Class 12 Important Questions: CBSE Maths Chapter 6 Application of Derivatives 2024-25

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

1. The side of a square is increasing at the rate of $\mathbf{0}\mathbf{.2~cm/sec}$. Find the rate of increase of the perimeter of the square.

Ans: It is given that the side of a square is increasing at the rate of $0.2~\text{cm}/\text{sec}$.

Let us consider the edge of the given cube be $x~\text{cm}$ at any instant.

According to the question,

The rate of side of the square increasing is,

$\dfrac{\text{dx}}{dt}=0.2~\text{cm}/\text{sec}\ldots $...(i)

Therefore the perimeter of the square at any time $t$ will be,

$P=4x~\text{cm}$

By applying derivative with respect to time on both sides, we get

$ \Rightarrow \dfrac{dP}{dt}=\dfrac{d(4x)}{dt} $ 

 $ \Rightarrow \dfrac{dP}{dt}=4\dfrac{dx}{dt} $ 

 $ \Rightarrow \dfrac{dP}{dt}=4\times 0.2=0.8~\text{cm/sec} $ 

Hence from equation (i). The rate at which the perimeter of the square will increase is $0.8\,\,\text{cm/sec}$.

2. The radius of the circle is increasing at the rate of $\mathbf{0}\mathbf{.7~cm/sec}$. What is the rate of increase of its circumference?

Ans: It is given that the radius of a circle is increasing at the rate of $0.7~\text{cm}/\text{sec}$.

Let us consider that the radius of the given circle be r $\text{cm}$ at any instant.

According to the question,

The rate of radius of a circle is increasing as,

$\dfrac{\text{dr}}{dt}=0.7~\text{cm}/\text{sec}$     ...(i)

Now the circumference of the circle at any time $t$ will be,

$\text{C}=2\pi \,\text{cm}$

By applying derivative with respect to time on both sides, we get

$ \Rightarrow \dfrac{dC}{dt}=\dfrac{d(2\pi r)}{dt} $ 

$ \Rightarrow \dfrac{dC}{dt}=2\pi \dfrac{dr}{dt} $ 

$ \Rightarrow \dfrac{dC}{dt}=2\pi \times 0.7=1.4\pi \,\text{cm}/\text{sec} $ 

From the equation (i). We can conclude that the rate at which the circumference of the circle will be increasing is $1.4\,\pi \,cm/\sec $

3. If the radius of a soap bubble is increasing at the rate of $\dfrac{\mathbf{1}}{\mathbf{2}}\mathbf{~cm/sec}$. At what rate its volume is increasing when the radius is $\mathbf{1~cm}$.

Ans: It is given that the radius of an air bubble is increasing at the rate of $0.5~\text{cm}/\text{sec}$.

Let us consider that the radius of the given air bubble be r $\text{cm}$ and let $\text{V}$ be the volume of the air bubble at any instant.

According to the question,

The rate at which the radius of the bubble is increasing is,

$\dfrac{dr}{dt}=0.5\,~\text{cm/sec}$    ... (i)

The volume of the bubble, i.e., volume of sphere is $V=\dfrac{4}{3}\pi {{r}^{3}}$

By applying derivative with respect to time on both sides,

$ \Rightarrow \dfrac{dV}{dt}=\dfrac{d\left( \dfrac{4}{3}\pi {{r}^{3}} \right)}{dt} $ 

$ \Rightarrow \dfrac{dV}{dt}=\dfrac{4}{3}\pi \dfrac{d\left( {{r}^{3}} \right)}{dt} $ 

$ \Rightarrow \dfrac{dV}{dt}=\dfrac{4}{3}\pi \times 3{{r}^{2}}\dfrac{dr}{dt} $ 

$ \Rightarrow \dfrac{dV}{dt}=4\pi {{r}^{2}}\times 0.5\,\,\,\,\,\,\,\,...\left( ii \right) $ 

When the radius is $1~\text{cm}$,

The above equation becomes

$ \Rightarrow \dfrac{\text{dV}}{\text{dt}}=4\pi \times {{(1)}^{2}}\times 0.5 $ 

$ \Rightarrow \dfrac{\text{dV}}{\text{dt}}=2\pi \,\text{c}{{\text{m}}^{\text{3}}}\text{/sec} $ 

Hence the volume of air bubble is increasing at the rate of $2\pi \,\text{c}{{\text{m}}^{\text{3}}}\text{/sec}$.

