MHT CET 2018 :: Mathematics :: General questions

• Mathematics - General questions

 The area of the region bounded by x² =4y, y=1, y=4 and the Y-axis lying  in the first quadrant is............. square units

Answer:

Explanation:

We have,

    x²=4y

    y=1

    y=4

 and the Y-axis lying  in the first quadrant.

$\therefore$ Required  area =  $\int_{1}^{4} \sqrt{4y}dy$

 =$2\left[\frac{2}{3}y^{3/2}\right]_{1}^{4}=\frac{4}{3}[4^{3/2}-1]$

$\frac{4}{3}(8-1)=\frac{4}{3}\times 7=\frac{28}{3}$

A coin is tossed three times if X denotes the absolute difference between the number of heads and the number of heads and the number of tails, then P(X =1) = 

Answer:

Explanation:

 Given, 

A coin is tossed three times

and X= absolute difference between the number of heads and number of tails

 Now, X= 1 when exactly two head or two  tail comes

 $\therefore$     $P(X=1)=\frac{3}{8}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4}$

           [ $\because$ probability of exactlly two head= $\frac{3}{8}$

                and probability of exactly  two tails= $\frac{3}{8}$]

The sum of the first 10 terms of the series 9+99+999+... is 

Answer:

Explanation:

Let ,

 S_n= 9+99+999+.........n terms

 $\Rightarrow$    $ S_{n}=(10-1)+(100-1)+(1000-1)+$.... n  terms

 $\Rightarrow$    $ S_{n}=(10+10^{2}+10^{3}+....... $   n terms -(1+1+..... n terms)

$\Rightarrow $     $S_{n}=\frac{10(10^{n}-1)}{10-1}-n$

$[a+ar+ar^{2}+..... ar^{n-1}= \frac{a(r^{n}-1}{r-1}, r>1]$

 $\Rightarrow $    $S_{n}= \frac{10}{9}(10^{n}-1)-n$

 put n=10

$\Rightarrow $    $S_{10}= \frac{10}{9}(10^{10}-1)-10$

 $= \frac{10}{9}(10^{10}-1-9)$

    $= \frac{10}{9}(10^{10}-10)=\frac{100}{9}(10^{9}-1)$

The equation of line passing through (3,-1,2) and perpendicular to the lines

 $\overrightarrow{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2\hat{i}-2\hat{j}+2\hat{k})$

 and $\overrightarrow{r}=(2\hat{i}+\hat{j}-3\hat{k})+\mu(\hat{i}-2\hat{j}+2\hat{k})$  is 

Answer:

Explanation:

 Direction ratio of line perpendicular to the lines  $\overrightarrow{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$

 and  $\overrightarrow{r}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}$ is $\alpha(\overrightarrow{b_{1}}\times\overrightarrow{b_{2}})$

 $\therefore$    Direction ratio of line perpendicular to the lines

   $\overrightarrow{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2\hat{i}-2\hat{j}+2\hat{k})$

and 

  $\overrightarrow{r}=(2\hat{i}+\hat{j}-3\hat{k})+\mu(\hat{i}-2\hat{j}+2\hat{k})$  is  

                                        $\alpha\begin{bmatrix}\hat{i} & \hat{j}&\hat{k} \\2 & -2&1 \\1 &-2 &2 \end{bmatrix}$

 = $\alpha [{(-4+2)\hat{i}-(4-1)\hat{j}+(-4+2)\hat{k}]}$

    = $\alpha [{-2\hat{i}-3\hat{j}-2\hat{k}]}$

  Now, equation of line passing through (3,-1,2) and parallel to ${-2\hat{i}-3\hat{j}-2\hat{k}}$

 $\overrightarrow{r}=3\hat{i}-\hat{j}-2\hat{k}+\beta(2\hat{i}+3\hat{j}+2\hat{k})$

 Hence, cartesian form of the above equation is 

$\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-2}{2}$

If  $A=\begin{bmatrix}1 & 2&3 \\-1 & 1&2\\ 1&2&4 \end{bmatrix}$ n , then (A²-5A)A^-1= 

Answer:

Explanation:

We have,

 $A=\begin{bmatrix}1 & 2&3 \\-1 & 1&2\\ 1&2&4 \end{bmatrix}$

 Now, (A²-5A)A^-1

= A². A^-1 -5A. A^-1 = A-5l

$=\begin{bmatrix}1 & 2&3 \\-1 & 1&2\\ 1&2&4 \end{bmatrix}-\begin{bmatrix}5 &0&0 \\0 & 5&0\\ 0&0&5 \end{bmatrix}$

  $=\begin{bmatrix}-4 & 2&3 \\-1 & -4&2\\ 1&2&-1 \end{bmatrix}$

• Mathematics - General questions