NCERT Grade 12-Application of Derivatives-Answers

NCERT Solutions for Class 12 Maths

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1.    Let x denote the area of the circle of variable radius r.

∵ Area of circle (x) = πr2

∴ Rate of change of area x w.r.t. r

(a) When r = 3 cm, then

(b) When r = 4 cm, then

2.    Let x cm be the edge of the cube and y be the surface area of the cube at any time t

Given: Rate of increase of volume of cube = 8 cm3/sec

Let y be the surface area of the cube, i.e., y = 6x2

Since  is positive, therefore surface area is increasing at the rate of  cm2/sec.

3.    Let r cm be the radius and A be the area of the circle at any time t.

Rate of increase of radius of circle = 3 cm/sec

Since  is positive, therefore surface area is increasing at the rate of 60π cm2/sec.

4.    Let x cm be the edge and y be the volume of the variable cube at any time t.

Rate of increase of edge = 3 cm/sec

Since  is positive, therefore volume of cube is increasing at the rate of 900 cm3/sec.

5.    Let x cm be the radius and y be the enclosed area of the circular wave at any time t.

Rate of increase of radius of circular wave = 5 cm/sec

Since  is positive, therefore area of circular wave is increasing at the rate of 80π cm2/sec.

6.    Let x cm be the radius and y be the circumference of the circle at any time t.

Rate of increase of radius of circle = 0.7 cm/sec

 is positive and = 0.7 cm/sec

 = 0.7 cm/sec   …(i)

∴ Rate of change of circumference of circle 

= 1.4 π cm/sec.

7.    Given: Rate of decrease of length x of rectangle is 5 cm/minute.

 is negative = –5 cm/minute

Also, Rate of increase of width y of rectangle is 4 cm/minute

 is positive

= 4 cm/minute

(a) Let z denotes the perimeter of rectangle.

∴ Perimeter of the rectangle is decreasing at the rate of 2 cm/min.

(b) Let z denotes the area of rectangle at any time t.

∴ Area of the rectangle is increasing at the rate of 2cm2/min

8.    Let x cm be the radius of the spherical balloon at time t.

∴ Radius of balloon is increasing at the rate of  cm sec.

9.    

Therefore, the volume is increasing at the rate of 400π cm3/sec.

10.  Let AB be the ladder and C is the junction of wall and ground, AB = 5 m

Let CA = x meters, CB = y meters

It is given that the bottom of the ladder is pulled along the ground away from the wall at the rate of 2 cm/sec

So x increases and y decreases

11.   Given: Equation of the curve 6y = x3 + 2  ……….(i)

Let (x, y) be the required point on curve (i)

Taking x = 4, 6y = 64 + 2

⇒ y = 11

∴ Required point is (4, 11).

Taking x = -4, 6y = -64 + 2

12.  Let x cm be the radius of the air bubble at time t

Therefore, required rate of increase of volume of air bubble is 2π cm3/sec.

13.  

14.  Let the height and radius of the sand-cone formed at time t second be y cm and x cm respectively.

15.  

Therefore, required Marginal cost is Rs 20.97

16.  

Now, when x = 7, MR = 26 × 7 + 26 = 208

Therefore, the required marginal revenue is Rs 208.

17.  Area of circle (A) = πr2

Therefore, option (B) is correct.

18.  Total revenue R(x) = 3x + 36x + 5

∴ Marginal revenue 

= 6 × 15 + 36 = 126

Therefore, option (D) is correct.

19.  Given: f(x) = 3x + 17

∴ f'(x) = 3(1) + 0 = 3 > 0 i.e., positive for all x ∈ R

Therefore, f(x) is strictly increasing on R.

20.  Given: f(x) = e2x

 i.e., positive for all x ∈ R

Therefore, f(x) is strictly increasing on R.

21.  Given: f(x) = sin x

∴ f'(x) = cos x

∴ f'(x) does not have the same sign in the interval (0, π).