4. A stone is dropped into a quiet lake and waves move in circles at a speed of $\mathbf{4~cm/sec}$. At the instant when the radius of the circular wave is \[\mathbf{10~cm}\], how fast is the enclosed area increasing?

Ans: It is given that when a stone is dropped into a quiet lake and waves are formed which moves in circles at a speed of $4~\text{cm}/\text{sec}$.

Let us consider that, r be the radius of the circle and $A$ be the area of the circle.

When a stone is dropped into the lake, waves are formed which move in a circle at speed of $4~\text{cm}/\text{sec}$.

Thus, we can say that the radius of the circle increases at a rate of,

$\dfrac{\text{dr}}{dt}=4~\text{cm/sec}$

Area of the circle is $\pi {{r}^{2}}$, therefore

$ \Rightarrow \dfrac{\text{dA}}{\text{dt}}=\dfrac{\text{d}\left( \pi {{\text{r}}^{2}} \right)}{\text{dt}} $ 

$ \Rightarrow \dfrac{\text{dA}}{\text{dt}}=\pi \dfrac{\text{d}\left( {{\text{r}}^{2}} \right)}{\text{dt}} $ 

$ \Rightarrow \dfrac{\text{dA}}{\text{dt}}=\pi \times 2\text{r}\dfrac{\text{dr}}{\text{dt}} $ 

$ \Rightarrow \dfrac{\text{dA}}{\text{dt}}=2\pi \text{r}\times 4\ldots \ldots .\text{ (ii)} $ 

Hence, when the radius of the circular wave is $10~\text{cm}$, the above equation becomes

$ \Rightarrow \dfrac{\text{dA}}{dt}=2\pi \times 10\times 4 $ 

 $ \Rightarrow \dfrac{\text{dA}}{dt}=80\pi \,\text{c}{{\text{m}}^{\text{2}}}\text{/sec} $ 

Thus, the enclosed area is increasing at the rate of $80\pi \,\text{c}{{\text{m}}^{\text{2}}}\text{/sec}$.

5. The total revenue in Rupees received from the sale of $\mathbf{x}$ units of a product is given by, $\mathbf{R(x)=13}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+26x+15}$. Find the marginal revenue when $\mathbf{x=7}$.

Ans: Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

Let us consider ‘MR’ to be the marginal revenue, therefore

$MR=\dfrac{dR}{dx}$

It is given that,

Total revenue, i.e., $R(x)=13{{x}^{2}}+26x+15$  …(1)

We need to find marginal revenue when $x=7$

i.e., MR when $x=7$

$ \Rightarrow \text{MR}=\dfrac{d(R(x))}{dx} $ 

$ \Rightarrow \text{MR}=\dfrac{d\left( 13{{x}^{2}}+26x+15 \right)}{dx} $ 

$ \Rightarrow \text{MR}=\dfrac{d\left( 13{{x}^{2}} \right)}{dx}+\dfrac{d(26x)}{dx}+\dfrac{d(15)}{dx} $ 

$ \Rightarrow \text{MR}=13\dfrac{d\left( {{x}^{2}} \right)}{dx}+26\dfrac{d(x)}{dx}+0 $ 

$ \Rightarrow \text{MR}=13\times 2x+26 $ 

$ \Rightarrow \text{MR}=26x+26 $ 

$ \Rightarrow MR=26(x+1) $ 

Taking $x=7$, we get

$ \Rightarrow MR=26\left( 7+1 \right) $ 

$ \Rightarrow MR=26\times 8 $ 

$ \Rightarrow MR=208 $ 

Therefore, the required marginal revenue is $\text{Rs}\,208$.

