MHT CET 2020 :: Mathematics :: General questions

• Mathematics - General questions

 If  $f(x)=\frac{2x+3}{3x-2},x\neq\frac{2}{3}$  , then the function f of  is 

Answer:

Explanation:

we have, 

$f(x)=\frac{2x+3}{3x-2}$

  $fof=f(f(x))=\frac{2f(x)+3}{3f(x)-2}$

$\Rightarrow fof=\frac{2(\frac{2x+3}{3x-2})+3}{3(\frac{2x+3}{3x-2})-2}$

$\Rightarrow $    $fof=\frac{4x+6+9x-6}{6x+9-6x+4}$

$\Rightarrow fof=\frac{13x}{13}=x= $ an identity  function.

 The p.d.f  of c.r.v X is given by  $f(x)=\frac{x+2}{18}$  , if -2 < x < 4=0, otherwise =0, then P[|x| <1]=

Answer:

Explanation:

$f(x)=\begin{cases}\frac{x+2}{18}, & -2<x<4 = 0\\0 & otherwise\end{cases}$

$P[|x| <1]=\int_{-1}^{1 }f(x) dx$

 $=\int_{-1}^{1 }\left(\frac{x+2}{18}\right)dx=\frac{1}{18}\left[\frac{x^{2}}{2}+2x\right]^{1}_{-1}$

$=\frac{1}{18}\left[\left(\frac{1}{2}+2\right)-\left(\frac{1}{2}-2\right)\right]=\frac{4}{18}=\frac{2}{9}$

The value of  $\sin^{-1}\left(\frac{1}{2}\right)+\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$  is

Answer:

Explanation:

 we have,

 $\sin^{-1}\left(\frac{-1}{2}\right)+\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

$-\sin^{-1}\left(\frac{1}{2}\right)+\pi-\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$

$=-\frac{\pi}{6}+\pi-\frac{\pi}{6}=\pi-\frac{\pi}{3}=\frac{2\pi}{3}=\cos^{-1}(\frac{-1}{2})$

 The minimum value of Z=5x+8y  subject to  $x+y\geq 5,0\leq x\leq4,y\geq2,x\geq0, y\geq0$   is 

Answer:

Explanation:

 Given minimise  z=5x+8y

  Subject to constants   $x+y\geq 5$

 $0\leq x\leq4,y\geq2,x\geq0, y\geq0$ 

 corner point        z=5x+8y

A (3,2)                  15+16=31

B(4,2)                   20+16=36

C(0,5)                   0+40=40

 $\therefore$   Minimum value=31

If   $\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0, x\neq y,$  then  $ (1+x)^{2}\frac{dy}{dx}=$

Answer:

Explanation:

We have

$\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0$

$\Rightarrow$     $ x\sqrt{1+y}+y\sqrt{1+x}=0$

$\Rightarrow $     $x\sqrt{1+y}=-y\sqrt{1+x}=0$

$\Rightarrow $    $x^{2}(1+y)=y^{2}(1+x)$

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

$\Rightarrow$     $ (x+y)(x-y)+xy(x-y)=0$

$\Rightarrow$     $ (x-y)(x+y+xy)=0$

$\Rightarrow$     $ y(1+x)=-x$    $  [\because x-y\neq0]$

$\Rightarrow$    y=   $ \frac{-x}{1+x}$

$\Rightarrow$    $\frac{dy}{dx}=-\left[\frac{(1+x)(1)-(x)}{(1+x)^{2}}\right]$

$\Rightarrow$    $(1+x)^{2}\frac{dy}{dx}=-1$

• Mathematics - General questions