2 Marks

Question 12 Marks

An unbiased dice is thrown twice. Let the event E be 'odd number on the first throw' and F the event 'odd number on the second throw'. Check the independence of events E and F .

Answer

If all the 36 elementary events of the experiment are considered to be equally likely, we have
$P(A)=\frac{18}{36}=\frac{1}{2}$ and
$P(B)=\frac{18}{36}=\frac{1}{2}$
Also, $P ( A \cap B )= P \quad$ (odd Number on both throws)
$\begin{array}{l}=\frac{9}{36} \\ =\frac{1}{4}\end{array}$
Now, $P(A) \cdot P(B)=\frac{1}{2} \times \frac{1}{2}$
$=\frac{1}{4}$
Cleary, $P ( A \cap B )= P ( A ) \cdot P ( B )$
Thus, A and B are independent events.

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Question 22 Marks

Given that $A$ and $B$ are events such that $P(A)=0.6, P(B)=0.3$ and $P(A \cap B)=0.2$, Find $P(A \mid B)$ and $P(B \mid A)$.

Answer

$\begin{array}{l} P ( E )=0.6 \\ P ( F )=0.3 \\ P ( E \cap F )=0.2\end{array}$
$\begin{aligned} P ( E \mid F ) & =\frac{ P ( E \cap F )}{ P ( F )} \\ & =\frac{0.2}{0.3} \\ & =\frac{2}{3}\end{aligned}$
$\begin{aligned} P(F \mid E) & =\frac{P(E \cap F)}{P(E)} \\ & =\frac{0.2}{0.6} \\ & =\frac{1}{3}\end{aligned}$

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Question 32 Marks

Find the vector and the Cartesian equations of the line through the point $(5,2,-4)$ and which is parallel to the vector $3 \hat{i}-2 \hat{j}+8 \hat{k}$.

Answer

We have $\vec{a}=5 \hat{i}+2 \hat{j}-4 \hat{k}$ yLku $\vec{b}=3 \hat{i}-2 \hat{j}+8 \hat{k} Au$. Therefore, the vector equation of the line is $\vec{r}=5 \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(3 \hat{i}-2 \hat{j}+8 \hat{k})$
Now, $\vec{r}$ is the position vector of any point $P (x, y, z)$ on the line.
Therefore, $x \hat{i}+y \hat{j}+z \hat{k}$
$\begin{array}{l}
=5 \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(3 \hat{i}+2 \hat{j}-8 \hat{k}) \\
=(5+3 \lambda) \hat{i}+(2+2 \lambda) \hat{j}+(-4-8 \lambda) \hat{k}
\end{array}$
Eliminating $\lambda$, we get
$\frac{x-5}{3}=\frac{y-2}{2}=\frac{z+4}{-8}$
Which is the equation of the line in cartesian form.

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Question 42 Marks

Find the vector and the Cartesian equation of the line passing through the point $(1,2,-4)$ and perpendicular to the two lines :
$\begin{array}{l}
\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7} \text { and } \\
\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}
\end{array}$

Answer

Line $L _1: \frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$
$\vec{r}=(8 \hat{i}-19 \hat{j}+10 \hat{k})+\lambda(3 \hat{i}-16 \hat{j}+7 \hat{k})$
Direction of line $\overrightarrow{b_1}=3 \hat{i}-16 \hat{j}+7 \hat{k}$
Line $L _2: \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$
$\overrightarrow{r_2}=(15 \hat{i}+29 \hat{j}+5 \hat{k})+\lambda(3 \hat{i}+8 \hat{j}-5 \hat{k})$
Direction of line $\overrightarrow{b_2}=3 \hat{i}+8 \hat{j}-5 \hat{k}$
$\begin{aligned}
\overrightarrow{b_1} \times \overrightarrow{b_2} & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -16 & 7 \\
3 & 8 & -5
\end{array}\right| \\
& =24 \hat{i}+36 \hat{j}+72 \hat{k} \\
& =12(2 \hat{i}+3 \hat{j}+6 \hat{k})
\end{aligned}$
$\therefore$ Direction of given line $\vec{b}=2 \hat{i}+3 \hat{j}+6 \hat{k}$ $A (\vec{a})=\hat{i}+2 \hat{j}-4 \hat{k}$ line on the line.
Vector equation of line :
$\begin{array}{l}
\frac{x-x_1}{l_1}=\frac{y-y_1}{l_2}=\frac{z-z_1}{l_3} \\
\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}
\end{array}$