6. Find the maximum and minimum values of function $\mathbf{f(x)=sin2x+5}$.

Ans: Given function is,

$f(x)=\sin 2x+5$

We know that,

$-1\le \sin \theta \le 1,\,\,\,\forall \theta \in R$

$-1\le \sin 2x\le 1$

Adding 5 on both sides,

$-1+5\le \sin 2x+5\le 1+5$

$4\le \sin 2x+5\le 6$

Therefore,

Max value of $f(x)=\sin 2x+5$ will be 6 and,

Min value of $f(x)=\sin 2x+5$ will be 4.

7. Find the maximum and minimum values (if any) of the function

$f(x)=-|x-1|+7\forall x\in R$

Ans: Given equation is $f(x)=-|x+1|+3$

$|x+1|>0$

$\Rightarrow -|x+1|<0$

Maximum value of $g(\text{x})=$ maximum value of $-|\text{x}+1|+7$

$\Rightarrow 0+7=7$

Maximum value of $f(x)=3$

There is no minimum value of $f(x)$.

8. Find the value of $a$ for which the function $\mathbf{f(x)=}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-2ax+6,x>0}$ is strictly increasing.

Ans: Given function is $f(x)={{x}^{2}}-2ax+6,x>0$

It will be strictly increasing when $f'\left( x \right)>0$.

$ f'\left( x \right)=2x-2a>0 $ 

$ \Rightarrow 2\left( x-a \right)>0 $ 

$ \Rightarrow x-a>0 $ 

$ \Rightarrow a<x $ 

But $x>0$

Therefore, the maximum possible value of $a$ is 0 and all other values of a  will be less than 0.

Hence, we get $a\le 0$.

9. Write the interval for which the function $\mathbf{f(x)=cosx,0\text{£}x\text{£}2\pi }$ is decreasing.

Ans: The given function is $f(x)=\cos x,0\le x\le 2\pi $.

It will be a strictly decreasing function when $f'\left( x \right)<0$.

Differentiating w.r.t. $x$, we get

$ f'\left( x \right)=-\sin x $ 

$ \text{Now,} $ 

$ f'\left( x \right)<0 $ 

$ \Rightarrow -\sin x<0 $ 

$ \Rightarrow \sin x>0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{i}\text{.e}\text{.,}\left( 0,\,\pi  \right) $ 

Hence, the given function is decreasing in $\left( 0,\,\pi  \right)$.

10. What is the interval on which the function $\mathbf{f(x)=}\dfrac{\mathbf{logx}}{\mathbf{x}}\mathbf{,x\hat{I}(0,\text{¥})}$ is increasing?

Ans: The given function is $f(x)=\dfrac{\log x}{x},x\in (0,\infty )$.

It will be a strictly increasing function when $f'\left( x \right)>0$.

$f(x)=\dfrac{\log x}{x}$

Therefore,

$ {{f}^{\prime }}(x)=\dfrac{1}{{{x}^{2}}}-\dfrac{\log x}{{{x}^{2}}} $ 

$ {{f}^{\prime }}(x)=\dfrac{1-\log x}{{{x}^{2}}} $ 

$ \because \,\,{{f}^{\prime }}(x)>0 $ 

$ \Rightarrow \dfrac{1-\log {{x}^{2}}}{{{x}^{2}}}>0 $ 

$ \Rightarrow 1-\log x>0 $ 

$ \Rightarrow 1>\log x $ 

$ \Rightarrow e>x $ 

Therefore, $\text{f}(\text{x})$ is increasing in the interval $\left( 0,\,e \right)$.

11. For which values of $\mathbf{x}$, the functions $\mathbf{y=}{{\mathbf{x}}^{\mathbf{4}}}\mathbf{-}\dfrac{\mathbf{4}}{\mathbf{3}}{{\mathbf{x}}^{\mathbf{3}}}$ is increasing?

Ans: The given function is $y={{x}^{4}}-\dfrac{4}{3}{{x}^{3}}$

It will be a strictly increasing function when $f'\left( x \right)>0$.

