MHT CET 2016 :: Mathematics :: General questions

• Mathematics - General questions

If $\int\frac{f(x)}{\log(\sin x)}dx=\log[\log \sin x]+c$ , then f(x)  is equal to

Answer:

Explanation:

Given,   $\int\frac{f(x)}{\log(\sin x)}dx=\log[\log \sin x]+c$

On differentiating both sides , we get

$\frac{f(x)}{\log (\sin x)}=\frac{1}{\log \sin x}\frac{d}{dx}(\log \sin x)+0$

 $\Rightarrow$    $\frac{f(x)}{\log (\sin x)}=\frac{1}{\log \sin x}\times\frac{1}{\sin x}\times \cos x$

 $\Rightarrow$    f(x)  = cot x

The area of the region bounded by the curve y=2x-x² and X- axis is

Answer:

Explanation:

Given equation of curve is y=2x-x²

 $\Rightarrow$       $x^{2}-2x=-y$

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

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

This is the equation of parabola having vertex (1,1) and open downward.

 The parabola intersect  the X-axis , put y=0 , we get

    $ 0=2x-x^{2}$

$\Rightarrow$    x(2-x)=0

$\Rightarrow$    x=0,2

 $\therefore$    Area of bounded region between the curve and X- axis

 $=\int_{0}^{2} y dx$

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

 $=\left[4-\frac{8}{3}-0-0\right]=\frac{4}{3}$  sq. units

The degree and order of the differential equation 

$\left[1+\left(\frac{dy}{dx}\right)^{3}\right]^{7/3}=7\left(\frac{d^{2}y}{dx^{2}}\right)$ respectively are

Answer:

Explanation:

Given , differential equation  is 

$\left[1+\left(\frac{dy}{dx}\right)\right]^{7/3}=7\left(\frac{d^{2}y}{dx^{2}}\right)$

 On cubing both sides, we get

$\left[1+\left(\frac{dy}{dx}\right)^{3}\right]^{7}=7^{3}\left(\frac{d^{2}y}{dx^{2}}\right)^{3}$

Here, we see that highest  order derivative is 2, whose degree is 3.

 Hence , degree =3 and order=2

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}  \log\left(\frac{2-\sin x}{2+\sin x}\right)dx$  is equal to 

Answer:

Explanation:

We have ,l=$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}  \log\left(\frac{2-\sin x}{2+\sin x}\right)dx$ 

Let   $f(x)=\log\left(\frac{2-\sin x}{2+\sin x}\right)$

Then,   $f(-x)=\log\left(\frac{2-\sin (-x)}{2+\sin (-x)}\right)$

$=\log\left(\frac{2+\sin x}{2-\sin x}\right)=\log\left(\frac{2-\sin x}{2+\sin x}\right)^{-1}$

 $=-\log\left(\frac{2-\sin x}{2+\sin x}\right)=-f(x)$

 then, f(x) is odd function.

$\therefore$    $\int_{-\pi/2}^{\pi/2}  f(x) dx=0$

   [ $\because $  f(x) is an odd function, then $\int_{-a}^{a}  f(x) dx=0$]

For what value of k, the function defined by 

$f(x)=\begin{cases}\frac{log(1+2x)\sin x^{0}}{x^{2}} &for x\neq0\\k & for x=0\end{cases}$

 is continuous at x=0?

Answer:

Explanation:

Given , p(x)$=\begin{cases}\frac{log(1+2x)\sin x^{0}}{x^{2}} &for x\neq0\\k & for x=0\end{cases}$

 =  $f(x)=\begin{cases}\frac{log(1+2x)\sin \frac{\pi x}{180}}{x^{2}} &for x\neq0\\k & for x=0\end{cases}$

 Since f(x) is continuous at x=0

 $\therefore$   LHL=f(0)

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

  =$\lim_{h \rightarrow 0}\frac{\log(1+2(0-h)\sin\frac{\pi}{180^{0}}(0-h))}{(0-h)^{2}}$

 =$\lim_{h \rightarrow 0}\frac{\log(1-2h)\left\{-\sin\frac{\pi h}{180}\right\}}{h^{2}}$

 =  $\lim_{h \rightarrow 0}(-2)\frac{\log(1-2h)}{-2h}\times(-)\lim_{h \rightarrow 0}\frac{\sin \frac{\pi h}{180}}{\frac{\pi h}{180}}\times\frac{\pi}{180}$

= $(-2)\times(-1)\times 1\times \frac{\pi}{180}$

$\left[\because \lim_{x \rightarrow 0}\log\frac{1+x}{x}=1 and    \lim_{x \rightarrow 0}\frac{\sin x}{x}=1\right]$

=$\frac{\pi}{90}$

• Mathematics - General questions