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Maths 3 Marks Question

Application of Derivatives · Uttarakhand Board (UBSE) STD 12 Science (Uttarakhand - English Medium). 20 questions.

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Question 13 Marks
Find the intervals in which the function f given by $f(x) = x ^ { 3 } + \frac { 1 } { x ^ { 3 } } , x \neq 0$ is $x (i)$ increasing. $(ii)$ decreasing.
Answer
$f(x) = x ^ { 3 } + \frac { 1 } { x ^ { 3 } } , x \neq 0$On differentiating both sides w.r.t. $x,$ we get,
$f'(x) = 3 x ^ { 2 } - \frac { 3 } { x ^ { 4 } }$
On putting $f'(x) = 0,$ we get,
$ 3 x ^ { 2 } - \frac { 3 } { x ^ { 4 } } = 0 \Rightarrow \frac { 3 x ^ { 6 } - 3 } { x ^ { 4 } } = 0$
$ \Rightarrow  3x^6 - 3 = 0   \Rightarrow 3x^6 = 3$
$ \Rightarrow  x^6 = 1   \Rightarrow  x =   \pm1$
Now, we find intervals in whch $f(x)$ is strictly increasing or strictly decreasing.
Interval $f'(x) = \frac { 3 x ^ { 6 } - 3 } { x ^ { 4 } }$ Sign of $f'(x)$
$x <-1$ $\frac { ( + ) } { ( + ) }$ $+ve$
$-1$ $\frac { ( - ) } { ( + ) }$ $-ve$
$x > 1$ $\frac { ( + ) } { ( + ) }$ $+v$
So, $f(x)$ is strictly increasing on intervals $(- \infty, -1)$ and $(1, \infty)$ and it is strictly decreasing on $(-1, 1)-\{0\}.$
Also, $f(x)$ is continuous at $x = 1 , -1.$
Hence, $f(x)$ is
  1. increasig on intervals $(- \infty, -1 ]$ and $[1, \infty).$
  2. decreasing on interval $[-1, 1]- \{0\}.$
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Question 23 Marks
Find the intervals in which the function f given by $f\left( x \right) = \frac{{4\sin x - 2x - x\cos x}}{{2 + \cos x}}$ is
  1. increasing
  2. decreasing
Answer
Given function is, $f\left( x \right) = \frac{{4\sin x - 2x - x\cos x}}{{2 + \cos x}}$
$= \frac{{4\sin x - x\left( {2 + \cos x} \right)}}{{2 + \cos x}}$
$= \frac{{4\sin x}}{{2 + \cos x}} - x$
$$Now,
$f'(x)=4\left({\frac{(2+cosx)cosx-sinx(0-sinx)}{(2+cosx)^2}}\right)-1$
$=4\left({\frac{2cosx+cos^2x+sin^2x}{(2+cosx)^2}}\right)-1$

$=4\left({\frac{8cosx+4-(2+cosx)^2}{(2+cosx)^2}}\right)$$f'(x)=4\left({\frac{cosx(4-cosx)}{(2+cosx)^2}}\right)$

${since\ - 1 \leqslant \cos x \leqslant 1} $
Hence
$\frac{{\cos x\left( {4 - \cos x} \right)}}{{{{\left( {2 + \cos x} \right)}^2}}} > 0\forall x \in \left( {0,\frac{\pi }{2}} \right)$ and $\left( {\frac{{3\pi }}{2},2\pi } \right)$
$\frac{{\cos x\left( {4 - \cos x} \right)}}{{{{\left( {2 + \cos x} \right)}^2}}} < 0\forall \in \left( {\frac{\pi }{2},\frac{{3\pi }}{2}} \right)$
So, f(x) is increasing in $\left( {0,\frac{\pi }{2}} \right)$ and $\left( {\frac{{3\pi }}{2},2\pi } \right)$
and if f(x) is decreasing in $\left( {\frac{\pi }{2},\frac{{3\pi }}{2}} \right)$

