MHT CET 2018 :: Mathematics :: General questions

• Mathematics - General questions

If a line makes angle $120^{0}$ and $60^{0}$ with the positive directions of X and Z -axes respectively, then the angle made by the line  with a positive Y-axis is 

Answer:

Explanation:

Let angles $\alpha, \beta, \gamma$  makes positive direction of X,Y and Z  axes, respectively,

$\therefore$     $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1$

Here, $\alpha$= $120^{0}$   and $\gamma$ =$60^{0}$

 $ \therefore$   $\cos^{2}120^{0}+\cos^{2}\beta+\cos^{2}60^{0}=1$

$\Rightarrow$   $(\frac{1}{2})^{2}+\cos^{2}\beta+(\frac{1}{2})^{2}=1$

$\Rightarrow$    $\cos^{2}\beta=1-\frac{1}{2}=\frac{1}{2}$

$\Rightarrow$     $\cos^{}\beta=\frac{\pm 1}{\sqrt{2}}$

$\Rightarrow$   $\beta$=$135^{0}$

The order of the differential equation of all parabolas, whose latus rectum is 4a and axis parallel to the X-axis, is 

Answer:

Explanation:

Equation  of all parabola , whose axis is parallel  to X- axis  and have latusrectum 4a , is (y-K)²  =4a(x-h)²

Here, we have two effective constants, K  and h. So, we have to differentiate it twice.

 Hence, the order of the differential equation is 2

The general solution of differential equation $\frac{dx}{dy}=cos(x+y)$  is

Answer:

Explanation:

We have ,   $\frac{dx}{dy}=cos(x+y)$

 Put , x+y=v

 $\Rightarrow$     $\frac{dx}{dy}+1=\frac{dv}{dy}$

$\Rightarrow$     $\frac{dx}{dy}=\frac{dv}{dy}-1$

$\therefore$   Equation  becomes

 $\frac{dv}{dy}-1= \cos v$

$\Rightarrow$  $\frac{dv}{dy}=1+\cos v$

$\Rightarrow$   $\frac{dv}{1+\cos v}+dy$

$\Rightarrow$   $\frac{dv}{2\cos ^{2}\frac{v}{2}}=dy$

 On integrating , we get

$\frac{1}{2}\int \sec^{2}\frac{v}{2}dv=\int dy$

  $\Rightarrow$     $\tan\frac{v}{2}=y+C$

 $\Rightarrow$    $\tan\left(\frac{x+y}{2}\right)=y+C$     [$\therefore$   v=x+y]

If  $X\sim B(n,p)$  with n=10 , p=0.4  , then E(X)²=

Answer:

Explanation:

Given , n=10  and p=0.4

 $\therefore$ q=1-p=1-0.4=0.6

 E(x) = np=10(0.4)=4

$\sigma^{2}$ = npq= 10(0.4)(0.6)=2.4

We know that,

$\sigma^{2}= E(X^{2})-(E(X))^{2}$

$\therefore$     $ E(X^{2})=\sigma^{2}+(E(X))^{2}$

$ \Rightarrow$   $ E(X^{2})=24+(4)^{2}$

  =2.4+16=18.4

The minimum value of the function f(x)= x log x is  

Answer:

Explanation:

We have ,

 f(x)= x log x

f'(x)= 1+log x

For maxima or minima , put f '(x)= 0

 $\therefore$    1+ log x=0

 $\Rightarrow$  logx =-1

 $\Rightarrow$  x = e^-1

 Now, f "(x) = $\frac{1}{x}$

 $\Rightarrow$    $ f^{''}(e^{-1})=\frac{1}{e^{-1}}, e >0$

 $\therefore$ f(x) is decreasing

 Hence, minimum value of f(x) at x= e^-1 is

 $\Rightarrow $     $f^{}(e^{-1})=e^{-1}\log e^{-1}=\frac{1}{e}(-log e)=-\frac{1}{e}$

• Mathematics - General questions