### Featured Posts

### Math Assignment Class XII Ch -06 | Application of Derivatives

- Get link
- Other Apps

## 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 ****x ^{2} = 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 cm ^{2}/s**

**Question 3**

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

**Answer: 3cm ^{2}/s**

**Question 4**

**For the curve y = 5x – 2x ^{3}. 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

∴

⇒ y = 2x

Differentiating on both side w.r.t. t we get

△ABE ~ △ CDE

∴

⇒ y = 2x

Differentiating on both side w.r.t. t we get

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 Î¸.

On the basis of the above information, answer the following questions :

(i) Express Î¸ in terms of height of the camera installed on the pole and x.

(ii) Find d Î¸ / 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 Î¸ = 5/x ⇒ Î¸ = tan^{-1}(5/x)

(ii)

(iii) (a)

(iii) (b)

**At x = 50**

**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) = ****x ^{2 }- 4x - 12**

**is (i) Strictly Increasing (ii) Strictly Decreasing**

**Question 3**

**Find the intervals in which the function f(x) = (x ^{4 }/ 4) – x^{3} – 5x^{2} + 24x + 12
is**

**i) Strictly increasing ii) Strictly decreasing**

**Answer: **

**Critical points are : -3, 2, 4**

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

**Question 4**

**Find the interval in which the function:**

**f(x) = sin ^{4}x + cos^{4}x, 0 ≤ x < Ï€/2**

**is strictly increasing or strictly decreasing.**

**Answer:**

**Strictly increasing in (Ï€/4, Ï€/2)**

**Strictly decreasing in (0, Ï€/4)**

**Question 5**

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

**Answer:**

**Strictly increasing in (0, Ï€/4) ⋃ (7Ï€/12, 11Ï€/12)**

**Strictly decreasing in (Ï€/4, 7Ï€/12) ⋃ (11Ï€/12, Ï€)**

**Question 6**

**Prove that f(x) = x**

^{2}-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) = 4x**

^{3}– 6x^{2}- 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) = 2x ^{3} – 12x^{2} + 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 :**

**Answer:**

**Strictly increasing in **

**Strictly Decreasing in **

**Question 10**

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

**Answer**

**Increasing on [0, Ï€/6]**

**Decreasing on [Ï€/6, Ï€/2]**

**Question 11 Show that f(x) = e**

Solution

∴ f is strictly increasing over its domain RQuestion 12: Find the intervals in which the function is strictly increasing or strictly decreasing.

^{x}– e^{–x}+ x – tan^{–1}x is strictly increasing in its domain.Solution

∴ f is strictly increasing over its domain RQuestion 12: Find the intervals in which the function 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:

For strictly increasing / decreasing, put f’(x) = 0 ⇒ x = e, x > 0

For strictly increasing, x ∈ (0, e ) and for strictly decreasing x

**Question 13: **Find the absolute maximum and absolute minimum values of the function f given by , on the interval [1, 2].

Answer: Absolute maximum value = 5/2

Absolute minimum value = 2

**Question 14: **

Show that the function , is strictly decreasing in .

Solution Hint

**Question 15:**

**The function f(x) = x ^{3} – 3x^{2} + 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(**

**x**

^{2}- 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 , then find the maximum value of f(x)**

**Ans: 4/3**

**Solution Hint:**

**f(x) is maximum if 4x ^{2}
+ 2x + 1 is minimum**

**Let g(x) = 4x ^{2} + 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 **

**Ans: 1/e**

**Solution Hint**

**Find the derivative of f(x) we get **

**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 **

**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 - x**

^{2}**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 = - x**

^{3}+ 3x^{2}+ 9x - 27 is maximum? Also find the maximum slope.**Answer: Maximum slope = m = 12 at x = 1**

**Solution Hint**

**y = - x**

^{3}+ 3x^{2}+ 9x - 27**Slope = m = dy/dx = -3x**

^{2}+ 6x + 9**Now using second derivative test to find the maxima and minima.**

**Question 7**

**Show that the function f(x) = 4x**

^{3}– 18x^{2}+ 27x – 7 has neither maxima nor minima.Solution: f ′ (x) = 12x^{2} − 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) = x**

^{4}– 62x^{2}+ 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) = 4x**

^{3}– 124x + aAs at x = 1, 'f' attains local maximum value, f’(1) = 0 ⇒ a = 120

Now, f’(x) = 4x^{3} – 124x + 120

= 4(x - 1)(x^{2} + x - 30)

= 4(x - 1)(x - 5)(x + 6)

Critical Points are x = - 6, 1, 5

f’’(x) = 12x^{2} – 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 = 2Ï€r = x

**Volume of cyllinder = Ï€r ^{2}h = **

**Differentiating w r t x**

**At x = 100 cm**

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

**Question 10:**

**Show that the function f(x) = 4x ^{3} – 18x^{2}
+ 27x – 7 has neither maxima nor minima.**

**Solution**

**f ′ (x) = 12x ^{2} − 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**

** 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**

Thankx for your visitplease comment below

- Get link
- Other Apps

## Comments

## Post a Comment