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Question 52 Marks

The points $A(2 \hat{i}-\hat{j}+\hat{k}), B(\hat{i}-3 \hat{j}-5 \hat{k})$,
$C (3 \hat{i}-4 \hat{j}-4 \hat{k})$ are the vertices of triangle. Decide the type of triangle.

Answer

Here,
$\begin{aligned}
\overrightarrow{AB} & =(1-2) \hat{i}+(-3+1) \hat{j}+(-5-1) \hat{k} \\
& =-\hat{i}-2 \hat{j}-6 \hat{k} \\
\overrightarrow{AB} & =\sqrt{1+4+36}=\sqrt{41} \\
\overrightarrow{BC} & =(3-1) \hat{i}+(-4+3) \hat{j}+(-4+5) \hat{k} \\
& =2 \hat{i}-\hat{j}+\hat{k} \\
\overrightarrow{BC} & =\sqrt{4+1+1}=\sqrt{6}
\end{aligned}$
and
$\begin{aligned}
\overrightarrow{CA} & =(2-3) \hat{i}+(-1+4) \hat{j}+(1+4) \hat{k} \\
& =-\hat{i}+3 \hat{j}+5 \hat{k} \\
\overrightarrow{CA} & =\sqrt{1+9+25}=\sqrt{35}
\end{aligned}$
Further, Note that
$|\overrightarrow{AB}|^2=41$
$\begin{array}{l}=6+35 \\ =|\overrightarrow{ BC }|^2+|\overrightarrow{ CA }|^2\end{array}$
Hence, the triangle is a right angled triangle.

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Question 62 Marks

Find general solution of differential equation :
$\sec ^2 x \tan y d x+\sec ^2 y \tan x d y=0 .$

Answer

$\begin{array}{l}\sec ^2 x \cdot \tan y d x+\sec ^2 y \tan x d y=0 \\ \therefore \sec ^2 x \tan y d x=-\sec ^2 y \tan x d y \\ \therefore \frac{\sec ^2 y}{\tan y} d y=\frac{-\sec ^2 x}{\tan x} d x\end{array}$
$\rightarrow$ Integrate both the sides,
$\int \frac{\sec ^2 y}{\tan y} d y=-\int \frac{\sec ^2 x}{\tan x} d x$
$\therefore \int \frac{\frac{d}{d y}(\tan y)}{\tan y} d y=-\int \frac{\frac{d}{d x}(\tan x)}{\tan x} d x$
$\begin{array}{ll}\therefore & \log |\tan y|=-\log |\tan x|+\log |c| \\ \therefore & \log |\tan y|=\log \left|\frac{c}{\tan x}\right|\end{array}$
$\begin{array}{rlrl}\therefore \tan y =\frac{c}{\tan x} \\ \therefore \tan x \cdot \tan y & =c\end{array}$
Which is required general solution of given differential equation.

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Question 72 Marks

Find the area of the region bounded by the line $y=3 x+2$, the X-axis and the ordinates $x=-1$ and $x=1$.

Answer

As shown in the fig., the line $y=3 x+2$, meets X -axis at $\left(-\frac{2}{3}, 0\right)$ and its graph lie below X-axis for $x \in\left(-1,-\frac{2}{3}\right)$ and above X -axis for $x \in\left(-\frac{2}{3}, 1\right)$.

