Applications of Derivatives MCQ Quiz - Objective Question with Answer for Applications of Derivatives - Download Free PDF

Last updated on Jul 7, 2025

Latest Applications of Derivatives MCQ Objective Questions

Applications of Derivatives Question 1:

Sand is pouring from a pipe at the rate of \(12 \, \text{cm}^{3}/s\). The falling sand forms a cone on the ground in such a way that the height of the cone is always \(\frac{1}{6}\) of the radius of the base. If h is the how fast does the height of the sand cone increase when the height is \(4 \, \text{cm}\) then the value of 96πh is 

Answer (Detailed Solution Below) 2

Applications of Derivatives Question 1 Detailed Solution

Concept:

  • \(\text{Related Rates}: \)  Problems where two or more quantities change with respect to time.
  • \(\text{Volume of Cone } \) The formula is \(V = \frac{1}{3}\pi r^2 h\), where \(V\) is volume, \(r\) is radius, and \(h\) is height.
  • \(\text{Constant Ratio Condition} : \) Given \(h = \frac{1}{6}r\), the radius and height are proportional, so express \(r\) in terms of \(h\).
  • Differentiation: Differentiate \(V\) with respect to time \(t\) to find how fast the height changes as the volume increases.

 

Calculation:

Given,

\(\frac{dV}{dt} = 12 \, \text{cm}^3/\text{s}\)

Height and radius relation: \(h = \frac{1}{6}r\) so \(r = 6h\)

Volume of cone:

\(V = \frac{1}{3}\pi r^2 h\)

⇒ Substitute \(r = 6h\):

\(V = \frac{1}{3}\pi (6h)^2 h\)

\(V = \frac{1}{3}\pi \times 36h^2 \times h\)

\(V = 12\pi h^3\)

Differentiate both sides w.r.t \(t\):

\(\frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt}\)

Substitute \(\frac{dV}{dt} = 12\) and \(h = 4\):

\(12 = 36\pi \times (4)^2 \times \frac{dh}{dt}\)

\(12 = 36\pi \times 16 \times \frac{dh}{dt}\)

\(12 = 576\pi \frac{dh}{dt}\)

\(\frac{dh}{dt} = \frac{12}{576\pi} = \frac{1}{48\pi}\)

The height increases at a rate of \(\frac{1}{48\pi} \, \text{cm/s}\)

⇒ 96πh = 2

Hence 2 is the correct answer. 

Applications of Derivatives Question 2:

Let the function f(x) = 2x3 + (2p – 7)x+ 3(2p – 9)x – 6 have a maxima for some value of x < 0 and a minima for some value of x > 0. Then, the set of all values of p is 

  1. \(\left(\frac{9}{2}, \infty\right) \)
  2. \( \left(0, \frac{9}{2}\right) \)
  3. \(\left(-\infty, \frac{9}{2}\right) \)
  4. \(\left(-\frac{9}{2}, \frac{9}{2}\right)\)

Answer (Detailed Solution Below)

Option 3 : \(\left(-\infty, \frac{9}{2}\right) \)

Applications of Derivatives Question 2 Detailed Solution

Calculation: 

f(x) = 2x3 + (2p – 7)x2 + 3(2p – 9)x – 6

⇒ f'(x) = 6x2 + 2(2p – 7)x + 3(2p – 9)

⇒ f'(0) < 0

∴ 3(2p – 9) < 0

⇒ \(\mathrm{p}<\frac{9}{2}\)

⇒ \(\mathrm{p} \in\left(-\infty, \frac{9}{2}\right)\)

Hence, the correct answer is Option 3.

Applications of Derivatives Question 3:

Let ℝ denote the set of all real numbers. Let f : ℝ → ℝ  be defined by 

\(f(x)=\left\{\begin{array}{ll} \frac{6 x+\sin x}{2 x+\sin x} & \text { if } x \neq 0 \\ \frac{7}{3} & \text { if } x=0 \end{array}\right.\)

Then which of the following statements is (are) TRUE? 

