JEE 2015 :: Advanced 2015 :: Mathematics 2015

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• Advanced 2015 - Mathematics 2015

Let n₁  and n₂  be the number of red and black balls. respectively in box I. Let  n₃ and n₄  be the number of red and black balls respectively in box II.

A ball is drawn at random from box  I and transferred to box II. If the probability  of drawing  a red ball from box I, after this transfer  is  $\frac{1}{3}$, then the correct option(s)  with the possible values of n₁ and n₂  is/are

Answer:

Explanation:

$\therefore$     P (drawing  red ball from B₁)=  $\frac{1}{3}$

  $\Rightarrow$                     $\left(\frac{n_{1}-1}{n_{1}+n_{2}-1}\right)\left(\frac{n_{1}}{n_{1}+n_{2}}\right)+\left(\frac{n_{2}}{n_{1}+n_{2}}\right)\left(\frac{n_{1}}{n_{1}+n_{2}-1}\right)=\frac{1}{3}$

$\Rightarrow$         $\frac{n_1^2+n_{1}.n_{2}-n_{1}}{(n_{1}+n_{2})(n_{1}+n_{2}-1)}=\frac{1}{3}$

   Clearly, options  (c) and (d)  satisfy

Let n₁  and n₂  be the number of red and black balls. respectively in box I. Let  n₃ and n₄  be the number of red and black balls respectively in box II.

One of the two boxes, box I and box II a was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability  that this red ball was drawn  from box II, is  $\frac{1}{3}$, then the correct option(s)  with the possible  values of n₁,n₂,n₃  and n₄ is/are

Answer:

Explanation:

Let A= Drawing red ball

  $\therefore$        $ P(A)=P(B_{1}).P(A/B_{1})+P(B_{2}).P(A/B_{2})$

$=\frac{1}{2}\left(\frac{n_{1}}{n_{1}+n_{2}}\right)+\frac{1}{2}\left(\frac{n_{3}}{n_{3}+n_{4}}\right)$

 Given      $ P(B_{2}/A)$=$\frac{1}{3}$

$\Rightarrow$             $\frac{P(B_{2}).P(B_{2}\cap A)}{P(A)}=\frac{1}{2}$

$\Rightarrow$         $ \frac{\frac{1}{2}\left(\frac{n_{3}}{n_{3}+n_{4}}\right)}{\frac{1}{2}\left(\frac{n_{1}}{n_{1}+n_{2}}\right)+\frac{1}{2}\left(\frac{n_{3}}{n_{3}+n_{4}}\right)}=\frac{1}{3}$

$\Rightarrow$           $\frac{n_{3}(n_{1}+n_{2})}{n_{1}(n_{3}+n_{4})+n_{3}(n_{1}+n_{2})}=\frac{1}{3}$

 Now, check options, then clearly options (a) and (b) satisfy.

 Let   $F:R\rightarrow R$  be a thrice  differentiable  function. Suppose that  F(1)=0,F(3)=-4  and F'(x)<0 for x ε (1,3)  , Let f(x)=xF(X) for all x ε R.

 If  $\int_{1}^{3}x^{2} F'(x)dx=-12$      and           $\int_{1}^{3} x^{3}F"(x)=dx=40$, then the correct expression(s) is/are

Answer:

Explanation:

