JEE 2015 :: Advanced 2015 :: Mathematics 2015

• Advanced 2015 - Physics 2015
• Advanced 2015 - Chemistry 2015
• Advanced 2015 - Mathematics 2015

Let X  and Y be two arbitrary , 3x3  , non zero, skew-symmetric  matrices and Z be an arbitary , 3 x3 , non -zero , symmetric matrix. Then , which of the following matrices is/are skew-symmetric?

Answer:

Explanation:

Given,

      $X^{T}=-X,Y^{T}=-Y,Z^{T}=Z$

(a) Let  P= $Y^{3}Z^{4}-Z^{4}Y^{3}$

   Then,   $P^{T}= (Y^{3}Z^{4})^{T}-(Z^{4}Y^{3})^{T}$

                     $=(Z^{T})^{4}(Y^{T})^{3}-(Y^{T})^{3}(Z^{T})^{4}$

                          $=-Z^{4}Y^{3}+Y^{3}Z^{4}=P$

$\therefore$      P is symmetric matrix.

  (b)   Let       $P=X^{44}+Y^{44}$

         Then,    $P^{T}=(X^{T})^{44}+(Y^{T})^{44}$

                               =   $X^{44}+Y^{44}=P$

 $\therefore$       P is skew-symmetric matrix.

    (c)     Let   $P=X^{4}Z^{3}-Z^{3}X^{4}$

  Then,     $P^{T}=(X^{4}Z^{3})^{T}-(Z^{3}X^{4})^{T}$

    =     $(Z^{T})^{3}(X^{T})^{4}-(X^{T})^{4}(Z^{T})^{3}$

                        = $Z^{3}X^{4}-X^{4}Z^{3}=-P$

$\therefore$    P is skew- symmetric matrix

(d)        Let   $P=X^{23}+Y^{23}$

   Then,            $P^{T}=(X^{T})^{23}+(Y^{T})^{23}$

   =   -X²³  - Y ²³

           =-P

$\therefore$    P is skew-symmetric matrix.

Let $\triangle PQR$  be a triangle . Let  a=QR, b= RP and  c= PQ . If  |a|=12, |b|= $4\sqrt{3}$ and  b.c=24, then which of the following is/are true?

Answer:

Explanation:

Given,  |a|= 12,   |b|= 4  $\sqrt{3}$

         a+b+c= 0

 $\Rightarrow$               a=-(b+c)

$\Rightarrow$        $|a|^{2}=|b+c|^{2}$

$\Rightarrow$                      $|a|^{2}=|b|^{2}+|c|^{2}+2 b.c$

$\Rightarrow$               $144=48+|c|^{2}+48$

$\Rightarrow$                   $|c|^{2}=48\Rightarrow |c|=4\sqrt{3}$

 Also,  $|c|^{2}=|a|^{2}+|b|^{2}+2.a.b$

$\Rightarrow$       48=144+48+2.a.b

$\Rightarrow$      a.b=-72

    $\therefore$  Option (d) is correct.

  Also,    a x b=c x a

$\Rightarrow$      a x b +c x a =2a x b

$\Rightarrow$          |a x b +c x a| =2|a x b|

                                                   =   $2\sqrt{|a|^{2}|b|^{2}-(a.b)^{2}}$

                  =   $2\sqrt{(144)(48)-(-72)^{2}}$

                =   $2(12)\sqrt{48-36}= 48\sqrt{3}$

. $\therefore$    Option (c) correct.

 Also,    $\frac{|c|^{2}}{2}-|a|=24-12=12$

$\therefore$    Option (a ) is  correct.

and 

            $\frac{|c|^{2}}{2}+|a|=24+12=36$

 $\therefore$    Option (b ) is  not   correct.

Let   $ f(x)= \sin\left[ \frac{\pi}{6}\sin(\frac{\pi}{2}\sin x)\right]$  for all x ε  R and   $g(x)= (\pi/2)\sin x4$ for all x ε  R . Let (fog)(x)  denotes f(g(x)) and (gof)(x) denotes g{f(x)} . Then, which of the following is /are true?

Answer:

Explanation:

(a)       $f(x)=\sin \left[ \frac{\pi}{6}sin\frac{\pi}{2}(\sin x)\right],x\epsilon R$

                      =  $\sin (\frac{\pi}{6}\sin \theta),\theta\epsilon\left[-\frac{\pi}{2},\frac{\pi}{2}\right],$

   where               $\theta =\frac{\pi}{2}\sin x$

                                      = $\sin \alpha,\alpha \epsilon\left[-\frac{\pi}{6},\frac{\pi}{6}\right],$

where   $\alpha= \frac{\pi}{6}\sin\theta$

$\therefore$        $f(x)\epsilon  \left[-\frac{1}{2},\frac{1}{2}\right]$

   Hence, range of      $f(x)\epsilon  \left[-\frac{1}{2},\frac{1}{2}\right]$

  So,  option (a) is correct.

 (b)    $f\left\{g(x)\right\}=f(t),t\epsilon \left[- \frac{\pi}{2},\frac{\pi}{2}\right]$

   $\Rightarrow$    $f(t)\epsilon  \left[-\frac{1}{2},\frac{1}{2}\right]$

$\therefore$      Option (b) is correct.