$ f'\left( x \right)>0\,\,\text{and,} $ 

$ f'\left( x \right)=4{{x}^{3}}-4{{x}^{2}} $ 

$ f'\left( x \right)=4{{x}^{2}}(x-1) $ 

$ 4{{x}^{2}}(x-1)>0 $ 

Now, 

$\dfrac{dy}{dx}=0\Rightarrow x=0,x=1$

Since ${{f}^{\prime }}(x)<0\forall x\in (-\infty ,0)\cup (0,1)$ and $f$ is continuous in $(-\infty ,0]$ and $[0,1]$. Therefore $f$ is decreasing in $(-\infty ,1]$ and $f$ is increasing in $[1,\infty )$.

Here $f$ is strictly decreasing in $(-\infty ,0)\cup (0,1)$ and is strictly increasing in $(1,\infty )$.

12. Write the interval for which the function $\mathbf{f(x)=}\dfrac{\mathbf{1}}{\mathbf{x}}$ is strictly decreasing.

Ans: The given equation is $\mathbf{f(x)=}\dfrac{\mathbf{1}}{\mathbf{x}}$.

It will be a strictly decreasing function when $f'\left( x \right)<0$.

$ f(x)=x+\dfrac{1}{x} $ 

$ \Rightarrow {{f}^{\prime }}(x)=1-\dfrac{1}{{{x}^{2}}} $ 

$ \Rightarrow {{f}^{\prime }}(x)=\dfrac{{{x}^{2}}-1}{{{x}^{2}}} $ 

$ \Rightarrow {{f}^{\prime }}(x)=0 $ 

$ \Rightarrow \dfrac{{{x}^{2}}-1}{{{x}^{2}}}=0 $ 

$ \Rightarrow {{x}^{2}}-1=0 $ 

$ \Rightarrow x=\pm 1 $ 

The intervals are $(-\infty ,-1),(-1,1),(1,\infty )$

${{f}^{\prime }}(0)<0$

$\therefore $ Strictly decreasing in $(-1,1)$

13. Find the sub-interval of the interval $\mathbf{(0,\pi /2)}$ in which the function $\mathbf{f(x)=sin3x}$ is increasing.

Ans: The given function is $f(x)=\sin 3x$

On differentiating the above function with respect to $x$, we get,

${{f}^{\prime }}(x)=3\cos 3x$

$f(x)$ will be increasing, when ${{f}^{\prime }}(x)>0$

Given that $x\in \left( 0,\dfrac{\pi }{2} \right)$

$\Rightarrow 3x\in \left( 0,\dfrac{3\pi }{2} \right)$

Cosine function is positive in the first quadrant and negative in the second quadrant.

case 1:

When $3x\in \left( 0,\dfrac{\pi }{2} \right)$

$ \Rightarrow \cos 3x>0 $ 

$ \Rightarrow 3\cos 3x>0 $ 

$ \Rightarrow {{f}^{\prime }}(x)>0\text{ for }0<3x<\dfrac{\pi }{2} $ 

$ \Rightarrow {{f}^{\prime }}(x)>0\text{ for }0<x<\dfrac{\pi }{6} $ 

$\therefore f(x)$ is increasing in the interval $\left( 0,\dfrac{\pi }{6} \right)$ 

case 2:

When $3x\in \left( \dfrac{\pi }{2},\dfrac{3\pi }{2} \right)$

$\Rightarrow \cos 3x<0$ 

$\Rightarrow 3\cos 3x<0$ 

$\Rightarrow {{f}^{\prime }}(x)<0$ for $\dfrac{\pi }{2}<3x<\dfrac{3\pi }{2}$ 

$\Rightarrow {{f}^{\prime }}(x)<0$ for $\dfrac{\pi }{6}<x<\dfrac{\pi }{2}$

$\therefore f(x)$ is decreasing in the interval $\left( \dfrac{\pi }{6},\dfrac{\pi }{2} \right)$

14. Without using derivatives, find the maximum and minimum value of $\mathbf{y=|3sinx+1|}$.

Ans: The given function is $y=|3\sin x+1|$

Maximum and minimum values of $\sin x=\{-1,1\}$ respectively.

Therefore, the value of the given function will be maximum and minimum at only these points.

Taking $\sin x=-1$

$y=|3\times (-1)+1|=>2$

Now, put $\sin x=1$

$y=|3\times 1+1|=>4$

The maximum and minimum values of the given function are 4 and 2 respectively.

