When x = 15, f’’(x) = -12(60-15)(30-15) = -12×45×15 < 0

Then, by second derivative test, x =15 is a point of local maxima of f.

Then, function xy³ is maximum when x =15 and y = 60 – 15 = 45.

Therefore, required numbers are 15 and 45.

Question 34.

Find two positive numbers x and y such that their sum is 35 and the product x² y⁵ is a maximum.

Answer:

Let one number be x. Then, the other number is y = (35-x)

Let P(x)= x² y⁵

Then, we get,

P(x) = 2x(35 –x)⁵ – 5x²(35-x)⁴

= x(35 –x)⁴ [2(35-x)– 5x]

= x(35 –x)⁴ [70-7x]

= 7x(35 –x)⁴(10-x)

Now, P’’(x) = 7(35 –x)⁴ (10-x)+7x[-(35 –x)⁴ – 4(35-x)³ (10-x)]

= 7(35 –x)⁴ (10-x) - 7x(35 –x)⁴ – 28x(35-x)³ (10-x)]

=7(35 –x)³ [(35-x)(10-x) - x(35 –x) – 4x(10-x)]

=7(35 –x)³ [350-45x+x²-35x+x²-40x+4x²]

=7(35 –x)³ [6x²-120x+350]

Now, P’(x) = 0

⇒ x = 0, 35, 10

When x = 0, 35 This will make the product x2y5 equal to 0.

Therefore, x= 0, 35 cannot be possible values of x.

And when x = 10

Then, we have,

P’’(x) =7(35 –10)³[6(10)²-120(10)+350]

= 7(25)³ [-250]<0

Then, by second derivative test,

x = 10 and y = 35 -10 = 25 is the point of local maxima of P.

Therefore, the required number are 10 and 25.

Question 35.

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 S(x) be the sum of these number. Then,

S(x) = x³ + (16-x)³

⇒ S’(x) = 3x² -3(16-x)²

⇒ S’’(x) = 6x + 6(16-x)

Now, S’(x) =0

⇒ 3x² -3(16-x)² = 0

⇒ x² -(16-x)² = 0

⇒ x² – 256 - x² + 32x = 0

⇒ x = 8

Now, S’’(8) = 6(8) + 6(16-8)

= 48 + 48 = 96 > 0

Then, by second derivative test, x = 8 is the point of local minima of S.

Therefore, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16-8 = 8.

Question 36.

A square piece of tin of side 18 cm is to be 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, then, the length and the breadth of the box will be (18 – x) cm each and the height of the box is x cm.

Then, the volume {V(x)} of the box is given by:

V(x) = x(18-x)²

⇒ V’(x) = (18-x)² - 2x(18-x)

= (18 - x)[18- x -2x]

= (18 - x)(18 - 3x)

Now, V’’(x) = (18 - x)(-3) + (18 - 3x)(-1)

= -3(18 - x) - (18 - 3x)

= -54 + 3x - 18 + 3x

= 6x - 72

Now, V’(x) = 0

⇒ x = 18 or 3

If x = 18 then breadth becomes 0 which is not possible

Therefore, x =3

V’’(3) = 6.3 - 72 = -ve

Then, by second derivative test, x= 3 is the point of maxima of V.

Therefore, If we remove a square of side 3cm 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 37.

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 maximum?

Answer:

Let the side of the square to be cut off be x, then, the height of the box is x and the length is 45-2x and the breadth is 24 – 2x.

Then, the volume {V(x)} of the box is given by:

V(x) = x(45-2x)(24-x)

= x(1080-90x-48x+4x²)

= 4x³ – 138x² + 1080x

⇒ V’(x) = 12x² – 276x + 1080

= 12(x² - 23x + 90)

=12(x – 18)(x – 5)

Now, V’’(x) = 24x – 276 = 12(2x-23)

Now, V’(x) = 0

⇒ x = 18 or 5

It is not possible to cut off a square of side 18cm from each corner of the rectangular sheet. So, x cannot be equal to1 8.

