3 Marks Question

Question 13 Marks

If $x \sqrt{1+y}+y \sqrt{1+x}=0$ and $x \neq y$, prove that $\frac{d y}{d x}=-\frac{1}{(x+1)^2}$.

Answer

According to the question, we have to prove that $\frac{d y}{d x}=-\frac{1}{(1+x)^2}$ if $x \sqrt{1+y}+y \sqrt{1+x}=0$
where $x \neq y$.
we shall first write $y$ in terms of $x$ explicitly i.e $y=f(x)$
Clearly,$x \sqrt{1+y}=-y \sqrt{1+x}$
Squaring both sides, we get,
$x^2(1+y)=y^2(1+x)$
$\Rightarrow x^2+x^2 y=y^2(1+x)$
$\Rightarrow x^2-y^2=y^2 x-x^2 y$
$\Rightarrow(x-y)(x+y)=-x y(x-y)$
$\Rightarrow(x-y)(x+y)+x y(x-y)=0$
$\Rightarrow(x-y)(x+y+x y)=0$
$\therefore $ Either, $x - y = 0$ or $x + y + xy = 0$
Now, $x - y = 0 \Rightarrow x = y$
But, it is given that $ x \neq y$
So, it is a contradiction
Therefore, $x - y = 0$ is rejected.
Now, consider $y + xy + x = 0$
$\Rightarrow y(1+x)=-x$
$\Rightarrow y=\frac{-x}{1+x}.....(i)$
Therefore, on differentiating both sides $w.r.t x$, we get,
$\frac{d y}{d x}=\frac{(1+x) \times \frac{d}{d x}(-x)-(-x) \times \frac{d}{d z}(1+x)}{(1+x)^2}[$ By using quotient rule of derivative $]$
$\Rightarrow \frac{d y}{d x}=\frac{(1+x)(-1)+x(1)}{(1+x)^2}$
$\Rightarrow \frac{d y}{d x}=\frac{-1-x+x}{(1+x)^2}$
$\therefore \frac{d y}{d x}=\frac{-1}{(1+x)^2}$

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Question 23 Marks

Solve graphically the following linear programming problem:
Maximise $z = 6x + 3y,$
Subject to the constraints:
$4 x+y \geq 80$
$3 x+2 y \leq 150$
$x+5 y \geq 115$
$x > 0, y \geq 0$

Answer

Subject to the constraints are
$4 x+y \geq 80$
$x+5 y \geq 115$
$3 x+2 y \leq 150$
and the non negative constraint $x, y \geq 0$
Converting the given inequations into equations,
we get $4x + y = 80 x + 5y = 115 3x + 2y = 150 x = 0$ and $y = 0$ These lines are drawn on the graph and the shaded region $\text{ABC}$ represents the feasible region of the given $LPP.$

It can be observed that the feasible region is bounded. The coordinates of the corner points of the feasible region are $A(2, 72) B(15, 20)$ and $C(40,15)$ The values of the objective function, $Z$ at these corner points are given in the following table:
Corner Point Value of the Objective Function $z = 6x + 3y$
$ A (2,72): Z=6 \times 2+3 \times 72=228$
$B(15,20): Z=6 \times 15+3 \times 20=150$
$C (40,15): Z=6 \times 40+3 \times 15=285$
From the table, $Z$ is minimum at $x = 15$ and $y = 20$ and the minimum value of $Z$ is $150$. Thus, the minimum value of $Z$ is $150$

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Question 33 Marks

Solve the differential equation: $\left( x ^3+ x ^2+ x +1\right) \frac{d y}{d x}=2 x ^2+ x$

