Case study (4 Marks)

Question 14 Marks

x and y are the sides of two squares such that y = x - x². Find the rate of change of the area of second square with respect to the area of first square.

Answer

Given x and y are the sides of two squares such that y = x - x².
$\therefore$ Area of the first square, A₁ = x²
and area of the second square, A₂ = y² = (x - x²)²
$\therefore\ \frac{\text{dA}_1}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{x}^2)=2\text{x}\cdot\frac{\text{dx}}{\text{dt}}$
and $\frac{\text{dA}_2}{\text{dt}}=\frac{\text{dt}}{\text{dx}}(\text{x}-\text{x}^2)^2$
$=2(\text{x}-\text{x}^2)(1-2\text{x})\frac{\text{dx}}{\text{dt}}$
$\therefore\ \frac{\text{dA}_2}{\text{dx}_1}=\frac{\frac{\text{dA}_2}{\text{dt}}}{\frac{\text{dA}_1}{\text{dt}}}=\frac{(2\text{x}-2\text{x}^2)(1-2\text{x})\frac{\text{dx}}{\text{dt}}}{2\text{x}\frac{\text{dx}}{\text{dt}}}$
$=\frac{(1-2\text{x})2\text{x}(1-\text{x})}{2\text{x}}=(1-2\text{x})(1-\text{x})=2\text{x}^2-3\text{x}+1$

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Question 24 Marks

A telephone company in a town has 500 subscribers on its list and collects fixed charges of Rs. 300/- per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of Re 1/- one subscriber will discontinue the service. Find what increase will bring maximum profit?

Answer

Consider that company increases the annual subscription by Rs. x
So, x subscribes will discontinue the service.
$\therefore$ Total revenue of company after the increment is given by
R(x) = (500 - x)(300 + x)
= -x² + 200x + 150000
⇒ R's) = -2x + 200
R'(x) = 0
⇒ 2x = 200 ⇒ x = 100
Also R''(x) = -2 < 0
So, R(x) is maximum when x = 100.
Hence, the company should increase the subscription fee Rs. 100, so that it has maximum profit.

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Question 34 Marks

Show that $\text{f(x)}=\tan^{-1}(\sin\text{x}+\cos\text{x})$ is an increasing function in $\Big(0,\frac{\pi}{4}\Big).$

Answer

We have, $\text{f(x)}=\tan^{-1}(\sin\text{x}+\cos\text{x})$
$\therefore\ \text{f}'(\text{x})=\frac{1}{1+(\sin\text{x}+\cos\text{x})^2}\cdot(\cos\text{x}-\sin\text{x})$
$=\frac{1}{1+\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}.\cos\text{x}}(\cos\text{x}-\sin\text{x})$
$=\frac{1}{(2+\sin2\text{x})}(\cos\text{x}-\sin\text{x})$
$\big[\because\sin2\text{x}=2\sin\text{x}\cos\text{x and }\sin^2\text{x}+\cos^2\text{x}=1\big]$
For $\text{f}'(\text{x})\geq0.$
$\frac{1}{(2+\sin2\text{x})}\cdot(\cos\text{x}-\sin\text{x})\geq0$
$\Rightarrow\ \cos\text{x}-\sin\text{x}\geq0$ $\Big[\because(2+\sin2\text{x})\geq0\text{ in }\Big(0,\frac{\pi}{4}\Big)\Big]$
$\Rightarrow\ \cos\text{x}\geq\sin\text{x}$
Which is true, if $\text{x}\in\Big(0,\frac{\pi}{4}\Big)$
Hence, f(x) is an increasing function in $\Big(0,\frac{\pi}{4}\Big).$

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Question 44 Marks

Show that for $\text{a}\geq1,\text{ f(x)}=\sqrt{3}\sin\text{x}-\cos\text{x}-2\text{ax}+\text{b}$ is decreasing in R.

