Math Assignment Class XII Ch -06 | Application of Derivatives

Math Assignment Class XII Ch - 06

Application of Derivatives

Extra questions of chapter 06 Applications of Derivatives, class XII with answers and hints to the difficult questions, strictly according to the CBSE Board . Important and useful math. assignment for the students of class XII

MATHEMATICS ASSIGNMENT OF CHAPTER 06

STRICTLY ACCORDING TO THE PREVIOUS CBSE SAMPLE QUESTION PAPERS AND CBSE BOARD PAPERS

RATE OF CHANGE OF QUANTITIES

Question 1

A particle moves along the curve x² = 2y. At what point, ordinate increases at the same rate as abscissa increases?

Answer (1, 1/2)

Solution Hint:

Differentiating the given equation w. r. t. t

Now putting dy/dt = dx/dt and find the value of x.

Putting the value of x in the given equation and find the value of y

Question 2

The side of an equilateral triangle is increasing at the rate of 2 cm/s. At what rate is its area increasing when the side of triangle is 20 cm?

Answer: 20√3 cm²/s

Question 3

Volume of sphere is increasing at the rate of 3cm³/s. Find the rate of increase of surface area, when radius is 2 cm

Answer: 3cm²/s

Question 4

For the curve y = 5x – 2x³. If x increases at the rate of 2 unit/sec., find the rate of change of the slope of the curve when x = 3.

Answer: -72 unit /sec

Question 5

A man 2m high walks at a uniform speed of 6km/h away from the lamp post 6m high. Find the rate at which the length of his shadow increases.

Answer : 3km/h

Solution Hint

AB = 6m, CD = 2m dy / dt = 6 km/h,

Let AC = y, CE = x, AE = x + y,

To find : dx/dt = ?

△ABE ~ △ CDE
∴ $\frac{AB}{AE}=\frac{CD}{CE}$
$\frac{6}{x+y}=\frac{2}{x}$
⇒ y = 2x
Differentiating on both side w.r.t. t we get

$\frac{dy}{dt}=2\frac{dx}{dt}$

$6=2\frac{dx}{dt}\Rightarrow \frac{dx}{dt}=3km/h$

Case Study Based Question
Question 17: (CBSE Sample Paper 2024 set-1)
The traffic police has installed Over Speed Violation Detection (OSVD) system at various locations in a city. These cameras can capture a speeding vehicle from a distance of 300 m and even function in the dark.
A camera is installed on a pole at the height of 5 m. It detects a car travelling away from the pole at the speed of 20 m/s. At any point, x m away from the base of the pole, the angle of elevation of the speed camera from the car C is θ.
On the basis of the above information, answer the following questions :
(i) Express θ in terms of height of the camera installed on the pole and x.
(ii) Find d θ / dx .
(iii) (a) Find the rate of change of angle of elevation with respect to time at an instant when the car is 50 m away from the pole.
OR
(iii) (b) If the rate of change of angle of elevation with respect to time of another car at a distance of 50 m from the base of the pole is 3/101 rad/s, then find the speed of the car.
Solution
(i) tan θ = 5/x  ⇒ θ = tan^-1(5/x)
(ii) $\frac{d\theta}{dx}=-\frac{5}{5^{2}+x^{2}}$
(iii) (a)    $\frac{d\theta}{dt}=\frac{d\theta}{dx}\times\frac{dx}{dt}=-\frac{5}{5^{2}+x^{2}}\times 20$
   $\left(\frac{d\theta}{dx}\right)_{x=50}=-\frac{100}{2525}=-\frac{4}{101}\;rad/sec$
(iii) (b) $\frac{d\theta}{dt}=\frac{d\theta}{dx}\times\frac{dx}{dt}$

$\frac{3}{101}=\frac{-5}{5^{2}+x^{2}}\times\frac{dx}{dt}$

At x = 50

$\frac{3}{101}=\frac{-5}{2525}\times\frac{dx}{dt}$

$\frac{dx}{dt}=-15m/sec$

Speed of car = 15 m/sec

INCREASING & DECREASING OF FUNCTIONS

Question 1

Let f(x) be a continuous function on [a, b] and differentiable on (a, b). Then, this function f(x) is strictly increasing in (a, b) if

(A) f’(x) < 0, ∀ x ∈ (a, b)

(B) f’(x) > 0, ∀ x ∈ (a, b)

(C) f’(x) = 0, ∀ x ∈ (a, b)

(D) f(x) > 0, ∀ x ∈ (a, b)

Answer B

Question 2

Find the interval in which the function f(x) = x²- 4x - 12 is (i) Strictly Increasing (ii) Strictly Decreasing

Question 3

Find the intervals in which the function f(x) = (x⁴/ 4) – x³ – 5x² + 24x + 12 is

i) Strictly increasing ii) Strictly decreasing

Answer:

Critical points are : -3, 2, 4

Strictly increasing in (-3, 2)⋃(4, ∞).

