Solve the Following Question.(2 Marks)

Question 12 Marks

If two coins are tossed simultaneously, write the probability distribution of the number of heads.

Answer

Self

Question 22 Marks

Solve: $1+\frac{d y}{d x}=\operatorname{cosec}(x+y)$; put $x+y=u$.

Answer

Put x + y = u.
$\therefore 1+\frac{ dy }{ d x}=\frac{ du }{ d x}$
Given differential equation becomes
$\frac{ du }{ d x}=\operatorname{cosec} u$
$\therefore \frac{ du }{\operatorname{cosecu}}= dx$
$\therefore$sin u.du = dx
Integrating both sides, we get
$\therefore \int \sin u \cdot d u=\int d x$
$\therefore-\cos u=x+c$
$\therefore x+\cos u+c=0$
$\therefore x+\cos (x+y)+c=0 \quad \ldots(\because x+y=u)$

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Question 32 Marks

Find the area of the region bounded by the curve $y=x^2$, and the lines $x=1, x=2$, and $y=0$.

Answer

Area of region $=\lambda_1^2 y dx$
$=\lambda_1^2 x ^2 dx$
$=\left[\frac{x^3}{3}\right]_1^2$
$=\frac{1}{3}\left[2^3-1^3\right]$
$=\frac{1}{3}[8-1]$
$=\frac{7}{3}$
Therefore, required Area $=\frac{7}{3}$ square units.

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Question 42 Marks

Evaluate the definite integral: $\int_0^{\frac{\pi}{2}} \cos ^2 x d x$

Answer

$\int_0^{\frac{\pi}{2}} \cos ^2 x d x$
$=\frac{1}{2} \int_0^{\frac{\pi}{2}}(1+\cos 2 x) d x$
$\because \cos 2 x=1-2 \cos ^2 x$
$\Rightarrow \cos ^2 x=\frac{1}{2}(1+\cos 2 x)$
$=\frac{1}{2}\left[x+\frac{\sin 2 x}{2}\right]_0^{\frac{\pi}{2}}$
$=\frac{1}{2}\left[\frac{\pi}{2}-\frac{\sin \pi}{2}-0-\frac{1}{2} \sin 0\right]$
$=\frac{1}{2}\left[\frac{\pi}{2}+0-0\right]$
$=\frac{\pi}{4}$

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Question 52 Marks

Evaluate: $\int \log x d x$

Answer

We can write log (x) = 1. log (x) so
$\int \log x d x=\int 1 \cdot \log (x) d x$
We can solve it by using integration by parts , For this we take log(x)as first function and 1 as second function.
$\int 1 \cdot \log (x) d x$
$\log ( x ) \int 1 . d x-\int\left(\frac{d}{d x}\right)(\log x) \int 1 . d x$
$=\log (x) \cdot x-\int \frac{1}{x} \times x d x$
$=\log (x) \cdot x-\int 1 \cdot d x$
= log (x) x - x + c
= x (log(x) - 1) + c
Hence, the value of $\int \log (x) d x$ is $(\log (x)-1)+c$

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Question 62 Marks

Find $\frac{d y}{d x}$, if $y=(\log x)^x$

Answer

$y=(\log x)^x$
$\log y=\log (\log x)^x$
$\log y=x \log (\log x)$
Differentiate $w.r.t$ to $ x$
$\frac{1}{y} \frac{d y}{d x}=x \times \frac{1}{\log x} \times \frac{1}{x}+\log \times 1$
$\frac{1}{y} \frac{d y}{d x}=\frac{1}{\log x}+\log (\log x)$
$\frac{d y}{d x}=y\left(\frac{1}{\log x}+\log (\log x)\right)$

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Question 72 Marks

Find the vector equation of the line passing through the point having position vector $4 \hat{i}-\hat{j}+2 \hat{k}$ and parallel to the vector $-2 \hat{i}-\hat{j}+\hat{k}$

Answer

The vector equation of a line passing through a point with position vector $\bar{a}$ and parallel to $\overline{ b }$ is $\overline{ r }=\overline{ a }+\lambda \overline{ b }$.
$\therefore$ The vector equation of the line passing through the point having position vector
$4 \hat{i}-\hat{j}+2 \hat{k}$ and parallel to the vector $-2 \hat{i}-\hat{j}+\hat{k}$ is
$\bar{r}=(4 \hat{i}-\hat{j}+2 \hat{k})+\lambda(-2 \hat{i}-\hat{j}+\hat{k})$

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Question 82 Marks

If $\bar{a}, \bar{b}, \bar{c}$ are the position vector of the points $A, B, C$ respectively and $5 \bar{a}-3 \bar{b}-2 \bar{c}=\overline{0}$ then find the ratio in which the point $C$ divides the line segment $B A$

Answer

$5 \bar{a}-3 \bar{b}=2 \bar{c}$
$\frac{5 \bar{a}-3 \bar{b}}{2}=\bar{c}$
$\frac{5 \bar{a}-3 \bar{b}}{5-3}=\bar{c}$
Point C divides segment BA externally in the ratio 5 : 3

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Question 92 Marks

Find $K,$ if the sum of the slopes of the lines represented by $x^2+k x y-3 y^2=0$ is twice their product.

Answer

Comparing the equation $x^2+k x y-3 y^2=0$ with $a x^2+2 h x y-b y^2=0$,
We get, $a = 1, 2h = k, b = −3.$
Let $m_1$ and $m_2$ be the slopes of the lines represented by $x^2+k x y-3 y^2=0$
$\therefore m _1+ m _2=\frac{-2 h}{ b }=-\frac{ k }{-3}=\frac{ k }{3}$
And $m_1 m_2=\frac{a}{b}=\frac{1}{-3}=-\frac{1}{3}$
Now, $m_1+m_2=2\left(m_1 m_2\right)($Given$)$
$\therefore \frac{ k }{3}=2\left(-\frac{1}{3}\right)$
$\therefore k =-2 .$

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Question 102 Marks

In $\triangle A B C$, if $a =18, b=24, c =30$ then find the values of $\sin \frac{A}{2}$.

Answer

Given: $a = 18, b = 24$ and $c = 30$
$\therefore 2s = a + b + c$
$= 18 + 24 + 30$
$= 72$
$\therefore S = 36$
$\sin \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{b c}}$
$=\sqrt{\frac{(36-24)(36-30)}{(24)(30)}}$
$=\sqrt{\frac{12 \times 6}{24 \times 30}}$
$=\sqrt{\frac{1}{10}}$
$=\frac{1}{\sqrt{10}} .$

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Question 112 Marks

Check whether the following matrix is invertible or not:$
\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]
$

Answer

Let $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$
Then, $|A|=\left|\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right|=\cos ^2 \theta+\sin ^2 \theta=1 \neq 0$
$\therefore A$ is a non-singular matrix.
Hence, A is invertible.

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Question 122 Marks

Construct the truth table for the statement pattern:
$[(p \rightarrow q) \wedge q] \rightarrow p$

Answer

p	q	$(p \rightarrow q)$	$[(p \rightarrow q) \wedge q]$	$[(p \rightarrow q) \wedge q] \rightarrow p$
T	T	T	T	T
T	F	F	F	T
F	T	T	T	F
F	F	T	F	T

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Maths Solve the Following Question.(2 Marks)

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