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)x2 + 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
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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 =
Answer (Detailed Solution Below)
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) = (2n2 - 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 |
Answer (Detailed Solution Below)
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) :
Answer (Detailed Solution Below)
Applications of Derivatives Question 6 Detailed Solution
Download Solution PDFConcept:
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) = 3x2
m = f'(1) = 3 × 12 = 3.
(y - b) = m(x - a)
⇒ (y - 1) = 3(x - 1)
⇒ y - 1 = 3x - 3
Find the minimum value of function f(x) = x2 - x + 2
Answer (Detailed Solution Below)
Applications of Derivatives Question 7 Detailed Solution
Download Solution PDFConcept:
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?
Answer (Detailed Solution Below)
Applications of Derivatives Question 8 Detailed Solution
Download Solution PDFConcept:
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
Answer (Detailed Solution Below)
Applications of Derivatives Question 9 Detailed Solution
Download Solution PDFConcept:
|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) = x2 - 2x is strictly increasing ?
Answer (Detailed Solution Below)
Applications of Derivatives Question 10 Detailed Solution
Download Solution PDFConcept:
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 = ________
Answer (Detailed Solution Below)
Applications of Derivatives Question 11 Detailed Solution
Download Solution PDFConcept:
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
Answer (Detailed Solution Below)
Applications of Derivatives Question 12 Detailed Solution
Download Solution PDFConcept:
- 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
Answer (Detailed Solution Below)
Applications of Derivatives Question 13 Detailed Solution
Download Solution PDFConcept:
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.
- 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:
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:
Answer (Detailed Solution Below)
Applications of Derivatives Question 14 Detailed Solution
Download Solution PDFConcept:
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 ________.
Answer (Detailed Solution Below)
Applications of Derivatives Question 15 Detailed Solution
Download Solution PDFConcept;
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
\(\rm \Rightarrow \frac{dV}{dr} = 12π \times 4 = 48π\)