The required area
$=$ Area of the region $ACBA +$ Area of the region ADEA
$\begin{array}{l}
=\left|\int_{-1}^{-\frac{2}{3}}(3 x+2) d x\right|+\int_{-\frac{2}{3}}^1(3 x+2) d x \\
=\left|\left(\frac{3}{2} x^2+2 x\right)_{-1}^{-\frac{2}{3}}\right|+\left(\frac{3}{2} x^2+2 x\right)_{-\frac{2}{3}}^1 \\
=\left|\left(\frac{3}{2}\left(\frac{4}{9}\right)-\frac{4}{3}\right)-\left(\frac{3}{2}(1)+2(-1)\right)\right|+\left(\frac{3}{2}(1)+2(1)\right) -\left(\frac{3}{2}\left(\frac{4}{9}\right)+2\left(-\frac{2}{3}\right)\right)
\end{array}$
$\begin{array}{l}=\left|\frac{2}{3}-\frac{4}{3}-\frac{3}{2}+2\right|+\frac{3}{2}+2-\frac{2}{3}+\frac{4}{3} \\ =\left|\frac{-2}{3}-\frac{3}{2}+2\right|+\frac{3}{2}+2+\frac{2}{3} \\ =\left|\frac{-4-9+12}{6}\right|+\frac{9+12+4}{6} \\ =\frac{1}{6}+\frac{25}{6} \\ =\frac{26}{6} \\ =\frac{13}{3} \text { sq. units. }\end{array}$

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Question 82 Marks

Find the area of the region bounded by the ellipse $9 x^2+16 y^2=144$.

Answer

$\begin{array}{l}9 x^2+16 y^2=144 \\ \therefore \quad \frac{9 x^2}{144}+\frac{16 y^2}{144}=1 \\ \quad \frac{x^2}{16}+\frac{y^2}{9}=1 \\ a^2=16, a=4(a>b) \\ b^2=9, b=3\end{array}$

$\begin{array}{l}
I=\int_0^4 \frac{3}{4} \sqrt{16-x^2} d x \\
I=\frac{3}{4} \int_0^4 \sqrt{16-x^2} d x \\
I=\frac{3}{4}\left[\frac{x}{2} \sqrt{16-x^2}+\frac{16}{2} \sin ^{-1}\left(\frac{x}{4}\right)\right]_0^4 \\
I=\frac{3}{4}\left[\left(\frac{4}{2}(0)+8 \sin ^{-1}(1)\right)-\left(0+\sin ^{-1}(0)\right)\right] \\
I=\frac{3}{4}\left(8 \cdot \frac{\pi}{2}\right) \\
I=3 \pi
\end{array}$
Now, $A=4|I|$
$=4|3 \pi|$
$\therefore \quad A=12 \pi$ sq. units

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Question 92 Marks

Evaluate : $\int \frac{1}{\sqrt{8+3 x-x^2}} d x$.

Answer

$\begin{aligned} I & =\int \frac{d x}{\sqrt{8+3 x-x^2}} \\ & =\int \frac{d x}{\sqrt{-\left(x^2-3 x-8\right)}} \\ & =\int \frac{d x}{\sqrt{-\left(x^2-2\left(\frac{3 x}{2}\right)+\frac{9}{4}-\frac{9}{4}-8\right)}} \\ & =\int \frac{d x}{\sqrt{-\left(\left(x-\frac{3}{2}\right)^2-\frac{41}{4}\right)}} \\ & =\int \frac{d x}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-\left(x-\frac{3}{2}\right)^2}}\end{aligned}$
$\begin{aligned} & =\sin ^{-1}\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right)+c \\ \therefore I & =\sin ^{-1}\left(\frac{2 x-3}{\sqrt{41}}\right)+c\end{aligned}$

