MHT CET 2016 :: Mathematics :: General questions

• Mathematics - General questions

If Rolle's theorm for f(x)=e^x(sin x-cos x) is verified on $\left[\frac{\pi}{4},\frac{5\pi}{4}\right]$, then the value of c is 

Answer:

Explanation:

 Given ,   f(x)=e^x(sin x-cos x) 

 On differentiating  both sides w.r.t x, we get

$ f'(x)= e^{x}\frac{d}{dx}(\sin x-\cos x)+(\sin x-\cos x )\frac{d}{dx}(e^{x})$

                       [ by using product rule of derivative]

$= e^{x}(\cos x+\sin x)+(\sin x-\cos x )e^{x}= 2e^{x} \sin x$

 We know that, if Rolle's theorm is verified,

then their exist   $c\in \left(\frac{\pi}{4},\frac{5 \pi}{4}\right)$ , such that f'(c)=0

 $\therefore$   $2e^{c} \sin c =0$ $\Rightarrow$ $\sin c=0$

$\Rightarrow$         $c=\frac{\pi}{2}$  $\in \left(\frac{\pi}{4},\frac{5 \pi}{4}\right)$

If $A=\begin{bmatrix}1 & 1&0 \\2 & 1&5\\1&2&1 \end{bmatrix}$ , then

$a_{11}A_{21}+a_{12}A_{22}+a_{13}A_{23}$  is equal to

Answer:

Explanation:

Given , $A=\begin{bmatrix}1 & 1&0 \\2 & 1&5\\1&2&1 \end{bmatrix}$

We know that, the sum of the product of element other than the corresponding cofactor is zero.

$\therefore$    $a_{11}A_{21}+a_{12}A_{22}+a_{13}A_{23}$ =0

The differential equation of the family of circles touching Y-axis at the origin is

Answer:

Explanation:

Let centre of circle on X-axis be (h,0) .The radius of circle will be h

 $\therefore$  The equation of circle having centre (h,0) and radius h is

 $(x-h)^{2}+(y-0)^{2}=h^{2}$

 $\Rightarrow$    $x^{2}+h^{2}-2hx+y^{2}=h^{2}$

$\Rightarrow$       $x^{2}-2hx+y^{2}$=0  ............(i)

 On differentiating both sides w.r.t x, we get

    $2x-2h+2y \frac{dy}{dx}=0 $  $\Rightarrow$    $h= x+y\frac{dy}{dx}$

On putting h=x+y $\frac{dy}{dx}$  in Eq.(i)  , we get

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

$\Rightarrow$      $-x^{2}+y^{2}-2xy\frac{dy}{dx}=0$

 $\Rightarrow$     $(x^{2}-y^{2})+2xy\frac{dy}{dx}=0$

Let $X\sim B(n,p),if E(x)=5,Var(X)=2.5,$   then $ p(X<1)$ is equal to 

Answer:

Explanation:

Given  , mean E(X)=5, and variance , Var(X)=2.5

$\therefore$    np=5 and npq=2.5

$\Rightarrow$  5q=2.5 $\Rightarrow$ q=$\frac{1}{2}$

$\Rightarrow$    p+q=1

$\therefore$     $p=1-\frac{1}{2}=\frac{1}{2}$

$\therefore$ np=5                  

$\Rightarrow$  $n\times\frac{1}{2}=5$                 $\left[\because p=\frac{1}{2}\right]$

$\Rightarrow$     n=10

$p(X<1)=p(X=0)=^{n}C_{r}p^{r}q^{n-r}$

= $p(X<1)=p(X=0)=^{10}C_{0}\left(\frac{1}{2}\right)^{0}\left(\frac{1}{2}\right)^{10-0}$

$=1\times1\times\left(\frac{1}{2}\right)^{10}=\left(\frac{1}{2}\right)^{10}$

• Mathematics - General questions