Therefore, x =5

V’’(5) = 12(10 – 23) = -156 < 0

Then, by second derivative test, x = 5 is the point of maxima of V.

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

Question 38.

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

Answer:

The figure is given below:

Let a rectangle of length l and breadth b be inscribed in the circle of radius a.

Then, the diagonal passes through the centre and is of length 2a cm.

Now, by Pythagoras theorem, we get,

(2a)² = l² + b²

⇒ b² = 4a² – l²

⇒ b = 

Therefore, Area of rectangle, A = 

.

.

.

Gives 4a² = 2l²

⇒ l = a

⇒ b = 

when l = 

Then,  = -4 < 0

Then, by second derivative test, when l =, then the area of the rectangle is the maximum.

Since, l = b =,

Therefore, the rectangle is square.

Hence proved.

Question 39.

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

Answer:

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

Then, the surface area (S) of the cylinder is given by:

S = 2πr² + 2πrh

⇒ h 

Let V be the volume of the cylinder. Then

V = πr²h

Now, 

If 

So, when  then <0

Then, by second derivative test, the volume is the maximum when 

Now, when . then h = 

Therefore, the volume is the maximum when the height is the twice the radius or height is equal to diameter.

Question 40.

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 be the radius and h be the height of the cylinder.

Let V be the volume of the cylinder. Then

V = πr²h = 100(given)

⇒ h = 

hen, the surface area (S) of the cylinder is given by:

S = 2πr² + 2πrh

Now, , <0

If 

So, when  then  > 0

Then, by second derivative test, the surface area is the minimum when 

Now, when  then h = cm.

Therefore, the dimensions of the can which has the minimum surface area are  and h cm.

Question 41.

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, 2πr = 28 – l

⇒ r = 

Therefore, the required area (a) is given by

A = (side of the square)2 + πr2

= 

Then, 

Now, if 

Then, 

⇒ (π + 4)l – 112 = 0

⇒ 

So, when 

Then, 

Then, by second derivative test, the area (A) is the minimum when .

Therefore, 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

28 -  = cm.

Question 42.

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 cone.

Then, V= 

And height of cone h = R + 

Now, 

Now, if, 

After solving this we get, 

So, when, , then < 0

Then, by second derivative test, the volume of the cone is the maximum when 

So, when, , h = R +

Therefore, V = 

Therefore, the volume of the largest cone that can be inscribed in the sphere is  the volume of the sphere.

Question 43.

Show that the right circular cone of least curved surface and given volume has an altitude equal to √2 time the radius of the base.

Answer:

Let r and h be the radius and height of the cone respectively.

Then, the volume (V) of the cone is given by:

V= 

The surface area (S) of the cone

S = πrl

= 

Then, 

Now, if 

So, when  then  > 0

Then, by second derivative test, the surface area of the cone is the least when  .

So when  then h = 

Therefore, for a given volume, the right circular cone of the least curved surface has an altitude equal to √2 times the radius of the base.

Question 44.

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan^-1 √2.

Answer:

Let Ɵ be the semi- vertical angle of the cone.

Let r, h and l be the radius, height and the slant height of the cone respectively.

It is given that slant height is constant.

Now, r = lsinƟ and h = lcosƟ

Then, the volume of the cone (V)

V = 

.

Now, if 

sin³Ɵ = 2sinƟcos²Ɵ

⇒ tan²Ɵ = 2

⇒ tanƟ = √2

⇒ 

Now, when, then tan²Ɵ = 2 or sin²Ɵ = 2cos²Ɵ.

Then, we get

= -4πl³cos³Ɵ < 0 for Ɵ ϵ 

Then, by second derivative test, the volume (V) is the maximum when 

Therefore, the semi-vertical angle of the cone of the maximum volume and of given slant height is .

Hence Proved.

Question 45.

Show that semi-vertical angle of right circular cone of given surface area and maximum volume is .

