Important Questions for Class 12 Maths Chapter 6 – Applications of Derivatives are provided here. The questions are taken as per the syllabus of the CBSE board. These important questions help the students to secure good marks in the class 12 board examination. The important questions provided here covers 1 mark, 2 marks, 4 marks, and 6 marks. Also, get the important questions for all the chapters of 12th Maths and practice more important problems at BYJU’S.
Class 12 chapter 6 – Application of Derivative covers the important concepts in Maths such as tangents and normals, rate of change, maxima and minima, increasing and decreasing functions, and some simple problems that illustrate the basic concept of derivative and its application in the real-life situations.
- Important 4 Marks Questions for CBSE Class 12 Maths
- Important 6 marks Questions for CBSE Class 12 Maths
Class 12 Chapter 6 Applications of Derivatives Important Questions with Solutions
Some of the important questions of chapter 6 – Application of Derivative class 12 Maths are provided below with step by step solutions. Students can score good marks in the final examination by practising these problems, as the below-given problems are important in the examination point of view.
For the given curve: y = 5x – 2x3, when x increases at the rate of 2 units/sec, then how fast is the slope of curve changes when x = 3?
Given that, y = 5x – 2x3
Then, the slope of the curve, dy/dx = 5-6x2
⇒d/dt [dy/dx]= -12x. dx/dt
= -72 units per second
Hence, the slope of the curve is decreasing at the rate of 72 units per second when x is increasing at the rate of 2 units per second.
Show that the function f(x) – tan x – 4x is strictly decreasing on [-π/3, π/3]
Given that, f(x) – tan x – 4x
Then, the differentiation of the function is given by:
f’(x)= sec2x – 4
When -π/3 <x π/3, 1<sec x <2
Then, 1<sec2x <4
Hence, it becomes -3 < (sec2x-4)<0
Hence, for -π/3 <x π/3, f’(x)<0
Therefore, the function “f” is strictly decreasing on [-π/3, π/3]
A stone is dropped into a quiet lake and waves move in the form of circles at a speed of 4cm/sec. At the instant, when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing?
We know that the area of a circle with radius “r” is given by A = πr2.
Hence, the rate of change of area “A’ with respect to the time “t” is given by:
dA/dt = (d/dt) πr2
By using the chain rule, we get:
(d/dr)(πr2). (dr/dt) = 2πr.(dr/dt)
It is given that, dr/dt = 4cm/sec
Therefore, when r = 10cm,
dA/dt = 2π. (10). (4)
dA.dt = 80 π
Hence, when r = 10 cm, the enclosing area is increasing at a rate of 80π cm2/sec.
What is the equation of the normal to the curve y = sin x at (0, 0)?
(a)x =0 (b) y=0 (c)x+y =0 (d)x-y=0
A correct answer is an option (c)
Given that, y = sinx
Hence, dy/dx = cosx
Thus, the slope of the normal = (-1/cosx)x =0 = -1
Therefore, the equation of the normal is y-0 = -1(x-0) or x+y=0
Hence, the correct solution is option c.
Determine all the points of local maxima and local minima of the following function: f(x) = (-¾)x4 – 8x3 – (45/2)x2 + 105
Given function: f(x) = (-¾)x4 – 8x3 – (45/2)x2 + 105
Thus, differentiate the function with respect to x, we get
f ′ (x) = –3x3 – 24x2 – 45x
Now take, -3x as common:
= – 3x (x2 + 8x + 15)
Factorise the expression inside the bracket, then we have:
= – 3x (x +5)(x+3)
f ′ (x) = 0
⇒ x = –5, x = –3, x = 0
Now, again differentiate the function:
f ″(x) = –9x2 – 48x – 45
Take -3 outside,
= –3 (3x2 + 16x + 15)
Now, substitue the value of x in the second derivative function.
f ″(0) = – 45 < 0. Hence, x = 0 is point of local maxima
f ″(–3) = 18 > 0. Hence, x = –3 is point of local minima
f ″(–5) = –30 < 0. Hence, x = –5 is point of local maxima.
A circular disc of radius 3 cm is being heated. Due to expansion, its radius increases at a rate of 0.05 cm per second. Find the rate at which its area is increasing if the radius is 3.2 cm.
Let us assume that “r” be the radius of the given disc and “A” be the area, then the area is given as:
A = πr2
By using the chain rule,
Then dA/dt = 2πr(dr/dt)
Thus, the approximate rate of increase of radius = dr = (dr/dt) ∆t = 0.05 cm per second
Hence, the approximate rate of increase in area is:
dA = (dA/dt)(∆t) = 2πr[(dr/dt) ∆t ]
= 2π (3.2) (0.05)
= 0.320π cm2 per second.
Therefore, when r= 3.2 cm, then the area is increasing at a rate of 0.320π cm2/second.
Practice Problems for Class 12 Maths Chapter 6
Solve the practice problem given below:
- Water is dropping out at a constant rate of 1 cubic cm/sec through a small hole at the vertex of the conical container, whose axis is vertical. If the slant height of water in the vessel is 4 cm, find the rate of decrease of slant height, where the vertical angle of the conical vessel is π/6.
- Show that the function f(x) = log x/x has maximum at x=e
- Determine the approximate variation in the surface area of a cube of side x metres caused by decreasing the side by 1%
- The radius of a sphere is estimated as 9 cm with an error of 0.03 cm, then calculate the approximate error in measuring its volume
- Calculate two positive numbers whose sum is 15 and the sum of whose squares is minimum.
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