JEE 2014 :: Main 2014 :: Mathematics 2014

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The statement    $\sim (p\leftrightarrow \sim q)$    is 

Answer:

Explanation:

 $\sim (p\leftrightarrow \sim q)$     is equivalent to  $ p\leftrightarrow  q$

 A bird is sitting on the top of vertical pole 20m high and its elevation from a point O on the ground 45° . It flies off horizontally straight away from the point O. After 1s,  the elevatiion  of the bird from O is reduced to 30° . Then the speed (in m/s)  of the bird is 

Answer:

Explanation:

In  $\triangle OA_{1}B_{1},$

    $\tan 45^{0}=\frac{A_{1}B_{1}}{OB_{1}}\Rightarrow \frac{20}{OB_{1}}=1$

  $\Rightarrow$      $ OB_{1}=20$

$\triangle OA_{2}B_{2},$

   $\tan 30^{0}=\frac{20}{OB_{2}}$

 $\Rightarrow$           $OB_{2}=20\sqrt{3}$

$\Rightarrow$                     $B_{1}B_{2}+OB_{1}=20\sqrt{3}$

 $\Rightarrow$                    $B_{1}B_{2}=20\sqrt{3}-20$            

 $\Rightarrow$               $B_{1}B_{2}=20(\sqrt{3}-1)m$

 Now, speed = $\frac{Distance}{Time}=\frac{20(\sqrt{3}-1)}{1}$

                              =   $20(\sqrt{3}-1)$  m/s

 If   $f_{k}(x)=1/k(\sin^{k}x+\cos^{k}x)$ , where x ε  R and  $k\geq 1$   , then  $f_{4}(x)-f_{6}(x)$   is equal to 

Answer:

Explanation:

Given     $f_{k}(x)=1/k(\sin^{k}x+\cos^{k}x)$ , where x ε  R and k>1

         $f_{4}(x)-f_{6}(x)$    =   $\frac{1}{4}(\sin^{4}x+\cos^{4}x)-\frac{1}{6}(\sin^{6}x+\cos^{6}x)$

               =   $\frac{1}{4}(1-2\sin^{2}x.\cos^{2}x-\frac{1}{6}(1-3\sin^{2}x.\cos^{2}x)$

   =  $\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$

 The variance of first 50 even natural numbers is 

Answer:

Explanation:

Here,   $\overline{X}=\frac{\sum    X_{i}}{n}$

                       =$\frac{2+4+6+8+...+100}{50}=\frac{50 \times 51}{50}=51$

                                                      [ $\because \sum{}2n=n(n+1), $  , here n=50]

 Variance  ,    $\sigma^{2} =\frac{1}{n}\sum{}Xi^{2}-(\overline{x})^{-2}$

   $\sigma^{2} =\frac{1}{50}(2^{2}+4^{2}+....+100^{2})-(51)^{2}=833$

 Let A and B be two events such that    $P(\overline{A\cup B})=\frac{1}{6}, P(A\cap B)=\frac{1}{4}$   and 

$P(\overline{A})=\frac{1}{4}$ , where  $\overline{A}$  stands  for the complement  of the event A. Then , the events A and B are

Answer:

Explanation:

 Given ,   $P(\overline{A\cup B})=\frac{1}{6}, P(A\cap B)=\frac{1}{4}$

  $P(\overline{A})=\frac{1}{4}$

 $\therefore$           $P(A\cup B)=1-P(\overline{A\cup B)}$

                       =$1-\frac{1}{6}=\frac{5}{6}$

 and                   $P(A)=1-P(\overline{A})$

                    =  $1-\frac{1}{4}=\frac{3}{4}$

   $P(A \cup B)=P(A)+P(B)-P(A\cap B)$

 $\Rightarrow$   $\frac{5}{6}=\frac{3}{4}+P(B)-\frac{1}{4}$

$P(B)=\frac{1}{3}\Rightarrow$  A and B are not equally  likely

   $P(A\cap B)= P(A).P(B)=\frac{1}{4}$

 So, events are independent.

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