Answer:

We know that total surface area of the cone = S = πr(l + r) …(1)

and Volume of the cone(V) = 

Then by (1), we get,

P = V²

Now, differentiating P with respect to r, we get,

Now, if, , then

S = 4πr²

Now again differentiating with respect to r, we get 

Therefore, P is maximum when S = 4πr²

And V is maximum when S = 4πr²

⇒ πr(l + r) = 4πr²

⇒ l = 3r

SinƟ = 

Therefore, semi-vertical angle of right circular cone of given surface area and maximum volume is .

Hence Proved.

Question 46.

The point on the curve x² = 2y which is nearest to the point (0, 5) is
A. (2√2,4)

B. (2√2,0)

C. (0, 0)

D. (2, 2)

Answer:

It is given that x² = 2y

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:

d’(x) = 0

⇒ x³ – 8x = 0

⇒ x(x²-8) =0

⇒ x = 0, 

And, .

So, now when x = 0, then d’’(x) =  < 0

And when, x = , d’’(x) >0

Then, by second derivative test, d(x) is minimum at 

So, when 

Therefore, the point on the curve x2 = 2y which is nearest to the point (0,5) is .

Question 47.

For all real values of x, the minimum value of  is
A. 0

B. 1

C. 3

D. 

Answer:

Let 

Then, f’(x) = 0

⇒ x² = 1

⇒ x = �1

Now, 

And, 

Also, f’’(-1) = -4 < 0

Then, by second derivative test, f is minimum at x = 1 and the minimum value is given by

Question 48.

The maximum value of  is
A. 

B. 

C. 1

D. 0

Answer:

Let f(x) = 

Now, if f’(x) = 0

Then, we evaluate the value of f at critical point  and at the end points of the interval [0, 1].

Therefore, we can conclude that the maximum value of f in the interval [0, 1] is 1.

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Miscellaneous Exercise

Question 1.

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

Answer:

Let us consider y = x^1/4 and  and 

Then, 

Therefore, 

Now, dy is approximately equal to Δy and is equal to

dy = 

Therefore, the approximate value of  is 

Question 2.

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

Answer:

Let us consider y =  and x=32 and 

Then, 

= 

Therefore, 

Now, dy is approximately equal to Δy and is equal to

Therefore, the approximate value of  is 

Question 3.

Show that the function given by  has maximum at x = e.

Answer:

It is given that f(x) = 

Then, f’(x) = 

Now, f’(x) = 0

⇒ 1 - logx =0

⇒ log x =1

⇒ log x = log e

⇒ x = e

Now, f’’(x) = 

Now, f’’(e)=

Therefore, by second derivative test, f is the maximum at x = e.

Question 4.

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.

Now, draw a perpendicular AD to BC. Then, we have

In ΔADC, by using Pythagoras theorem,

AD = 

Then, Area of triangle (A) = 

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 3cm per second.

Then 

And when a = b we have,

Therefore, if the two sides are equal to the base, then the area of the triangle is decreasing at the rate of  cm²/s.

Question 5.

Find the equation of the normal to curve x2 = 4y which passes through the point (1, 2).

Answer:

It is given that curve is y² = 4x.

Differentiating with respect to x, we get,

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

Therefore, Equation of the normal at (1,2) is y - 2 = - 1(x – 1)

⇒ y - 2 = - x + 1

⇒ x + y - 3 = 0

Question 6.

Show that the normal at any point θ to the curve

x = a cosθ + a θ sinθ, y = a sinθ – aθ cosθ is at a constant distance from the origin.

Answer:

We have x = a cosθ + a θ sinθ,

⇒ 

And y = a sinθ – aθ cosθ

⇒ 

So, 

Then, Slope of the normal at any point θ is .

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

y - a sinθ + aθ cosθ = (x - a cosθ - a θ sinθ)

⇒ ysinθ – asin²θ + aθ sinθ cosθ = - x cosθ + acos²θ + aθ sinθ cosθ

⇒ xcosθ + ysinθ – a(sin²θ + cos²θ ) = 0

⇒ xcosθ + ysinθ –a = 0

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

, which is independent of θ .

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

Question 7.

Find the intervals in which the function f given by



is (i) strictly increasing (ii) strictly decreasing.

