Solve the Following Question.(2 Marks)

Question 12 Marks

Find the particular solution of : $r \frac{d r}{d \theta}+\cos \theta=5$ at $r=\sqrt{2}$ and $\theta=0$.

Answer

$r \frac{d r}{d \theta}+\cos \theta=5$
$r \frac{d r}{d \theta}=5-\cos \theta$
$r \cdot d r=(5-\cos \theta) \cdot d \theta$
Integrating both sides, we get,
$\int r \cdot d r=\int(5-\cos \theta) \cdot d \theta$
$\frac{r^2}{2}=5 \theta-\sin \theta+c\ldots\ldots (1)$
Put, $\theta=0$ and $r=\sqrt{2}$, we get
$\frac{2^2}{2}=0-\sin 0+c$
$\therefore c=1$
Put $c=1$ in equation $(1),$ we get
$\therefore \frac{r^2}{2}=5 \theta-\sin \theta+1$ is the required particular solution.

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Question 22 Marks

The probability distribution of $X$ is as follows:

$X=x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$0.1$	$K$	$2K$	$2K$	$K$

Find $(i) \ K$ $(ii) \ P(X < 2)$

Answer

$(i)$ We know that,
$P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=1$
$\therefore 0.1+K+2 k+2 k+K=1$
$\therefore 6 K=0.9$
$\therefore K=\frac{0.9}{6}$
$\therefore K=\frac{3}{20}$
$(ii)\ P(X<2)=P(X=0)+P(X=1)$
$\therefore \quad=0.1+K$
$\therefore \quad=0.1+\frac{3}{20}$
$P(X<2)=\frac{5}{20}$

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Question 32 Marks

Evaluate :
$\underset{\overset{\int}{0}}{{\frac{\pi}{4}}} \sqrt{1+\sin 2 x} \cdot d x$

Answer

Let$\quad$$I=\underset{\overset{\int}{0}}{{\frac{\pi}{4}}} \sqrt{1+\sin 2 x} \cdot d x$
$\therefore \quad=\underset{\overset{\int}{0}}{{\frac{\pi}{4}}} \sqrt{(\cos x+\sin x)^2} \cdot d x \quad \ldots \ldots \ldots \ldots .\left[\because 1+\sin 2 \theta=(\cos \theta+\sin \theta)^2\right]$
$\therefore \quad=\underset{\overset{\int}{0}}{{\frac{\pi}{4}}}(\cos x+\sin x) \cdot d x$
$\therefore \quad=[\sin x-\cos x]_0^{\frac{\pi}{4}}$
$\therefore \quad=\left[\left(\sin \frac{\pi}{4}-\cos \frac{\pi}{4}\right)-(\sin 0-\cos 0)\right]$
$\therefore \quad=\left[\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)-(0-1)\right]$
$\begin{array}{ll}\therefore & =1\end{array}$
$\therefore \quad \underset{\overset{\int}{0}}{{\frac{\pi}{4}}} \sqrt{1+\sin 2 x} \cdot d x=1$

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Question 42 Marks

The displacement of a particle at a time $t$ is given by $s=2 t^3-5 t^2+4 t-3$. Find the time when acceleration is $14 ft / sec ^2$.

Answer

$s=2 t^3-5 t^2+4 t-3$
Differentiate w.r.t. t, we get
Velocity, $v=\frac{d s}{d t}=\frac{d}{d t}\left(2 t^3-5 t^2+4 t-3\right)$
$\therefore \quad v=6 t^2-10 t+4$
Differentiate w.r.t. t, again, we get
Acceleration, $a=\frac{d v}{d t}=\frac{d}{d t}\left(6 t^2-10 t+4\right)$
$\therefore \quad a=12 t-10$
Given, acceleration is $14 ft / sec ^2$
$\begin{array}{ll}\therefore & 14=12 t-10 \\
\therefore & 12 t=24 \\
\therefore & t=2 sec \end{array}$

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Question 52 Marks

Evaluate: $ \int \frac{e^x(1+x)}{\sin ^2\left(x e^x\right)} d x$

Answer

$I=\int \frac{e^x(1+x)}{\sin ^2\left(x e^x\right)} d x$
Put $x e^x=t$
Differentiating both sides,
$x \cdot e^x+e^x=\frac{d t}{d x}$
$e^x(1+x) \cdot d x=d t$
$\therefore I=\int \frac{1}{\sin ^2(t)} d t$
$\therefore \quad=\int \operatorname{cosec}^2(t) d t$
$\therefore I=-\cot t+c$$\qquad$$\ldots\ldots\left[\because \int \operatorname{cosec}^2 x \cdot d x=-\cot x+c\right]$
$\therefore \int \frac{e^x(1+x)}{\sin ^2\left(x e^x\right)} d x$$=-\cot \left(x e^x\right)+c$

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Question 62 Marks

Find the points on the curve given by $y=x^3-6 x^2+x+3$ where the tangents are parallel to the line $y=x+5$.

Answer

Comparing $y=x+5$ with $y=m x+c$, we get
Slope, $m=1$.
Now, $y=x^3-6 x^2+x+3\ldots\ldots (1)$
Differentiate w.r.t. $x$, we get
$\therefore \quad \frac{d y}{d x}=3 x^2-12 x+1$
As the tangents are parallel to the line $y=x+5$.
$\therefore \quad \frac{d y}{d x}=m$
$\therefore 3 x^2-12 x+1=1$
$\therefore 3 x^2-12 x=0$
$\therefore x^2-4 x=0$
$\therefore x(x-4)=0$
$\therefore x=0$ or $x-4=0$
$\therefore x=0 $ or $x=4$
Now, put $x=0$ in equation (1), we get,
$y=0-0+0+3=3$
put $x=4$ in equation (1), we get,
$y=4^3-6 \times 4^2+4+3=64-96+7=-25$
Hence, The required points on the curve where the tangents are parallel to the line $y=x+5$ are $(0,3)$ and $(4,-25)$.

