JEE 2015 :: Advanced 2015 :: Mathematics 2015

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Let   $f(x)= 7\tan^{8}x+7\tan^{6}x-3\tan^{4}x$ $x-3\tan^{2}x$   for all   $x\epsilon (-\frac{\pi}{2},\frac{\pi}{2})$  , Then the correct expression (s) is /are

Answer:

Explanation:

Here,      $f(x)= 7\tan^{8}x+7\tan^{6}x-3\tan^{4}x-3\tan^{2}x$

   For all x   $(\frac{-\pi}{2},\frac{\pi}{2})$

 $\therefore$          $f(x)= 7\tan^{6}x \sec^{2}x-3\tan^{2}x \sec^{2}x$

= $(7\tan^{6}x-3\tan^{2}x)\sec^{2}x$

 Now,     $\int_{0}^{\frac{\pi}{4}} x f(x) dx=\int_{0}^{\frac{\pi}{4}} x(7\tan^{6}x-3\tan^{2}x)\sec^{2}x dx$

                      = $\left[ x(\tan^{7}x-\tan^{3}x)\right]_0^{\pi/4}$

                                                         - $\int_{0}^{\pi/4} 1(tan^{7}x-\tan^{3}x)dx$

  = 0-  $\int_{0}^{\pi/4} tan^{3}x(tan^{4}x-1)dx$

     =   $-\int_{0}^{\pi/4} tan^{3}x(tan^{2}x-1)\sec^{2}xdx$

Put tan x=t   $\Rightarrow \sec^{2}x dx=dt$

    $\therefore$           $\int_{0}^{\pi/4}x f(x)=-\int_{0}^{1} t^{3}(t^{2}-1)dt$

   $=\int_{0}^{1} (t^{b}-t^{5})dt=\left[\frac{t^{4}}{4}-\frac{t^{5}}{5}\right]_0^1=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$

 Also, 

         $\int_{0}^{\pi/4} f(x) dx= \int_{0}^{\pi/4} (7\tan^{6}x-3\tan^{2}x)\sec^{2}x dx$

            =    $\int_{0}^{1}   (7t^{6}-3t^{2}) dt=\left[t^{7}-t^{3}\right]_0^1=0$

                                      -

 Let f.g:[-1,2]→  R,  be continuous  functions which are twice differentiable  on the interval (-1,2) . Let the values  of f and g at the points -1,0 and 2 be as given in the following table .

 In each of the intervals (-1,0) and (0,2), the function (f-3g)" never  vanishes. Then the correct statement(s) is/are

Answer:

Explanation:

Let F(x)=f(x)-3g(x)

   $\therefore$    F(-1)=3, F(0)=3 and F(2)=3

 So, F'(x) will vanish atleast twice in   $(-1,0)\cup (0,2)$

$\therefore$     $F"(x)>0$  or $<0,\forall x \epsilon(-1,0)\cup(0,2)$

 Hence, f'(x) -3g'(x)=0 has exactly  one solution in (-1,0) and one solution in (0,2)

 For any integer k, let $\alpha_{k}=\cos \left(\frac{k\pi}{7}\right)+i\sin \left(\frac{k\pi}{7}\right)$ , where  $i=\sqrt{-1}$ . The value of the expression  $\frac{\sum_{k=1}^{12}|\alpha_{k+1}-\alpha_{k}|}{\sum_{k=1}^{3}|\alpha_{4k-1}-\alpha_{4k-2}|}=\frac{12(a)}{3(a)}$  is 

Answer:

Explanation:

Given       $\alpha_{k}=\cos \left(\frac{k\pi}{7}\right)+i\sin \left(\frac{k\pi}{7}\right)$

                              $=\cos \left(\frac{2k\pi}{14}\right)+i\sin \left(\frac{2k\pi}{14}\right)$

 $\therefore$    $\alpha_{k}$   are  vertices of regular  polygon having 14 sides.

      Let the side length of regular  polygon be a.

  $\therefore$    $|\alpha_{k+1}-\alpha_{k}|$    = length  ofa side of the regular polygon =a ........(i)

 and  $|\alpha_{4k-1}-\alpha_{4k-2}|$  = length of a side of the regular  polygon=a

$\therefore$          $\frac{\sum_{k=1}^{12}|\alpha_{k+1}-\alpha_{k}|}{\sum_{k=1}^{3}|\alpha_{4k-1}-\alpha_{4k-2}|}=\frac{12(a)}{3(a)}=4$

Suppose  that p,q, and r  three non-coplanar vectors  in R³. Let  the components of a vector  s along p,q and r be 4,3 and 5  , respectively. If the components  of this vector  s along    (-p+q+r)  (p-q+r)  and (-p-q+r)  are x,y and z respectively, then the value of 2x+y+z is 

Answer:

Explanation:

 Here, s = 4p+3q+5r              ........(i)

  and s=   (-p+q+r)x+(p-q+r)y

                                  +(-p-q+r)z        .........(ii)

 $\therefore$    4p+3q+5r = p(-x+y-z)+q(x-y-z)+r(x+y+z)

 On comparing both sides, we get

             -x+y-z=4, x-y-z=3

         and x+y+z=5

  On solving above equations, we get

  $x=4,y=\frac{9}{2},z=\frac{-7}{2}$

 $\therefore$     $2x+y+z=8+\frac{9}{2}-\frac{7}{2}=9$

 Let f:R→ R be a continuous  odd function, which vanishes exactly  at one point   $f(1)=\frac{1}{2}$ . Suppose that $F(x)=\int_{-1}^{x} f(t) dt$    for all x ε [-1,2] and    $G(x)=\int_{-1}^{x}t|f\left\{ f(t)\right\}|dt$ for all  x ε [-1,2] If   $\lim_{x \rightarrow 1}\frac{F(x)}{G(x)}=\frac{1}{14}$ , then the value of    $f(\frac{1}{2})$  is

Answer:

Explanation:

Here,   $\lim_{x \rightarrow1}\frac{F(x)}{G(x)}=\frac{1}{14}$

     $\Rightarrow$      $\lim_{x \rightarrow1}\frac{F^{'}(x)}{G^{'}(x)}=\frac{1}{14}$

                                                   [ Using  L' Hospital's rule ]...........(i)

As                  $F(x)=\int_{-1}^{x} f(t)dt$

 $\Rightarrow$   F'(x)= f(x)     .........(ii)

and      $G(x)=\int_{-1}^{x}t|f\left\{ f(t)\right\}|dt$

$\Rightarrow$            $G'(x)=x|f\left\{f(x)\right\}|$            ...........(iii)

    $\therefore$      $\lim_{x \rightarrow 1}\frac{F(x)}{G(x)}=\lim_{x \rightarrow 1}\frac{F'(x)}{G'(x)}=\lim_{x \rightarrow 1}\frac{f(x)}{x|f\left\{f(x)\right\}|}$

                              =   $\frac{f(1)}{1|f\left\{f(1)\right\}|}=\frac{1/2}{|f(1/2)|}$  ........(iv)

 Given,     $\lim_{x \rightarrow 1}\frac{F(x)}{G(x)}=\frac{1}{14}$

  $\therefore$    $\frac{\frac{1}{2}}{|f\left(\frac{1}{2}\right)|}=\frac{1}{14}\Rightarrow |f\left(\frac{1}{2}\right)|=7$

• Advanced 2015 - Physics 2015
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