MHT CET 2016 :: Mathematics :: General questions

• Mathematics - General questions

$\frac{tan^{-1})(\sqrt{3})-\sec^{-1}(-2)}{cosec^{-1}(-\sqrt{2})+\cos^{-1}(-\frac{1}{2})}$ is equal to

Answer:

Explanation:

$\frac{tan^{-1})(\sqrt{3})-\sec^{-1}(-2)}{cosec^{-1}(-\sqrt{2})+\cos^{-1}(-\frac{1}{2})}$

 =   $\frac{tan^{-1}(\sqrt{3})-(\pi-\sec^{-1}(-2))}{-cosec^{-1}(\sqrt{2})+\pi -\cos^{-1}(\frac{1}{2})}$

 =  $\frac{\frac{\pi}{3}-\left(\pi-\frac{\pi}{3}\right)}{-\frac{\pi}{4}+\pi-\frac{\pi}{3}}=\frac{\frac{2 \pi}{3}-\pi}{\pi-\frac{\pi}{4}-\frac{\pi}{3}}$

 =  $\frac{-\frac{\pi}{3}}{\frac{12\pi-3\pi-4\pi}{12}}=\frac{-\frac{ \pi}{3}}{\frac{5\pi}{12}}=-\frac{4}{5}$

$\int\left(\frac{4e^{x}-25}{2e^{x}-5}\right)dx= Ax+B\log(2e^{x}-5)+c$, then

Answer:

Explanation:

Let   l=$\int\left(\frac{4e^{x}-25}{2e^{x}-5}\right)dx$

 =  $\int\frac{4e^{x}}{2e^{x}-5}dx-\int\frac{25}{2e^{x}-5}dx$

 $=4\int\frac{e^{x}}{2e^{x}-5}dx-25\int\frac{e^{-x}}{2-5e^{-x}}dx$

 Put $2e^{x}-5=u$ and 2-5$e^{-x}$=v

$\Rightarrow$    $ 2e^{x} dx$=du    and  5$e^{-x}dx= du$

$\Rightarrow e^{x}dx=\frac{du}{2} $  and    $ e^{-x}dx=\frac{dv}{5}$

$\therefore$    $l=4\int_{}^{} \frac{du}{2u}-25\int\frac{du}{5v}$

  = 2 log u-5 log v+c

 $=2\log(2e^{x}-5)-5\log(2-5e^{x})+c$

$=2\log(2e^{x}-5)-5\log\left(\frac{2e^{x}-5}{e^{x}}\right)+c$

 $=2\log(2e^{x}-5)-5\log(2e^{x}-5)+5\log e^{x}+c$

$=-3\log(2e^{x}-5)+5x+c$

 $\Rightarrow l=5x-3\log(2e^{x}-5)+c$

 But it is given l=Ax+Blog (2e^x -5)+c

$\therefore$   A=5  and B=-3

If the function f(x) defined by

$f(x)=\begin{cases}x \sin\frac{1}{x} ,&for  x \neq 0\\k, & for x=0 \end{cases}$

 is continuous at x=0 , then k is equal to

Answer:

Explanation:

Given 

$f(x)=\begin{cases}x \sin\frac{1}{x} ,&for  x \neq 0\\k, & for x=0 \end{cases}$

 iSince , f(x) is continuous at x=0

, $\therefore$    LHL=RHL=f(0)

 .......(i)

 Now, LHL$=\lim_{x \rightarrow 0-}f(x)=\lim_{h \rightarrow 0}f(0-h)$

   $=\lim_{h \rightarrow 0}(0-h)\sin \frac{1}{(0-h)}=\lim_{h \rightarrow 0}-h\sin\left(-\frac{1}{h}\right)$

 $=\lim_{h \rightarrow 0}h \sin \frac{1}{h}=0\times$ finite value =0

                                         $\left[\because \lim_{h \rightarrow 0}\sin \frac{1}{h}=finite  value\right]$

 and f(0)=k

$\therefore$   From Eq.(i)   , LHL= RHL

$\Rightarrow$    0=k

$\Rightarrow$    k=0

The point of the curve  $6y= x^{3}+2$  at which  y-coordinate is changing 8 times as fact  as x-coordinate is 

Answer:

Explanation:

Given , $6y= x^{3}+2$...........(i)

 and $\triangle y=8 \triangle x$

 On  differentiating both sides of Eq.(i)  w.r.t  x. we get

 $\frac{6dy}{dx}=3x^{2}\Rightarrow \frac{dy}{dx}=\frac{1}{2} x^{2}$

$\because$  $\triangle y=\frac{dy}{dx}\triangle x\Rightarrow 8\triangle x=\frac{1}{2}x^{2} \triangle x$

 $\Rightarrow$    $   x^{2}=16 \Rightarrow x=\pm 4$

 When x=4 , 6y=$(4)^{3}$+2

 $\Rightarrow$     6y=66  $\Rightarrow$    y=11

 Hence , required point is (4,11)

 The joint equation of bisectors of angles between lines x=5 and y=3 is

Answer:

Explanation:

The equation of the  bisector of the angle between the lines

 (x-5) and (y-3) is

$\frac{(x-5)}{\sqrt{1^{2}}}=\pm\frac{y-3}{\sqrt{1^{2}}}\Rightarrow\frac{x-5}{1}=\pm\frac{y-3}{1}$

$\Rightarrow$     x-5=+(y-3) and x-5 =-(y-3)

$\Rightarrow$  (x-y-2)=0 and (x+y-8)=0

 $\therefore$  combined equation  of bisector of angle between the lines is

            (x-y-2)(x+y-8)=0

$\Rightarrow x^{2}+xy-8x-xy-y^{2}+8y-2x-2y-16=0$

$\Rightarrow x^{2}-y^{2}-10x+6y+16=0$

• Mathematics - General questions