JEE 2014 :: Main 2014 :: Mathematics 2014

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• Main 2014 - Mathematics 2014

 The integral  $\int_{}^{} (1+x-\frac{1}{x})e^{x+\frac{1}{x}}dx$  id equal to 

Answer:

Explanation:

$\int_{}^{} (1+x-\frac{1}{x})e^{x+\frac{1}{x}}dx$

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

 =  $\int_{}^{} e^{x+1/x}dx+xe^{x+1/x}-\int_{}^{} \frac{d}{dx}(x)e^{x+1/x} dx$

    =  $\int_{}^{} e^{x+1/x}dx+xe^{x+1/x}-\int_{}^{} e^{x+\frac{1}{x}}dx$

   $\left[\because \int_{}^{} \left(1-\frac{1}{x^{2}}\right)e^{x+1/x}dx=e^{x+1/x}\right]$

    $=\int_{}^{} e^{x+1/x}dx+xe^{x+1/x}-\int_{}^{} ex^{+1/x}dx$

  =  $xe^{x+1/x}+C$

If x=-1 and x=2  are extreme points of   $f(x)=\alpha\log|x|+\beta x^{2}+x$ , then

Answer:

Explanation:

Here , x=-1 and x=2 are extreme points of $f(x)=\alpha\log|x|+\beta x^{2}+x$, then

$f '(x)=\frac{\alpha}{x}+2\beta x+1$

$f '(-1)=-\alpha-2\beta+1=0$            ........(i)

                     [ At extreme point, f '(x0=0]

$f'(2)=\frac{\alpha}{2}+4\beta+1=0$                  .......(ii)

 On solving  Eqs. (i) and (ii) , we get

     $\alpha=2,\beta=-1/2$

 If f and g are differentiable functions in (0,1)  satisfying f(0)=2=g(1), g(0)=0  and f(1)=6 , then for some c ε ]0,1[

Answer:

Explanation:

Given  , f(0)= 2=g(1),g(0)=0  and f(1)=6

  f and g are differentiable  in (0,1)

Let h(x)= f(x)-2g(x)  .......(i)

   h(0)=f(0)-2g(0)

 h(0)=2-0

    h(0)=2

 and h(1)=f(1)-2g(1)=6-2(2)

  h(1)=2, h(0)=h(1)=2

Hence, using Rolle's theorem.

    h'(c)=0,  such that c ε  (0,1)

 Differentiating Eq.(i) at, we get

 $\Rightarrow$    f'(c)-2 g'(c)=0

 $\Rightarrow$  f '(c) =2 g '(c)

 If g is the inverse of  a function f and   $f'(x)=\frac{1}{1+x^{5}}$  , then g'(x) is equal to

Answer:

Explanation:

Here 'g' is the inverse of f(x)

    $\Rightarrow$    fog(x)= x

On differentiating w.r.t x , we get

 $f'\left\{g(x)\right\}\times g'(x)=1$

$g'(x)=\frac{1}{f'(g(x))}=\frac{1}{\frac{1}{1+\left\{g(x)\right\}^{5}}}$

                                       $ [ \because f'(x)=\frac{1}{1+x^{5}}]$

                   $g'(x)=1+\left\{g(x)\right\}^{5}$

$\lim_{x \rightarrow 0}\frac{\sin(\pi\cos^{2}x)}{x^{2}}$   is equal to 

Answer:

Explanation:

  $\lim_{x \rightarrow 0}\frac{\sin(\pi\cos^{2}x)}{x^{2}}$

   = $\lim_{x \rightarrow 0}\frac{\sin\pi(1-\sin^{2}x)}{x^{2}}$

     = $\lim_{x \rightarrow 0}\frac{\sin(\pi-\pi\sin^{2}x)}{x^{2}}$

  =$\lim_{x \rightarrow 0}\frac{\sin(\pi\sin^{2}x)}{x^{2}}$

                   [ $\because$   $\sin(\pi-\theta)=\sin\theta$ ]

               $=\lim_{x \rightarrow0}\frac{\sin\pi\sin ^{2}x}{\pi\sin^{2}x}\times (\pi)\left[\frac{\sin^{2}x}{x^{2}}\right]$

  =      $\pi$                     [$\lim_{x \rightarrow0}\frac{\sin\theta}{\theta}=1$]

• Main 2014 - Physics 2014
• Main 2014 - Chemistry 2014
• Main 2014 - Mathematics 2014