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Question 33 Marks
Show that the function given by f(x) = $\frac{\log x}{x}$ has maximum at x = e.
Answer
It is given that f(x) = $\frac{\log x}{x}$
Then, f'(x) = $\frac{\mathrm{x}\left(\frac{1}{\mathrm{x}}\right)-\log \mathrm{x}}{\mathrm{x}^{2}}=\frac{1-\log \mathrm{x}}{\mathrm{x}^{2}}$
Now, f'(x) = 0
$\Rightarrow$ 1 - log x = 0
$\Rightarrow$ log x = 1
$\Rightarrow$ log x = log e
$\Rightarrow$ x = e
Further, $f^{\prime \prime}(x)=\frac{x^{2}\left(-\frac{1}{x}\right)-(1-\log x)(2 x)}{x^{4}}$
= $\frac{-x-2 x(1-\log x)}{x^{4}}$
= $\frac{-3+2 \log x}{x^{3}}$
Now, $f^\prime{^\prime}$(e) = $\frac{-3+2 \log e}{e^{3}}=\frac{-3+2}{e^{3}}=\frac{-1}{e^{3}}<0$
Therefore, by second derivative test, f is the maximum at x = e.
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Question 43 Marks
Find the local maxima and local minima of function,
$f(x)=x \sqrt{1-x}, \quad 0<x<1$
Find also the local maximum and the local minimum value.
Answer
Here,
$f(x)=x \sqrt{1-x}, 0<x<1$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\sqrt{1-\mathrm{x}}+\mathrm{x} \frac{1}{2 \sqrt{1-\mathrm{x}}}(-1)$
$=\sqrt{1-\mathrm{x}}-\frac{\mathrm{x}}{2 \sqrt{1-\mathrm{x}}}$
$=\frac{2(1-\mathrm{x})-\mathrm{x}}{2 \sqrt{1-\mathrm{x}}}$
f'(x) = 0
$\Rightarrow \frac{2-3 x}{2 \sqrt{1-x}}=0$
$\Rightarrow$ 2 - 3x = 0
$\Rightarrow \mathrm{x}=\frac{2}{3}$
$f^{\prime \prime}(x)=\frac{1}{2}\left[\frac{\sqrt{1-x}(-3)-(2-3 x)\left(\frac{-1}{2 \sqrt{1-x}}\right)}{1-x}\right]$
$=\left[\frac{\sqrt{1-\mathrm{x}}(-3)+(2-3 \mathrm{x})\left(\frac{-1}{2 \sqrt{1-\mathrm{x}}}\right)}{2(1-\mathrm{x})}\right]$
$=\left[\frac{-6(1-\mathrm{x})+(2-3 \mathrm{x})}{2(1-\mathrm{x})^{\frac{3}{2}}}\right]$
$=\frac{3 x-4}{4(1-x)^{\frac{3}{2}}}$
Now,
$\mathrm{f}^{\prime \prime}\left(\frac{2}{3}\right)=-\frac{2}{4( \frac13)^ \frac32}$ < 0,
Therefore, the second derivative test, x = $\frac23$ is the point of local maxima and the local maximum value
of 'f ' at x = $\frac23$ is
f($\frac23$) = $\frac{2}{3\sqrt3}$
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Question 53 Marks
Find the local maxima and local minima of function
$g(x)=\frac{1}{x^{2}+2}$
Find also the local maximum and the local minimum value.
Answer
Here,
$g(x)=\frac{1}{x^{2}+2}$
$\Rightarrow g^{\prime}(x)=\frac{-(2 x)}{\left(x^{2}+2\right)^{2}}=0$
$\Rightarrow$ x = 0
Now, for values close to x = 0 and to the left of 0
g'(x) > 0
Also, for values close to x = 0 and to the right of 0,
g'(x) < 0
Then, by first derivative test,
$\Rightarrow$ x = 0 is point of local maxima and local maximum of g at x = 0 is
$g(0)=\frac{1}{0+2}=\frac{1}{2}$
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Question 63 Marks
Find the local maxima and local minima, of function
$g(x)=\frac{x}{2}+\frac{2}{x}, \quad x>0$
Find also the local maximum and the local minimum value.
Answer
Here,
$g(x)=\frac{x}{2}+\frac{2}{x}, x>0$
$\Rightarrow g^{\prime}(x)=\frac{1}{2}-\frac{2}{x^{2}}$
Now, $g'(x) = 0$
$\Rightarrow \frac{1}{2}-\frac{2}{x^{2}}=0$
$\Rightarrow \frac{2}{x^{2}}=\frac{1}{2}$
$\Rightarrow x^2 = 4$
$\Rightarrow x = 2, -2.$
Since $x > 0,$ we take $x = 2$
$g^{\prime \prime}(x)=\frac{4}{x^{3}}$
$g^{\prime \prime}(2)=\frac{4}{2^{3}}=\frac{1}{2}>0$
