3 Marks Question

Question 13 Marks

The feasible region for a LPP is shown in Figure. Evaluate Z = 4x + y at each of the corner points of this region. Find the minimum value of Z, if it exists.

Answer

From the shaded region, it is clear that feasible region is unbounded with the corner points A(4, 0), B(2, 1) and C(0, 3). Also, we have Z = 4x + y
(Since, x + 2y = 4 and x + y = 3y =1 and x = 2)

Corner Points	Corresponding value of Z
(4,0)	16
(2, 1)	9
(0, 3)	3 (minimum)

Now, we see that 3 is the smallest value of Z at the corner point (0, 3). Note that here we see that the region is unbounded, therefore 3 may or may not be the minimum value of Z.
To decide this issue, we graph the inequality 4x + y < 3 and check whether the resulting open half plan has no point in common with feasible region otherwise, Z has no minimum value
From the shown graph above, it is clear that there is no point in common with feasible region and hence Z has minimum value of 3 at (0, 3).

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Question 23 Marks

Evaluate the integral: $\int(x+1) \sqrt{x^2+x+1} d x$

Answer

Let the given integral be,
$ I =\int(x+1) \sqrt{x^2+x+1} d x$
Also$, x +1=\lambda \frac{d}{d x}\left(x^2+x+1\right)+\mu$
$\Rightarrow x +1=\lambda(2 x +1)+\mu$
$\Rightarrow x +1=(2 \lambda) x +\lambda+\mu$
Eqyating coeffient of like terms
$2 \lambda=1$
$\Rightarrow \lambda=\frac{1}{2}$
And
$\lambda+\mu=1$
$\Rightarrow \frac{1}{2}+\mu=1$
$\therefore \mu=\frac{1}{2}$
$\therefore I =\frac{1}{2} \int(2 x+1) \sqrt{x^2+x+1} d x+\frac{1}{2} \int \sqrt{x^2+x+1} d x$
$=\frac{1}{2} \int(2 x+1) \sqrt{x^2+x+1} d x+\frac{1}{2} \int \sqrt{x^2+x+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1} d x$
$=\frac{1}{2} \int(2 x+1) \sqrt{x^2+x+1} d x+\frac{1}{2} \int \sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} d x$
Let $x ^2+ x +1= t$
$\Rightarrow(2 x +1) dx = dt$
Then,
$I=\frac{1}{2} \int \sqrt{t} d t+\frac{1}{2}\left[\frac{x+\frac{1}{2}}{2} \sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}+\frac{3}{8} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\right|\right]+C$
$=\frac{1}{2} \times \frac{2}{3} t^{\frac{3}{2}}+\frac{1}{2}\left[\left(\frac{2 x+1}{4}\right) \sqrt{x^2+x+1}+\frac{3}{8} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^2+x+1}\right|\right]+C$
$=\frac{1}{3}\left(x^2+x+1\right)^{\frac{3}{2}}+\frac{1}{2}\left[\left(\frac{2 x+1}{4}\right) \sqrt{x^2+x+1}+\frac{3}{8} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^2+x+1}\right|\right]+C$

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Question 33 Marks

Evaluate: $\int x \sin ^3 x \cos xdx$.

Answer

We can write it as $\int x \sin ^2 x \sin x \cos x d x$
We also know that $2 \sin x \cdot \cos x=\sin 2 x$
$\int x \sin ^2 x \sin x \cos x d x=\frac{1}{2} \int x \sin ^2 x \sin 2 x d x$
We also know that $\sin ^2 x=\frac{1-\cos 2 x}{2}$
$\frac{1}{2} \int x \sin ^2 x \sin 2 x d x=\frac{1}{2} \int x \cdot\left(\frac{1-\cos 2 x}{2}\right) \sin 2 x d x$
$=\frac{1}{2}\left[\left(\int \frac{x \sin 2 x}{2} d x-\int \frac{x \cos 2 x \sin 2 x}{2} d x\right)\right]$
Here $\operatorname{Sin} 4 x =2 \sin 2 x \cdot \cos 2 x$
$=\frac{1}{2}\left[\left(\int \frac{x \sin 2 x}{2} d x-\frac{1}{4} \int x \sin 4 x d x\right)\right]$
Using $\text{BY PART METHOD.}$
Here $x$ is first function and $\operatorname{Sin} 2 x$ and $\sin 4 x$ as the second function.
$\int \text { a.b.dx }=a \int b . dx-\int\left[\frac{d a}{d x} \cdot \int b d x\right] d x$
$=\frac{1}{2}\left[\left(\frac{1}{2}\left\{x \int \sin 2 xdx-\int\left(\frac{dx}{dx} \cdot \int \sin 2 xdx\right) dx\right\}\right)-\left(\frac{1}{4}\left\{x \int \sin 4 x-\int\left(\frac{dx}{dx} \cdot \int \sin 4 xdx\right) dx\right\}\right)\right]$
$=\frac{1}{2}\left[\left(\frac{1}{2}\left\{-x \frac{\cos 2 x}{2}+\int \frac{\cos 2 x}{2} d x\right\}\right)-\left(\frac{1}{4}\left\{-x \frac{\cos 4 x}{4}+\int \frac{\cos 4 x}{4} d x\right\}\right)\right]$
$=\frac{1}{2}\left[\left(\frac{1}{2}\left\{-x \frac{\cos 2 x}{2}+\frac{\sin 2 x}{4}\right\}\right)-\left(\frac{1}{4}\left\{-x \frac{\cos 4 x}{4}+\frac{\sin 4 x}{16}\right\}\right)\right]+c$
$=\frac{-x \cos 2 x}{8}+\frac{\sin 2 x}{16}+\frac{x \cos 4 x}{32}-\frac{\sin 4 x}{128}+c$