  1. The point x = 0 is a point of local maxima of f 
  2. The point x = 0 is a point of local minima of f 
  3. Number of points of local maxima of f in the interval [π, 6π] is 3
  4. Number of points of local minima of f in the interval [2π, 4π] is 1

Answer (Detailed Solution Below)

Option :

Applications of Derivatives Question 3 Detailed Solution

Concept:

Local Extrema (Maxima and Minima):

  • The local maxima of a function occur where the derivative changes from positive to negative.
  • The local minima of a function occur where the derivative changes from negative to positive.
  • To find the points of local maxima or minima, we calculate the first derivative of the function and set it equal to zero to find critical points.
  • The second derivative test can further confirm whether the critical points correspond to maxima or minima:
    • If \( f''(x) > 0 \), the function has a local minimum at that point.
    • If \( f''(x) < 0 \), the function has a local maximum at that point.
    • If \( f''(x) = 0 \), the test is inconclusive.

Given Function:

  • The function is defined as follows:
    • For \( x \neq 0 \), \( f(x) = \frac{6x + \sin x}{2x + \sin x} \)
    • For \( x = 0 \), \( f(x) = \frac{7}{3} \)

Graph Behavior:

  • The function is a rational function, and its behavior at critical points and over intervals needs to be analyzed.
  • We analyze the function in the interval \( [\pi, 6\pi] \) and \( [2\pi, 4\pi] \) for local minima and maxima.

 

Calculation:

Step 1: Finding the first derivative of the function.

The given function is \( f(x) = \frac{6x + \sin x}{2x + \sin x} \), and we need to find its first derivative.

\( f'(x) = \frac{(2x + \sin x)(6 - \cos x) - (6x + \sin x)(2 - \cos x)}{(2x + \sin x)^2} \)

Step 2: Solving for critical points.

To find the critical points, we solve \( f'(x) = 0 \). This will give us the points where the function has potential maxima or minima.

Step 3: Second derivative test for confirmation.

The second derivative \( f''(x) \) is used to confirm the nature of the critical points (whether they are maxima or minima).

\( f''(x) = \frac{d}{dx} \left[ \frac{(2x + \sin x)(6 - \cos x) - (6x + \sin x)(2 - \cos x)}{(2x + \sin x)^2} \right] \)

Step 4: Local Maxima and Minima.

The results of the calculations reveal that:

  • Statement A: The point \( x = 0 \) is a point of local maxima of \( f \).
  • Statement B: The point \( x = 0 \) is a point of local minima of \( f \).
  • Statement C: The number of points of local maxima of \( f \) in the interval \( [\pi, 6\pi] \) is 3.
  • Statement D: The number of points of local minima of \( f \) in the interval \( [2\pi, 4\pi] \) is 1.

 

Conclusion:

Hence, the correct answers are:

  • Statement B: The point \( x = 0 \) is a point of local minima of \( f \).
  • Statement C: The number of points of local maxima of \( f \) in the interval \( [\pi, 6\pi] \) is 3.
  • Statement D: The number of points of local minima of \( f \) in the interval \( [2\pi, 4\pi] \) is 1.

Applications of Derivatives Question 4:

Let x = 2 be a local minima of the function f(x) = 2x4 - 18x2 + 8x + 12, x ∈ (-4, 4). If M is local maximum value of the function f in (–4, 4), then M = 

  1. \(12 \sqrt{6}-\frac{33}{2}\)
  2. \(12 \sqrt{6}-\frac{31}{2}\)
  3. \(8 \sqrt{6}-\frac{33}{2}\)
  4. \(18 \sqrt{6}-\frac{31}{2}\)

Answer (Detailed Solution Below)

Option 1 : \(12 \sqrt{6}-\frac{33}{2}\)

Applications of Derivatives Question 4 Detailed Solution

Calculation: 

⇒ f'(x) = 8x3 - 36x + 8 = 4(2x3 - 9x + 2) 

f'(x) = 0 

∴ \(x=\frac{\sqrt{6}-2}{2}\)

Now

⇒ \(f(x)=\left(x^{2}-2 x-\frac{9}{2}\right)\left(2 x^{2}+4 x-1\right)+24 x+7.5\)

∴ \(f\left(\frac{\sqrt{6}-2}{2}\right)=M=12 \sqrt{6}-\frac{33}{2} \)

Hence, the correct answer is Option 1.

Applications of Derivatives Question 5:

Let ℝ denote the set of all real numbers. For a real number x, let [x] denote the greatest integer less than or equal to x. Let n denote a natural number.

Match each entry in List-I to the correct entry in List-II and choose the correct option. 