Strictly Decreasing in (-∞, -3)⋃(2, 4)

Question 4

Find the interval in which the function:

f(x) = sin⁴x + cos⁴x, 0 ≤ x < π/2

is strictly increasing or strictly decreasing.

Answer:

Strictly increasing in (π/4, π/2)

Strictly decreasing in (0, π/4)

Question 5

Find the intervals in which f(x) = sin3x - cos3x, 0 < x < π, is strictly increasing or strictly decreasing.

Answer:

Strictly increasing in (0, π/4) ⋃ (7π/12, 11π/12)

Strictly decreasing in (π/4, 7π/12) ⋃ (11π/12, π)

Question 6

Prove that f(x) = x² -x + 1 is neither increasing nor decreasing strictly on (-1, 1)

Question 7

Find the interval in which the following functions are strictly increasing and strictly decreasing : f(x) = 4x³ – 6x² - 72x + 30

Answer:

Strictly increasing in (-∞, -2)⋃(3, ∞).

Strictly Decreasing in (-2, 3)

Question 8

Find the interval in which the following functions are strictly increasing and strictly decreasing : f(x) = 2x³ – 12x² + 18x + 5

Answer:

Strictly increasing in (-∞, 1)⋃(3, ∞).

Strictly Decreasing in (1, 3)

Question 9:

Find the interval in which the following functions are strictly increasing and strictly decreasing : $f(x)=\frac{4x^{2}+1}{x}$

Answer:

Strictly increasing in $\left ( -\infty ,-\frac{1}{2} \right )\cup \left ( \frac{1}{2},\infty \right )$

Strictly Decreasing in $\left ( -\frac{1}{2},\frac{1}{2} \right )$

Question 10

Find the intervals in which the function given by f(x) = sin3x, x∈ [0, π/2] is increasing and decreasing.

Answer

Increasing on [0, π/6]

Decreasing on [π/6, π/2]

Question 11 Show that f(x) = e^x – e^–x + x – tan^–1 x is strictly increasing in its domain.
Solution
$f'(x)=e^{x}+e^{-x}+1-\frac{1}{1+x^{2}}$
$f'(x)=e^{x}+\frac{1}{e^{x}}+\frac{x^{2}}{1+x^{2}}>0\;\;for\;all\;x\epsilon R$
∴ f is strictly increasing over its domain RQuestion 12: Find the intervals in which the function $f(x)=\frac{logx}{x}$ is strictly increasing or strictly decreasing.

(A) strictly decreasing on R

(B) strictly increasing on R

(C) neither strictly increasing nor strictly decreasing on R

Solution: $f'(x)=\frac{1-logx}{x^{2}}$
For strictly increasing / decreasing, put f’(x) = 0 ⇒ x = e, x > 0

For strictly increasing, x ∈ (0, e ) and for strictly decreasing x ∈ (e, ∞ )
Question 13: Find the absolute maximum and absolute minimum values of the function f given by $f(x)=\frac{x}{2}+\frac{2}{x}$ , on the interval [1, 2].
Answer: Absolute maximum value = 5/2
Absolute minimum value = 2
Question 14:
Show that the function $\mathbf{ f(x)=\frac{16sinx}{4+cosx}-x}$ , is strictly decreasing in $\left(\frac{\pi}{2},\pi\right)$ .
Solution Hint
Question 15:

The function f(x) = x³ – 3x² + 12x – 18 is :

(A) strictly decreasing on R

(B) strictly increasing on R

(C) neither strictly increasing nor strictly decreasing on R

(D) strictly decreasing on (– ∞, 0)

Ans: (B)

Solution Hint:

Find f'(x) = 3(x² - 2x + 4)

= 3[(x-1)² + 3] ≥ 3, for all x ∊ R

⇒ f'(x) > 0, for all x ∊ R

⇒ f(x) is strictly increasing for all x ∊ R

MAXIMA AND MINIMA

Question 1

If $f(x)=\frac{1}{4x^{2}+2x+1}$ , then find the maximum value of f(x)

Ans: 4/3

Solution Hint:

f(x) is maximum if 4x² + 2x + 1 is minimum

Let g(x) = 4x² + 2x + 1

g'(x) = 8x + 2

For critical point g'(x) =0 ⇒ x = -1/4

g''(x) = 8 > 0

⇒ g(x) is minimum at x = -1/4

⇒ f(x) is maximum at x = -1/4

Maximum value of f(x) = 4/3 at x = -1/4

Question 2

Find the maximum value of $f(x)=\frac{logx}{x}$

Ans: 1/e

Solution Hint

Find the derivative of f(x) we get $f'(x)=\frac{1-logx}{x^{2}}$

f ' (x) = 0 ⇒ x = 0 and e (critical points)

But at x = 0 logx is not defined so we have x = e only

Now find f '' (x) we get

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

At x = e, f ''(x) < 0 ⇒ f(x) is maximum at x = e

Max. value of f(x) = f(e) = 1/e (∵ Loge = 1)

Question 3

Find the least value of f(x) = e^x + e^-x

Answer: 2

Solution Hint

Find f'(x) and find critical point.

x = 0 is the critical point.

f(x) is minimum at x = 0

Question 4

Without using the derivatives, find the maximum and minimum values if any for the function f(x) = sin2x + 5.