Answer

We have,$\text{ f(x)}=\sqrt{3}\sin\text{x}-\cos\text{x}-2\text{ax}+\text{b}$
$\Rightarrow\ \text{f(x)}=\sqrt{3}\cos\text{x}-(-\sin\text{x})-2\text{a}$
$=\sqrt{3}\cos\text{x}+\sin\text{x}-2\text{a}$
$=2\Big[\frac{\sqrt{3}}{2}\cdot\cos\text{x}+\frac{1}{2}\cdot\sin\text{x}\Big]-2\text{a}$
$=2\Big[\cos\frac{\pi}{6}\cdot\cos\text{x}+\sin\frac{\pi}{6}\cdot\sin\text{x}\Big]-2\text{a}$
$=2\cos\Big(\frac{\pi}{6}-\text{x}\Big)-2\text{a}$ $\big[\therefore\cos(\text{A}-\text{B})=\cos\text{A}\cdot\text{B}+\sin\text{A}\cdot\sin\text{B}\big]$
$=2\Big[\cos\Big(\frac{\pi}{6}-\text{x}\Big)-\text{a}\Big]$
Since, $\cos\text{x}\in[-1,1]\text{ and a}\geq1$
$\therefore\ 2\Big[\cos\Big(\frac{\pi}{6}-\text{a}\Big)-\text{a}\Big]\leq0$
We know tha if $\text{f(x)}\leq0,$ then f(x) is a decreasing function.
Thus, f(x) is a decreasing function in R.

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Question 54 Marks

A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is propotional to the surface. Prove that the radius is decreasing at a constant rate.

Answer

We have, rate of decrease of the volume of spherical ball of salt at any instant is surface. Let the radius of the spherical ball of the salt be r.
$\therefore$ Volume of the ball $(\text{V})=\frac{4}{3}\pi\text{r}^3$
and surface area $(\text{S})=4\pi\text{r}^2$
$\because\ \frac{\text{dV}}{\text{dT}}\propto\text{S}$
$\Rightarrow\ \frac{\text{d}}{\text{dt}}\Big(\frac{4}{3}\pi\text{r}^3\Big)\propto4\pi\text{r}^2$
$\Rightarrow\ \frac{4}{3}\pi3\text{r}^2\frac{\text{dr}}{\text{dt}}\propto4\pi\text{r}^2$
$\Rightarrow\ \frac{\text{dr}}{\text{dt}}\propto\frac{4\pi\text{r}^2}{4\pi\text{r}^2}$
$\Rightarrow\ \frac{\text{dr}}{\text{dt}}=\text{k.1}$ [where, k is the proportionality constant]
$\Rightarrow\ \frac{\text{dr}}{\text{dt}}=\text{k}$
Hence, the radius of ball is decreasing at a constant rate.

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Question 64 Marks

If the area of a circle increases at a uniform rate, then prove that perimeter varies inversely as the radius.

Answer

Let the radius of circle at any time t is r.
Then area of the circle at any time t is $\text{A}=\pi\text{r}^2$
$\therefore\ \frac{\text{d}}{\text{dt}}\text{A}=\frac{\text{d}}{\text{dt}}(\pi\text{r}^2)$
$\Rightarrow\ \frac{\text{dA}}{\text{dt}}=2\pi\text{r}\cdot\frac{\text{dr}}{\text{dt}}\ \ \dots(\text{i})$
Since, the area of a circle increases at a uniform rate, we have
$\frac{\text{dA}}{\text{dt}}=\text{k},$ Where, k is a constant ...(ii)
From (i) and (ii), we get
$2\pi\text{r}\cdot\frac{\text{dr}}{\text{dt}}=\text{k}$
$\Rightarrow\ \frac{\text{dr}}{\text{dt}}=\frac{\text{k}}{2\pi\text{r}}=\frac{\text{k}}{2\pi}\cdot\Big(\frac{1}{\text{r}}\Big)$
$\Rightarrow\ 2\pi\frac{\text{dr}}{\text{dt}}=\frac{\text{k}}{\text{r}}$
$\Rightarrow\ \frac{\text{d}(2\pi\text{r})}{\text{dt}}=\frac{\text{k}}{\text{r}}$
$\Rightarrow\ \frac{\text{dP}}{\text{dt}}=\frac{\text{k}}{\text{r}},$ where $\text{P}=2\pi\text{r}$
$\Rightarrow\ \frac{\text{dP}}{\text{dt}}\propto\frac{1}{\text{r}}$
Thus perimeter varies inversely as time radius.

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Maths Case study (4 Marks)

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