Answer

The given differential equation is,
$\left( x ^3+ x ^2+ x +1\right) \frac{d y}{d x}=2 x ^2+ x$
$\Rightarrow \frac{d y}{d x}=\frac{2 x^2+x}{x^3+x^2+x+1}$
$\Rightarrow d y=\frac{2 x^2+x}{(x+1)\left(x^2+1\right)} d x$
Integrating both sides, we get
$\int d y=\int\left\{\frac{2 x^2+x}{(x+1)\left(x^2+1\right)}\right\} d x$
$\Rightarrow y=\int\left\{\frac{2 x^2+x}{(x+1)\left(x^2+1\right)}\right\} d x$
Let $\frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^2+1}$
$\Rightarrow 2 x^2+x=A x^2+A+B x^2+B x+C x+C$
$\Rightarrow 2 x^2+x=(A+B) x^2+(B+C) x+(A+C)$
Comparing the coefficients on both sides, we get
$A + B = 2..... (i)$
$B + C = 1 ...(ii)$
$A + C = 0.... (iii)$
Solving $(i), (ii)$ and $(iii),$ we get
$A =\frac{1}{2}, B=\frac{3}{2}, C =-\frac{1}{2}$
$\therefore y=\frac{1}{2} \int \frac{1}{(x+1)} d x+\int \frac{\frac{3}{2} x-\frac{1}{2}}{x^2+1} d x$
$=\frac{1}{2} \int \frac{1}{(x+1)} d x+\frac{1}{2} \int \frac{3 x}{x^2+1} d x-\frac{1}{2} \int \frac{1}{x^2+1} d x$
$=\frac{1}{2} \int \frac{1}{(x+1)} d x+\frac{3}{4} \int \frac{2 x}{x^2+1} d x-\frac{1}{2} \int \frac{1}{x^2+1} d x$
$=\frac{1}{2} \log |x+1|+\frac{3}{4} \log \left|x^2+1\right|-\frac{1}{2} \tan ^{-1} x+C$
Hence, $y =\frac{1}{2} \log | x +1|+\frac{3}{4} \log \left| x ^2+1\right|-\frac{1}{2} \tan ^{-1} x + C$ is the solution to the given differential equation.

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Question 43 Marks

There are two boxes, namely $\text{Box-I}$ and $\text{Box-II}. \text{Box-I}$ contains $3$ red and $6$ black balls. $\text{Box-II}$ contains $5$ red and $5$ black balls. One of the two boxes, is selected at random and a ball is drawn at random. The ball drawn is found to be red. Find the probability that this red ball comes out from $\text{Box-II}$

Answer

Let $E_1$ be the event when box $I$ is selected
$E_2$ be the event when box $II$ is selected
A be the event of getting red ball in any box
$P\left(E_1\right)=\frac{1}{2}, P\left(E_2\right)=\frac{1}{2}$
$P\left(\frac{A}{E_1}\right)=\frac{3}{9}=\frac{1}{2}$
$P\left(\frac{A}{E_2}\right)=\frac{5}{10}=\frac{1}{2}$
Now, required probability =
$P\left(\frac{E_2}{A}\right)=\frac{P\left(\frac{A}{E_2}\right) P\left(E_2\right)}{P\left(\frac{A}{E_1}\right) P\left(E_1\right)+P\left(\frac{A}{E_2}\right) P\left(E_2\right)}$
$P\left(\frac{E_2}{A}\right)=\frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{3}+\frac{1}{2} \times \frac{1}{2}}=\frac{\frac{1}{4}}{\frac{5}{12}}$
$P\left(\frac{E_2}{A}\right)=\frac{3}{5}$

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Question 53 Marks

Minimise $Z = 3x + 5y$ subject to the constraints:
$x+2 y \geqslant 10$
$x+y \geqslant 6$
$3 x+y \geqslant 8$
$x, y \geqslant 0$

Answer

We first draw the graphs of $x + 2y = 10 x + y = 6 3x + y = 8$ The shaded region $\text{ABCD}$ is the feasible region $(R)$ determined by the above constraints. The feasible region is unbounded. Therefore, minimum of $Z$ may or may not occur. If it occurs, it will be on the corner point.

Corner Point	Value of Z
$A(0, 8)$	$40$
$B(1, 5)$	$28$
$C(2,4)$	$26 ($ smallest $)$
$D(10, 0)$	$30$

Let us draw the graph of $3 x+5 y < 26$ as shown in Fig by dotted line.
We see that the open half plane determined by $3 x+5 y < 26$ and $R$ do not have a point in common. Thus, $26$ is the minimum value of $Z.$

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Question 63 Marks

Solve: $2 y e ^{ x / y } d x+\left(y-2 x e ^{ x / y }\right) dy =0$

Answer

We have, $2 y e^{x / y} d x+\left(y-2 x e^{x / y}\right) d y=0$
$\Rightarrow \frac{d x}{d y}=\frac{2 x e^{x / y}-y}{2 y e^{x / y}} \ldots (i)$
Clearly, the given differential equation is a homogeneous differential equation.
As the right hand side of $(i)$ is expressible as a function of $\frac{x}{y}$.
So, we put $X =$ vy and $\frac{d x}{d y}=v+y \frac{d v}{d y}$ to get
$v+y \frac{d v}{d y}=\frac{2 v e^v-1}{2 e^v}$
$\Rightarrow y \frac{d v}{d y}=\frac{2 v e^v-1}{2 e^v}-v$
$\Rightarrow y \frac{d v}{d y}=-\frac{1}{2 e^v}$
$\Rightarrow 2 ye ^{ v } dv =- dy$
$\Rightarrow 2 e ^{ v } dv =-\frac{1}{y} dy $
Integrating both sides,
$\Rightarrow 2 \int e^v d v=-\int \frac{1}{y} d y$
$\Rightarrow 2 e ^{ v }=-\log | y |+\log C$
$\Rightarrow 2 e ^{ v }=\log \left|\frac{c}{y}\right|$
$\Rightarrow 2 e ^{ x / y }=\log \left|\frac{c}{y}\right|$
Hence, 2 $e ^{ x / y }=\log \left|\frac{c}{y}\right|$ gives the general solution of the given differential equation.