Answer:

(i) It is given that f(x) = 

Now, if f’(x) =0

⇒ cos x = 0 or cosx = 4

But, cosx = 4 is not possible

Therefore, cosx =0

⇒ x = 

Now, x = divides (0,2π) into three disjoints intervals

In the intervals and, f’(x)>0

Therefore, f(x) is increasing for 0< x <  and  < x < 2π.

In interval, f’(x)<0

Therefore, f(x) is decreasing for  < x < .

Question 8.

Find the intervals in which the function f given by  is

(i) Increasing (ii) decreasing.

Answer:

It is given that f(x) =

Then, f’(x) =0

⇒ 3x⁶ - 3 = 0

⇒ x⁶ = 1

⇒ x = 1

Now, the points x =1 and x = - 1 divide the real line into three disjoint intervals

( - ∞, - 1), ( - 1,1) and (1,∞).

In interval ( - ∞, - 1) and (1,∞) when x < - 1 and x > 1 then f’(x) >0

Therefore, when x < - 1 and x > 1, f is increasing.

And, in interval ( - 1,1) when - 1< x < 1 then f’(x) < 0.

Therefore, when - 1 < x < 1, f is decreasing.

Question 9.

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

Answer:

It is given that ellipse 

Let the major axis be along the x – axis.1).

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

Since, the ellipse is symmetrical w.r.t. x - axis and y - axis, we can assume the coordinates of A to be ( - x₁,y₁) and the coordinates of B to be ( - x₁, - y₁).

Now, we have y1 = ± 

Therefore, Coordinates of A  and the coordinates of B

As the point(x₁,y₁) lies on the ellipse, the area of triangle ABC (A) is given by:

A = 

……..(1)

Now, 

But, x₁ cannot be equal to a.

⇒ x₁ = 

y₁ = 

Now, 

Also, when x₁ = , then,

 < 0

Then, the area is the maximum when x₁ = .

Therefore, Maximum area of the triangle is given by:

A = 

Question 10.

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 be the length, breadth and height of the tank respectively.

then, we have h = 2m

Volume of the tank = 8 m³

Volume of the tank = l × b × h

⇒ 8 = l × b × 2

⇒ lb = 4

⇒ b = 

Now, area of the base = lb = 4

Area of the 4 walls (A) = 2h(l + b)

Now, 

⇒  = 0

⇒ l² = 4

⇒ l = ±2

Since, length cannot be negative therefore l =2.

⇒ b = 2

Now, 

When l =2, 

Then, by second derivative test, the area is the minimum when l =2.

We have, l =b=h=2

Therefore, Cost of building the base = Rs 70 × (lb) = Rs 70 (4) = Rs 280.

Cost of building the walls = Rs 2h (l + b) × 45 = Rs 90(2)(2 + 2)

= Rs 8(90) = Rs 720.

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

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

Question 11.

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, 2πr + 4a = k (where k is constant)

The sum of the areas of the circle and square (A) is

= πr² + a² = πr² + 

Now, 

8r = k - 2πr

⇒ (8 + 2π)r = k

⇒ r = 

Now,  > 0

Therefore, When r = , > 0

⇒ The sum of the area is least when r = 

So, when r = 

Then a = 

Therefore, it is proved that the sum of their areas is least when the side of the square is double the radius of the circle.

Question 12.

A window is in the form of a rectangle surmounted by a semi - circular 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 semi - circular opening = 

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

⇒ x + 2y + 

Therefore, Area of the window (A) is given by

= 

Now,  , then

=0

Then, when x =  then  < 0.

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

x =  m.

Now, y = 

Therefore, the required dimensions of the window to admit maximum light is given by length =  m and breadth =m.

Question 13.

A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle.

Show that the maximum 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 .

Now, we have,

Ac = 

Now, PC = b cosecθ

And AP = a secθ

⇒ AC = AP + PC

⇒ AC = a secθ + b cosecθ

Now, if 

⇒ asecθ tanθ = bcosecθ cotθ

⇒ 

⇒ asin³θ =bcos³θ

⇒ 

⇒ 

………..(1)

So, it is clear that  < 0 when 

Therefore, by second derivative test, the length of the hypotenuse is the maximum when 

Now, when , we get,

Ac = 

Therefore, the maximum length of the hypotenuses is .