Therefore, f(x) is neither increasing nor decreasing in (0, π).

22.  Given: f(x) = 2x2 – 3x

23.  (a) Given: f(x) = 2x3 – 3x2 – 36x + 7

⇒ f'(x) = 6x2 – 6x – 36 = 6 (x – x – 6)

⇒ f'(x) = 6 (x + 2) (x – 3)  ….(i)

Now 6(x + 2) (x – 3) = 0

⇒ x + 2 = 0 or x – 3 = 0

x = -2 or x = 3

Therefore, we have three disjoint sub-intervals (-∞, -2), (-2, 3) and (3, ∞).

For interval (-∞, -2), taking x = -3 (say), from eq.(i),

f'(x) = (+) (-) (-) = (+) > 0

Therefore, f is strictly increasing in (-∞, -2).

For interval (-2, -3), taking x = 2 (say), from eq.(i),

f'(x) = (+) (+) (-) = (-) < 0

Therefore, f is strictly increasing in (-2, 3).

For interval (3, ∞), taking x = 4 (say), from eq.(i),

f'(x) = (+) (+) (+) = (+) > 0

Therefore, f is strictly increasing in (3, ∞).

Hence, (a) f is strictly increasing in (-∞, -2) and (3, ∞).

(b) f is strictly decreasing in (-2, 3)

24.  (a) Given: f(x) = x + 2x – 5

⇒ f'(x) = 2x + 2 = 2(x + 1)   ….(i)

Now 2(x + 1) = 0

⇒ x = -1

Therefore, we have two sub-intervals (-∞, -1) and (-1, ∞).

For interval (-∞, -1) taking x = -2 (say), from eq.(i) f'(x) = (-) < 0

Therefore, f is strictly decreasing.

For interval (-1, ∞) taking x = 0 (say), from eq.(i) f'(x) = (+) > 0

Therefore, f is strictly increasing.

(b) Given: f(x) = 10 – 6x – 2x2

f'(x) = (+) (-) (-) = (+) > 0

Therefore, f is strictly increasing.

For interval  taking x = -1 (say), from eq.(i)

f'(x) = (-) (+) = (-) < 0

Therefore, f is strictly decreasing.

(c) Given: f(x) = -2x – 9x – 12x + 1

⇒ f'(x) = -6x – 18x – 12

⇒ f'(x) = -6(x + 3x + 2)

= -6(x + 1)(x + 2) …..(i)

Now -6(x + 1)(x + 2) = 0

⇒ x = -1 or x = -2

Therefore, we have three disjoint intervals (-∞, -2), (-2, -1) and (-1, ∞).

For interval (-∞, -2), from eq.(i),

f'(x) = (-) (-) (-) = (-) < 0

Therefore, f is strictly decreasing.

For interval (-2, -1), from eq. (i),

f'(x) = (-) (-) (+) = (+) > 0

Therefore, f is strictly increasing.

For interval (-1, ∞), from eq. (i),

f'(x) = (-) (+) (+) = (-) < 0

Therefore, f is strictly decreasing.

(e) Given: f(x) = (x + 1)3 (x – 3)3

⇒ f'(x) = (x + 1).3(x – 3)2 + (x – 3)3 .3(x + 1)2

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

⇒ f'(x) = 3(x + 1)(x – 3)2 (2x – 2)

⇒ f'(x) = 6(x + 1)(x – 3)2 + (x – 1)

Here, factors (x + 1) and (x – 3) are non-negative for all x

Therefore, f(x) is strictly increasing if f'(x) > 0

⇒ x – 1 > 0

⇒ x > 1

And f(x) is strictly decreasing if f'(x) < 0

⇒ x – 1 < 0

⇒ x < 1

Hence, f is strictly increasing in (1, ∞) and f is strictly decreasing in (-∞, 1).

25.  

Domain of the given function is given to be x > -1

⇒ x + 1 > 0

Also (2 + x)2 > 0 and x2 ≥ 0

∴ From eq.(i)  for all x in domain x > -1 and f is increasing function.