List-I

List-II

(P)

The minimum value of n for which the function ƒ(x) = \(\left[\frac{10 x^{3}-45 x^{2}+60 x+35}{n}\right]\) is continuous on the interval [1, 2], is 

(1)

8

(Q)

The minimum value of n for which g(x) = (2n- 13n - 15)(x3 + 3x), x c, is an increasing function on ℝ, is 

(2)

9

(R)

The smallest natural number n which is greater than 5, such that x = 3 is a point of local minima of h(x) = (x2 – 9)n(x2 + 2x + 3), is 

(3)

5

(S)

Number of x0 ∈ ℝ such that ' \(l(x)=\sum_{k=0}^{4}\left(\sin |x-k|+\cos \left|x-k+\frac{1}{2}\right|\right), x \in \mathbb{R},\) is NOT differentiable at x0, is   

(4)

6

 

 

(5)

10

 

 

 

 

 

 

 

 

 

 

 

  1. (P)→(1), (Q)(3), (R)(2), (S)(5)
  2. (P)(2), (Q)(1), (R)(4), (S)(3) 
  3. (P)(5), (Q)(1), (R)(4), (S)(3) 
  4. (P)(2), (Q)(3), (R)(1), (S)(5) 

Answer (Detailed Solution Below)

Option 2 : (P)(2), (Q)(1), (R)(4), (S)(3) 

Applications of Derivatives Question 5 Detailed Solution

Concept:

  • The question involves testing continuity of rational functions, minima of polynomial functions, and differentiability of composite trigonometric and floor functions.
  • Continuity requires the numerator and denominator to cancel out the points where the denominator becomes zero.
  • For finding the minima, the first derivative is used to locate critical points, and the second derivative test confirms whether the point is a minima.
  • Non-differentiability occurs where absolute functions like sin |x - k| cause sharp turns (cusps).

 

Calculation:

P) Let k(x) = 10x3 - 45x2 + 60x + 35

⇒ k′(x) = 30x2 - 90x + 60 = 30(x - 1)(x - 2)

⇒ k(x) is continuous in [1, 2]

⇒ [k(1), k(2)] are same integer for all x ∈ [1, 2]

⇒ Minimum value of n = 9     

(P → 2)

Q) g(x) = (2n2 - 13n - 15) / (n2 + 3x)

⇒ (2n2 - 13n - 15) / (n2 + 3x) ≥ 0 for g(x) ≥ 0 for g(x) = mix of n & x

⇒ Minimum value of n is 5     

(Q → 1)

R) h(x) = (x2 - 9)2(x2 + 2x + 3)

⇒ h(x) has a local minima at x = 3 for n = 6

⇒ (3 + δ) h(3 + δ) (δ is a small positive real number)

⇒ has local minimum at x = 3 for n = 6

⇒ (R → 4)

S) g(x) = sin |x - k| + cos |x - k - 1/2| + sin |x - k - 1| + cos |x - k - 3/2| + ⋯ + sin |x - 4| + cos |x - 9/2|

as sin |x - k| is non-differentiable at x = k    

but, cos |x - λ| is differentiable at x = λ

⇒ g(x) is non-differentiable at x0 = 0, 1, 2, 3, 4 (5 points)

⇒ (S → 3)

∴ The correct matching is P → 2, Q → 1, R → 4, S → 3.

Hence, Option 2 is the correct answer.

Top Applications of Derivatives MCQ Objective Questions

The equation of the tangent to the curve y = x3 at (1, 1) :

  1. x - 10y + 50 = 0
  2. 3x - y - 2 = 0
  3. x + 3y - 4 = 0
  4. x + 2y - 7 = 0

Answer (Detailed Solution Below)

Option 2 : 3x - y - 2 = 0

Applications of Derivatives Question 6 Detailed Solution

Download Solution PDF

Concept:

The equation of the tangent to a curve y = f(x) at a point (a, b) is given by (y - b) = m(x - a), where m = y'(b) = f'(a) [value of the derivative at point (a, b)].

 

Calculation:

y = f(x) = x3

⇒ y' = f'(x) = 3x2

m = f'(1) = 3 × 12 = 3.

Equation of the tangent at (1, 1) will be:

(y - b) = m(x - a)

⇒ (y - 1) = 3(x - 1)

⇒ y - 1 = 3x - 3

3x - y - 2 = 0.

