3 Marks Question

Question 13 Marks

A random variable $X$ can take all non negative integral values and the probability that $X$ takes the value is proportional to $5^{-r}$. Find $P(X<3)$

Answer

Given $P(X=r) \alpha \frac{1}{5^r}$
$P(X=r)=k \frac{1}{5^r}($ where $k$ is a non$-$zero constant$)$
$P(r=0)=k \cdot \frac{1}{5^0}$
$P(r=1)=k \cdot \frac{1}{5^1}$
$P(r=2)=k \cdot \frac{1}{5^2}$
$P(r=3)=k \cdot \frac{1}{5^3}$
We have, $P(X=0)+P(X=1)+P(X=2)+\ldots \ldots \ldots \ldots=1$
$\Rightarrow k\left(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\ldots \ldots \ldots\right)=1$
$\Rightarrow k\left(\frac{1}{1-\frac{1}{5}}\right)=1$
$\Rightarrow k=\frac{4}{5} $
So, $P(X<3)=P(X=0)+P(X=1)+P(X=2)$
$=\frac{4}{5}\left(1+\frac{1}{5}+\frac{1}{5^2}\right)$
$=\frac{4}{5}\left(\frac{25+5+1}{25}\right)$
$=\frac{124}{125} .$

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Question 23 Marks

Evaluate : $\int_0^1 x(1-x)^n d x$; $($ where $n \in N)$

Answer

$\int_0^1 x(1-x)^n d x$
$\begin{array}{c}=\int_0^1(1-x)\{1-(1-x)\}^n d x,\left(a s, \int_0^a f(x) d x=\int_0^a f(a-x) d x\right) \\ =\int_0^1 x^n(1-x) d x \\ =\int_0^1 x^n d x-\int_0^1 x^{n+1} d x \\ =\frac{1}{n+1}\left[x^{n+1}\right]_0^1-\frac{1}{n+2}\left[x^{n+2}\right]_0^1 \\ =\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}\end{array}$

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Question 33 Marks

Find the vector and the cartesian equation of the line that passes through through (-1, 2, 7) and is perpendicular to the lines $\vec{r}=2 \hat{\imath}+\hat{\jmath}-3 \hat{ k }+\lambda(\hat{\imath}+2 \hat{\jmath}+5 \hat{ k })$ and $\vec{r}=3 \hat{\imath}+3 \hat{\jmath}-7 \hat{ k }+\mu(3 \hat{\imath}-2 \hat{\jmath}+5 \hat{ k })$.

Answer

Line perpendicular to the lines $\vec{r}=2 \hat{\imath}+\hat{\jmath}-3 \hat{k}+\lambda(\hat{\imath}+2 \hat{\jmath}+5 \hat{k}) \text { and } \vec{r}=3 \hat{\imath}+3 \hat{\jmath}-7 \hat{k}+\mu(3 \hat{\imath}-2 \hat{\jmath}+5 \hat{k}) \text {. }$
has a vector parallel it is given by $\vec{b}=\overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 2 & 5 \\ 3 & -2 & 5\end{array}\right|=20 \hat{\imath}+10 \hat{\jmath}-8 \hat{k}$
$\therefore$ equation of line in vector form is $\vec{r}=-\hat{\imath}+2 \hat{\jmath}+7 \hat{k}+a(10 \hat{i}+5 \hat{\jmath}-4 \hat{k})$
and equation of line in cartesian form is $\frac{x+1}{10}=\frac{y-2}{5}=\frac{z-7}{-4}$

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Question 43 Marks

The probability that it rains today is 0.4. If it rains today, the probability that it will rain tomorrow is 0.8. If it does not rain today, the probability that it will rain tomorrow is 0.7. If
P1: denotes the probability that it does not rain today.
P2: denotes the probability that it will not rain tomorrow, if it rains today.
P3: denotes the probability that it will rain tomorrow, if it does not rain today.
P4: denotes the probability that it will not rain tomorrow, if it does not rain today.
(i) Find the value of $P_1 \times P_4-P_2 \times P_3$
(ii) Calculate the probability of raining tomorrow.

Answer

Since the event of raining today and not raining today are complementary events so if the probability that it rains today is 0.4 then the probability that it does not rain today is $1-0.4=0.6 \Rightarrow P_1=0.6$
If it rains today, the probability that it will rain tomorrow is 0.8 then the probability that it will not rain tomorrow is 1-0.8-0.2.
If it does not rain today, the probability that it will rain tomorrow is 0.7 then the probability that it will not rain tomorrow is 1-0.7-0.3

(i) $P_1 \times P_4-P_2 \times P_3=0.6 \times 0.3-0.2 \times 0.7=0.04$.
(ii) Let $E_1$ and $E_2$ be the events that it will rain today and it will not rain today respectively.
$P\left(E_1\right)=0.4 \& P\left(E_2\right)=0.6$$A$ be the event that it will rain tomorrow. $P\left(\frac{A}{E_1}\right)=0.8 \& P\left(\frac{A}{E_2}\right)=0.7$
We have, $P(A)=P\left(E_1\right) P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right)=0.4 \times 0.8+0.6 \times 0.7=0.74$.
The probability of rain tomorrow is $0 . 7 4$.