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Question 102 Marks

Find $\frac{d y}{d x}$ of the function : $x^y+y^x=1$

Answer

Suppose, $u=x^y$ and $v=y^x$
$\therefore u+v=1$
Now, differentiate w.r.t. $x$,
$\frac{d u}{d x}+\frac{d v}{d x}=0\ldots(1)$
Here, $u=x^y$
Take both the sides $\log$,
$\log u=y \log x$
Now, differentiate w.r.t. $x$,
$\begin{aligned}
\frac{d u}{d x} \frac{1}{u} & =y \frac{d}{d x} \log x+\log x \frac{d}{d x} y \\
\therefore \quad \frac{d u}{d x} \frac{1}{u} & =y \cdot \frac{1}{x}+\log x \frac{d y}{d x} \\
\therefore \quad \frac{d u}{d x} & =u\left[\frac{y}{x}+\log x \frac{d y}{d x}\right] \\
& =x^y\left[\frac{y}{x}+\log x \frac{d y}{d x}\right] \\
\therefore \quad \frac{d u}{d x} & =x^{y-1} y+x^y \log \frac{d y}{d x}\ldots(2)
\end{aligned}$
Now, $v=y^x$
Take both the side $\log$,
$\log v=x \log y$
Now, differentiate w.r.t. $x$,
$\begin{aligned}
\frac{1}{v} \frac{d v}{d x} & =x \frac{d}{d x} \log y+\log y \frac{d}{d x} x \\
\therefore \quad \frac{1}{v} \frac{d v}{d x} & =x \cdot \frac{1}{y} \frac{d y}{d x}+\log y \\
\therefore \quad \frac{d v}{d x} & =v\left[\frac{x}{y} \frac{d y}{d x}+\log y\right]
\end{aligned}$
$=y^x\left[\frac{x}{y} \frac{d y}{d x}+\log y\right]\ldots(3)$
Put the value of equation (2) and (3) in equation (1),
$\begin{array}{l}
x^{y-1} y+x^y \log x \frac{d y}{d x}+y^{x-1} x \frac{d y}{d x}+y^x \log y=0 \\
\frac{d y}{d x}\left[x^y \log x+y^{x-1} x\right]=-y^x \log y-x^{y-1} y \\
\frac{d y}{d x}=-\frac{\left[y^x \log y+x^{y-1} y\right]}{\left[x^y \log x+y^{x-1} x\right]}
\end{array}$

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Question 112 Marks

Prove that : $\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)=2 \sin ^{-1} x$, $-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$.

Answer

$\begin{array}{l}\text { L.H.S. }=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right) \\ \text { Suppose, } x=\sin \theta\end{array}$
$\begin{aligned} \therefore \theta & =\sin ^{-1} x, \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ & =\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^2 \theta}\right) \\ & =\sin ^{-1}(2 \sin \theta \cos \theta) \\ & =\sin ^{-1}(\sin 2 \theta)\end{aligned}$
$\begin{array}{l}\text { Here, }-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} \\ \therefore-\sin \frac{\pi}{4} \leq \sin \theta \leq \sin \frac{\pi}{4} \\ \therefore \sin \left(-\frac{\pi}{4}\right) \leq \sin \theta \leq \sin \frac{\pi}{4} \\ \therefore-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} \\ \therefore-\frac{\pi}{2} \leq 2 \theta \leq \frac{\pi}{2} \\ \therefore 2 \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \ldots(1)\\ \therefore \sin ^{-1}(\sin 2 \theta)=2 \theta(\because \text { From equation }(1))\end{array}$
$\begin{array}{l}=2 \sin ^{-1} x \\ =\text { R.H.S. }\end{array}$

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Question 122 Marks

Prove that : $\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}$

Answer

$\begin{array}{l}\text { L.H.S. }=\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13} \\ \text { Take, } \cos ^{-1} \frac{4}{5}=\alpha, \quad \cos ^{-1} \frac{12}{13}=\beta \\ \therefore \quad \cos \alpha=\frac{4}{5} \quad, \quad \cos \beta=\frac{12}{13}\end{array}$

$\begin{array}{l}\therefore \quad \sin \alpha=\frac{3}{5} \quad, \quad \sin \beta=\frac{5}{13} \\ \text { Here, } \cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta\end{array}$
$\begin{array}{l}=\left(\frac{4}{5} \times \frac{12}{13}\right)-\left(\frac{3}{5} \times \frac{5}{13}\right) \\ =\frac{48}{65}-\frac{15}{65}\end{array}$
$\cos (\alpha+\beta)=\frac{33}{65}$
$\begin{array}{cc}\therefore & \alpha+\beta=\cos ^{-1}\left(\frac{33}{65}\right) \\ \therefore & \cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}\end{array}$

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Maths 2 Marks

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