Then, by second derivative test,
$\Rightarrow x = 2$ is point of local minima and local minimum value of $g$ at $x = 2$ is
$g(2)=\frac{2}{2}+\frac{2}{2}=1+1=2$
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Question 73 Marks
Find the local maxima and local minima of the function
$f(x) = x^3 - 6x^2 + 9x + 15$
Find also the local maximum and the local minimum value.
Answer
Here,
$f(x) = x^3 - 6x^2 + 9x + 15$
$\Rightarrow f'(x) = 3x^2 - 12x + 9$
Now, $f'(x) = 0$
$\Rightarrow 3x^2 - 12x + 9 = 0$
$\Rightarrow 3(x - 1)(x - 3) = 0$
$\Rightarrow$ x = 1, 3
$f ''(x) = 6x - 12 = 6(x - 2)$
Now, $f ''(1) = 6(1 - 2) = -6 < 0$ and $f ''(3) = 6(3 - 2) = 6 > 0$
Then, by second derivative test,
$\Rightarrow x = 1$ is point of local maxima and local maximum value of f at $x = 1$ is
$f(1) = 13 - 6 + 9 + 15 = 19$
Also,
$x = 3$ is point of local minima and local minimum value of $f$ at $x = 3$ is
$f(3) = 27 - 54 + 27 + 15 = 15$
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Question 83 Marks
Find the local maxima and local minima of the function. Find also the local maximum and the local minimum value: f(x) = sinx - cosx, 0 < x < 2$ \pi$.
Answer
We have, $ f ( x ) = \sin x - \cos x , 0 < x < 2 \pi$
On differentiating both sides w.r.t. x ,we get,
f'(x) = cosx + sinx..........(i)
For local maximum and local minimum,
Put f'(x) = 0,
i.e. $ \cos x + \sin x = 0 \Rightarrow \cos x = - \sin x$
$ \Rightarrow \quad \tan x = - 1 \Rightarrow x = \pi - \frac { \pi } { 4 } \text { or } 2 \pi - \frac { \pi } { 4 }$
$ \Rightarrow \quad x = \frac { 3 \pi } { 4 } \text { or } \frac { 7 \pi } { 4 }$
Again, on differentiating both sides of Eq. (i)
w.r.t. x, we get,
f'' (x)=-sin x + cos x
When $x = \frac { 3 \pi } { 4 } ,$then
$f''\left( \frac { 3 \pi } { 4 } \right) = - \sin \frac { 3 \pi } { 4 } + \cos \frac { 3 \pi } { 4 }$
$ = - \sin \left( \pi - \frac { \pi } { 4 } \right) + \cos \left( \pi - \frac { \pi } { 4 } \right)$
$= - \sin \frac { \pi } { 4 } - \cos \frac { \pi } { 4 } < 0$
When $ x = \frac { 7 \pi } { 4 } , \text { then } f ^ { \prime \prime } \left( \frac { 7 \pi } { 4 } \right) = - \sin \frac { 7 \pi } { 4 } + \cos \frac { 7 \pi } { 4 }$
$ = - \sin \left( 2 \pi - \frac { \pi } { 4 } \right) + \cos \left( 2 \pi - \frac { \pi } { 4 } \right)$
$ = \sin \frac { \pi } { 4 } + \cos \frac { \pi } { 4 } > 0$
Thus, $ x = \frac { 3 { \pi } } { 4 }$ is a point of local maxima and $ x = \frac { 7 \pi } { 4 }$ is a point of local minima.
Now, the local maximum value,
$ f \left( \frac { 3 \pi } { 4 } \right) = \sin \frac { 3 \pi } { 4 } - \cos \frac { 3 \pi } { 4 }$
$ = \sin \left( \pi - \frac { \pi } { 4 } \right) - \cos \left( \pi - \frac { \pi } { 4 } \right)$
$ = \sin \frac { \pi } { 4 } + \cos \frac { \pi } { 4 } = \frac { 1 } { \sqrt { 2 } } + \frac { 1 } { \sqrt { 2 } } = \frac { 2 } { \sqrt { 2 } } = \sqrt { 2 }$
and the local minimum value,
$ f \left( \frac { 7 \pi } { 4 } \right) = \sin \frac { 7 \pi } { 4 } - \cos \frac { 7 \pi } { 4 }$
$ = \sin \left( 2 \pi - \frac { \pi } { 4 } \right) - \cos \left( 2 \pi - \frac { \pi } { 4 } \right)$
$ = - \sin \frac { \pi } { 4 } - \cos \frac { \pi } { 4 } = - \frac { 1 } { \sqrt { 2 } } - \frac { 1 } { \sqrt { 2 } }$

