JEE 2015 :: Advanced 2015 :: Mathematics 2015

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

Let y(x) be a soilution of the differential equation $(1+e^{x})y^{'}+ye^{x}=1,$.If y(0)=2 , then which of the following statement(s) is/are true?

Answer:

Explanation:

Here , $(1+e^{x})y^{'}+ye^{x}=1,$

$\Rightarrow$                         $\frac{dy}{dx}+e^{x}.\frac{dy}{dx}+ye^{x}=1$

$\Rightarrow$                     $dy+e^{x}dy+ye^{x}dx=dx$

$\Rightarrow$                     $dy+d(e^{x}y) =dx$

  On integrating both sides, we get

                           $y+e^{x}y=x+C$

  Given      ,      y(0)=2

$\Rightarrow$      $2+e^{0}.2=0+C$

$\Rightarrow$      C=4

$\therefore$                       $y(1+e^{x})=x+4$

$\Rightarrow$                       $y=\frac{x+4}{1+e^{x}}$

  Now at x=-4,         $y=\frac{-4+4}{1+e^{-4}}=0$

  $\therefore$              y(-4)=0  ......(i)

  For  critical points,   $\frac{dy}{dx}=0$

i.e,   $\frac{dy}{dx}=\frac{(1+e^{x}).1-(x+4)e^{x}}{(1+e^{x})^{2}}=0$

$\Rightarrow$                      $e^{x}(x+3)-1=0$

 or                       $e^{-x}=(x+3)$

  Clearly, the intersecting  point lies between (-1,0)

  $\therefore$     y(x) has a critical point in the interval (-1,0)

Let  $f:R\rightarrow R$ be a  function defined by   $f(x)=\begin{cases}[x], & x \leq 2\\0 & x > 2\end{cases}$ , where [x] denotes the greatest integer less than or equal to x. If   $I= \int_{-1}^{2} \frac{x f(x^{2})}{2+f(x+1)}dx, $, then the value of (4l-1) is 

Answer:

Explanation:

Here     ,  $f(x)=\begin{cases}[x] ,& x \leq 2\\0, & x > 2\end{cases}$

            $I=\int_{-1}^{2} \frac{xf(x^{2})}{2+f(x+1)}dx$

                         $I=\int_{-1}^{0} \frac{xf(x^{2})}{2+f(x+1)}dx$ + $\int_{0}^{1} \frac{xf(x^{2})}{2+f(x+1)}dx$+  $\int_{1}^{\sqrt{2}} \frac{xf(x^{2})}{2+f(x+1)}dx$

                                                 +   $\int_{\sqrt{2}}^{\sqrt{3}} \frac{xf(x^{2})}{2+f(x+1)}dx$ +   $\int_{\sqrt{3}}^{2} \frac{xf(x^{2})}{2+f(x+1)}dx$

              =   $\int_{-1}^{0} odx+\int_{0}^{1} 0 dx+\int_{1}^{\sqrt{2}} \frac{x.1}{2+0}dx+ \int_{\sqrt{2}}^{\sqrt{3}} o dx+\int_{\sqrt{3}}^{2}o.dx $  

                                                 {    $\because -1<x<0\Rightarrow 0<x^{2}<1\Rightarrow [x^{2}]=0, $

                                                       $ o<x<1\Rightarrow 0<x^{2}<1\Rightarrow[x^{2}]=0,  $

                                                          $ 1< x<\sqrt{2}\Rightarrow $   $\begin{cases}1<x^{2}<2 & \Rightarrow[x^{2}]=1\\2<x+1<1+\sqrt{2}      &\Rightarrow f(x+1)=0\end{cases}$

                                           $\Rightarrow$    f(x²)= 0,

                                                   and  $\sqrt{3}<x<2\Rightarrow3<x^{2}<4$

                                                $\Rightarrow$    f(x² )=0 }

  $\Rightarrow$      $I= \int_{1}^{\sqrt{2}} \frac{x}{2}dx=\left[\frac{x^{2}}{4}\right]^{\sqrt{2}}_{1}$

                           $=\frac{1}{4}(2-1)=\frac{1}{4}$

                   $\therefore$          4l =1    $\Rightarrow$    4l-1=0

A cylindrical container is to be made from certain solid material with the following constraints: It has a fixed inner volume of V mm³, has a 2mm thick solid wall and is open at the top. The bottom of the container is a solid circular disc of thickness 2mm and is of radius equal to the outer radius of the container. If the volume of the material used to make the container is minimum, when  the inner radius  of the container is 10mm, then the value of  $\frac{V}{250\pi}$  is

Answer:

Explanation:

Here, volume of cylindrical container,

       V=  $\pi r^{2}h$           .............(i)

 and let volume of the material used be T.

