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Maths Solve the Following Question.(3 Marks)

MARCH 2024 · Maharashtra Board STD 12 Science (MSBSHSE - English Medium). 12 questions.

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Solve the Following Question.(3 Marks)

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12 questions · timed · auto-graded

Question 13 Marks
A die is thrown 6 times. If ‘getting an odd number’ is a success, find the probability of 5 successes.
Answer
Let X = number of successes, i.e. number of odd numbers.
p = probability of getting an odd number in a single throw of a die
$\therefore p =\frac{3}{6}=\frac{1}{2}$ and $q =1- p =1-\frac{1}{2}=\frac{1}{2}$
Given: n = 6
$\therefore X \sim B\left(6, \frac{1}{2}\right)$
The p.m.f. of X is given by
$p( X = x )={ }^n C_x p^x q^{n-x}$
i.e. $p ( x )={ }^6 C_x\left(\frac{1}{2}\right)^x\left(\frac{1}{2}\right)^{6-x}$
$={ }^6 C_x\left(\frac{1}{2}\right)^6, x =0,1,2, \ldots, 6$
P(5 successes) = P[X = 5]
$=p(5)={ }^6 C_5\left(\frac{1}{2}\right)^6$
$={ }^6 C_1 \times \frac{1}{64} \quad \ldots \ldots .\left[\because{ }^n C_x={ }^n C_{n-x}\right]$
$=\frac{6}{64}=\frac{3}{32}$
Hence, the probability of 5 successes is $\frac{3}{32}$
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Question 23 Marks
Find $k$, if
$f(x)=k x^2(1-x)$, for $0 < x < 1$,
$=0$
otherwise
is the p.d.f. of random variable $X$.
Answer
Since f(x) is p.d.f. of r.v.X, we get
$\int_{-\infty}^{\infty} f(x) d x=1$
$\therefore \int_{-\infty}^0 f(x) d x+\int_0^1 f(x) d x+\int_1^{\infty} f(x) d x=1$
$\therefore 0+\int_0^1 k x^2(1-x) d x+0=1$
$\therefore k \int_0^1\left(x^2-x^3\right) d x=1$
$\therefore k\left[\frac{x^3}{3}-\frac{x^4}{4}\right]_0^1=1$
$\therefore k\left[\left(\frac{1}{3}-\frac{1}{4}\right)-0\right]=1$
$\therefore k\left(\frac{1}{12}\right)=1$
$\therefore$K=12