$ = - \frac { 2 } { \sqrt { 2 } } = - \sqrt { 2 }$

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Question 93 Marks
Find the local maxima and local minima, of function. Find also the local maximum and the local minimum value, as the case may be:
h(x) = sin x + cos x, $0<x<\frac{\pi}{2}$
Answer
Here,
h(x) = sin x + cos x, $0<x<\frac{\pi}{2}$
h'(x) = cos x - sin x
Now, h'(x) = 0
$\Rightarrow$ cos x - sin x = 0
$\Rightarrow$ cos x = sin x
$\Rightarrow$ tan x = 1
$\Rightarrow \mathrm{x}=\frac{\pi}{4}$
h''(x) = -sin x - cos x = -(sin x + cos x)
$h^{\prime \prime}\left(\frac{\pi}{4}\right)=-\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=-\frac{2}{\sqrt{2}}=-\sqrt{2}<0$
Then, by second derivative test,
$\Rightarrow \mathrm{x}=\frac{\pi}{4}$ is point of local maxima and local maximum value of h at $x=\frac{\pi}{4}$ is
$h\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$
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Question 103 Marks
Find the local maxima and local minima, of function. Find also the local maximum and the local minimum value, as the case may be: $g(x) = x^3 - 3x$
Answer
Here, $g(x) = x^3 - 3x$
$\Rightarrow g'(x) = 3x^2 - 3$
Now, $g'(x) = 0$
$\Rightarrow 3x^2 - 3 = 0$
$\Rightarrow 3x^2 = 3$
$\Rightarrow x = 1, -1$ are the critical points.
Further, $g''(x) = 6x$
Now, $g''(1) = 6 > 0$
and $g''(-1) = -6 < 0$
Then, by second derivative test,
$\Rightarrow x = 1$ is point of local minimum of $g$ and at $x = 1,$ we have
$g(1) = 1^3 - 3 = 1 - 3 = -2$
Moreover,
$x = -1$ is point of local maximum value of $g$ and at $x = -1,$ we have
$g(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2$
So, the local maxima and minima are $-1$ and $1$ respectively.
The local minimum and maximum values are $-2$ and $2$ respectively.
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Question 113 Marks
Find two positive numbers whose sum is $16$ and the sum of whose cubes is minimum.
Answer
Let one number be x . Then, the other number is $(16-\mathrm{x})$.
Let $S(x)$ be the sum of these number. Then,
$S(x)=x^3+(16-x)^3$
$\Rightarrow S^{\prime}(x)=3 x^2-3(16-x)^2$
$\Rightarrow S^{\prime \prime}(x)=6 x+6(16-x)$
Now, $S^{\prime}(x)=0$
$\Rightarrow 3 x^2-3(16-x)^2=0$
$\Rightarrow x^2-(16-x)^2=0$
$\Rightarrow x^2-256-x^2+32 x=0$
$\Rightarrow x=8$
Now, $S^{\prime \prime}(8)=6(8)+6(16-8)$
$=48+48=96>0$
Then, by second derivative test, $x=8$ is the point of local minima of $S$.
Therefore, the sum of the cubes of the numbers is the minimum when the numbers are 8 and $16-8=8$.
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Question 123 Marks
Find two positive numbers $x$ and $y$ such that their sum is $35$ and the product $x^2y^5$ is a maximum.
Answer
Let one number be $x$. Then, the other number is $y = (35 - x)$
Let $P(x) = x^2y^5$ 
$\Rightarrow P(x) = x^2(35 - x)^5$
Then,
$P'(x) = 2x(35 - x)^5 - 5x^2(35 - x)^4$
$= x(35 - x)^4 [2(35 - x) - 5x]$
$= x(35 - x)^4 [70 - 7x]$
$= 7x(35x - x)^4 (10 - x)$
Now, $P''(x) = 7(35 - x)^4 (10 - x) + 7 \times [-(35 - x)^4 - 4(35 - x)^3 (10 - x)]$
$=7(35-x)^{4}(10-x)-7 x(35-x)^{4}-28 x(35-x)^{3}(10-x)$
$= 7(35-x)^{3}[(35-x)(10-x)-x(35-x)-4 x(10-x)]$
$= 7(35-x)^{3}\left[350-45 x+x^{2}-35 x+x^{2}-40 x+4 x^{2}\right]$
$= 7(35 - x)^3 [6x^2 - 120x + 350]$
Now, $P'(x) = 0$
$\Rightarrow x = 0, 35, 10$
When $x = 0, 35$ This will make the product $x^2y^5$ equal to 0.
Therefore, $x = 0, 35$ cannot be possible values of $x.$
And when $x = 10$
Then, we have,
$P''(x) = 7(35 - 10)3[6(10)2 - 120(10) + 350]$
$= 7(25)3 [-250] < 0$
Then, by second derivative test,
$x = 10$ and $y = 35 - 10 = 25$ is the point of local maxima of $P.$
Therefore, the required number are $10$ and $25.$
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Question 133 Marks
Find two positive integers x and y such that $x + y = 60$ and $x{y^3}$ is maximum.
Answer
Given: $x + y = 60,x > 0,y > 0$…(i)