26.  

27.  

28.  

Therefore, f(x) is strictly increasing on (0, ∞).

29.  Given: f(x) = x2 – x + 1

⇒ f'(x) = 2x – 1

f(x) is strictly increasing if f'(x) > 0

hence, f(x) is neither strictly increasing nor decreasing on the interval (-1, 1).

30.  

31.  Given: f(x) = x100 + sin x – 1

⇒ f'(x) = 100x99 + cos x

(A) on (0, 1), x > 0 therefore 100x99 > 0

And for cos x

⇒ (0, 1 radian) = (0.57° nearly) > 0

Therefore, f(x) is strictly increasing on (0, 1).

32.  f(x) = x2 + ax + 1

⇒ f'(x) = 2x + a

Since f(x) is strictly increasing on (1, 2) therefore f'(x) = 2x + a > 0 for all x in (1, 2)

∴ On (1, 2) 1 < x < 2

⇒ 2 < 2x < 4

⇒ 2 + a < 2x + a < 4 + a

∴ Minimum value of f'(x) is 2 + a and maximum value is 4 + a.

Since f'(x) > 0 for all x in (1, 2)

∴ 2 + a > 0 and 4 + a > 0

⇒ a> -2 and a > -4

Therefore leas value of a is -2.

33.  

34.  

35.  

36.  Give: f(x) = x3 – 3x2 + 3x – 100

⇒ f'(x) = 3x2 – 6x + 3 = 3(x2 – 2x + 1)

⇒ f'(x) = 3(x – 1)2 ≥ 0 for all x in R.

Therefore, f(x) is increasing on R.

37.  

38.  Given: Equation of the curve y = 3x – 4x    …(i)

Slope of the tangent to the curve = value of  at the point (x, y)

∴ Slope of the tangent at point x = 4 to the curve (i)

= 12(4)2 – 4 = 768 – 4 = 764.

39.  

40.  Given: Equation of the curve y = x3 – x + 1   ……….(i)

∴ Slope of the tangent at point x = 2 to the curve (i)

= 3(2)2 – 1 = 12 – 1 = 11.

41.  Given: Equation of the curve y = x3 – 3x + 2   ……….(i)

∴ Slope of the tangent at point x = 3 to the curve (i)

= 3(3)2 – 3 = 27 – 3 = 24

42.  Given: Equation of the curves are x = a cos3 θ, y = a sin3 θ

43.  Given: Equation of the curves are x = 1 – a sin θ and y = b cos3 θ

44.  Given: Equation of the curve y = x3 – 3x2 – 9x + 7   …(i)

Therefore, the required points are (3, -20) and (-1, 12).

45.  Let the given points are A(2, 0) and B(4, 4).

Slope of the chord AB = 

Equation of the curve is y = (x – 2)2

∴ Slope of the tangent at (x, y)

If the tangent is parallel to the chord AB, then Slope of tangent = Slope of chord

⇒ 2(x – 2) = 2

⇒ x = 3

∴ y = (3 – 2)2 = 1

Therefore, the required point is (3, 1).

46.  Given: Equation of the curve y = x3 – 11x + 5    ….(i)

Equation of the tangent y = x – 11   ……….(ii)

⇒ x – y – 11 = 0

From eq. (i), 

= Slope of the tangent at (x, y)

But from eq. (ii), the slope of tangent = 

∴ 3x2 – 11 = 1

⇒ x2 = 4

⇒ x = ±2

From eq. (i), when x = 2, y = 8 – 22 + 5 = -9

And when x = -2, y = -8 + 22 + 5 = 19

Since (-2, 19) does not satisfy eq. (ii), therefore the required point is (2, -9).

47.  

∴ Points of contact are (2, 1) and (0, -1).

And Equation of two tangents are y – 1 = -1 (x – 2)

= x + y – 3 = 0 and

y – (-1) = -1(x – 0) = x + y + 1 = 0.