   (c)   $\lim_{x \rightarrow 0}\frac{f(x)}{g(x)}$

                                        $=\lim_{x \rightarrow 0}\frac{\sin\left[\frac{\pi}{6}\sin(\frac{\pi}{2}\sin x)\right]}{\frac{\pi}{2}(\sin x)}$

                          $=\lim_{x \rightarrow 0}\frac{\sin\left[\frac{\pi}{6}\sin(\frac{\pi}{2}\sin x)\right]}{\frac{\pi}{6}\sin(\frac{\pi}{2}\sin x)}$

                                                      $\frac{\frac{\pi}{6}\sin(\frac{\pi}{2}\sin x)}{(\frac{\pi}{2}\sin x)}$

                              =  $1\times \frac{\pi}{6} \times 1=\frac{\pi}{6}$

  $\therefore$    Option (c) is correct.

  (d) g{ f(x)}=1

$\Rightarrow$                  $\frac{\pi}{2}\sin\left\{f(x)\right\}=1$

$\Rightarrow$               $\sin\left\{f(x)\right\}=\frac{2}{\pi}$                    .........(i)

 But              $ f(x)\epsilon \left[-\frac{1}{2},\frac{1}{2}\right]\subset\left[ -\frac{\pi}{6},\frac{\pi}{6}\right]$

$\therefore$   $\sin \left\{  f(x)\right\}\epsilon \left[-\frac{1}{2},\frac{1}{2}\right]$ ..............(ii)

$\Rightarrow$      $\sin \left\{ f(x)\right\}\neq\frac{2}{\pi}$

    [from Eqs(i) and  (ii) ]

 i.e, no solution

 $\therefore$    Option (d) is not correct. 

 Let  $g:R\rightarrow R$  be a differentiable  function with g(0)=0, g'(0)=0 and g'(1)≠ 0,$f(x)= \begin{cases}\frac{x}{|x|}g(x) & x \neq 0\\0 & x = 0\end{cases}$  and $h(x)=e^{|x|}$   For all x ε R. Let (foh)(x) denotes  f{h(x)}  and (hof)(x) denotes h{f(x)} . Then which of the following is/are true?

Answer:

Explanation:

 Here  ,   $f(x)=\begin{cases}g(x) ,& x > 0\\0, & x = 0\\-g(x),   & x<0\end{cases}$

              $f'(x)=\begin{cases}g'(x) ,& x \geq 0\\-g(x), & x < 0\end{cases}$

  $\therefore$     Option (a) is correct

 (b)           $h(x)=e^{|x|}=\begin{cases}e^{x} ,& x \geq 0\\e^{-x}, & x < 0\end{cases}$

   $\Rightarrow$      $h'(x)=\begin{cases}e^{x} ,& x \geq 0\\-e^{-x}, & x < 0\end{cases}$

 $\Rightarrow$     h' (0⁺ )=1

 and     h'(o^-)=-1

  So, h(x)  is not differentiable at x=0,

 $\therefore$   Option (b) is not correct.

  (c)    $(foh)(x)=f\left\{h(x)\right\},$  as     $h(x)>0$

                        $= \begin{cases}g(e^{x} ),& x \geq 0\\g(e^{-x}), & x < 0\end{cases}$

 $\Rightarrow$     $(foh)'(x)= \begin{cases}e^{x}g^{'}(e^{x}), & x \geq 0\\-e^{-x}g^{'}(e^{-x}) ,& x < 0\end{cases}$

$\Rightarrow$          $(foh)'(0^{+})=g'(1),(foh)'(0^{-})=-g'(1) $

             So, (foh)(x)  is not differentiable at x=0

   $\therefore$    Option (c) is not correct.

  (d)    $ (hof)(x)=e^{|f(x)| }=\begin{cases}e^{|g(x)|}, & x \neq 0\\e^{0}=1 & x = 0\end{cases}$

 Now,         $  (hof)'(0)=\lim_{h \rightarrow0}\frac{e^{|g(x)|}-1}{x}$

                        $=\lim_{h \rightarrow0}\frac{e^{|g(x)|}-1}{|g(x)|}.\frac{|g(x)|}{x}$

                $=\lim_{h \rightarrow0}\frac{e^{|g(x)|}-1}{|g(x)|}.\lim_{h \rightarrow 0}\frac{|g(x)-0|}{|x|}.\lim_{h \rightarrow 0}\frac{|x|}{x}$

                            = $1.g'(0).\lim_{h \rightarrow 0}\frac{|x|}{x}$

    = 0, as g'(0)=0

     $\therefore$      Option  (d) is correct.                

 Consider the family of all circles whose centres lie on the straight line y=x, If this family of circles is represented by the differential equation   $Py"+Qy'+1=0$ , where P,Q are the functions of x, y and y'   (here,    $y'=\frac{dy}{dx},y''=\frac{d^{2}y}{dx^{2}})$, then which of the following statement (s) is /are true?

Answer:

Explanation:

 Since the centre lies on y=x.

      $\therefore$  Equation of circle is

                    $x^{2}+y^{2}-2ax-2ay+c=0$

  On differentiating , we get

          $2x+2yy'-2a-2ay'=0$

   $\Rightarrow$       $x+yy'-a-ay'=0$

  $\Rightarrow$    $a=\frac{x+yy'}{1+y'}$

 Again differentiating , we get

         0  $= \frac{(1+y')[1+yy'+(y')^{2}]-(x+yy').(y")}{(1+y')^{2}}$

$\Rightarrow$    $ (1+y')[1+(y')^{2}+yy"]- (x+yy')(y")=0$

$\Rightarrow$    $1+y'[(y')^{2}+y'+1]+y"(y-x)=0$

 On comparing  with   $Py"+Qy'+1=0,$ we get 

 P=y-x  and $Q= (y')^{2}+y'+1$

• Advanced 2015 - Physics 2015
• Advanced 2015 - Chemistry 2015
• Advanced 2015 - Mathematics 2015