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Question 33 Marks
Solve the following differential equation:
$x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0$
Answer
$ x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0 $ .....(1)
Put y = vx
$\therefore \frac{ dy }{ dx }= V + x \frac{ dV }{ dx }$ and $\frac{ y }{ x }= V$
$\therefore$ (1) becomes, $x \left( V + x \frac{ dV }{ dx }\right)- Vx + x \sin V=0$
$\therefore V x+x^2 \frac{d V}{d x}-V x+x \sin V=0$
$\therefore x ^2 \frac{ dV }{ dx }+ x \sin V =0$
$\therefore \frac{1}{\sin V} d V+\frac{1}{x} d x=0$
Integrating, we get
$\therefore \int \operatorname{cosec} VdV +\int \frac{1}{ x } dx = c _1$
$\therefore \log |\operatorname{cosec} V -\cot V |+\log | x |=\log c$, where $c _1=\log c$
$\therefore \log | x (\operatorname{cosec} V -\cot V )|=\log c$
$\therefore x\left(\frac{1}{\sin V}-\frac{\cos V}{\sin V}\right)=c$
$\therefore x(1-\cos V)=c \sin V$
$\therefore x\left[1-\cos \left(\frac{y}{x}\right)\right]=c \sin \left(\frac{y}{x}\right)$
This is the general solution.
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Question 43 Marks
Prove that $\int \frac{1}{a^2-x^2} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+c$
Answer
$\begin{aligned} & \int \frac{1}{a^2-x^2} d x=\int \frac{1}{(a-x)(a+x)} d x \\ = & \frac{1}{2 a} \int \frac{(a-x)+(a+x)}{(a-x)(a+x)} d x \\ = & \frac{1}{2 a} \int\left(\frac{1}{a+x}+\frac{1}{a-x}\right) d x \\ = & \frac{1}{2 a}\left[\int \frac{1}{a+x} d x+\int \frac{1}{a-x} d x\right] \\ = & \frac{1}{2 a}\left[\log |a+x|+\frac{\log |a-x|}{-1}\right]+C=\frac{1}{2 a}[\log |a+x|-\log |a-x|]+C \\ = & \frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\end{aligned}$
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Question 53 Marks
Find the approximate value of $\tan ^{-1}$ (1.002). [Given: $\pi=3.1416$ ]
Answer
x = 1. 002 = 1 + 0.002
x = a + b
a = 1, h = 0.002
$f(x)=\tan ^{-1}(x)$
$f^{\prime}(x)=\frac{1}{1+x^2}$
$f(1)=\tan ^{-1}(1)=\frac{\pi}{4}$
$f(1)=\frac{1}{1+1^2}$
$f(1)=\frac{1}{2}$
$f(a+h) \doteqdot f(a)+h f^{\prime}(a)$
$\doteqdot \frac{\pi}{4}+0.002 \times \frac{1}{2}$
$\doteqdot \frac{3.1416}{4}+0.001$
= 0.7864
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Question 63 Marks
If $y =\sin ^{-1} x$, show that $\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=0$
Answer
$y=\sin ^{-1} x$
Differentiating w.r.t. x,
$\frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}$
$\therefore \sqrt{1-x^2} \frac{d y}{d x}=1$
Squaring, we get
$\therefore\left(1-x^2\right)\left(\frac{d y}{d x}\right)^2=1$
Again differentiating w.r.t. x
$\left(1-x^2\right) 2 \frac{d y}{d x} \cdot \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2(-2 x)=0$
Dividing by $2 \frac{d y}{d x}$, we get
$\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=0$
Hence proved.
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Question 73 Marks
Find the angle between the line $\bar{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})$ and the plane $\bar{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=8$
Answer
$\overline{ b }=\hat{ i }+\hat{ j }+\hat{ k } $ and $ \overline{ n }=2 \hat{ i }-\hat{ j }+\hat{ k }$
$\sin \theta=\left|\frac{\overline{ b } \cdot \overline{ n }}{|\overline{ b }| \cdot|\overline{ n }|}\right|$
$=\left|\frac{(1)(2)+(1)(-1)+(1)(1)}{\sqrt{1^2+1^2+1^2} \sqrt{2^2+(-1)^2+1^2}}\right|$
$\therefore \sin \theta=\left|\frac{2-1+1}{\sqrt{3} \sqrt{6}}\right|=\frac{\sqrt{2}}{3}$
$\Rightarrow \theta=\sin ^{-1}\left(\frac{\sqrt{2}}{3}\right)$
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Question 83 Marks
Find the shortest distance between the lines $\bar{r}=(4 \hat{i}-\hat{j})+\lambda(\hat{i}+2 \hat{j}-3 \widehat{k})$ and $\bar{r}=(\hat{i}-\hat{j}-2 \widehat{k})+\mu(\hat{i}+4 \hat{j}-5 \widehat{k})$
Answer
We know that the shortest distance between the skew lines
$\bar{r}=\bar{a}_1+\lambda \bar{b}$ and $\bar{r}=\bar{a}_2+\mu \bar{b}_2$ is given by
$d =\left|\frac{\left(\bar{a}_2-\bar{a}_1\right) \cdot\left(\bar{b}_1 \times \bar{b}_2\right)}{\left|\bar{b}_1 \times \bar{b}_2\right|}\right|$
Here, $\bar{a}_1=4 \hat{i}-\hat{j}_1$
$\bar{a}_2=\hat{i}-\hat{j}-2 \widehat{k}$
$\bar{b}_1=\hat{i}+2 \hat{j}-3 \widehat{k}$
$\bar{b}_2=\hat{i}+4 \hat{j}-5 \widehat{k}$
$\therefore \bar{b}_1 \times \bar{b}_2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \widehat{k} \\ 1 & 2 & -3 \\ 1 & 4 & -5\end{array}\right|$
$=(-10+12) \hat{i}-(-5+3) \hat{j}+(4-2) \widehat{k}$
$=2 \hat{i}+2 \hat{j}+2 \widehat{k}$
and $\bar{a}_2-\bar{a}_1=(\hat{i}-\hat{j}-2 \widehat{k})-(4 \hat{i}-\hat{j})$
$=-3 \hat{i}-2 \widehat{k}$
$\therefore\left(\bar{a}_2-\bar{a}_2\right) \cdot\left(\bar{b}_1 \times \bar{b}_2\right)=(-3 \hat{i}-2 \widehat{k}) \cdot(2 \hat{i}+2 \hat{j}+2 \widehat{k})$
$=-3(2)+0(2)-2(2)$
$=-6+0-4$
$=-10$
and $\left|\bar{b}_1 \times \bar{b}_2\right|=\sqrt{2^2+2^2+2^2}$
$=\sqrt{4+4+4}$
$=2 \sqrt{3}$
$\therefore$Required shortest distance between the given lines
$=\left|\frac{-2}{2 \sqrt{3}}\right|$
$=\frac{1}{\sqrt{3}}$ units.