48.  

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

49.  

50.  

(i) If tangent is parallel to x-axis, then slope of tangent = 0

Therefore, the points on curve (i) where tangents are parallel to x-axis are (0, ±4).

(ii) If the tangent parallel to y-axis.

⇒ Slope of the tangent = ±∞

Therefore, the points on curve (i) where tangents are parallel to y-axis are (±3, 0).

51.  (i) Equation of the curve y = x4 – 6x3 + 13x2 – 10x + 5

(ii) Equation of the curve y = x4 – 6x3 + 13x2 – 10x + 5

(iii) Equation of the curve y = x3 ….(i)

(iv) Equation of the curve y = x2   ….(i)

= 2 × 0 = 0 = m (say)

∴ Equation of the tangent at (0, 0) is y – 0 = 0(x – 0)

⇒ y = 0 which is equation of x-axis

thus normal at (0, 0) is y-axis.

∴ equation of normal is x = 0.

(v) Equation of the curves are x = cos t, y = sin t

52.  Given: Equation of the curve y = x2 – 2x + 7    …(i)

∴ Slope of tangent 

(a) Slope of the live 

∴ Slope of tangent parallel to this line is also = 2

∴ From eq.(ii), 2x – 2 = 2

⇒ x = 2

∴ From eq.(ii), y = 4 – 4 + 7 = 7

Therefore, point of contact is (2, 7).

∴ Equation of the tangent at *2, 7) is y – 7 = 2(x – 2)

⇒ y – 7 = 2x – 4

⇒ y – 2x – 3 = 0

53.  Given: Equation of the curve y = 7x3 + 11

∴ Slope of tangent at 

At the point x = 2, Slope of the tangent = 21(2)2 = 21 × 4 = 84

At the point x = -2, Slope of the tangent = 21(-2)2 = 21 × 4 = 84

Since, the slopes of the two tangents are equal.

Therefore, tangents at x = 2 and x = -2 are parallel.

54.  Given: Equation of the curve y = x3   ………(i)

∴ Slope of tangent at (x, y)

According to question, Slope of the tangent = y-coordinate of the point

∴ 3x2 = x3

⇒ 3x2 – x = 0

⇒ x2(3 – x) = 0

⇒ x2 = 0 or 3 – x = 0

⇒ x = 0 or x = 3

∴ From eq. (i), at x = 0, y = 0 The point is (0, 0).

And From eq. (i), at x = 3, y = 27 The point is (3, 27).

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

55.  Given: Equation of the curve y = 4x3 – 2x5   ……….(i)

∴ Slope of the tangent at (x, y) passing through origin (0, 0)

Substituting this value of y in eq.(i), we get,

∴ From eq. (i), at x = 0, y = 0

From eq. (i), at x = 1, y = 4 – 2 = 2

From eq. (i), at x = -1, y = -4 + 2 = -2

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

56.  Equation of the curve x2 + y2 – 2x – 3 = 0    …(i)

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

57.  Given: Equation of the curve ay2 = x3    …(i)

58.  Given: Equation of the curve y = x3 + 2x + 6    …(i)

∴ Slope of the tangent at (x, y)

∴ From eq.(i), at x = 2, y = 8 + 4 + 6 = 18

at x = -2, y = -8 – 4 + 6 = -6

Therefore, the points of contact are (2, 18) and (-2, -6).

59.  Given: Equation of the parabola y2 = 4ax ……….(i)

Differentiating eq.(i) w.r.t x

60.  Given: Equations of the curves are x = y2    …(i) and xy = k  …(iii)

Substituting the value of x in eq.(ii), we get y2.y = k

61.  

62.  

63.  Given: Equation of the curve y = 2x2 + 3 sin x   ……….(i)

∴ Slope of the tangent at point (x, y) is 

∴ Slope of the tangent at x = 0, 4(0) + 3 cos 0 = 3 = m(say)

∴ Slope of the normal 

Therefore, option (D) if correct.