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Question 53 Marks

Consider the following Linear Programming Problem:
Minimise Z = x + 2y
Subject to $2 x+y \geq 3, x+2 y \geq 6, x, y \geq 0$
Show graphically that the minimum of Z occurs at more than two points

Answer

The feasible region determined by the constraints$2 x+y \geq 3, x+2 y \geq 6, x \geq 0, y \geq 0$ is as shown.

The corner points of the unbounded feasible region are A(6, 0) and B(0, 3) The values of Z at these corner points are as follows:

Corner point	Value of the objective function z = x + 2y
A(6, 0)	6
B(0, 3)	6

We observe the region x + 2y < 6 have no points in common with the unbounded feasible region. Hence the minimum value of z = 6
It can be seen that the value of Z at points A and B is same. If we take any other point on the line x + 2y = 6 such as (2,2) on line x+ y = 6 then Z = 6
Thus, the minimum value of Z occurs for more than 2 points, and is equal to 6.

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Question 63 Marks

Evaluate: $\int\left\{\frac{1}{\log x}-\frac{1}{(\log x)^2}\right\} d x; ($where $\left.x>1\right)$.

Answer

$ \int\left\{\frac{1}{\log _e x}-\frac{1}{\left(\log _e x\right)^2}\right\} d x$
$=\int \frac{d x}{\log _e x}-\int \frac{1}{\left(\log _e x\right)^2} d x=\frac{1}{\log _e x} \int d x-\int\left\{\frac{d}{d x}\left(\frac{1}{\log _e x}\right) \int d x\right\} d x-\int \frac{1}{\left(\log _e x\right)^2} d x$
$=\frac{x}{\log _e x}+\int \frac{1}{\left(\log _e x\right)^2} \frac{1}{x} \cdot x \cdot d x-\int \frac{1}{\left(\log _e x\right)^2} d x$
$=\frac{x}{\log _e x}+\int \frac{1}{\left(\log _e x\right)^2} d x-\int \frac{d x}{\left(\log _e x\right)^2}=\frac{x}{\log _e x}+c $where' $c$ 'is any arbitary constant of integration.

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Question 73 Marks

An ant is moving along the vector $\overrightarrow{l_1}=\hat{\imath}-2 \hat{\jmath}+3 \hat{k}$ Few sugar crystals are kept along the vector $\overrightarrow{l_2}=3 \hat{\imath}-2 \hat{j}+\hat{k}$ which is inclined at an angle $\theta$ with the vector $\overrightarrow{l_1}$. Then find the angle $\theta$. Also find the scalar projection of $\overrightarrow{l_1}$ on $\overrightarrow{l_2}$

Answer

$(i) \theta=\cos ^{-1}\left(\frac{\overrightarrow{l_1} \cdot \overrightarrow{l_2}}{\left|\overrightarrow{l_1}\right| \cdot\left|\overrightarrow{l_2}\right|}\right)$
$=\cos ^{-1}\left(\frac{(\hat{i}-2 \hat{\jmath}+3 \hat{k}) \cdot(3 \hat{i}-2 \hat{\jmath}+\hat{k})}{|(\hat{i}-2 \hat{\jmath}+3 \hat{k})||(3 \hat{i}-2 \hat{\jmath}+\hat{k})|}\right)$
$=\cos ^{-1}\left(\frac{3+4+3}{\sqrt{1+4+9} \sqrt{9+4+1}}\right)$
$=\cos ^{-1}\left(\frac{10}{14}\right)$
$=\cos ^{-1}\left(\frac{5}{7}\right)$
$(ii)$ Scalar projection of $\overrightarrow{l_1}$ on $\overrightarrow{l_2}$
$=\frac{\overrightarrow{l_1} \cdot \overrightarrow{l_2}}{\left|\overrightarrow{l_2}\right|}$
$=\frac{(\hat{\imath}-2 \hat{\jmath}+3 \hat{ k }) \cdot(3 \hat{i}-2 \hat{\jmath}+\hat{k})}{|(3 \hat{\imath}-2 \hat{\jmath}+\hat{k})|}$
$=\frac{3+4+3}{\sqrt{9+4+1}}=\frac{10}{\sqrt{14}}$.

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Question 83 Marks

According to a psychologist, the ability of a person to understand spatial concepts is given by
$A=\frac{1}{3} \sqrt{t}$, where $t$ is the age in years, $t \in[5,18]$. Show that the rate of increase of the ability to understand spatial concepts decreases with age in between $5$ and $18.$

Answer

$A=\frac{1}{3} \sqrt{t}: \frac{d A}{d t}=\frac{1}{6} t^{-\frac{1}{2}}$
$=\frac{1}{6 \sqrt{t}} ; \forall t \in(5,18)$
$\frac{d A}{d t}=\frac{1}{6 \sqrt{t}} $
$\therefore \frac{d^2 A}{d t^2}=-\frac{1}{12 t \sqrt{t}}$
$\text { So, } \frac{d^2 A}{d t^2}<0, \forall t \in(5,18)$
This means that the rate of change of the ability to understand spatial concepts decreases $($slows down$)$ with age.

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Question 93 Marks

A kite is flying at a height of 3 metres and 5 metres of string is out. If the kite is moving away horizontally at the rate of 200 cm/s, find the rate at which the string is being released.

Answer

$x^2+3^2=y^2$
When $y=5$ then $x=4$, now $2 x \frac{d x}{d t}=2 y \frac{d y}{d t}$
$4(200)=5 \frac{d y}{d t} \Rightarrow\frac{d y}{d t}=160 cm / s$

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Maths 3 Marks Question

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