Application of Derivatives MCQ with Solutions

Application of Derivatives: Chapter 6 Links

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6.1 | 6.2 | 6.3 | Misc | QRTs

Application of Derivatives MCQ with answers and solutions from NCERT Class 12 Maths. Revise all 9 important questions step by step with clear explanations. You can also watch the video solutions for better understanding. This is in continuation of the previous topic on Inverse Trigonometric Functions MCQs.

Ready? Okay, let’s begin with the first question!

Application of Derivatives MCQ#1 (NCERT Exercise 6.1 – Question 17)

📍The rate of change of the area of a circle with respect to its radius r at r = 6 cm is

Options:

(A) $10\pi$
(B) $12\pi$
(C) $8\pi$
(D) $11\pi$

Answer:
✅ Correct option: (B)

Explanation:
Area of a circle is
$A=\pi r^{2}$

Differentiate w.r.t. r:
$\dfrac{dA}{dr}=2\pi r$

At r = 6 cm:
$\dfrac{dA}{dr}=2\pi \times 6 = 12\pi$

Hence, the rate of change is $12\pi$ .

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Let me tell you all of these 9 Application of Derivatives MCQ are very simple and easy to understand and therefore scoring too. But you must practice them regularly.

So, my suggestion to you is to solve these NCERT Application of Derivatives MCQ a good number of times and feel confident. And if you ever need any clarification, just drop a comment — I’ll be more than happy to help.

Application of Derivatives MCQ#2 (NCERT Exercise 6.1 – Question 18)

📍The total revenue in rupees received from the sale of x units of a product is given by
R(x) = 3x² + 36x + 5. The marginal revenue, when x = 15, is

Options:

(A) $116$
(B) $96$
(C) $90$
(D) $126$

Answer:
✅ Correct Option: (D)

Explanation:

Revenue function:
$R(x)=3x^{2}+36x+5$

Differentiate to get marginal revenue:
$R'(x)=6x+36$

At x = 15:
$R'(15)=6\cdot 15 + 36 = 126$

Hence, marginal revenue is $126$ .

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Application of Derivatives MCQ#3 (NCERT Exercise 6.2 – Question 12)

📍Which of the following functions are decreasing on $(0, \dfrac{\pi}{2})$ ?

Options:

(A) $\cos x$
(B) $\cos 2x$
(C) $\cos 3x$
(D) $\tan x$

Answer:
✅ Correct Options: (A) and (B)

Explanation:

A function is decreasing where its derivative is negative.

1️⃣ For $\cos x:$
$\dfrac{d}{dx}(\cos x) = -\sin x$ which is $< 0$ on $(0, \dfrac{\pi}{2})$
So cos x is decreasing on the whole interval.

2️⃣ For $\cos 2x:$
$\dfrac{d}{dx}(\cos 2x) = -2\sin 2x$ ; on $(0,\dfrac{\pi}{2})$ we have $2x \in (0,\pi)$ so $\sin 2x>0$ and derivative is $<0$ .
So cos 2x is decreasing on the whole interval.

3️⃣ For $\cos 3x:$
$\dfrac{d}{dx}(\cos 3x) = -3\sin 3x$ . But for $x \in (0,\dfrac{\pi}{2})$ the sign of $\sin 3x$ changes: it is positive for $x\in(0,\dfrac{\pi}{3})$ and negative for $x\in(\dfrac{\pi}{3},\dfrac{\pi}{2})$ . Hence $\cos 3x$ is not decreasing on the whole interval.

4️⃣ For $\tan x:$
$\dfrac{d}{dx}(\tan x)=\sec^{2}x$ which is $>0$ on the interval, so tan x is increasing.

Hence only (A) and (B) are correct.

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Did you know, there are about 100 MCQs in NCERT textbooks for Class 12 (both Part 1 and Part 2)? I shall be taking one Chapter at a time and show you the explanations for each of the questions.

Moreover, I have also covered these solutions in the video form and they are freely available on my YouTube channel, @Mathsbetter. And here also, in each of the 9 Application of Derivatives MCQ on this page, I have provided the direct links to each and every MCQ for you to understand the solutions in a very simple and easy manner.

Application of Derivative MCQ#4 (NCERT Exercise 6.2 – Question 13)

📍On which of the following intervals is the function given by
$f(x) = x^{100} + \sin x - 1$ decreasing?

Options:

(A) $(0,1)$
(B) $\left(\dfrac{\pi}{2},\pi\right)$
(C) $\left(0,\dfrac{\pi}{2}\right)$
(D) None of these

Answer:
✅ Correct Option: (D) None of these

Explanation:

Differentiate:
$f'(x)=100x^{99}+\cos x$

For the function to be decreasing we need $f'(x) < 0$ .