64.  Given: Equation of the curve y2 = 4x    ….(i)

Therefore, required point is (1, 2).

Therefore, option (A) is correct.

65.  

66.  Let f(x) = y = 4x + 5x + 2   …(i)

⇒ dy = (8x + 5)dx

= (8x + 5)Δx   ….(ii)

Changing x  to x + Δx and y to y + Δy in eq.(i)

y + Δy = f(x + Δx)

= f(2.01) = f(2 + 0.01)   ….(iii)

Here, x = 2 and Δx = 0.01

∴ From eq.(iii), f(2.01) = y + Δy

Since, Δx and Δy is approximately equal to dx and dy respectively.

∴ From eq.(i) and (ii), f(2.01) = (4x2 + 5x + 2) + (8x + 5)Δx

⇒ f(2.01) = 4(4) + 5(2) + 2 + (8 × 2 + 5) (0.01) = 28.21

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

67.  Let f(x) = y = x3 – 7x2 + 15   …(i)

⇒ dy = (3x2 – 14x)dx = (3x2 – 14x)Δx     ….(ii)

Changing x  to x + Δx and y to y + Δy in eq.(i)

y + Δy = f(x + Δx)

= f(5.001) = f(5 + 0.001)   ….(iii)

Here, x = 5 and Δx = 0.001

∴ From eq.(iii), f(5.001) = y + Δy

Since, Δx and Δy is approximately equal to dx and dy respectively.

∴ From eq.(i) and (ii),

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

⇒ f(5.001) = 125 – 175 + 15 + (75 – 70) (0.001)

= -35 + 0.005 = -34.995

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

68.  

69.  

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

Then, according to the question, r = 7 m and Δr = 0.02 m

∴ Approximate error in calculating the volume = Approximate value of ΔV

Therefore, the approximate error in calculating volume is 12.32 m3.

71.  Let r be the radius of the sphere.

∴ Surface area of the sphere (S) = 4πr2

According to question r = 9m and Δr = 0.3 m

ds = 8π(9) (0.3)

= 2.16 π square meters.

72.  Let f(x) = y = 3x2 + 15x + 5   …(i)

⇒ dy = (6x – 15) dx = (6x + 15)Δx     ….(ii)

Changing x  to x + Δx and y to y + Δy in eq.(i)

y + Δy = f(x + Δx)

= f(3.02) = f(3 + 0.02)   ….(iii)

Here, x = 3 and Δx = 0.02

∴ From eq.(iii), f(3.02) = y + Δy

Since, Δx and Δy is approximately equal to dx and dy respectively.

∴ From eq.(i) and (ii),

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

⇒ f(3.02) = 3(9) + 15(3) + 5 + (6 × 3 + 15) (0.02)

= 77 + 0.66 = 77.66

Therefore, option (D) is correct.

73. 

74.  

75.  

76.  Let BC = b be the fixed base and AB = AC = x be the two equal sides of given isosceles triangle.

 

77.  Equation of the curve is y2 = 4x ……….(i)

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

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

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

⇒ y – 2 = -x + 1

⇒ x + y = 3

78.  The parametric equations of the curve are

79.  

80.  

Here, 3 (x4 + x2 + 1) is positive for all real x ≠ 0

∴ x + 1 = 0 or x – 1 = 0 [Turning points]

Therefore, x = -1 or x = 1 divide the real line into three dsijoint sub intervals (-∞, -1), (-1, 1) and (1, ∞)

For (-∞, -1), from eq.(i) at x = -2 (say),

f'(x) = (+) (-) (-) = (+)

Therefore, f(x) is increasing at (-∞, -1)

For (-1, 1) from eq.(i) at  (say)

f'(x) = (+) (+) (-) = (-)

Therefore, f(x) is increasing at (-1, 1)

For (1, ∞), from eq.(i) at x = 2 (say),

f'(x) = (+) (+) (+) = (+)

Therefore, f(x) is increasing at (1, ∞),

Therefore, f(x) is (i) an increasing function for x ≤ -1 and for x ≥1 and (ii) decreasing function for -1 ≤ x ≤ 1.