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Question 73 Marks

Evaluate: $\int \frac{x^3+x^2+2 x+1}{x^2-x+1} d x$

Answer

Let $I =\int \frac{x^3+x^2+2 x+1}{x^2-x+1} d x$
Here we have,
$\frac{x^3+x^2+2 x+1}{x^2-x+1}= x +2+\frac{3 x-1}{x^2-x+1} \ldots (i)$
$\text { Let } 3 x -1=A \frac{d}{d x}( x - x +1)+ B$
$\Rightarrow 3 x -1= A (2 x +1)+ B$
$\Rightarrow 3 x -1=(2 A) x + B - A $
Equating Coefficients of like terms, we get
$2 A=3 \Rightarrow A=\frac{3}{2}$
$\text { Also, } B - A =-1$
$\Rightarrow B-\frac{3}{2}=-1$
$\Rightarrow B=\frac{1}{2}$
So, $\int\left(\frac{x^3+x^2+2 x+1}{x^2-x+1}\right) d x=\int( x +2) dx +\int\left(\frac{\frac{3}{2}(2 x-1)+\frac{1}{2}}{x^2-x+1}\right)$
$=\int( x +2) dx +\frac{3}{2} \int\left(\frac{2 x-1}{x^2-x+1}\right) d x+\frac{1}{2} \int \frac{d x}{x^2-x+1}$
$=\int( x +2) dx +\frac{3}{2} \int \frac{(2 x-1) d x}{x^2-x+1}+\frac{1}{2} \int \frac{d x}{x^2-x+\frac{1}{4}-\frac{1}{4}+1}$
$=\int( x +2) dx +\frac{3}{2} \int \frac{(2 x-1) d x}{x^2-x+1}+\frac{1}{2} \int \frac{d x}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}$
$=\frac{x^2}{2}+2 x +\frac{3}{2} \log \left| x ^2- x +1\right|+\frac{1}{2} \times \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C$
$=\frac{x^2}{2}+2 x +\frac{3}{2} \log \left| x ^2- x +1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+C$

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Question 83 Marks

Evaluate: $\int \frac{1}{\cos x-\sin x} d x$

Answer

Let the given integral be,
$ I =\int \frac{1}{\cos x-\sin x} d x$
$\text { Putting } \cos x =\frac{1-\tan ^2\left(\frac{x}{2}\right)}{1+\tan ^2\left(\frac{x}{2}\right)} \text { and } \sin x =\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}$
$\Rightarrow I=\int \frac{d x}{\frac{1-\tan ^2\left(\frac{x}{2}\right)}{1+\tan ^2\left(\frac{x}{2}\right)}-\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}$
$=\int \frac{\sec ^2\left(\frac{x}{2}\right) d x}{1-\tan ^2\left(\frac{x}{2}\right)-2 \tan \left(\frac{x}{2}\right)}$
$\text { Let } \tan \left(\frac{x}{2}\right)= t$
$\Rightarrow \frac{1}{2} \sec ^2\left(\frac{x}{2}\right) dx = dt$
$\sec ^2\left(\frac{x}{2}\right) dx =2 dt $
$\therefore I=\int \frac{2 d t}{1-t^2-2 t}$
$=\int \frac{-2 d t}{t^2+2 t-1}$
$=\int \frac{-2 d t}{t^2+2 t+1-2}$
$=-\int \frac{2 d t}{(t+1)^2-(\sqrt{2})^2}$
$=\int \frac{2 d t}{(\sqrt{2})^2-(t-1)^2}$
$=\frac{2}{2 \sqrt{2}} \ln \left|\frac{\sqrt{2}+t+1}{\sqrt{2}-t-1}\right|+C$
$=\frac{1}{\sqrt{2}} \ln \left|\frac{\sqrt{2}+\tan \frac{x}{2}+1}{\sqrt{2}-\tan \frac{x}{2}-1}\right|+C$

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Question 93 Marks

Evaluate: $\int_0^{\pi / 2} \frac{\sin ^2 x}{\sin x+\cos x} d x$

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