On $(0,1):$ we have $100x^{99} \ge 0$ and $\cos x > 0$ , so $f'(x) > 0$ .

On $\left(0,\dfrac{\pi}{2}\right):$ cosine is positive and $100x^{99}$ is non-negative, so $f'(x) > 0$ .

And On $\left(\dfrac{\pi}{2},\pi\right):$ cosine is negative, but $100x^{99}$ is large positive for $x \ge \dfrac{\pi}{2}\approx1.57$ (in fact $100x^{99} \gg 1$ ), so $f'(x) > 0$ there as well.

Therefore $f'(x) > 0$ on all listed intervals, so the function is not decreasing on any of them.

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Application of Derivatives MCQ#5 (NCERT Exercise 6.2 – Question 19)

📍The interval in which
$y = x^{2} e^{-x}$
is increasing is

Options:

(A) $(-\infty,\infty)$
(B) $(-2,0)$
(C) $(2,\infty)$
(D) $(0,2)$

Answer:
✅ Correct Option: (D)

Explanation:

Differentiate:
$y' = \dfrac{d}{dx}\big(x^{2}e^{-x}\big) = e^{-x}(2x - x^{2})$
Rewrite:
$y' = e^{-x},x(2 - x)$

Since $e^{-x}$ is always positive, the sign of $y'$ is determined by $x(2-x)$ .

$x(2-x) > 0$ when $0 < x < 2$ .

Therefore $y' > 0$ on $(0,2)$ , so the function is increasing on $(0,2)$ .

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If you have been following this website as a resource for some of the important questions. For example, how to Integrate Square Root of tan x, or may be a complete guide to look at using Matrix Method in solving a system of linear equations.

Then trust me, similarly, this series on NCERT Class 12 MCQs is going to help you in many ways.

Moving onto the next Application of Derivatives MCQ in this part of the series.

Applications of Derivatives MCQ#6 (NCERT Exercise 6.3 – Question 27)

📍The point on the curve
$x^{2} = 2y$
which is nearest to the point
$(0,5)$ is

Options:

(A) $(2\sqrt{2},4)$
(B) $(2\sqrt{2},0)$
(C) $(0,0)$
(D) $(2,2)$

Answer:
✅ Correct option: (A)

Explanation:
From the curve $x^{2}=2y$ , we have
$y=\dfrac{x^{2}}{2}$

Distance squared from $(0,5)$ to a point on the curve is
$D^{2}=(x-0)^{2}+\left(\dfrac{x^{2}}{2}-5\right)^{2}$

Minimise $D^{2}$ :

$D^{2}=x^{2}+\dfrac{x^{4}}{4}-5x^{2}+25$
$=\dfrac{x^{4}}{4}-4x^{2}+25$

Differentiate:
$\dfrac{d(D^{2})}{dx}=x^{3}-8x$

Set equal to zero:
$x(x^{2}-8)=0$

Critical points:
$x=0,\ \pm2\sqrt{2}$

Corresponding points on curve:
$(0,0),\ (2\sqrt{2},4)$

Comparing distances,
$(2\sqrt{2},4)$ is nearest to $(0,5)$ .

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Application of Derivatives MCQ#7 (NCERT Exercise 6.3 – Question 28)

📍For all real values of x, the minimum value of
$\dfrac{1-x+x^{2}}{1+x+x^{2}}$ is

Options:

(A) $0$
(B) $1$
(C) $3$
(D) $\dfrac{1}{3}$

Answer:
✅ Correct Options: (D)

Explanation:
Let
$f(x)=\dfrac{1-x+x^{2}}{1+x+x^{2}}$

Differentiate using quotient rule:
$f'(x)=\dfrac{(2x-1)(1+x+x^{2})-(1-x+x^{2})(2x+1)}{(1+x+x^{2})^{2}}$

Simplifying numerator:
$f'(x)=\dfrac{2x^{2}-2}{(1+x+x^{2})^{2}}$

Set derivative zero:
$2x^{2}-2=0 \Rightarrow x=\pm1$

Check values:
$f(1)=\dfrac{1}{3},\quad f(-1)=3$

Hence minimum value is
$\dfrac{1}{3}$ .

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Applications of Derivative MCQ#8 (NCERT Exercise 6.3 – Question 29)

📍The maximum value of
$[x(x-1)+1]^{\tfrac{1}{3}}$
for $0\le x\le1$ is

Options:

(A) $\left(\dfrac{1}{3}\right)^{\tfrac{1}{3}}$
(B) $\dfrac{1}{2}$
(C) $1$
(D) $0$

Answer:
✅ Correct Option: (C)

Explanation:
Let
$f(x)=\big[x(x-1)+1\big]^{1/3}$

Since cube-root is an increasing function, maximum of $f(x)$ occurs when
$x(x-1)+1$ is maximum.