81.  

∴ Any point on ellipse is P (a cos θ, b sin θ).

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

∴ OM = a cos θ and PM = b sin θ

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

82.  Given: Depth of tank = 2 m

Let x m be the length and y m be the branch of the base of the tank.

Volume of tank = 8 cubic meters

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

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

45(x.2 + x.2 + y.2 + y.2)

= ` (180x + 180y)

Let z be the total cost of building the tank.

83.  Let x be the radius of the circle and y be the side of square.

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

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

84.  Let x m be the radius of the semi-circular opening of the window. Then one side of rectangle part of window is 2x and y m be the other side of rectangle.

∴ Perimeter of window

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

85.  Let P be a point on the hypotenuse AC of a right triangle ABC such that PL ⊥ AB = a and PM ⊥ BC = b and let ∠BAC = ∠MPC = θ, then in right angled ΔALP, 

86.  

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.

Therefore, 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.

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

Now as the values of x varies through -1, f'(x) does not change its sign. Therefore, x = -1 is the point of inflexion.

87.  

88.  Let x be the radius of base of cone and y be the height of the cone inscribed in a sphere of radius r.

∴ OD = AD – AO = y – r

In right angled triangle OBD,

OD2 + BD2 = OB2

⇒ (y – r)2 + x2 = r2

⇒ y2 + r2 – 2ry + x2 = r2

⇒ x2 = 2ry – y2       …(i)

89.  Let I be the interval (a, b)

Given: f'(x) > 0 for all x in an interval I. Let x1, x2 ∈ I with x1 < x2

By Lagrange’s Mean Value theorem, we have,

Now x1 < x2

⇒ x2 – x1 > 0    ….(i)

Also, f'(x) > 0 for all x in interval I

⇒ f'(c) > 0

∴ From eq.(i), f(x2) – f(x1) > 0

⇒ f(x1) < f(x2)

Thus, for every pair of points x1, x2 ∈ I with x1 < x2

⇒ f(x1) < f(x2)

Therefore, f(x) is strictly increasing in I.

90.  Let x be the radius and y be the height of the cylinder inscribed in a sphere having centre O and radius R. (x > 0, y > 0)

91.  Let r be the radius of the right circular cone of height h. Let the radius of the inscribed cylinder be x and height  y.

92.  Let y be the depth of the wheat in the cylindrical tank of radius 10 m at time t.

∴ V = Volume of wheat in cylindrical tank at time t = π(10)2 y = 100πy cu. m

Therefore, option (A) is correct.

93.  Equation of the curves are x = t2 + 3t – 8 ….(i) and y = 2t2 – 2t – 5   …..(ii)

∴ Slope of the tangent to the given curve at point 

At the given point (2, -1), x = 2 and y = -1

At x = 2, from eq.(i), 2 = t2 + 3t – 8

⇒ t2 + 3t – 10 = 0

⇒ (t + 5) (t – 2) = 0

⇒ t = -5, t = 2

At y = -1, from eq.(ii), -1 = 2t2 – 2t – 5

⇒ 2t2 – 2t – 4 = 0

⇒ t2 – t – 2 = 0

⇒ (t – 2) (t + 1) = 0

⇒ t = 2, t = -1

Here, common value of t in the two sets of values is t = 2

∴ From eq.(iii),

Slope of the tangent to the given curve at point 

Therefore, option (B) is correct.

94.  Equation of the curve is y2 = 4x    …….(i)

Therefore, option (A) is correct.

95.  Equation of the given curve is 2y + x2 = 3    ….(i)

Therefore, option (B) is correct.

96.  Equation of the curve is x2 = 4y    …(i)

Therefore, option (A) is correct.

97.  Equation of the curve is 9y2 = x3 …..(i)

Therefore, option (A) is correct.

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