Applications of Derivatives MCQ Quiz in मल्याळम - Objective Question with Answer for Applications of Derivatives - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Apr 14, 2025
Latest Applications of Derivatives MCQ Objective Questions
Top Applications of Derivatives MCQ Objective Questions
Applications of Derivatives Question 1:
It is given that at x = 2, the function x3 - 12x2 + kx - 8 attains its maximum value, on the interval [0, 3]. Find the value of k
Answer (Detailed Solution Below)
Applications of Derivatives Question 1 Detailed Solution
Concept:
Following steps to finding maxima and minima using derivatives.
- Find the derivative of the function.
- Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.
- Now we have to find the second derivative.
- f``(x) is less than 0 then the given function is said to be maxima
- If f``(x) Is greater than 0 then the function is said to be minima
Calculation:
Let f(x) = x3 - 12x2 + kx – 8
Differentiating with respect to x, we get
⇒ f’(x) = 3x2 – 24x + k
It is given that function attains its maximum value of the interval [0, 3] at x = 2
∴ f’(2) = 0
⇒ 3 × 22 – (24 × 2) + k = 0
∴ k = 36
Applications of Derivatives Question 2:
Find the interval in which the function f(x) = x2 - 4x is strictly increasing ?
Answer (Detailed Solution Below)
Applications of Derivatives Question 2 Detailed Solution
Concept:
If f′(x) > 0 at each point in an interval, then the function is said to be strictly increasing.
Calculations:
Given , f(x) = x2 - 4x
Differentiating, we get
f'(x) = 2x - 4
f(x) is strictly increasing function
∴ f'(x) > 0
⇒ 2x - 4 > 0
⇒ x > 2
∴ x ∈ (2, ∞)
Mistake Points
If a, b ∈ R and a
- Open interval is indicated by (a, b) = {x : a
- Closed interval is indicated by [a, b] = {x : a ≤ x ≤ b}, i.e. 'a' and 'b' are included.
- [a, b) = {x : a ≤ x
- (a, b ] = { x : a
Since '2' is not included here, Option (1) will not be correct.
Applications of Derivatives Question 3:
The equation of the tangent line to the curve y = 2x sin x at the point
Answer (Detailed Solution Below)
Applications of Derivatives Question 3 Detailed Solution
Concept:
The equation of tangent at point (x1 , y1) with slope m is given by
Calculations:
Given curve is y = 2x sin x
Taking derivative on both side, we get
Put x =
The equation of tangent at point (x1 , y1) with slope m is given by
y = 2x
Hence, the equation of the tangent line to the curve y = 2x sin x at the point
Applications of Derivatives Question 4:
The derivative of
Answer (Detailed Solution Below)
Applications of Derivatives Question 4 Detailed Solution
From the question, the derivative given is:
Now,
∵
From the question, let’s assume,
From the question, we need to find
Now,
Now,
Applications of Derivatives Question 5:
The maximum value of
Answer (Detailed Solution Below)
Applications of Derivatives Question 5 Detailed Solution
Concept:
Differentiation formula
Quotient rule:
Calculation:
f(x) =
⇒ f'(x) =
⇒ f′(x) = 0 at x = e
⇒ f′(x) > 0 when x
⇒ f′(x) e
So, f(x) is maxima at x = e
⇒ f(e) =
∴ The maximum value of
Applications of Derivatives Question 6:
Find the value of x for which f(x) = x - ex is an increasing function
Answer (Detailed Solution Below)
Applications of Derivatives Question 6 Detailed Solution
Concept:
- If f′(x) > 0 then the function is said to be increasing.
- If f′(x) decreasing.
Calculation:
Given:
f(x) = x - ex
Differentiating with respect to x, we get
⇒ f’(x) = 1 - ex
For increasing function,
f'(x) > 0
⇒ 1 – ex > 0
⇒ ex
⇒ ex 0
∴ x
So, x ∈ (-∞, 0)
Applications of Derivatives Question 7:
Let
Answer (Detailed Solution Below)
Applications of Derivatives Question 7 Detailed Solution
Concept:
Monotonic function: If a function is differentiable on the interval (a, b) and if the function is increasing/strictly increasing or decreasing/strictly decreasing, then the function is known as monotonic function.
First derivative test:
- If
for all x in (a, b) then, f(x) is an increasing function in (a, b). - If
for all x in (a, b) then, f(x) is an decreasing function in (a, b).
For strictly increasing or decreasing:
0\) for strictly increasing. for strictly decreasing.
Calculation:
Given:
Differentiating w.r.t x
So, for x ∈ (0, 1), f’(x)
Hence, f(x) is decreases for x ∈ (0, 1).
Applications of Derivatives Question 8:
The radius of the circle is increasing at the rate of 0.8 cm/sec. What is the rate of increase of its circumference ?
Answer (Detailed Solution Below)
Applications of Derivatives Question 8 Detailed Solution
Concept:
Circumference of circle = 2πr, where r is the radius of the circle.
Calculation:
Given:
Circumference of circle (C) = 2πr
Differentiating both the sides with respect to t, we get:
⇒
By, substituting the values of
⇒
Applications of Derivatives Question 9:
The radius of a cylinder is increasing at the rate of 3 m / s and its altitude is decreasing at the rate of 4 m / s. The rate of change of volume when radius is 4 m and altitude is 6 m is:
Answer (Detailed Solution Below)
Applications of Derivatives Question 9 Detailed Solution
CONCEPT:
- If y = f(x), then dy/dx denotes the rate of change of y with respect to x its value at x = a is denoted by:
- Decreasing rate is represented by negative sign whereas increasing rate is represented by positive sign.
- Volume of cylinder is given by: πr2h where r is the radius and h is the altitude.
CALCULATION:
Given: dr/dt = 3m/s and dh/dt = - 4m/s
As we know that, volume of cylinder is given by: πr2h where r is the radius and h is the altitude.
Let V = πr2h
Now by differentiating v with respect to t we get,
Now by substituting r = 4m, h = 6m, dr/dt = 3m/s and dh/dt = - 4m/s we get,
Hence, option D is the correct answer.
Applications of Derivatives Question 10:
Find the approximate value of f (3.01), where f(x) = 3x2 + 3.
Answer (Detailed Solution Below)
Applications of Derivatives Question 10 Detailed Solution
Concept:
Let small charge in x be Δx and the corresponding change in y is Δy.
Now that Δy = f(x + Δx) - f(x)
Therefore, f(x + Δx) = f(x) + Δy
Calculation:
Given:f(x) = 3x2 + 3
Let x + Δx = 3.01 = 3 + 0.01
Therefore, x = 3 and Δx = 0.01
f(x + Δx) = f(x) + Δy
= f(x + Δx) = f(x) + f'(x)Δx
= f(3.01) = 3x2 + 3 + (6x)Δx
= f(3.01) = 3(3)2 + 3 + (6⋅3)(0.01)
= f(3.01) = 30 + 0.18
= f(3.01) = 30.18