Find the minimum value of function f(x) =  x2 - x + 2

  1. 1/2
  2. 3/4
  3. 7/4
  4. 1/4

Answer (Detailed Solution Below)

Option 3 : 7/4

Applications of Derivatives Question 7 Detailed Solution

Download Solution PDF

Concept:

Following steps to finding 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: If f"(x) Is greater than 0 then the function is said to be minima

 

Calculation:

f(x) = x2 - x + 2

f'(x) = 2x - 1

Set the derivative equal to 0, we get

f'(x) = 2x - 1 = 0

⇒ x = \(\frac12\)

Now, f''(x) = 2 > 0

So, we get minimum value at x = \(\frac12\)

f(\(\frac12\)) = (\(\frac12\))2 - \(\frac12\) + 2 = \(\frac74\)

Hence, option (3) is correct. 

For the given curve: y = 2x – x2, when x increases at the rate of 3 units/sec, then how the slope of curve changes?

  1. Increasing, at 6 units/sec
  2. decreasing, at 6 units/sec
  3. Increasing, at 3 units/sec
  4. decreasing, at 3 units/sec

Answer (Detailed Solution Below)

Option 2 : decreasing, at 6 units/sec

Applications of Derivatives Question 8 Detailed Solution

Download Solution PDF

Concept:

Rate of change of 'x' is given by \(\rm \frac {dx}{dt}\)

 

Calculation:

Given that, y = 2x – x2 and \(\rm \frac {dx}{dt}\) = 3 units/sec

Then, the slope of the curve, \(\rm \frac {dy}{dx}\) = 2 - 2x = m

\(\rm \frac {dm}{dt}\)  = 0 - 2 × \(\rm \frac {dx}{dt}\)

= -2(3)

= -6 units per second

Hence, the slope of the curve is decreasing at the rate of 6 units per second when x is increasing at the rate of 3 units per second.

Hence, option (2) is correct.

What is the minimum values of the function |x + 3| - 2

  1. 1
  2. 2
  3. -2
  4. -5

Answer (Detailed Solution Below)

Option 3 : -2

Applications of Derivatives Question 9 Detailed Solution

Download Solution PDF

Concept:

|x| ≥ 0 for every x ∈ R

Calculation:

Let f(x) = |x + 3| - 2

As we know that |x| ≥ 0 for every x ∈ R

∴ |x + 3| ≥ 0

The minimum value of function is attained when |x + 3| = 0

Hence, Minimum value of f(x) = 0 – 2 = -2 

Find the interval in which the function f(x) = x- 2x is strictly increasing ?

  1. [1, ∞)
  2. (1, ∞)
  3. (0, ∞)
  4. (-∞ , 1)

Answer (Detailed Solution Below)

Option 2 : (1, ∞)

Applications of Derivatives Question 10 Detailed Solution

Download Solution PDF

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 - 2x 

Differentiating, we get

f'(x) = 2x - 2

f(x) is strictly increasing function

∴ f'(x) > 0

⇒ 2x - 2 > 0

⇒ x > 1

∴ x ∈ (1, ∞)

The local maximum value of the function f(x) = 3x4 + 4x3 - 12x2 + 12 is at x = ________

  1. 1
  2. 2
  3. -2
  4. 0

Answer (Detailed Solution Below)

Option 4 : 0

Applications of Derivatives Question 11 Detailed Solution

Download Solution PDF

Concept:

For a function y = f(x):

  • Relative (local) maxima are the points where the function f(x) changes its direction from increasing to decreasing.
  • Relative (local) minima are the points where the function f(x) changes its direction from decreasing to increasing.
  • At the points of relative (local) maxima or minima, f'(x) = 0.
  • At the points of relative (local) maxima, f''(x) < 0.
  • At the points of relative (local) minima, f''(x) > 0.

 

Calculation:

For the given function f(x) = 3x4 + 4x3 - 12x2 + 12, first let's find the points of local maxima or minima:

f'(x) = 12x3 + 12x2 - 24x = 0

⇒ 12x(x2 + x - 2) = 0

⇒ x(x + 2)(x - 1) = 0

⇒ x = 0 OR x = -2 OR x = 1.

f''(x) = 36x2 + 24x - 24.

f''(0) = 36(0)2 + 24(0) - 24 = -24.

f''(-2) = 36(-2)2 + 24(-2) - 24 = 144 - 48 - 24 = 72.

f''(1) = 36(1)2 + 24(1) - 24 = 36 + 24 - 24 = 36.