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Question 43 Marks

Find the maximum value of $Z=7 x+7 y$ subject to the constraints $x \geq 0, y \geq 0, x+y \geq 2$ and $2 x+3 y \leq 6$

Answer

Given $Z=7 x+7 y$ subject to the constraints $x \geq 0, y \geq 0, x+y \geq 2$ and $2 x+3 y \leq 6$ Now, draw the line $x+y=2$ and $2 x+3 y=6$

And shaded region satisfied by above inequalities here the feasible region is bounded. The value of Z. at A(3, 0) Z = 7 x 2 + 7 x 0 = 14 at B(3, 0) Z = 7 x 3 + 7 x 0 = 21 and at C(0,2) Z = 7 x 0 + 7 x 2 = 14 Therefore, the maximum value of Z is 21, this is the required solution which occurs at B(3)

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Question 53 Marks

If $x ^{ x }+ y ^{ x }=1$, prove that $\frac{d y}{d x}=-\left\{\frac{x^x(1+\log x)+y^x \cdot \log y}{x \cdot y^{(x-1)}}\right\}$

Answer

$\text { ATQ, } x^{x}+y^{x}=1$
$\Rightarrow e^{\log x^x}+e^{\log y^x}=1\left\{\text { As } e^{Log a}=a\right\}$
$\Rightarrow e^{x \log x}+e^{x \log y}=1$
Differentiating with respect to $x$ using chain rule,
$\frac{d}{d x}\left(e^{x \log x}\right)+\frac{d}{d x}\left(e^{x \log y}\right)=\frac{d}{d x}(1)$
$\Rightarrow e^{x \log x} \frac{d}{d x}(x \log x)+e^{x \log y} \frac{d}{d x}(x \log y)=0$
$\Rightarrow e^{x \log x}\left[x \frac{d}{d x}(\log x)+\log x \frac{d}{d x}(x)\right] e^{x \log x}+e^{\log y^x}\left[x \frac{d}{d x}(\log y)+\log y \frac{d}{d x}(x)\right]=0$
$\Rightarrow x^{x}\left[x\left(\frac{1}{x}\right)+\log x(1)\right]+y^{x}\left[x\left(\frac{1}{x}\right) \frac{d y}{d x}+\log y(1)\right]=0$
$\Rightarrow x^{x}[1+\log x]+y^{x}\left(\frac{d y}{d x}+\log y\right)=0$
$\Rightarrow y^x \times \frac{x}{y} \frac{d y}{d x}=-\left[x^{x}(1+\log x)+y^{x} \log y\right]$
$\Rightarrow\left(xy^{x-1}\right) \frac{d y}{d x}=-\left[x^{x}(1+\log x)+y^{x} \log y\right]$
$\Rightarrow \frac{d y}{d x}=-\left[\frac{x^x(1+\log x)+y^x \log y}{x y^{x-1}}\right]$
$\text{LHS = RHS}$
Hence Proved.

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Question 63 Marks

Solve the differential equation $\left(1+y^2\right)(1+\log x) d x+x d y=0$ given that when $x=1, y=1$

Answer

We have,
$\left(1+y^2\right)(1+\log x) d x+x d y=0$
$\Rightarrow(1+\log x )\left(1+ y ^2\right) dx =- x dy$
$\Rightarrow \frac{(1+\log x)}{x} d x=-\frac{1}{1+y^2} d y$
$\Rightarrow \int \frac{1+\log x}{x} d x=-\int \frac{1}{1+y^2} d y . \ ($Integrating both sides$)$
$\Rightarrow \int t d t=-\int \frac{1}{1+y^2} d y,$ where $1+\log x = t$
$\Rightarrow \frac{t^2}{2}=-\tan ^{-1} y + C$
$\Rightarrow \frac{1}{2}(1+\log x )^2=-\tan ^{-1} y + C $
It is given that when $x=1, y=1$.
So, putting $x=1, y=1$ in $(i)$, we obtain
$\frac{1}{2}(1+\log 1)^2=-\tan ^{-1} 1+C$
$\Rightarrow \frac{1}{2}=-\frac{\pi}{4}+C $
$\Rightarrow C=\frac{1}{2}+\frac{\pi}{4}$
Putting $C=\frac{1}{2}+\frac{\pi}{4}$ in $(i),$ we obtain
$\frac{1}{2}(1+\log x)^2=-\tan ^{-1} y+\frac{1}{2}+\frac{\pi}{4}$
$\Rightarrow \tan ^{-1} y=\frac{\pi}{4}+\frac{1}{2}-\frac{1}{2}(1+\log x)^2$
$\Rightarrow y=\tan \left\{\frac{\pi}{4}+\frac{1}{2}-\frac{1}{2}(1+\log x)^2\right\}$, which is the solution of the given differential equation.