15. If $\mathbf{f(x)=ax+cosx}$ is strictly increasing on $\mathbf{R}$, find $\mathbf{a}$.

Ans: It is given that the function $f(x)=ax+\cos x$ is strictly increasing on $R$

Here function, $f(x)=ax+\cos x$

Differentiating $\text{f}(\text{x})$ with respect to $\text{x}$ we get,

${{f}^{\prime }}(x)=a+(-\sin x)=a-\sin x$

for strictly increasing, ${{f}^{\prime }}(x)>0$

Therefore,

$a-\sin x>0$ it will be correct for all real value of $x$ only when $a\in (-1,1)$

Hence the value of a belongs to $(-1,1)$.

16. Write the interval in which the function $\mathbf{f(x)=}{{\mathbf{x}}^{\mathbf{9}}}\mathbf{+3}{{\mathbf{x}}^{\mathbf{7}}}\mathbf{+64}$ is increasing.

Ans: The given function is $f(x)={{x}^{9}}+3{{x}^{7}}+64$.

For it to be a increasing function $f'\left( x \right)>0$

On differentiating both sides with respect to x, we get

$ \text{f}(\text{x})={{\text{x}}^{9}}+3{{\text{x}}^{7}}+64 $ 

$ \Rightarrow {{\text{f}}^{\prime }}(\text{x})=9{{\text{x}}^{8}}+21{{\text{x}}^{6}} $ 

$ \Rightarrow f'\left( x \right)=3{{\text{x}}^{6}}\left( 3{{\text{x}}^{2}}+7 \right) $ 

$\because$ function is increasing.

$3{{x}^{6}}\left( 3{{x}^{2}}+7 \right)>0$

$\Rightarrow $ function is increasing on $\text{R}$.

17. What is the slope of the tangent to the curve $\mathbf{f}\left( \mathbf{x} \right)\mathbf{=}{{\mathbf{x}}^{\mathbf{3}}}\mathbf{-5x+3}$ at the point whose $\mathbf{x}$ co-ordinate is 2?

Ans: The given equation of the curve is $f={{x}^{3}}-5x+3$   …(1)

When $x=2$,

$ y={{2}^{3}}-5.2+3 $ 

$ y=8-10+3 $ 

$ y=1 $ 

Therefore, the point on the curve is $\left( 2,\,\,1 \right)$.

Differentiating equation (1) with respect to x, we get

$\dfrac{dy}{dx}=3{{x}^{2}}-5$

Slope of tangent $\dfrac{dy}{dx}$

Since $x=2$,

$ \Rightarrow {{3.2}^{2}}-5 $ 

$ \Rightarrow 12-5 $ 

$ \Rightarrow 7 $ 

Hence the slope of tangent is 7.

18. At what point on the curve \[\mathbf{y=}{{\mathbf{x}}^{\mathbf{2}}}\] does the tangent make an angle of $\mathbf{4}{{\mathbf{5}}^{\mathbf{{}^\circ }}}$ with positive direction of the $x$-axis?

Ans: The given equation of the curve is $y={{x}^{2}}$

Differentiating the above with respect to $x$,

$ \Rightarrow \dfrac{dy}{dx}=2{{x}^{2-1}} $ 

$ \Rightarrow \dfrac{dy}{dx}=2x\,\,...\left( 1 \right) $ 

So,

$\dfrac{dy}{dx}=$ The slope of tangent $=\tan \theta $

The tangent makes an angle of ${{45}^{{}^\circ }}$ with $x$-axis

$\dfrac{dy}{dx}=\tan {{45}^{{}^\circ }}=1\ldots (2)$

Because the $\tan {{45}^{{}^\circ }}=1$

From the equation (1) $ (2), we get

$ \Rightarrow 2x=1 $ 

$ \Rightarrow x=\dfrac{1}{2} $ 

Substitute $x=\dfrac{1}{2}$ in $y={{x}^{2}}$

$ \Rightarrow y={{\left( \dfrac{1}{2} \right)}^{2}} $ 

$ \Rightarrow y=\dfrac{1}{4} $ 

Hence, the required point is $\left( \dfrac{1}{2},\dfrac{1}{4} \right)$.

