JEE 2011 :: General Questions :: Mathematics

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Let M and N be two 3x3 non-singular skew-symmetric matrices such that MN = NM. If P^T denotes the transpose of P, then  $M^{2}N^{2}(M^{T}N)^{-1}(MN^{-1})^{T}$ is  equal to 

Answer:

Explanation:

Given    $M^{T}=-M,N^{T}=-N$

 and MN=NM

 $\therefore$    $M^{2}N^{2}(M^{T}N)^{-1}(MN^{-1})^{T}$

$= M^{2}N^{2}N^{-1}(M^{T})^{-1}(N{-1})^{T}.M^{T}$

=$M^{2}N(NN^{-1})(-M)^{-1}(N^{T})^{-1}(-M)$

=$M^{2}NI(-M^{-1})(-N)^{-1}(-M)$

=$-M.(MN)M^{-1}N^{-1}M$

=$-MN(NM^{-1})N^{-1}M$

 =$-M(NN^{-1})M=-M^{2}$

Note Here, non-singular word should not be used, since there is no non-singular 3 x 3 skew-symmetric matrix

Let $\alpha$ and $\beta$  be the roots of $x^{2}-6x-2=0$ with $\alpha > \beta$ , if $a_{n}=\alpha^{n}-\beta^{n}$ for $\geq$1, then the value of $ \frac{ a_{10}-2a_{8}}{2a_{9}}$ is 

Answer:

Explanation:

$\frac{a_{10}-2a_{8}}{2a_{9}}=\frac{(\alpha^{10}-\beta^{10})-2(\alpha^{8}-\beta^{8})}{2(\alpha^{9}-\beta^{9})}$

=$\frac{\alpha^{8}(\alpha^{2}-2)-\beta^{8}(\beta^{2}-2)}{2(\alpha^{9}-\beta^{9})}$

 [ $\therefore$  is root of  $x^{2}-6x-2=0$

                                            $\alpha^{2}-2=6\alpha]$

Also,$\beta$ is the root of $x^{2}-6x-2=0$

$\rightarrow$  $\beta^{2}-2=6 \beta$

$= \frac{\alpha^{8}(6\alpha)-\beta^{8}(6\beta)}{2(\alpha^{9}-\beta^{9})}=\frac{6(\alpha^{9}-\beta^{9})}{2(\alpha^{9}-\beta^{9})}=3$

Let  $P=( \theta: \sin \theta -\cos \theta= \sqrt{2} \cos \theta)$ and  $Q=({ \theta :\sin \theta+\cos \theta= \sqrt{2} \sin \theta })$ be two  sets,  Then, 

Answer:

Explanation:

$P=( \theta: \sin \theta-\cos \theta= \sqrt{2} \cos \theta)$

$\Rightarrow$  $\cos \theta ( \sqrt{2}+1)= \sin \theta$

$\Rightarrow$  $\tan \theta = \sqrt{2}+1$ ...(i)

$Q=( \theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta)$

$\Rightarrow$   $\sin \theta(\sqrt{2}-1)=\cos \theta$

$\Rightarrow$  $\tan \theta= \frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}$

   $= (\sqrt{2}+1)$....(ii)

$\therefore$ P=Q

 Let ($x_{0},y_{0})$ be the solution of the following equations  $(2x)^{\log 2}=(3y)^{\log 3}$  $3^{\log x}=2 ^{\log y}$, then $x_{0}$ is equal to 

Answer:

Explanation:

Taking log on both  sides

 $\log 2.\log(2x)= \log 3(\log 3y)$

$\Rightarrow$   $\log 2(\log 2+\log x)$

= $\log 3(\log 3+\log y)$....(i)

and $\log x.\log 3= \log y \log 2$

$\log y= \frac{ \log x. \log 3}{ \log 2}$...(ii)

 from Eq.(i) and (ii) we get

 $\log 2( \log 2+\log x)= $

                      $\log 3.\left\{ \log 3+\frac{\log x.\log 3}{\log 2}\right\}$

 $\Rightarrow$  $(\log 2)^{2}+\log 2.\log x=$

 $(\log 3)^{2}+\frac{ (\log 3)^{2}}{ (\log 2)}. \log x$

$\Rightarrow$  $ \log x.\left\{\frac{(\log 3)^{2}}{\log 2}-\log 2\right\}=  (\log 2)^{2}-(\log 3)^{2}$

$\Rightarrow$   $\log x.\left\{\frac{(\log 3)^{2}-(\log 2)^{2}}{\log 2}\right\}=  (\log 2)^{2}-(\log 3)^{2}$

$\Rightarrow$   $\log x=-\log 2= \log 2^{-1}$

$\therefore$  $x= \frac{1}{2}$

The value  of $\int_{\sqrt{\log 2}}^{\sqrt{\log 3}} \frac{x \sin x^{2}}{\sin x^{2}+ \sin (\log 6-x^{2})}dx$ is 

Answer:

Explanation:

  put $x^{2}=t \Rightarrow x dx= \frac{dt}{2}$

 $\therefore$ $I=\int_{\log 2}^{\log 3} \frac{ \sin t \frac{dt}{2}}{ \sin t+\sin ( log 6-t)}$ ...(i)

 Using $\int_{a}^{b} f(x) dx= \int_{a}^{b}  f(a+b-x)dx$

$I=\frac{1}{2}\int_{\log 3}^{\log 2} \frac{\sin(\log 2+\log 3-t)}{\sin(  \log 2+\log 3-t)+\sin (\log 6-(\log 2+\log 3-t)) }dt$

 $I=\frac{1}{2}\int_{\log 2}^{\log 3} \frac{\sin(\log 6-t)}{\sin(  \log 6-t)+\sin  (t)}dt$

$\therefore$   $I=\frac{1}{2}\int_{\log 2}^{\log 3} \frac{\sin(\log 6-t)}{\sin(  \log 6-t)+\sin  (t)}dt$......(i)

On adding Eqs.(i) and (ii)  we get

 $2l=I=\frac{1}{2}\int_{\log 2}^{\log 3} \frac{\sin t+ \sin(log 6-t)}{\sin(  \log 6-t)+\sin  (t)}dt$

 $\therefore$   $2I=\frac{1}{2}(t)_{\log2}^{\log 3}= \frac{1}{2} (\log 3-\log 2)$

$\Rightarrow$    $I= \frac{1}{4} \log (\frac{3}{2})$

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