JEE 2013 :: Main 2013 :: Mathematics 2013

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The term independent of x in expansion of 

$\left(\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{x-1}{x-x^{\frac{1}{2}}}\right)^{10} is$

Answer:

Explanation:

$\therefore$    $\left[\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{x-1}{x-x^{\frac{1}{2}}}\right]^{10} $

= $\left[\frac{(x^{\frac{1}{3}})^{3}+1^{3}}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{((\sqrt{x}))^{2}-1}{\sqrt{x}(\sqrt{x}-1)}\right]^{10} $

  = $\left[\frac{(x^{\frac{1}{3}}+1)(x^{2/3}+1-x^{1/3}}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{((\sqrt{x}))^{2}-1}{\sqrt{x}(\sqrt{x}-1)}\right]^{10} $

   =   $\left[(x^{1/3}+1)-\frac{(\sqrt{x}+1)}{\sqrt{x}}\right]^{10}$

       = $(x^{1/3}-x^{-1/2})^{10}$

  $ \therefore$ The general term is

        $T_{r+1}=^{10}C_{r}(x^{1/3})^{10-r}(-x^{-1/2})^{r}$

        =  $^{10}C_{r}(-1)^{r}x^{\frac{10-r}{3}-\frac{r}{2}}$

 For independent of x, put

$\frac{10-r}{3}-\frac{r}{2}=0\Rightarrow 20-2r-3r=0$

 $\Rightarrow$    $  20=5r\Rightarrow r=4$

  $\therefore$      $T_{5}=^{10}C_{4}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210$

A multiple-choice examination has 5 questions, Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is 

Answer:

Explanation:

The probability of guessing a correct answer, p=1/3 and probability of guessing

wrong answer ,q$=\frac{2}{3}$

    $\therefore$ The probability of guessing a 4 or more correct answer

    = $^{5}C_{4}\left(\frac{1}{3}\right)^{4}.\frac{2}{3}+^{5}C_{5}\left(\frac{1}{3}\right)^{5}$

$=5.\frac{2}{3^{5}}+\frac{1}{3^{5}}=\frac{11}{3^{5}}$

The equation  of the circle passing through the foci of the ellipse 

$\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$  and having centre at (0,3) is 

Answer:

Explanation:

Given  equation of ellipse is  $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$  

 Here, a=4,b=3,   $e=\sqrt{1-\frac{9}{16}}\Rightarrow\frac{\sqrt{7}}{4}$

 $\therefore$  Foci is (±,0)= $\left(\pm4\times \frac{\sqrt{7}}{4},0\right)$

$= \left(\pm \sqrt{7},0\right)$

$\therefore$   Radius of the circle,

           $r= \sqrt{(ae)^{2}+b^{2}}=\sqrt{7+9}=\sqrt{16}=4$

 Now, equation of circle is 

                      $(x-0)^{2}+(y-3)^{2}=16$

  $\therefore$   $x^{2}+y^{2}-6y-7=0$

 The value of the integral

Statement I

 $\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\tan x}}$     is equal to   $ \pi/6$ 

 Statement II

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

Answer:

Explanation:

Let I=   $\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{tan x}}$   .........(i)

 $\therefore$     I =$\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{tan (\frac{\pi}{2}-x)}}$

   = $\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{cot x}}$

   $\Rightarrow$    $ I= \int_{\pi/6}^{\pi/3} \frac{\sqrt{\tan x}dx}{1+\sqrt{\tan x}}$........(ii)

          On adding Eqs.(i) and (ii)  , we get

    $2 I= \int_{\pi/6}^{\pi/3} dx\Rightarrow2I=[x]^{\pi/3}_{\pi/6}$

 $\Rightarrow I=\frac{1}{2}\left[\frac{\pi}{3}-\frac{\pi}{6}\right]=\frac{\pi}{12}$

 Statement I  false

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

 true statement by property of definite integrate

  $\lim_{x \rightarrow 0}\frac{(1-\cos 2x)(3+\cos x)}{x\tan 4x}$   is equal to 

Answer:

Explanation:

Let I=   $\lim_{x \rightarrow 0}\frac{(1-\cos 2x)}{x^{2}}\frac{(3+\cos x)}{1}.\frac{x}{\tan 4x}$

  = $\lim_{x \rightarrow 0}\frac{2\sin^{2}x}{x^{2}}.\frac{3+\cos x}{1}.\frac{x}{\tan 4x}$

      =  $2\lim_{x \rightarrow 0}\left(\frac{\sin^{}x}{x^{}}\right)^{2}\lim_{x \rightarrow 0}.(3+\cos x).\lim_{x \rightarrow 0}\frac{4x}{4\tan 4x}$

   = $2.(1)^{2}.(3+\cos 0^{o}).\frac{1}{4}(1)$

                  =   $2.1.(3+1).\frac{1}{4}=2.4.\frac{1}{4}=2$

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