JEE 2013 :: Main 2013 :: Mathematics 2013

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All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given?

Answer:

Explanation:

If initially  all marks were x₁, then 

$\sigma^{2}_{1}=\frac{\sum(x_{i}-\overline{x})^{2}}{N}$

 Now, each is increased by 10

   $\therefore$      $\sigma^{2}_{1}=\frac{\sum(x_{i}+10)-(\overline{x}+10))^{2}}{N}=\sigma^{2}_{1}$

  So, variance will not change whereas mean, median and mode will increase by 10

The expression   $\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$   can be written as

Answer:

Explanation:

Given expression is 

$\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$

 = $\frac{\sin A}{\cos A}\times\frac{\sin A}{\sin A-\cos A}+\frac{\cos A}{\sin A}\times\frac{\cos A}{\cos A-\sin A}$

     = $\frac{1}{\sin A-\cos A}\left\{\frac{\sin^{3}A-\cos^{3} A}{\cos A\sin A}\right\}$

   = $\frac{\sin^{2}A+\sin A \cos A+\cos^{2} A}{\cos A\sin A}$

     =   $\frac{1-\sin A \cos A}{\cos A\sin A}$

       =1+ sec A cosec A

If $y=\sec(\tan^{-1}x)$   then $\frac{dy}{dx}$  at x=1 is equal to 

Answer:

Explanation:

Given , y= sec {tan^-1   x)

   Let   $\tan^{-1}x=\theta\Rightarrow x=\tan\theta$

   $\therefore$     $y=\sec\theta=\sqrt{1+x^{2}}$

  On differentiating w.r.t x, we get

        $\frac{dy}{dx}=\frac{1}{2\sqrt{1+x^{2}}}2x$

   At x=1,    $\frac{dy}{dx}=\frac{1}{\sqrt{2}}$

Given A circle , $2x^{2}+2y^{2}=5$  and a parabola , $y^{2}=4\sqrt{5}x$

  Statement I , An equation of common tangne to these curves is $y=x+\sqrt{5}$

 Statement II, If the line  $y=mx+\frac{\sqrt{5}}{m}(m\neq0)$   is the common tangent, then m satisfies   $m^{4}-3m^{2}+2=0$.

Answer:

Explanation:

Equation of circle can be rewritten as 

$x^{2}+y^{2}=\frac{5}{2}$,

    Centre → (0,) and radius   $\rightarrow\sqrt{\frac{5}{2}}$

 Let common tangent be

                        $y=mx+\frac{\sqrt{5}}{m}$

  $\Rightarrow m^{2}x-my+\sqrt{5}=0$

 The perependicular from centre to the tangent is equal to radius of the cirtcle

$\therefore$                         $\frac{\frac{\sqrt{5}}{m}}{\sqrt{1+m^{2}}}=\sqrt{\frac{5}{2}}$

             $\Rightarrow  $     $ m\sqrt{1+m^{2}}=\sqrt{2}$

  $\Rightarrow$     $   m^{2}(1+m^{2})=2$

  $\Rightarrow $    $  m^{4}+m^{2}-2=0$

   $\Rightarrow $    $  (m^{2}+2)(m^{2}-1)=0$

$\Rightarrow $    $  m\pm1$

                                      ($\because m^{2}+2\neq0,$  as   $m\in R)$

   $\therefore$      $y=\pm(x+\sqrt{5})$  both statements are correct as m=$\pm$ 1 satisfies the given equation of statement II.

The intercepts on x-axis made by tangents to the curve,   $y=\int_{0}^{x} |t| dt,x\in R$   , which are parallel to the line y=2x, are equal to

Answer:

Explanation:

Given,  $y=\int_{0}^{x} |t| dt$

  $\therefore$    $\frac{dy}{dx}=|x|.1-0=|x|$

 (by Leibnitz rule)

 $\because$ Tangent to the curve

   $y=\int_{0}^{x} |t| dt,x\in R$   are parallel to the line y=2x

 $\therefore$ slope of both are equal

 $\Rightarrow x=\pm2$

 $\therefore$ points =   $y=\int_{0}^{\pm2} |t|dt=\pm2$

 $\therefore$   Equation of tangent is

   y-2=2(x-2)

and y+2=2(x+2)

 For x- intercept put y=0 , we get

 0-2=2(x-2)

 and 0 +2=2(x+2)= x$\pm$1

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