JEE 2013 :: Main 2013 :: Mathematics 2013

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If P=  $\begin{bmatrix}1 & \alpha &3 \\1 & 3&3 \\2&4& 4 \end{bmatrix}$   is the adjoint of a 3x 3  matrix A and |A| =4,  then $\alpha$ is equal  to

Answer:

Explanation:

Given , P=  $\begin{bmatrix}1 & \alpha &3 \\1 & 3&3 \\2&4& 4 \end{bmatrix}$ 

  $\therefore$     |P| = 1(12-12)- $\alpha$ (4-6) +3(4-6)

  = 2$\alpha$-6

$\because$  P=adj(A)           (given)

$\therefore$            |P|=|adj A|=|A|², =16

                                   ($\because$   |adj  A|=|A|^n-1)

 $\therefore$       2$\alpha$-6=16

   $\Rightarrow$      $2\alpha=22\Rightarrow\alpha=11$

ABCD is a trapezium such that AB and CD are parallel and   $BC\perp CD$ . If $\angle ADB=\theta$ , BC=p and CD= q , then AB is equal to   

Answer:

Explanation:

Let AB =x

 In $\triangle$ DAM,     $\tan  (\pi-\theta-\alpha)=\frac{p}{x-q}$

  $\Rightarrow$   $\tan  (\theta+\alpha)=\frac{p}{q-x}$

$\Rightarrow$    $q-x=p\cot(\theta+\alpha)$

$\Rightarrow$   $x=q-p \cot(\theta+\alpha)$

    $=q-p\left(\frac{\cot\theta\cot\alpha-1}{\cot\alpha+\cot\theta}\right)$

                                                    $\left(\because \cot\alpha=\frac{q}{p}\right)$

   =$q-p\left(\frac{\frac{q}{p}\cot\theta-1}{\frac{q}{p}+\cot\theta}\right)$

$=q-p\left(\frac{q\cot\theta-p}{q+p\cot\theta}\right)$

$=q-p\left(\frac{q\cos\theta-p\sin\theta}{q\sin\theta+p\cos\theta}\right)$

  $\Rightarrow \frac{q^{2}\sin\theta+pq\cos\theta-pq\cos\theta+p^{2}\sin\theta}{p\cos\theta+q\sin\theta}$

    $\Rightarrow AB=\frac{(p^{2}+q^{2})\sin\theta}{p\cos\theta+q\sin\theta}$

If z is a  complex number of unit modulus and argument $\theta$ , the  $arg\left[\frac{1+z}{1+\overline{z}}\right]$  is equal to

Answer:

Explanation:

Given  , |z|=1, arg z= $\theta$      $\therefore$  z = $e^{i\theta}$

  But   $\overline{z}=\frac{1}{z}$

   $\therefore$     $arg\left[\frac{1+z}{1+\frac{1}{z}}\right]=arg(z)=\theta$

Let T_n be the number of all possible triangles formed by joining vertices of an n -sided regular polygon. If T_n+1-T_n=10, then the value of n is

Answer:

Explanation:

Given ,   $T_{n}=^{n}C_{3}$

   $T_{n+1}=^{n+1}C_{3}$

 $\therefore$     $T_{n+1}-T_{n}=^{n+1}C_{3}-^{n}C_{3}=10(given)$

  $\Rightarrow$    $^{n}C_{2}+^{n}C_{3}-^{n}C_{3}=10$

                                        $(\because ^{n}C_{r}+^{n}C_{r+1}=^{n+1}C_{r+1})$

  $\Rightarrow$    $^{n}C_{2}=10\Rightarrow n=5$

The area (in square units) bounded by the curves     y= $\sqrt{x}$, 2y-x+3=0, x-axis and lying in the first quadrant is 

Answer:

Explanation:

Given curves are   $y=\sqrt{x}$   ........(i)

   and 2y-x+3=0      .......(ii)

 On solving Eqs(i) and (ii) , we get

   $2\sqrt{x}-(\sqrt{x})^{2}+3=0$

  $\Rightarrow$    $(\sqrt{x})^{2}-2\sqrt{x}-3=0$

$\Rightarrow$   $(\sqrt{x}-3)(\sqrt{x}+1)=0$

  $\Rightarrow$    $\sqrt{x}=3$

  ( $\therefore$   $\sqrt{x}=-1$   is not possible)

 $\therefore$  y=3

   $\therefore$  Required area =   $\int_{0}^{3} (x_{2}-x_{1})dy$

  $\int_{0}^{3} ((2y+3)-y^{2})dy$

      = $\left[y^{2}+3y-\frac{y^{3}}{3}\right]_{0}^{3}$

   =9+9-9=9

• Main 2013 - Physics 2013
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