Answer: Max. value = 6, Min Value = 4

Question 5

Of all the rectangles each of which has perimeter 40 meters, find one which has maximum area. Also find the maximum area?

Solution Hint:

Let sides of rectangle = x and y

Perimeter = 2(x + y)

ATQ : 2(x + y) = 40 ⇒ x + y = 20 ⇒ y = 20 - x

A = Area = xy = x(20 - x) ⇒ 20x - x²

A' = 20 - 2x

A' = 0 ⇒ 20 - 2x = 0 ⇒ x = 10

A'' = -2 < 0 ⇒ A is maximum at x = 10

If x = 10 then y = 10 ⇒ Rectangle is a square

Maximum area = xy = 10 x 10 = 100 sq m

Question 6

At what point, the slope of the curve y = - x³ + 3x² + 9x - 27 is maximum? Also find the maximum slope.

Answer: Maximum slope = m = 12 at x = 1

Solution Hint

y = - x³ + 3x² + 9x - 27

Slope = m = dy/dx = -3x² + 6x + 9

Now using second derivative test to find the maxima and minima.

Question 7 Show that the function f(x) = 4x³ – 18x² + 27x – 7 has neither maxima nor minima.

Solution: f ′ (x) = 12x² − 36x + 27

= 3 (2 x -3)² ≥ 0 for all x ∈ R

∴ f is increasing on R .

Hence f(x) does not have maxima or minima.

Question 8: It is given that function f(x) = x⁴ – 62x² + ax + 9 attains local maximum value at x = 1. Find the value of ‘a’, hence obtain all other points where the given function f(x) attains local maximum or local minimum values.

Solution: f’(x) = 4x³ – 124x + a

As at x = 1, 'f' attains local maximum value, f’(1) = 0 ⇒ a = 120
Now, f’(x) = 4x³ – 124x + 120
= 4(x - 1)(x² + x - 30)
= 4(x - 1)(x - 5)(x + 6)
Critical Points are x = - 6, 1, 5
f’’(x) = 12x² – 124
f’’(-6) > 0, f’’(5) > 0 and f’’(1) < 0
So f attains local maximum value at x = 1 and local minimum value at x = - 6, 5
Question 9:
The perimeter of a rectangular metallic sheet is 300 cm. It is rolled along one of its sides to form a cylinder. Find the dimensions of the rectangular sheet so that volume of cylinder so formed is maximum.
Solution: Let length of rectangle be x cm and breadth be (150 – x) cm.
Let r be the radius of cyllander
Circumference of base of cyllander = 2πr = x
$\Rightarrow\;r=\frac{x}{2\pi}$

Volume of cyllinder = πr²h =

$=\pi\times\left(\frac{x}{2\pi}\right)^{2}(150-x)$
$=\frac{75x^{2}}{2\pi}-\frac{x^{3}}{4\pi}$

Differentiating w r t x

$\frac{dV}{dx}=\frac{150x}{2\pi}-\frac{6x}{4\pi}$

$Let\;\;\frac{dV}{dx}=0\Rightarrow x=100\;cm$

$\frac{\partial^2 V}{\partial x^2}=\frac{150}{2\pi}-\frac{6x}{4\pi}=\frac{150-3x}{2\pi}$

At x = 100 cm

$\frac{\partial^2 V}{\partial x^2}=-\frac{75}{\pi}<0$

⇒ V is maximum when x = 100 cm and breadth of rectangle is 50 cm

Question 10:

Show that the function f(x) = 4x³ – 18x² + 27x – 7 has neither maxima nor minima.

Solution

f ′ (x) = 12x² − 36x + 27

= 3 (2 x -3)² ≥ 0 for all x ∈ R

Also x = 3/2 is the critical point, and at x = 3/2 f'(x) does not change its sign as we move from left to right at x = 3/2

Therefore f(x) neither have maximum value nor have minimum value. It is called the point of inflection.

Hence f(x) does not have maxima or minima.

Question 11:

If M and m denote the local maximum and local minimum values of the function

$f(x)=x+\frac{1}{x}\;\;(x\neq 0)$ respectively, find the value of (M – m).

Solution Hint

Find f'(x) and then find critical points -1 and 1 by putting f'(x) = 0.

Find f''(x)

f''(-1) = -2 < 0 ⇒ f(x) is maximum at x = -1

Local Maximum value = f(-1) = -2 = M

f''(1) = 2 > 0 ⇒ f(x) is mimimum at x = 1

Local Minimum value = f(1) = 2 = m

M - m = -2 -2 = - 4

Search This Blog

Featured Posts

Mathematics Class 09 Lab Manual | 17 Lab Activities