Question 14.

Find the points at which the function f given by f (x) = (x – 2)⁴ (x + 1)³ has

(i) local maxima (ii) local minima

(iii) point of inflexion

Answer:

It is given that function is f (x) = (x – 2)⁴ (x + 1)³

⇒ f’(x) = 4(x - 2)³ (x + 1)³ + 3(x + 1)²(x - 2)⁴

=(x - 2)³(x + 1)²[4(x + 1) + 3 (x - 2)]

=(x - 2)³(x + 1)²(7x - 2)

Now, f’(x) =0

⇒ x = - 1 and x =  or x = 2

Now, for values of x close to  and to the left of 

f’(x) > 0.

Also, for values of x close to  and to the right of , f’(x) < 0.

Then, x =  is the point of local maxima.

Now, for values of x close to 2 and to the left of 2, f’(x) < 0.

Also, for values of x close to 2 and to the right of 2. f’(x) > 0.

Then, x = 2 is the point of local minima.

Now, as the value of x varies through - 1, f’(x) does not changes its sign.

Then, x = - 1 is the point of inflexion.

Question 15.

Find the absolute maximum and minimum values of the function f given by

f (x) = cos² x + sin x, x ∈ [0, π]

Answer:

It is given that f (x) = cos² x + sin x, x ∈ [0, π]

f’(x) = 2cosx( - sinx) + cosx

= - 2sinxcosx + cosx

Now, if f’(x) = 0

⇒ 2sinxcosx = cosx

⇒ cosx(2sinx - 1)=0

⇒ sin x =  or cosx = 0

⇒ x = 

Now, evaluating the value of f at critical points x = and x = and at the end points of the interval [0,π], (ie, at x = 0 and x =π), we get,

f

f(0)=

f(π)=cos²π + sinπ = ( - 1)² + 0 =1

f

Therefore, the absolute maximum value of f is  occurring at x =  and the absolute minimum value of f is 1 occuring at x =1,  and π.

Question 16.

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

Answer:

Let R and h be the radius and the height of the cone respectively.

The volume (V) of the cone is given by;

V = 

Now, from the right triangle BCD, we get,

BC = 

V = 

Now, if , then,

Now, 

Now, when , it can be shown that  < 0.

Therefore, the volume is the maximum when .

When,

Height of the cone = r + .

Therefore, it can be seen that the altitude of the circular cone of maximum volume that can be inscribed in a sphere of radius r is.

Question 17.

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:

Since, f’(x) > 0 on (a,b)

Then, f is a differentiating function (a,b)

Also, every differentiable function is continuous,

Therefore, f is continuous on [a,b]

Let x1, x2 ϵ (a,b) and x2 > x1 then by LMV theorem, there exists c ϵ (a,b) s.t.

f’(c) = 

Therefore, f is an increasing function.

Question 18.

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:

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

Now, h = 

The volume (V) of the cylinder is given by:

V = πr²h =2 πr²

Now, if 

Now, 

Now, we can see that at, ,  < 0.

Therefore, the volume is the maximum when.

When , the height of the cylinder is .

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

Hence Proved.

Question 19.

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 is 

Answer:

The figure is given below:

Let VAB be a given cone of height h, semi-vertical angle α and let x b ethe radius of the base of the cylinder A'B'DC which is inscribed in the cone VAB.
In triangle VO'A',

VO' = x cotα
OO' = VO - VO' = h - x cotα
Let V be the volume of the cylinder. Then,
V = π(O'B')² (OO')
V = πx²(h - x cotα)
Differentiating with respect to x, we get, 

Now, putting dV/dx = 0, for maxima or minima, we get,

Putting the value of x, we get,

Therefore, there is maxima at x = 2h/3 tanα
Hence, putting the value of x, in formula of volume, we get,

Hence, Proved.

Question 20.