19. Find the point on the curve $\mathbf{y=3}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-12x+9}$ at which the tangent is parallel to $\mathbf{x}$-axis.

Ans: The given equation of the curve is $y=3{{x}^{2}}-12x+9$.

Differentiating the above equation with respect to x, we get

$ \dfrac{dy}{dx}=6x-12 $ 

$ m\,\,=6x-12 $ 

$\dfrac{dy}{dx}=$ The slope of tangent $=\tan \theta $

If the tangent is parallel to x-axis.

$ m=0 $ 

$ \Rightarrow 6x-12=0 $ 

$ \Rightarrow x=2 $ 

When $x=2$, then

$ y={{3.2}^{2}}-12.2+9 $ 

$ y=12-24+9 $ 

$ y=-3 $ 

Hence, the required point $\left( x,y \right)=\left( 2,\,-3 \right)$.

20. What is the slope of the normal to the curve $\mathbf{y=5}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-4sinx}$ at $\mathbf{x=0}$.

Ans: The given equation of the curve is $y=5{{x}^{2}}-4\sin x$.

Differentiating the above equation with respect to x, we get

$\dfrac{dy}{dx}=10x-4\cos x$

$\dfrac{dy}{dx}=$ The slope of tangent $=\tan \theta $

Thus, slope of tangent at $x=0$ is,

$ \Rightarrow 10\times 0-4\cos 0 $ 

$ \Rightarrow 0-4=4 $ 

Hence, slope of normal at the same point is,

$ \because \,\,{{m}_{1}}\times {{m}_{2}}=-1 $ 

$ \Rightarrow 4\times {{m}_{2}}=-1 $ 

$ \Rightarrow {{m}_{2}}=\dfrac{-1}{4} $

Important Formulas from Class 12 Maths Chapter 6 Application of Derivatives

These theorems form the foundation for understanding the properties of continuous and differentiable functions. Each theorem has specific applications in solving problems and proving concepts in calculus.

Tips to Study Class 12 Maths Chapter 6: Application of Derivatives 

1. Solidify Your Understanding of Derivatives: Before delving into applications, ensure you have a strong grasp of differentiation techniques, including the product, quotient, and chain rules. This foundational knowledge is essential for tackling more complex problems.

2. Explore Real-World Applications: Study how derivatives are applied to determine rates of change, find tangents and normals to curves, and identify increasing or decreasing functions. Understanding these applications will help you appreciate the practical utility of derivatives.

3. Practice Problem-Solving Regularly: Engage with a variety of problems from your textbook and additional resources to reinforce your understanding. Regular practice will enhance your problem-solving skills and build confidence.

4. Utilize Quality Study Materials: Refer to comprehensive notes and solutions that cover key concepts and provide step-by-step explanations. Resources like Vedantu offer free PDF downloads of important questions and revision notes tailored to the CBSE syllabus.

5. Clarify Doubts Promptly: If you encounter difficulties, seek clarification from teachers, peers, or reliable online platforms. Addressing doubts early prevents misconceptions and solidifies your understanding.

Benefits of Important Questions for Class 12 Maths Chapter 6 By Vedantu

• Comprehensive Coverage: These questions encompass all key topics, ensuring a thorough understanding of areas under curves and between lines and curves.

• Exam-Oriented Preparation: Curated by experts, the questions align with the latest CBSE syllabus and exam patterns, aiding in effective exam preparation.

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• Time Management: Practicing these questions helps students develop efficient time management skills, ensuring they can complete exam sections within the allotted time.

Conclusion 

The collection of important questions for CBSE Class 12 Maths Chapter 6 - "Application of Derivatives" for the academic year 2024-25 serves as a valuable tool for students preparing for their board examinations. These questions are strategically curated to cover key concepts and applications within the chapter, allowing students to focus their revision efforts effectively. By working through these questions, students can reinforce their understanding of calculus principles and problem-solving skills, ensuring they are well-prepared for the rigorous CBSE board exams. These important questions not only aid in achieving academic success but also in still confidence and competence in tackling real-world scenarios where calculus plays a crucial role.

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