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Question 73 Marks

A card from a pack of $52$ cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond

Answer

$E_1$ : lost card is diamond
$E_2$ : lost card is not diamond
let $A$ : two cards drawn from the remaining pack are diamonds.
$P\left(E_1\right)=\frac{13}{52}=\frac{1}{4}, P\left(E_2\right)=\frac{39}{52}=\frac{3}{4}$
$P\left(\frac{A}{E_1}\right)=\frac{12 C_2}{51 C_2}=\frac{12 \times 11}{51 \times 50}$
$P\left(\frac{A}{E_2}\right)=\frac{13 C_2}{51 C_2}=\frac{13 \times 12}{51 \times 50}$
$P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right) P\left(\frac{A}{E_1}\right)}{P\left(E_1\right) P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right)}$
$=\frac{\frac{13}{52} \times \frac{12 \times 11}{51 \times 50}}{\frac{13}{52} \times \frac{12 \times 11}{51 \times 50}+\frac{3}{4} \times \frac{13 \times 12}{51 \times 50}}$
$=\frac{11}{50}$

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Question 83 Marks

Evaluate: $\int \frac{(2 x+3)}{\sqrt{x^2+x+1}} d x$

Answer

Formula to be used $-\int \frac{d x}{\sqrt{x^2 \pm a^2}}=\log \left(x+\sqrt{x^2 \pm a^2}\right)+c$ where $c$ is the integrating constant
$\therefore \int \frac{(2 x+3)}{\sqrt{x^2+x+1}} d x$
$=\int \frac{(2 x+1)+2}{\sqrt{x^2+x+1}} d x$
$=\int \frac{(2 x+1)}{\sqrt{x^2+x+1}} d x+\int \frac{2}{\sqrt{x^2+x+1}} d x$
Put$, x ^2+ x +1= a ^2,(2 x +1) dx =2 ada$
$\therefore \int \frac{(2 x+1)}{\sqrt{x^2+x+1}} dx$
$=\int \frac{2 d a}{a}$
$=\int 2 da$
$=2 a + c _1$
$=2 \sqrt{x^2+x+1}+c_1$
For $2^{nd}$ part of integral.
$\therefore \int \frac{2}{\sqrt{x^2+x+1}} d x$
$=2 \int \frac{d x}{\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}$
$=2 \log \left|\left(x+\frac{1}{2}\right)+\sqrt{ x ^2+ x +1}\right|+ c _2$
$\therefore \int \frac{(2 x +1)}{\sqrt{ x ^2+ x +1}} dx +\int \frac{2}{\sqrt{x^2+x+1}} d x$
$=2 \sqrt{x^2+x+1}+2 \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^2+x+1}\right|+c, c$ is the integrating constant

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Question 93 Marks

The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was $20,000$ in $1999$ and $25000$ in the year $2004,$ what will be the population of the village in $2009$?

Answer

Let the population at any instant $(t)$ be $y.$
Now it is given that the rate of increase of population is proportional to the number of inhabitants at any instant.
$\therefore \frac{d y}{d t} \alpha y$
$\Rightarrow \frac{d y}{d t}=k y\ (k$ is constant$)$
$\Rightarrow \frac{d y}{y}=k d t$
Now, integrating both sides, we get,
$\log y=k t+C$
According to given conditions,
In the year $1999, t =0$ and $y =20000$
$\Rightarrow \log 20000=C$
Also, in the year $2004, t=5$ and $y=25000$
$\Rightarrow \log 25000=k \cdot 5+C$
$\Rightarrow \log 25000=5 k+\log 20000$
$\Rightarrow 5 k=\log \left(\frac{25000}{2000}\right)=\log \left(\frac{5}{4}\right)$
$\Rightarrow k=\frac{1}{5} \log \left(\frac{5}{4}\right) \ldots \ldots . \text { (iii) }$
Also, in the year $2009, t =10$
Now, substituting the values of $t , k$ and $c$ in equation $(i),$ we get
$\log y=10 \times \frac{1}{5} \log \left(\frac{5}{4}\right)+\log (20000)$
$\Rightarrow \log y=\log \left[20000 \times\left(\frac{5}{4}\right)^2\right]$
$\Rightarrow y=20000 \times \frac{5}{4} \times \frac{5}{4}$
$\Rightarrow y=31250$
Therefore, the population of the village in $2009$ will be $31250$ .

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