MHT CET 2020 :: Mathematics :: General questions

• Mathematics - General questions

 The c.d. f  F(x) associated  with p.d.f.

$f(x) =3(1-2x^{2})$. If  0 < x <1. is  $k\left( x-\frac{2x^{3}}{k}\right)$  , then value of k is 

Answer:

Explanation:

 we have , 

$f(x)=\begin{cases}3(1-2x^{2}), & 0<x<1\\0, & otherwise\end{cases}$

$F(x=n)=P(X\leq x)=\int_{0}^{x}f(x) dx $

 =$\int_{0}^{x}3(1-2x^{2}) dx$

 =  $\left[3\left(x-\frac{2x^{3}}{3}\right)\right]_{0}^{x}=3\left(x-\frac{2x^{3}}{3}\right)$

$\therefore$    k=3

 The negation of the statement pattern $\sim p \vee (q\rightarrow\sim r)$  is

Answer:

Explanation:

 We have 

  $\sim p \vee (q\rightarrow\sim r)$

 $\sim p \vee (\sim q\vee\sim r)$      $[\because p\rightarrow q=\sim p \vee q]$

 $\sim p \vee \sim (q\wedge r)$

$\sim (p \wedge  (q\wedge r))$

 $\therefore$     $\sim (\sim (p \wedge  (q\wedge r))$

            $p\wedge (q\wedge r)$

 The quadratic  equation whose roots are the numbers having arithmetic mean 34 and geometric mean 16 is

Answer:

Explanation:

 Let  $\alpha$ and $\beta $ are roots of quadratic equatuion

 Given,             $\frac{\alpha+\beta}{2}=34$

   $\alpha+\beta=68$

 and    $\sqrt{\alpha\beta}=16$

$\alpha\beta=256$

 $\therefore$ Required  quadratic equation is,

$x^{2}-(\alpha+\beta)x+\alpha\beta=0$

$x^{2}-68x+256=0$

$\int_{0}^{1} \tan^{-1}\left(\frac{2x-1}{1+x-x^{2}}\right)dx=$

Answer:

Explanation:

 We have,

l=$\int_{0}^{1} \tan^{-1}\left(\frac{2x-1}{1+x-x^{2}}\right)dx$

$l=\int_{0}^{1} \tan^{-1}\left(\frac{2(1-x)-1}{1+(1-x)-(1-x)^{2}}\right)dx$

                                   $\left[ \because\int_{0}^{a} f(x)dx=\int_{0}^{a} f(a-x)dx\right]$

$l=\int_{0}^{1} \tan^{-1}\left(\frac{1-2x}{1+x-x^{2}}\right)dx$

$l=\int_{0}^{1}- \tan^{-1}\left(\frac{2x-1}{1+x-x^{2}}\right)dx$

 2l=0

 l=0

 With usual notations, if the angles A,B,C of a  $\triangle$ABC  are in AP  and b:c= $\sqrt{3}:\sqrt{2}$

Answer:

Explanation:

In  $\triangle$ABC,

$\angle A,\angle B, \angle C $ are in A.P

 $\therefore$           2$\angle B $=  $\angle A+ \angle C $.............(i)

 and   $\angle A+\angle B+ \angle C =180^{0}$..........(ii)

  From eqs.(i) and (ii) , B= 60°

 $\therefore$     $\frac{b}{\sin B}=\frac{c}{\sin C}$

 $\Rightarrow\frac{\sin B}{\sin C}=\frac{b}{c}$

 $\Rightarrow\frac{\sin 60^{0}}{\sin C}=\frac{\sqrt{3}}{\sqrt{2}}$     $[\because$  given   $b:c=\sqrt{3}:\sqrt{2}]$

$\Rightarrow \sin C=\frac{\sqrt{2}\times\sqrt{3}}{\sqrt{3}\times2}=\frac{1}{\sqrt{2}}$

 $\therefore$        sin C= sin  45°

 $\Rightarrow$       C=45°

 $\angle A=2\angle B-\angle C=120^{0}-45^{0}=75^{0}$

• Mathematics - General questions