Let $P = x{y^3}$ [To be maximized] …(ii)

Putting from eq. (i), $x = 60 - y$ in eq. (ii),

$P = \left( {60 - y} \right){y^3} = 60{y^3} - {y^4}$

$\Rightarrow \frac{{dP}}{{dy}} = 180{y^2} - 4{y^3} = 4{y^2}\left( {45 - y} \right)$ …(iii)

Now $\frac{{dP}}{{dy}} = 0$

$\Rightarrow 4{y^2}\left( {45 - y} \right) = 0$

$\Rightarrow y = 0,45$

It is clear that $\frac{{dP}}{{dy}}$ changes sign from positive to negative as y increases through 45.

Therefore, P is maximum when $y = 45.$

Hence, $x{y^3}$ is maximum when $x = 60 - 45 = 15$ and y = 45.

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Question 143 Marks
Find two numbers whose sum is $24$ and whose product is as large as possible.
Answer
Let the two numbers be $x$ and $y$. According to the question, $x + y = 24$
$\Rightarrow y = 24 - x …(i)$
And let $z$ is the product of $x$ and $y.$
$\Rightarrow z = xy$
$\Rightarrow z = x(24 - x) [$From eq. $(i)]$
$\Rightarrow z = 24x - x^2$
$\Rightarrow \frac{{dz}}{{dx}} = 24 - 2x$ and $\frac{{{d^2}z}}{{d{x^2}}} = - 2$
Now to find turning point, $\frac{{dz}}{{dx}} = 0$
$\Rightarrow 24 - 2x = 0 \Rightarrow x = 12$
At $x = 12,\frac{{{d^2}z}}{{d{x^2}}} = - 2$ [Negative]
$\therefore x = 12$ is a point of local maxima and z is maximum at $x = 12.$
$\therefore $ From eq. $(i), ​​​​​​​ y = 24 - 12 = 12$
Therefore, the two required numbers are $12$ and $12.$
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Question 153 Marks
Prove that $y = \frac{{4\sin \theta }}{{\left( {2 + \cos \theta } \right)}} - \theta $ is an increasing function of $\theta$ in $\left[ {0,\frac{\pi }{2}} \right]$
Answer
Given: $y = \frac{{4\sin \theta }}{{\left( {2 + \cos \theta } \right)}} - \theta $

