Important Questions for CBSE Class 12 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 Related Links for CBSE Class 12 Maths