 Given,    $\int_{1}^{3}x^{2} F'(x)dx=-12$ 

      $\Rightarrow$             $ [x^{2}F(x)]_1^3-\int_{1}^{3}2x.F(x) dx=-12 $

                           $\Rightarrow 9F(3)-F(1)-\int_{1}^{3}f(x) dx=-12 $

                                          $[\therefore xF(x)=f(x), given]$

   $\Rightarrow$                   $ -36-0-2\int_{1}^{3} f(x)dx=-12$

                $\therefore$              $\int_{1}^{3} f(x) dx=-12$

 and         $\therefore$             $\int_{1}^{3} x^{3} F"(x) dx=40$

$\Rightarrow$          $[x^{3} F'(x)]_1^3-\int_{1}^{3} 3x^{2}F'(x)dx=40$

             $\Rightarrow$       $ [x^{2}(xF'(x)]_1^3-3\times(-12)=40$

            $\Rightarrow \left\{x^{2}.[f'(x)-F(x)]\right\}_1^3=4$

$\Rightarrow $   $9[f'(3)-F(3)]-[f'(1)-F(1)]=4$

$\Rightarrow$                   $9[f'(3)+4]-[f'(1)-0]=4$

$\Rightarrow$            $ 9f'(3)-f'(1)=-32$

 Let   $F:R\rightarrow R$  be a thrice  differentiable  function. Suppose that  F(1)=0,F(3)=-4  and F'(x)<0 for x ε (1,3)  , Let f(x)=xF(X) for all x ε R.

The correct statement(s)  is/are

Answer:

Explanation:

According to the given data,

   $F(x)<0,\forall x\epsilon(1,3)$

 We have, f(x)= x F(x)

$\Rightarrow$            $f'(x)=F(x)+xF'(x)$        .......(i)

$\Rightarrow$      $ f'(1)=F(1)+F'(1)<0$

         [given F(1)=0 and F'(x)<0]

 Also, f(2)=2F(2)<0

                   [using   $F(x)<0,\forall x \epsilon (1,3)$  ]

Now,  f'(x)=F(x)+x F'(x)<0

                 [using   $F(x)<0,\forall x \epsilon (1,3)$  ]

       $\Rightarrow f'(x)<0$

 The options (s) with the values of a and L that  satidfy the equation  $\frac{\int_{0}^{4\pi} e^{t}(\sin^{6}at+\cos^{4}at)dt}{\int_{0}^{\pi} e^{t}(\sin^{6}at+\cos^{4}at)dt} =L$   is\are

Answer:

Explanation:

Let    $I_{1}=\int_{0}^{4\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt$

 =$\int_{0}^{\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt$

+  $\int_{\pi}^{2\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt$

+  $\int_{2\pi}^{3\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt$

+$\int_{3\pi}^{4\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt$

    $\therefore$      $ I_{1}=I_{2}+I_{3}+I_{4}+I_{5}$       ..........(i)

Now,    $I_{3}=\int_{\pi}^{2\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt$

 Put     $t=\pi+t\Rightarrow dt=dt$

          $\therefore$      $I_{3}=\int_{0}^{\pi} e^{\pi+t}(\sin^{6}at+\cos^{6}at)dt$

=  $e^{t}.I_{2}$            .........(ii)

Now,

          I₄= $\int_{2\pi}^{3\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt$

               Put     $t=2\pi+t\Rightarrow dt=dt$

                        $\therefore$      $I_{4}=\int_{0}^{\pi} e^{2\pi+t}(\sin^{6}at+\cos^{6}at)dt$

=  $e^{2 \pi}.I_{2}$       ..........(iii)

and I₅=   $\int_{3\pi}^{4\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt$

Put            $t=3\pi+t\Rightarrow dt=dt$

                        $\therefore$      $I_{5}=\int_{0}^{\pi} e^{3\pi+t}(\sin^{6}at+\cos^{6}at)dt$

=  $e^{3 \pi}.I_{2}$ .........(iv)

 From Eqs. (i), (ii) ,(iii) and (iv)  , we get

              $I_{1}=I_{2}+e^{\pi}I_{2}+e^{2\pi}I_{2}+e^{3\pi}I_{2}$

     $=I_{2}(1++e^{\pi}+e^{2\pi}+e^{3\pi})$

 $L= \frac{\int_{0}^{4\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt}{\int_{0}^{\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt}$

=     $(1+e^{\pi}+e^{2\pi}+e^{3\pi})$

                = $\frac{1-(e^{4\pi}-1)}{e^{\pi}-1} $   for    $a \epsilon R$

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