$ \Rightarrow \frac{{dy}}{{d\theta }} = \frac{{\left( {2 + \cos \theta } \right).4\cos \theta - 4\sin \theta \left( { - \sin \theta } \right)}}{{{{\left( {2 + \cos \theta } \right)}^2}}} - 1$

$= \frac{{8\cos \theta + 4{{\cos }^2}\theta + 4{{\sin }^2}\theta }}{{{{\left( {2 + \cos \theta } \right)}^2}}} - 1$

$ \Rightarrow \frac{{dy}}{{d\theta }} = \frac{{8\cos \theta + 4\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) - {{\left( {2 + \cos \theta } \right)}^2}}}{{{{\left( {2 + \cos \theta } \right)}^2}}}$

$= \frac{{8\cos \theta + 4 - {{\left( {2 + \cos \theta } \right)}^2}}}{{{{\left( {2 + \cos \theta } \right)}^2}}}$

$\Rightarrow \frac{{dy}}{{d\theta }} = \frac{{\left( {8\cos \theta + 4} \right) - \left( {4 + 4\cos \theta + {{\cos }^2}\theta } \right)}}{{{{\left( {2 + \cos \theta } \right)}^2}}}$

$= \frac{{4\cos \theta - {{\cos }^2}\theta }}{{{{\left( {2 + \cos \theta } \right)}^2}}}$

$= \frac{{\cos \theta \left( {4 - \cos \theta } \right)}}{{{{\left( {2 + \cos \theta } \right)}^2}}}$

Since $0 \leqslant \theta \leqslant \frac{\pi }{2}$ and we have $0 \leqslant \cos \theta \leqslant 1$, therefore $4 - \cos \theta > 0$.

$\therefore \frac{{dy}}{{dx}} \geqslant 0$ for $0 \leqslant \theta \leqslant \frac{\pi }{2}$

Hence, y is an increasing function of $\theta $ in $\left[ {0,\frac{\pi }{2}} \right]$

