MHT CET 2017 :: Mathematics :: General questions

• Mathematics - General questions

If the volume of spherical ball is increasing at the rate of 4$\pi$ cm³/s, then the rate of change of its surface area when the  volume is 288 $\pi$ cm³ , is 

Answer:

Explanation:

Let V and r be the volume and radius of spherical ball , respectively,

 Volume of spherical ball = $\frac {4}{3} \pi r^{3}$

 $\Rightarrow$       $ 288 \pi$ =$\frac {4}{3} \pi r^{3}$    [ given, V=288 cm³]

$\Rightarrow$       288 =$\frac {4}{3}  r^{3}$

$\Rightarrow$         r³  =72 x 3=8 x 27

$\Rightarrow$      $ r^{3} = 2^{3} \times 3^{3}$

                                  [taking cube roots both sides]

$\Rightarrow$     r=6

On  differentiating Eq.(i)  w.r.t 't' , we get

                           $\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}$

$\therefore$    $4 \pi=4\pi r^{2}\frac{dr}{dt}$ 

                                            [ given $\frac{dV}{dt}=4\pi$ cubic cm/s]

$\Rightarrow$           $1=(6)^{2}\frac{dr}{dt}$     [$\because$ r=6]

$\Rightarrow$      $\frac{dr}{dt}=\frac{1}{36}$

Now, surface  area of spherical ball (s) = $4\pi r^{2}$

$\Rightarrow$   s= $ 4\pi r^{2}$

 On differentiating  both sides, w.r.t 't', we get

 $\frac{ds}{dt}=4\times 2\pi r\frac{dr}{dt}$

  $=8\times \pi \times6\times\frac{1}{36}$

                                       $\left[\because r= 6 and \frac{dr}{dt}=\frac{1}{36}\right]$

$\Rightarrow$                      $\frac{ds}{dt}=\frac{4\pi}{3}cm^{2}/s$

A box contains 6 pens , 2 of which are defective . Two pens are taken randomly  from the box. if r.v.X: number of defective pens obtained , then standard deviation of X= 

Answer:

Explanation:

 Let x be the number of defective pens.

 Two pens are taken randomly from the box,

 $\therefore$     x can take values 0,1,2

  $p(x=0)=\frac{^{4}C_{2}}{^{6}C_{2}}=\frac{4\times3}{6 \times 5}=\frac{2}{5}=\frac{6}{15}$

$p(x=1)=\frac{^{2}C_{1} \times^{4}C_{1}}{^{6}C_{2}}=\frac{2\times4\times2 \times 1}{6 \times 5}=\frac{8}{15}$

$p(x=2)=\frac{^{2}C_{2}}{^{6}C_{2}}=\frac{1\times2\times1 }{6 \times 5}=\frac{1}{15}$

E(x)= $\frac{10}{15}$  and =$\frac {2}{3}$

   E(x² )= $\frac{12}{15}=\frac{4}{5}$

 Standard deviation$=\sqrt{E(x^{2})-[E(x)]^{2}}$

 Standard deviation = $\sqrt{\left(\frac{4}{5}\right)-\left(\frac{2}{3}\right)^{2}}$

    = $\sqrt{\frac{4}{5}-\frac{4}{9}}=\sqrt{\frac{36-20}{45}}$

 $=\sqrt{\frac{16}{45}}=\frac{4}{3\sqrt{5}}$

 Tjhe area of the region bounded by the lines y=2x+1, y=3x+1 and x=4

Answer:

Explanation:

 Table for line y=2x+1

and table for line y=3x+1

 Area of shaded region = $\frac{1}{2}\begin{bmatrix}0 & 1&1 \\4 & 9&1\\4&13&1 \end{bmatrix}$

                        $\left[\because   Area of triangle=\frac{1}{2}\begin{bmatrix}x_{1} & y_{1}&1 \\x_{2} & y_{2} &1\\x_{3}&y_{3}&1\end{bmatrix}\right]$

$=\frac{1}{2}[0(9-13)-1(4-4)+1(4\times 13-4 \times 9)]$

  $=\frac{1}{2}[0-0+4(13-9)]=\frac{1}{2}\times 4\times 4=8 sq.units$

If $\int\sqrt{\frac{x-5}{x-7}}dx=A\sqrt{x^{2}-12x+35}+\log|x-6+\sqrt{x^{2}-12x+35|}+C,$  then A= 

Answer:

Explanation:

Let l=$\int\sqrt{\frac{x-5}{x-7}}dx$

 =$\int\sqrt{\frac{(x-5)(x-5)}{(x-7)(x-7)}}dx$   [rationlising]

=$ \int \frac{x-5}{\sqrt{x^{2}-5x-7x+35}}dx$

=$\int \frac{x-5}{\sqrt{x^{2}-12x+35}}dx$

$=\frac{1}{2}\int \frac{2x-10}{\sqrt{x^{2}-12x+35}}dx$

$=\frac{1}{2}\int \frac{2x-12+2}{\sqrt{x^{2}-12x+35}}dx$

$=\sqrt{x^{2}-12x+35}+\int \frac{1}{\sqrt{(x^{2}-12x+36-1)}}dx+C$

  $[\because $ Let $ x^{2}-12x+35=t\Rightarrow(2x-12)dx=dt]$

$=\sqrt{x^{2}-12x+35}+\int \frac{1}{\sqrt{(x-6)^{2}-1^2)}}dx+C$

$l=\sqrt{x^{2}-12x+35}+\log|x-6+\sqrt{x^{2}-12x+35}|+C$

$\therefore $   $l=A\sqrt{x^{2}-12x+35}+\log |x-6+\sqrt{x^{2}-12x+35}+C$

$ =1.\sqrt{x^{2}-12x+35}+\log| x-6+\sqrt{x^{2}-12x+35}+C$

 on comparing both sides , we get A=1

If $x=a\left(t-\frac{1}{t}\right),y=a\left(t+\frac{1}{t}\right)$   , where t is the parameter , then $\frac{dy}{dx}$= ?

Answer:

Explanation:

Given , $x=a\left(t-\frac{1}{t}\right),y=a\left(t+\frac{1}{t}\right)$

Now,  $y^{2}-x^{2}=a^{2}\left[\left(t+\frac{1}{t}\right)^{2}-a^{2}\left(t-\frac{1}{t}\right)^{2}\right]$

 = $a^{2}\left[t^{2}+\frac{1}{t^{2}}+2-t^{2}-\frac{1}{t^{2}}+2\right]$

$\Rightarrow $  $y^{2}-x^{2}=4a^{2}$

On differentiating  both sides w.r.t 'x' , we get

    2y $\frac{dy}{dx}$-2x=0

$\Rightarrow$    $ 2\left( y\frac{dy}{dx}-x\right)=0$

$\therefore$     $\frac{dy}{dx}=\frac{x}{y}$

• Mathematics - General questions