$\therefore  T=\pi \left[(r+2)^{2}-r^{2}\right]h+\pi(r+2)^{2}\times2$

 $ T=\pi \left[(r+2)^{2}-r^{2}\right]\frac{V}{\pi r^{2}}+2\pi(r+2)^{2}$

                                    $\left[\because  V=\pi r^{2}h\Rightarrow h=\frac{V}{\pi r^{2}}\right]$

                   $\Rightarrow T=V\left(\frac{r+2}{r}\right)^{2}+2\pi (r+2)^{2}-V$

 On differentiating  w.r.t , we get

                  $\frac{dT}{dr}=2V.\left(\frac{r+2}{r}\right).\left(\frac{-2}{r^{2}}\right)+4\pi (r+2)$

 At,            r=10,    $\frac{dT}{dr}=0$

 Now,   $0=(r+2).4\left(\pi-\frac{V}{r^{3}}\right)$

$\Rightarrow$                 $\frac{V}{r^{3}}=\pi, $      where r=10

$\Rightarrow$         $\frac{V}{1000}=\pi, or \frac{V}{250\pi}=4$

 Let  $F(X)=\int_{c}^{x^{2}+\frac{\pi}{6}} 2\cos^{2}tdt$ for all xε R and  $f:\left[0,\frac{1}{2}\right]\rightarrow (0,\infty)$ be a continuous function, For $a \epsilon \left[ 0,\frac{1}{2}\right]$ , if F'(a)+2 is the area of the region bounded by x=0, y=0, y=f(x) and x=a then f(0) is

Answer:

Explanation:

Since , F'(a)+2 is the area bounded by

 x=0, y=0,y=f(x)  and x=a

$\therefore$    $\int_{0}^{a} f(x) dx=F'(a)+2$

  Using Newton- Leibnitz formula

               $f(a)=F"(a)$

and f(0) =F" (0)                .........(i)

 Given, $f(x)=\int_{x}^{x^{2}+\frac{\pi}{6}} 2\cos^{2}t dt$

 On differentiating,

                  $F'(x)=2\cos^{2}\left(x^{2}+\frac{\pi}{6}\right).2x-2\cos^{2}x.1$

 Again differentiating,

            $F"(x)=4\left\{\cos ^{2}\left(x^{2}+\frac{\pi}{6}\right)-2xcos\left(x^{2}+\frac{\pi}{6}\right)\sin\left(x^{2}+\frac{\pi}{6}\right)2x\right\}+\left\{4\cos x.\sin x\right\}$

            $=4\left\{\cos ^{2}\left(x^{2}+\frac{\pi}{6}\right)-4x^{2}\cos \left( x^{2}+\frac{\pi}{6}\right) \sin\left(x^{2}+\frac{\pi}{6}\right)\right\}+2\sin 2x$

          $\therefore F"(0)=4\left\{\cos^{2}\left(\frac{\pi}{6}\right)\right\}=3$

                    $\therefore f(0)=3$

 The number of distinct solutions of the equation $ \frac{5}{4}\cos^{2}2x+\cos^{4}x+\sin^{4}x+\cos^{6}x+\sin^{6}x=2$ in the interval [0,2π) is 

Answer:

Explanation:

Here,     $\frac{5}{4}\cos^{2}2x+(\cos^{4}x+\sin^{4}x)+(\cos^{6}x+\sin^{6}x)=2$

               $\Rightarrow \frac{5}{4}\cot^{}2x+[(\cos^{2}x+\sin^{2}x)^{2}-2\sin^{2}x\cos^{2}x]$

                            $+(\cos^{2}x+\sin^{2}x)[(\cos^{2}x+\sin^{2}x)^{2}-3\sin^{2}x\cos^{2}x]=2$

             $\Rightarrow \frac{5}{4}\cos^{2}2x+(1-2\sin^{2}x\cos^{2}x)+(1-3\cos^{2} x \sin^{2}x)=2$

          $\Rightarrow \frac{5}{4}\cos^{2}2x-5\sin^{2}x\cos^{2}x=0$

  $\Rightarrow$                $\frac{5}{4}\cos^{2}2x-\frac{5}{4}\sin^{2}2x=0$

 $\Rightarrow$                   $\frac{5}{4}\cos^{2}2x-\frac{5}{4}+\frac{5}{4}\cos^{2}2x=0$

 $\Rightarrow$           $ \frac{5}{2}\cos^{2}2x=\frac{5}{4}$

 $\Rightarrow$            $\cos^{2}2x=\frac{1}{2}\Rightarrow 2\cos^{2}2x=1$

 $\Rightarrow$             1+cos 4x=1

 $\Rightarrow$     cos 4x=0, as  $0\leq x\leq 2\pi$

  $\therefore$

             $4x=\left\{\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2},\frac{9\pi}{2},\frac{11\pi}{2},\frac{13\pi}{2},\frac{15\pi}{2}\right\}$

            as   $0\leq 4x\leq8\pi$

    $\Rightarrow x=\left\{\frac{\pi}{8},\frac{3\pi}{8},\frac{5\pi}{8},\frac{7\pi}{8},\frac{9\pi}{8},\frac{11\pi}{8},\frac{13\pi}{8},\frac{15\pi}{8}\right\}$

             Hence, the total number of solutions is 8.

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