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Question 163 Marks
Find the value$(s)$ of $x$ for which $y = [x(x - 2)]^2$ is an increasing function.
Answer
Given function is $y = [x(x - 2)]^2= [x^2 - 2x]^2.$
Therefore,on differentiating both sides w.r.t x, we get,
$\frac { d y } { d x } = 2 \left( x ^ { 2 } - 2 x \right) \frac { d } { d x } \left( x ^ { 2 } - 2 x \right) $
$= 2(x^2 - 2x) (2x - 2)$
$= 4x(x - 2) (x - 1)$
Therefore,on putting $\frac { d y } { d x }$= 0, we get,
$4x(x - 2)(x - 1) = 0 \Rightarrow x = 0, 1$ and $2.$
Now, we find interval in which f(x) is strictly increasing or strictly decreasing.
Interval $f'(x) = 12x(x + 1)(x - 2)$ Sign of $f'(x)$
$(-\infty,0)$ $(-)(-)(-)$ $-ve$
$(0, 1)$ $(+)(-)(-)$ $+ve$
$(1, 2)$ $(+)(-)(+)$ $-ve$
$(2,\infty)$ $(+)(+)(+)$ $+ve$
Therefore, y is strictly increasing in $( 0, 1 )$ and $(2, \infty).$
Also, $y$ is a polynomial function, so it continuous at $x = 0, 1$ and $2.$
Hence, $y$ is increasing in $[ 0, 1] \cup  [2,\infty).$
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Question 173 Marks
Show that $y = \log \left( {1 + x} \right) - \frac{{2x}}{{2 + x}},x > - 1$ is an increasing function of x throughout its domain.
Answer
Given: $y = \log \left( {1 + x} \right) - \frac{{2x}}{{2+x}}$
$\therefore \frac{{dy}}{{dx}} = \frac{1}{{1 + x}}\frac{d}{{dx}}\left( {1 + x} \right) - \left[ {\frac{{\left( {2 + x} \right)\frac{d}{{dx}}\left( {2x} \right) - 2x\frac{d}{{dx}} \left( {2 + x} \right)}}{{{{\left( {2 + x} \right)}^2}}}} \right]$
$= \frac{1}{{1 + x}} - \frac{{\left( {4 + 2x - 2x} \right)}}{{{{\left( {2 + x} \right)}^2}}}$
$= \frac{1}{{1 + x}} - \frac{4}{{{{\left( {2 + x} \right)}^2}}}$
$\Rightarrow \frac{{dy}}{{dx}} = \frac{{\left( {2 + {x}} \right)^2 - 4\left( {1 + x} \right)}}{{\left( {1 + x} \right){{\left( {2 + x} \right)}^2}}}$
$=\frac{{{x^2}}}{{\left( {1 + x} \right)\left( {2 + {x^2}} \right)}}$ ...(i)
Domain of the given function is given to be $x > - 1$
$\Rightarrow x + 1 > 0$
Also ${\left( {2 + x} \right)^2} > 0$ and ${x^2} \geqslant 0$
$\therefore$ From eq. (i), $\frac{{dy}}{{dx}} \geqslant 0$ for all x in domain $x > - 1$ and f is an increasing function.
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Question 183 Marks
A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is $10 cm$.
Answer
Since, $V = \frac{4}{3}\pi {x^3}$
$\therefore \frac{{dV}}{{dx}} = \frac{d}{{dt}}\left( {\frac{4}{3}\pi {r^3}} \right)$
$= \frac{4}{3}\pi .3{r^2} = 4\pi {r^2}$
At x = 10 cm
$\Rightarrow \frac{{dV}}{{dx}} = 4\pi {\left( {10} \right)^2} = 400\pi $
Therefore, the volume is increasing at the rate of $400\pi\ cm^3/\sec$.
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Question 193 Marks
A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Answer
Let V be the volume of sphere
$V = \frac{4}{3}\pi {r^3}$
$\frac{{dv}}{{dt}} = 900c{m^3}/s$ (given)
Now , $\frac{{dv}}{{dt}} = \frac{4}{3}.\pi .3{r^2}.\frac{{dr}}{{dt}}$
$\implies900 = 4\pi {r^2}.\frac{{dr}}{{dt}}$
$\implies900 = 4\pi \times {\left( {15} \right)^2}.\frac{{dr}}{{dt}}$ [r = 15 cm]
$\implies\frac{{900}}{{4\pi \times 225}} = \frac{{dr}}{{dt}}$
$\therefore \frac{1}{\pi }cm/s = \frac{{dr}}{{dt}}$
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Question 203 Marks
The length $x$ of a rectangle is decreasing at the rate of $5$ cm/minute and the width $y$ is increasing at the rate of $4$ cm/minute. When $x = 8\ cm$ and $y = 6\ cm$, find the rates of change of $(a)$ the perimeter, and $(b)$ the area of the rectangle.
Answer
Given: Rate of decrease of length $x$ of rectangle is $5$ cm/minute.
$ \Rightarrow \frac{{dx}}{{dt}}$ is negative $= –5$ cm/minute
$ \Rightarrow \frac{{dx}}{{dt}} = – 5$ cm/minute $….(i)$
Also, Rate of increase of width y of rectangle is $4$ cm/minute
$\Rightarrow \frac{{dy}}{{dt}}$ is positive= 4 cm/minute
$\Rightarrow \frac{{dy}}{{dt}} = 4$ cm/minute $….(ii)$
let $z$ denotes the perimeter of rectangle at any time $t$
$\therefore x = 2x + 2y$
$\Rightarrow \frac{{dz}}{{dt}} = 2\frac{{dx}}{{dt}} + 2\frac{{dy}}{{dt}}$ (from (i) and (ii) )
$= 2\left( { - 5} \right) + 2\left( 4 \right) = - 2$ which is negative.
$\therefore$ Perimeter of the rectangle is decreasing at the rate of $2$ cm/min.
(b) Let $z$ denotes the area of rectangle at any time $t$
$\Rightarrow z = xy$
$ \Rightarrow \frac{{dz}}{{dt}} = x\frac{{dy}}{{dt}} + y\frac{{dx}}{{dt}}$
$= 8\left( 4 \right) + 6\left( { - 5} \right) = 2$ which is positive. (from (i) and (ii))
$\therefore$ Area of the rectangle is increasing at the rate of $2\ cm^2/min$
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