Define
$g(x)=x^{2}-x+1$

Differentiate:
$g'(x)=2x-1$

Set zero:
$x=\dfrac{1}{2}$

Check values:
$g(0)=1,\ g(1)=1,\ g\left(\dfrac{1}{2}\right)=\dfrac{3}{4}$

Maximum value of $g(x)$ is $1$ .

Hence maximum value of $f(x)$ is
$1$ .

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Application of Derivatives MCQ#9 (NCERT Miscellaneous Chapter 6 – Question 16)

📍A cylindrical tank of radius
$10\ \text{m}$
is being filled with wheat at the rate of
$314\ \text{m}^{3}!/\text{h}$ .
Then the depth of the wheat is increasing at the rate of

Options:

(A) $1\ \text{m/h}$
(B) $0.1\ \text{m/h}$
(C) $1.1\ \text{m/h}$
(D) $0.5\ \text{m/h}$

Answer:
✅ Correct Option: (A)

Explanation:
Volume of wheat in the cylindrical tank is
$V=\pi r^{2}h$

Here,
$r = 10\ \text{m}$
$\dfrac{dV}{dt} = 314\ \text{m}^3/\text{h}$

Volume of cylinder:
$V = \pi r^2 h = 100\pi h$

Differentiate w.r.t. time:
$\dfrac{dV}{dt} = 100\pi \dfrac{dh}{dt}$

Substitute values:
$314 = 100\pi \dfrac{dh}{dt}$

So,
$\dfrac{dh}{dt} = \dfrac{314}{100\pi}$

Since $\pi \approx 3.14$ ,
$\dfrac{dh}{dt} = \dfrac{314}{314} = 1\ \text{m/h}$

Hence, the depth of wheat is increasing at the rate
$1\ \text{m/h}$ .

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Quick Answer Key

Q1 → (B)
Q2 → (D)
Q3 → (A) and (B)
Q4 → (D)
Q5 → (D)
Q6 → (A)
Q7 → (D)
Q8 → (C)
Q9 → (A)

Closing Note

That completes all the important NCERT Class 12 Maths Application of Derivatives MCQ from chapter 6.
I hope the explanations and video solutions made things clearer and gave you more confidence. Stay tuned for the next chapter, where we’ll cover all the important Integrals MCQ (Chapter 7 from NCERT) with the same clarity and video support.

👉 Make sure to practice these questions again to strengthen your concepts.
👉 Watch the video solutions for a clearer and faster revision.

Keep practicing, and you’ll master the application of derivatives easily! 🚀

Application of Derivatives MCQ with Solutions

Application of Derivatives: Chapter 6 Links

Application of Derivatives MCQ#1 (NCERT Exercise 6.1 – Question 17)

Application of Derivatives MCQ#2 (NCERT Exercise 6.1 – Question 18)

Application of Derivatives MCQ#3 (NCERT Exercise 6.2 – Question 12)

Application of Derivative MCQ#4 (NCERT Exercise 6.2 – Question 13)

Application of Derivatives MCQ#5 (NCERT Exercise 6.2 – Question 19)

Applications of Derivatives MCQ#6 (NCERT Exercise 6.3 – Question 27)

Application of Derivatives MCQ#7 (NCERT Exercise 6.3 – Question 28)

Applications of Derivative MCQ#8 (NCERT Exercise 6.3 – Question 29)

Application of Derivatives MCQ#9 (NCERT Miscellaneous Chapter 6 – Question 16)

Quick Answer Key

Closing Note

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Application of Derivatives: Chapter 6 Links

Application of Derivatives MCQ#1 (NCERT Exercise 6.1 – Question 17)

Application of Derivatives MCQ#2 (NCERT Exercise 6.1 – Question 18)

Application of Derivatives MCQ#3 (NCERT Exercise 6.2 – Question 12)

Application of Derivative MCQ#4 (NCERT Exercise 6.2 – Question 13)

Application of Derivatives MCQ#5 (NCERT Exercise 6.2 – Question 19)

Applications of Derivatives MCQ#6 (NCERT Exercise 6.3 – Question 27)

Application of Derivatives MCQ#7 (NCERT Exercise 6.3 – Question 28)

Applications of Derivative MCQ#8 (NCERT Exercise 6.3 – Question 29)

Application of Derivatives MCQ#9 (NCERT Miscellaneous Chapter 6 – Question 16)

Quick Answer Key

Closing Note

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