MHT CET 2020 :: Mathematics :: General questions

• Mathematics - General questions

$\int\frac{dx}{\cos 2x-\cos^{2}x}=$

Answer:

Explanation:

 We have,

   l= $\int\frac{dx}{\cos 2x-\cos^{2}x}$

 $l=\int\frac{dx}{2\cos ^{2}x-1-\cos^{2}x}$

 $l=\int\frac{dx}{-\sin^{2}x}$

 $l=-\int cosec^{2}x dx$

l= cotx +c

 The negation  of the statement ' he is poor but happy ' is

Answer:

Explanation:

 The negation of the statement

 " He is poor but happy "

is " He is not poor or not happy"

 The value of 

$\tan^{-1}\left(\frac{1}{3}\right)+\tan^{-1}\left(\frac{1}{5}\right)+\tan^{-1}\left(\frac{1}{7}\right)+\tan^{-1}\left(\frac{1}{8}\right)$  is 

Answer:

Explanation:

 We have,

$\tan^{-1}\left(\frac{1}{3}\right)+\tan^{-1}\left(\frac{1}{5}\right)+\tan^{-1}\left(\frac{1}{7}\right)+\tan^{-1}\left(\frac{1}{8}\right)$ 

   = $\tan^{-1}\left(\frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{3}\times\frac{1}{5}}\right)+\tan^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7}\times\frac{1}{8}}\right)$

 =$\tan^{-1}\left(\frac{8}{14}\right)+\tan^{-1}\left(\frac{15}{55}\right)$

  = $\tan^{-1}\left(\frac{4}{7}\right)+\tan^{-1}\left(\frac{3}{11}\right)$

   = $\tan^{-1}\left(\frac{\frac{4}{7}+\frac{3}{11}}{1-\frac{4}{7}\times\frac{3}{11}}\right)$

 = $\tan^{-1}\left(\frac{65}{65}\right)=\tan^{-1} 1=\frac{\pi}{4}$

 The eccentricity of the ellipse y²+4x²-12x+6y+14=0 is

Answer:

Explanation:

 y²+4x²-12x+6y+14=0

 4x²-12x+y²+6y+14=0

 $4(x^{2}-3x+\frac{9}{4})+y^{2}+6y+9=-14+9+9$

  $4(x-\frac{3}{2})^{2}+(y+3)^{2}=4$

  $\frac{(x-3/2)^{2}}{1}+\frac{(y+3)^{2}}{4}=1$

Here, a²=1, b²=4

$\therefore$    $e=\sqrt{1-\frac{a^{2}}{b^{2}}}$

  $e=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$

 If the equation ax²+2hxy+by²+2gx+2fy=0 has one line as the bisector of the angle  between co-ordinate axes, then 

Answer:

Explanation:

Given , equation of line

ax²+2hxy+by²+2gx+2fy=0

 Let the slope of the line m₁  and m₂

 $\therefore$    $m_{1}+m_{2}=-\frac{2h}{b}$

$m_{1}m_{2}=\frac{a}{b}$

Given,  one line is bisector of angle between  coordinate axes.

  $\therefore$     $m_{1}=\tan\frac{\pi}{4}=1$

 $\Rightarrow$    1+m₂=-2h/b   .........(i)

 $m_{2}=\frac{a}{b}$  ......(ii)

 From eqs. (i) and (ii) , we get

 $1+\frac{a}{b}=\frac{-2h}{b}$

 $\Rightarrow$   a+b=-2h

 (a+b)²  =4 h²

• Mathematics - General questions