JEE 2015 :: Advanced 2015 :: Mathematics 2015

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If   $\alpha=\int_{0}^{1} e^{(9x+3\tan^{-1}x)}\left(\frac{12+9x^{2}}{1+x^{2}}\right)dx$, where tan^-1 x takes only principal values, then the value of   $\log_{e}|1+\alpha|-\frac{3\pi}{4}$  is 

Answer:

Explanation:

Here , 

                    $\alpha=\int_{0}^{1} e^{(9x+3\tan^{-1}x)}\left(\frac{12+9x^{2}}{1+x^{2}}\right)dx$

 Put         $9x+3\tan^{-1}x=1$

    $\Rightarrow$                          $\left(9+\frac{3}{1+x^{2}}\right)dx=dt$

   $\therefore$                     $\alpha=\int_{0}^{9+3\pi/4} e^{t}dt= \left[ e^{t}\right]_0^{9+3\pi/4}$

                                                  = $e^{9+3\pi/4}-1$

 $\Rightarrow$               $\log_{e}|1+\alpha|=9+\frac{3\pi}{4}$

   $\Rightarrow$     $\log_{e}|\alpha+1|-\frac{3\pi}{4}=9$

Let m and n be two positive integers greater than 1. If $\lim_{\alpha \rightarrow0}\left(\frac{e^{\cos(\alpha^{n})}-e}{\alpha^{m}}\right)=-\left(\frac{e}{2}\right)$ ,then the value of $\frac{m}{n}$ is

Answer:

Explanation:

Given      $\lim_{\alpha \rightarrow0}\left(\frac{e^{\cos(\alpha^{n})}-e}{\alpha^{m}}\right)=-\left(\frac{e}{2}\right)$

$\Rightarrow$         $\lim_{\alpha \rightarrow0}\frac{e[e^{\cos(\alpha^{n})-1}-1].\cos (\alpha^{n})-1}{\cos(\alpha^{n})-1 \alpha^{m}}=\frac{-e}{2}$

$\Rightarrow$               $\lim_{\alpha \rightarrow 0}e\left\{\frac{e^{\cos(\alpha^{n})-1}-1}{\cos(\alpha^{n})-1}\right\}$  .

                                                         $\lim_{\alpha \rightarrow 0}\frac{-2\sin^{2}\frac{\alpha^{n}}{2}}{\alpha^{m}}$

                                                            =-e/2

$\Rightarrow$        $e\times1\times(-2)\lim_{\alpha \rightarrow 0}\frac{\sin^{2}\left(\frac{\alpha^{n}}{2}\right)}{\frac{\alpha^{2n}}{4}}.\frac{\alpha^{2n}}{4\alpha^{m}}=\frac{-e}{2}$

  $\Rightarrow$     $e\times1\times(-2)\times 1\lim_{\alpha \rightarrow 0}\frac{\alpha^{2n-m}}{4}=\frac{-e}{2}$

 For this to exist.   

 2n-m=0   $\Rightarrow$    $\frac{m}{n}=2$

 Suppose that the foci of the ellipse   $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$  are (f₁,0) and (f₂, o) where f₁>0 and f₂<0 . Let P₁   and P₂ be two parabolas with a common vertex at (0,0) with foci at (f₁,0) and (2f₂,0)  respectively  .Let T₁ be a tangent to P₁ which passes through (2 f₂,o) and T₂ be  a tangent to P₂. P2 which passes through (f₁,0) . If m₁ is the slop of T₁  and m₂ is the slope of T₂, then the value of  $\left( \frac{1}{m_1^2}+m_2^2\right)$  is

Answer:

Explanation:

Tangent to P₁ passes through

   (2f₂,0)i.e, (-4,0)

  $\therefore  T_{1}:y=m_{1}x+\frac{2}{m_{1}}$

 $\Rightarrow$              $0=-4m_{1}+\frac{2}{m_{1}}$

$\Rightarrow$        $m_{1}^{2}=1/2$     ..........(i)

Also, tangent to  P₂   passes through (f₁,0) i.e, (2,0)

     $\Rightarrow$                $ T_{2}:y=m_{2}x+\frac{(-4)}{m_{2}}$

$\Rightarrow$                     $0=2m_{2}-\frac{4}{m_{2}}$

$\Rightarrow$               $m_2^2=2$

  $\therefore$    $\frac{1}{m_1^2}+m_2^2=2+2=4$

 The coefficient of x⁹   in the expansion of  $(1+x)(1+x^{2})+(1+x^{3}).......(1+x^{100})$  is

Answer:

Explanation:

The coefficient of x⁹   in the expansion of 

      $(1+x)(1+x^{2})+(1+x^{3}).......(1+x^{100})$  = Terms having x⁹

 =  $[I^{99}.x^{9},I^{98}.x.x^{8},I^{98}.x^{2}.x^{7},I^{98}.x^{3}.x^{6}I^{98}.x^{4}.x^{5}$

          $I^{97}.x.x^{2}.x^{6}I^{97}.x.x^{3}.x^{5}I^{97}.x^{2}.x^{3}.x^{4}]$

  $\therefore$ coefficient of x⁹=8

Suppose that all the term of an arithmetic progression are natural numbers. If the ratio of the sum of first seven terms to the sum of the first eleven terms  is 6:11  and the seventh term lies in between 130 and 140, then the common difference of this AP is

Answer:

Explanation:

Given , $\frac{S_{7}}{S_{11}}=\frac{6}{11}$    and 130< i₇ <140

  $\Rightarrow \frac{\frac{7}{2}[2a+6d]}{\frac{11}{2}[2a+10d]}=\frac{6}{11}$

          $\Rightarrow \frac{{7}{}[2a+6d]}{[2a+10d]}={6}$

$\Rightarrow $  a=9d        ........(i)

                   Also, 130 <i₇ <140

$\Rightarrow $     130<a+6d<140

$\Rightarrow $     130<9d+6d<140               [from Eq.(i)]

$\Rightarrow $     130<15d<140

$\Rightarrow $           $\frac{26}{3} <d<\frac{28}{3}$

                           [since , d is a natural  number]

$\therefore$    d=9

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