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Question 93 Marks
Prove by vector method, that the angle subtended on semicircle is a right angle.
Answer
Let seg AB be a diameter of a circle with centre C and P be any point on the circle other than A and B.
Then $\angle A P B$ is an angle subtended on a semicircle.
Let $\overline{ AC }=\overline{ CB }=\overline{ a }$ and $\overline{ CP }=\overline{ r }$
Then $|\overline{ a }|=|\overline{ r }|$ ....(1)
Image
$\overline{ AP }=\overline{ AC }+\overline{ CP }$
$=\overline{ a }+\overline{ r }$
$=\bar{r}+\bar{a}$
$\overline{ BP }=\overline{ BC }+\overline{ CP }$
$=-\overline{ CB }+\overline{ CP }$
$=-\bar{a}+\bar{r}$
$\therefore \overline{ AP } \cdot \overline{ BP }=(\overline{ r }+\overline{ a }) \cdot(\overline{ r }-\overline{ a })$
$=\overline{ r } \cdot \overline{ r }-\overline{ r } \cdot \overline{ a }+\overline{ a } \cdot \overline{ r }-\overline{ a } \cdot \overline{ a }$
$=|\overline{ r }|^2-|\overline{ a }|^2$
$=0 \quad \ldots(\because \bar{r} \cdot \bar{a}=\bar{a} \cdot \bar{r})$
$\therefore \overline{ AP } \perp \overline{ BP }$
$\therefore \angle APB$ is a right angle.
Hence, the angle subtended on a semicircle is the right angle.
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Question 103 Marks
In $\triangle A B C$, prove the following:
$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^2+b^2+c^2}{2 a b c}$
Answer
LHS $=\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}$
$=\frac{\left(\frac{b^2+c^2-a^2}{2 b c}\right)}{a}+\frac{\left(\frac{c^2+a^2-b^2}{2 c a}\right)}{b}+\frac{\left(\frac{a^2+b^2-c^2}{2 a b}\right)}{c}$
$=\frac{b^2+c^2-a^2}{2 a b c}+\frac{c^2+a^2-b^2}{2 a b c}+\frac{a^2+b^2-c^2}{2 a b c}$
$=\frac{b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2}{2 a b c}$
$=\frac{a^2+b^2+c^2}{2 a b c}$
= RHS
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Question 113 Marks
Prove the following:
$\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{4}$
Answer
L.H.S. $=\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{3}\right)$
$=\tan ^{-1}\left[\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}\right]$
$=\tan ^{-1}\left(\frac{3+2}{6-1}\right)$
$=\tan ^{-1}(1)$
$=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)$
$=\frac{\pi}{4}$
= R.H.S.
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Question 123 Marks
Express the following switching circuit in the symnolic from of logic,. Construct the switching table :
Image
Answer
Let $p, q$ denote the statements that the switches $S_1, S_2$ are closed respectively.
Let $\sim p, \sim q$ denote the statements that switches $ 'S_1\ '$ and $'S_2\ '$are closed respectively.
The symbolic form of the given switching circuit is $(p \vee q) \wedge(\sim p \wedge \sim q) \equiv 1$
The switching table for the given switching circuit is as follows:
        $ \text{RBCs.}$    
$p$ $q$ $\sim p$ $\sim q$ $p \ Vq$ $\sim p \wedge \sim q$ $A \wedge B$
$1$ $1$ $0$ $0$ $1$ $0$ $0$
$1$ $0$ $0$ $1$ $1$ $0$ $0$
$0$ $1$   $0$ $1$ $0$ $0$
$0$ $0$   $1$ $0$ $1$ $0$
In the last column of the switching table, each entry is zero, i.e. it is a contradiction.
Hence, no current will flow in the circuit irrespective of the positions of the switches.
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