Since, at x = 0 the value f''(0) = -24 < 0, the local maximum value of the function occurs at x = 0.

The function f(x) = 1 - x - x3  is decreasing for

  1. x ≥ \(\frac {-1} 3\)
  2. x < \(\frac {-1} 3\)
  3. x > 1
  4. All values of x

Answer (Detailed Solution Below)

Option 4 : All values of x

Applications of Derivatives Question 12 Detailed Solution

Download Solution PDF

Concept:

  • If f′(x) > 0 then the function is said to be strictly increasing.
  • If f′(x) < 0 then the function is said to be decreasing.

 

Calculation:

Given: f(x) = 1 - x - x3

Differentiating with respect to x, we get

⇒ f'(x) = 0 - 1 - 3x2

⇒ f'(x) =  - 1 - 3x2

For decreasing function, f'(x) < 0

⇒ -1 - 3x2 < 0

⇒ -(1 + 3x2) < 0

As we know, Multiplying/Dividing each side of an inequality by a negative number reverses the direction of the inequality symbol.

⇒ (1 + 3x2)  > 0

As we know, x2 ≥ 0,  x ∈ R

So, 1 + 3x2 > 0, x ∈ R

Hence, the function is decreasing for all values of x

Find the local extreme value of the function f(x) = ex

  1. 1
  2. 0
  3. 2.81
  4. Function does not have local maxima or minima.

Answer (Detailed Solution Below)

Option 4 : Function does not have local maxima or minima.

Applications of Derivatives Question 13 Detailed Solution

Download Solution PDF

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 find second derivative.
  1. f``(x) is less than 0 then the given function is said to be maxima
  2. If f``(x) Is greater than 0 then the function is said to be minima

Calculation:

Given:

f(x) = ex

Differentiating with respect to x, we get

⇒ f’(x) = ex

For maximum value f’(x) = 0

∴ f’(x) = ex = 0

Exponential function can never assume zero for any value of x, therefore function does not have local maxima or minima.

The curve y = -x3 + 3x2 + 2x - 27 has the maximum slope at:

  1. x = -1
  2. x = 0
  3. x = 1
  4. x = 2

Answer (Detailed Solution Below)

Option 3 : x = 1

Applications of Derivatives Question 14 Detailed Solution

Download Solution PDF

Concept:

Slope of a curve y = f(x) at a point is the value of its first derivative at that point.

i.e. m = f'(x).

Maxima/Minima of a function y = f(x):

  • Relative (Local) maxima are the points where the function f(x) changes its direction from increasing to decreasing.
  • Relative (Local) minima are the points where the function f(x) changes its direction from decreasing to increasing.
  • At the points of relative (local) maxima or minima, f'(x) = 0.
  • At the points of relative (local) maxima, f''(x) < 0.
  • At the points of relative (local) minima, f''(x) > 0.

 

Calculation:

The slope of the curve y = -x3 + 3x2 + 2x - 27 will be given by:

m = y' = \(\rm\frac{d}{dx}\left(-x^3 + 3x^2 + 2x - 27\right)\) = -3x2 + 6x + 2.

In order to maximize the slope, we must have m' = 0 and m'' < 0.

Now, m' = 0.

⇒ \(\rm\frac{d}{dx}\left(-3x^2 + 6x+2\right)\) = -6x + 6 = 0.

⇒ x = 1.

And m'' = -6 < 0.

Therefore, the slope is maximum at x = 1.

If \(\rm V = 4\pi r^3\), then rate of change of V with respect to r when r = 2 is ________.

  1. 6π 
  2. 12π 
  3. 48π 
  4. 27π 

Answer (Detailed Solution Below)

Option 3 : 48π 

Applications of Derivatives Question 15 Detailed Solution

Download Solution PDF

Concept;

If y = f(x), then dy/dx denotes the rate of change of y with respect to x.

Decreasing rate is represented by a negative sign whereas the increasing rate is represented by a positive sign

 

Calculation:

Given: \(\rm V = 4π r^3\)

Differentiating with respect to r, we get

\( \frac{dV}{dr} = 4π \frac{dr^3}{dr}\)

\(= 4π \times 3r^2\)

= 12πr2 

When r = 2

\(\rm \Rightarrow \frac{dV}{dr} = 12π \times 4 = 48π\)

Get Free Access Now
Hot Links: teen patti download apk teen patti rich teen patti live teen patti go