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Question 72 Marks

If $\tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}$ then find the value of $x,$ where $0 < 3 x < 1$.

Answer

$\tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}$
$\therefore \tan ^{-1}\left(\frac{2 x+3 x}{1-2 x \times 3 x}\right)=\frac{\pi}{4} \ldots . . .\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \text {, if } x>0, y>0, x y<1\right]$
$\therefore \frac{5 x}{1-6 x^2}=\tan \left(\frac{\pi}{4}\right)$
$\therefore \frac{5 x}{1-6 x^2}=1$
$\therefore 5 x=1-6 x^2$
$\therefore 6 x^2+5 x-1=0$
Comparing $6 x^2+5 x-1=0$ with $a x^2+b x+c=0$, we get,
$a=6, b=5, c=-1$
$\therefore x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$\therefore x=\frac{-5 \pm \sqrt{5^2-4(6)(-1)}}{2 \times 6}$
$\therefore x=\frac{-5 \pm \sqrt{25+24}}{12}$
$\therefore x=\frac{-5 \pm 7}{12}$
$\therefore x=\frac{-5+7}{12}$ Or $x=\frac{-5-7}{12}$
$\therefore x=\frac{1}{6} $ Or $ x=-1$
As per the given condition, $0<3 x<1$,
therefore we reject $x=-1$.
$\therefore x=\frac{1}{6}$

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Question 82 Marks

Using the truth table, show that the statement pattern $p \rightarrow(q \rightarrow p)$ is a tautology.

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Question 92 Marks

Find the acute angle between the lines represented by $x y+y^2=0$.

Answer

Comparing $x y+y^2$ with $a x^2+2 h x y+b y^2=0$, we get,
$a=0,2 h=1, b=1$
$\begin{array}{ll}\therefore & \theta=\tan ^{-1}\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right| \\
\therefore & \theta=\tan ^{-1}\left|\frac{2 \sqrt{\left(\frac{1}{2}\right)^2-0}}{0+1}\right|=\tan ^{-1}\left|2 \sqrt{\frac{1}{4}}\right|=\tan ^{-1}\left|\frac{2}{2}\right| \\ \therefore & \theta=\tan ^{-1}|1| \\
\therefore & \theta=\frac{\pi}{4}\end{array}$
Thus, the acute angle between the lines is $\frac{\pi}{4}$.

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Question 102 Marks

Find the polar co-ordinates of the point whose Cartesian co-ordinates are $(1,-\sqrt{3})$.

Answer

$x=1, y=-\sqrt{3}$
$\begin{array}{ll}\therefore & r=\sqrt{x^2+y^2}=\sqrt{1^2+(-\sqrt{3})^2}=\sqrt{1+3}=\sqrt{4}=2
\\ \therefore & \tan \theta=\frac{y}{x}\end{array}$
$\begin{array}{ll}\therefore & \tan \theta=\frac{-\sqrt{3}}{1} \\
\therefore & \tan \theta=-\sqrt{3} \\
\therefore & \tan \theta=-\tan \frac{\pi}{3} \\
\therefore & \tan \theta=\tan \left(2 \pi-\frac{\pi}{3}\right) \\
\therefore & \tan \theta=\tan \left(\frac{5 \pi}{3}\right) \\
\therefore & \theta=\frac{5 \pi}{3}\end{array}$
$\therefore \quad$ The required polar co-ordinated are $\left(2, \frac{5 \pi}{3}\right)$.

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Question 112 Marks

Find the inverse of matrix $A$ by elementary row transformation, where $A=\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$.

Answer

$A.A^{-1}=I$
$\therefore \quad\left[\begin{array}{cc}2 & -3 \\-1 & 2\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]$
Using $R_1 \rightarrow R_1+R_2$
$\begin{array}{l}\therefore \quad\left[\begin{array}{cc} \ \ \ 1\ \ \ - \ 1 \\-1 \ \ \ \ \ \ 2\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\end{array}$
Using $ R_2 \rightarrow R_2+R_1$
$\therefore \quad\left[\begin{array}{cc}1 & -1 \\0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 1 \\1 & 2
\end{array}\right]$
Using $R_1 \rightarrow R_1+R_2$
$\begin{array}{lrl}\therefore & {\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]}\end{array} $
$ \therefore A^{-1}=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$

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Question 122 Marks

If the statement p, q are the statements and r, s are false statements, then determine the truth value of the statement pattern :
$(q \wedge r) \vee(\sim p \wedge s)$

Answer

$(q \wedge r) \vee(\sim p \wedge s) \equiv(T \wedge F) \vee(\sim T \wedge F)$
$\therefore \quad(q \wedge r) \vee(\sim p \wedge s) \equiv F \vee(F \wedge F)\ldots\ldots$[Involution law]
$\begin{array}{ll}\therefore & (q \wedge r) \vee(\sim p \wedge s) \equiv F \vee(F \wedge F) \\
\therefore & (q \wedge r) \vee(\sim p \wedge s) \equiv F \vee F \\
\therefore & (q \wedge r) \vee(\sim p \wedge s) \equiv F\end{array}$
Hence truth value is $F$.

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Maths Solve the Following Question.(2 Marks)

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