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:

Let V be the volume of the cylinder

V = πr²h

= π(10)²h

⇒ V =100πh

Differentiating w.r.t. t we get,

The tank is being filled with wheat at the rate of 314 cubic meter per hour.

Then, we have,

314 = 

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

Question 21.

The slope of the tangent to the curve x = t² + 3t – 8, y = 2t² – 2t – 5 at the point (2,– 1) is
A. 

B. 

C. 

D. 

Answer:

It is given that curve x = t² + 3t – 8, y = 2t² – 2t – 5

Then, 

The given points is (2, - 1)

At x = 2, we get

t² + 3t – 8 = 2

⇒ t² + 3t – 10 = 0

⇒ (t – 2)(t + 5) = 0

⇒ t = 2 and - 5

At y = - 1, we get

2t² - 2t – 5 = - 1

⇒ 2t² - 2t – 4 = 0

⇒ 2(t² - t – 2) = 0

⇒ (t – 2)(t + 1) = 0

⇒ t = 2 and - 1

Therefore, the common value of t is 2.

Hence, the slope of the tangent to the given curve at point (2, - 1) is

Question 22.

The line y = mx + 1 is a tangent to the curve y² = 4x if the value of m is
A. 1

B. 2

C. 3

D. 

Answer:

It is given that the equation of the tangent to the given curve is

y = mx + 1

Now, substituting the value of y in y² = 4x, we get

⇒ (mx + 1)² = 4x

⇒ m²x² + 1 + 2mx - 4x =0

⇒ m²x² + x(2m - 4) + 1 = 0………………..(1)

Since, a tangent touches the curve at one point, the root of equation (1) must be equal.

Thus, we get

Discriminant = 0

(2m - 4)² – 4(m²)(1) = 0

⇒ 4m² + 16 - 16m - 4m² =0

⇒ 16 – 16m = 0

⇒ m =1

Therefore, the required value of m is 1.

Question 23.

The normal at the point (1,1) on the curve 2y + x² = 3 is
A. x + y = 0

B. x – y = 0

C. x + y + 1 = 0

D. x – y = 1

Answer:

It is given that the equation of curve is 2y + x² = 3

Differentiating w.r.t. x, we get,

The slope of the normal to the given curve at point (1,1) is

Then, the equation of the normal to the curve at (1,1) is

⇒ y – 1 =1(x - 1)

⇒ y - 1 = x – 1

⇒ x – y = 0

Question 24.

The normal to the curve x² = 4y passing (1,2) is
A. x + y = 3

B. x – y = 3

C. x + y = 1

D. x – y = 1

Answer:

It is given that the equation of curve is x² = 4y

Differentiating w.r.t. x, we get,

The slope of the normal to the given curve at point (h,k) is

Then, the equation of the normal to the curve at (h,k) is

⇒ y – k =

Now, it is given that the normal passes through the point (1,2)

Thus, we get,

⇒ 2 – k =

⇒ k = ………………(1)

Since (h,k) lies on the curve x² = 4y, we have h² = 4k

⇒ k = 

Now putting the value of of k in (1), we get

⇒ h³ = 8

⇒ h = 2

Therefore, the equation of the normal is given as:

⇒ y – 1 = 

⇒ y - 1 = - (x - 2)

⇒ x + y = 3

Question 25.

The points on the curve 9y² = x³, where the normal to the curve makes equal intercepts with the axes are
A. 

B. 

C. 

D. 

Answer:

It is given that the equation of the curve is 9y² = x³

Differentiating w.r.t. x, we get,

The slope of the normal to the given curve at point (x₁,y₁) is

Then, the equation of the normal to the curve at (x₁,y₁)is

⇒ y – y₁ =

⇒ 

⇒ 

⇒ 

⇒ 

Now, it is given that the normal makes equal intercepts with the axes.

Thus, we get,

⇒ 

⇒ 

⇒ …………………….(1)

Since, the point (x₁,y₁)lies on the curve …………(2)

From (1) and (2), we get

Now putting the value of x₁ in (2), we get

Therefore, the required points are