JEE 2018 :: Mathematics (2018) :: Solved questions (2018)

• Mathematics (2018) - Solved questions (2018)

For each t ε R, let [t] be the greatest integer less than or equal to t. Then,

$\lim_{x \rightarrow 0}x([\frac{1}{x}+[\frac{2}{x}]+......+[\frac{15}{x}])$

Answer:

Explanation:

Key Idea Use   property of greatest integer  function [x]= x - {x}

We have,

        $\lim_{x \rightarrow 0}x([\frac{1}{x}+[\frac{2}{x}]+......+[\frac{15}{x}])$

we know,    [x]= x - {x}

$\therefore$       $[\frac{1}{x}]=\frac{1}{x}-\left\{\frac{1}{x}\right\}$

Similarly,      $[\frac{n}{x}]=\frac{n}{x}-\left\{\frac{n}{x}\right\}$

$\therefore$     Given limit  

   $=\lim_{x \rightarrow 0}x \left(\frac{1}{x}-\left\{\frac{1}{x}\right\}+\frac{2}{x}-\left\{\frac{2}{x}\right\}+   ....\frac{15}{x}-\left\{\frac{15}{x}\right\}\right)$

    $=\lim_{x \rightarrow 0}(1+2+3+...+15) -x\left(\left\{\frac{1}{x}\right\}+\left\{\frac{2}{x}\right\}+.....+\left\{\frac{15}{x}\right\}\right)$

   =120-0=120

$\begin{bmatrix}\because 0\leq\left\{\frac{n}{x}\right\}<1,therefore \\0\leq x \left\{\frac{n}{x}\right\}< x \Rightarrow\lim_{x \rightarrow 0} x \left\{\frac{n}{x}\right\}=0 \end{bmatrix}$

Let y= y(x) be the solution  of the different equation 

$\sin x\frac{dy}{dx}+ycosx=4x,x\epsilon (0,\pi)$. If  $y(\frac{\pi}{2})=0$, then $y(\frac{\pi}{6})$ is equal to

Answer:

Explanation:

We have

    $\sin x\frac{dy}{dx}+y\cos x=4x$

$\Rightarrow$          $\frac{dy}{dx}+y\cot x=4x\times cosec x$

This is a linear differential equation of form

         $\frac{dy}{dx}+Py=Q$

where P= cot x, Q= 4x cosec x

Now,    IF= $e^{\int_{}^{} Pdx}=e^{\cot x dx}=e^{\log_{}{} \sin x}=\sin x$

Solution of the differential equation is

                      $y. \sin x= \int_{}^{} 4x cosecx\sin xdx +C$

$\Rightarrow$      $y \sin x=\int_{}^{} 4xdx+C=2x^{2}+C$

 Put   $x=\frac{\pi}{2},y=0,$ we get

        $C= \frac{-\pi^{2}}{2}$

$\Rightarrow$     $y \sin x=2x^{2}-\frac{\pi^{2}}{2}$

Put       $x=\frac{\pi}{6}$

  $y(\frac{1}{2})=2(\frac{\pi^{2}}{36})-\frac{\pi^{2}}{2}$

$\Rightarrow$     $y=\frac{\pi^{2}}{9}-\pi^{2}$

$\Rightarrow$      $y=-\frac{8\pi^{2}}{9}$

Alternative method

We have ,  $\sin x\frac{dy}{dx}+y\cos x=4x$ ,

which can be written as $\frac{d}{dx}(\sin x.y)=4x$

On integrating both sides, we get

      $\int_{}^{}\frac{d}{dx}(\sin x.y)dx= \int_{}^{} 4x.dx$

$\Rightarrow$     $y.\sin x=\frac{4x^{2}}{2}+C$

$\Rightarrow$      $y.\sin x=2x^{2}+C$

Now, as y=0 when $x=\frac{\pi}{2}$

$\therefore$  C= $-\frac{\pi^{2}}{2}$

$\Rightarrow$   $y.\sin x=2x^{2}-\frac{\pi^{2}}{2}$

Now , putting  $ x=\frac{\pi}{6}$, we get

$y(\frac{1}{2})=2(\frac{\pi^{2}}{36})-\frac{\pi^{2}}{2}$

  $y= \frac{\pi^{2}}{9}-\pi^{2}=-\frac{8\pi^{2}}{9}$

If sum of all the solutions of the equation

$8\cos x.( \cos ( \frac{\pi}{6}+x).\cos ( \frac{\pi}{6}-x)-\frac{1}{2})=1$ in [0,π ] is kπ , then k is equal to

Answer:

Explanation:

Key Idea Apply the  identity 

$\cos ( x+y)\cos ( x-y)=\cos^{2}x-\sin^{2}y$ and   $\cos 3x=4\cos^{3}x-3\cos x$

We have,

$8\cos x (\cos ( \frac{\pi}{6} +x)\cos ( \frac{\pi}{6}-x)-\frac{1}{2})=1$

$\Rightarrow$ $8\cos x( \cos^{2}\frac{\pi}{2}-\sin^{2}x-\frac{1}{2})=1$

$\Rightarrow$  $8\cos x( \frac{3}{4}-\sin^{2}x-\frac{1}{2})=1$

$\Rightarrow$  $8\cos x( \frac{3}{4}-\frac{1}{2}-1+\cos^{2}x)=1$

$\Rightarrow$  $8\cos x(\frac{-3+4\cos^{2}x}{4})=1$

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

$\Rightarrow$ $2\cos3x=1\Rightarrow\cos 3x=\frac{1}{2}$

$\Rightarrow$  \[3x=\frac{\pi}{3},\frac{5\pi}{3},\frac{7\pi}{3}\]   [0≤ 3x ≤ 3π ]

$\Rightarrow$   $x= \frac{\pi}{9},\frac{5\pi}{9},\frac{7\pi}{9}$

Sum  $= \frac{\pi}{9}+\frac{5\pi}{9}+\frac{7\pi}{9}=\frac{13\pi}{9}\Rightarrow k\pi =\frac{13\pi}{9}$

Hence    k= $\frac{13}{9}$

The Integral 

$\int_{}^{} \frac{\sin^{2}x cos^{2}x}{(\sin^{5}x+\cos^{3}x\sin^{2}x+\sin^{3}x\cos^{2}x+\cos^{5}x)^{2}}dx$

is equal to    (where C is constant of integration)

Answer:

Explanation:

We have

I= $\int_{}^{} \frac{\sin^{2}x cos^{2}x}{(\sin^{5}x+\cos^{3}x\sin^{2}x+\sin^{3}x\cos^{2}x+\cos^{5}x)^{2}}dx$

$\int_{}^{}\frac{\sin^{2}x cos^{2}x}{( \sin^{3}x( \sin^{2}x+\cos^{2}x)+\cos^{3}x( \sin^{2}x+\cos^{2}x))^{2}}dx$

= $\int_{}^{}\frac{\sin^{2}x cos^{2}x}{( \sin^{3}x+\cos^{3}x)^{2}}dx$

= $\int_{}^{}\frac{\sin^{2}x cos^{2}x}{\cos^{6}x( 1+\tan^{3}x)^{2}}dx$

= $\int_{}^{} \frac{\tan^{2}x\sec^{2}x}{( 1+\tan^{3}x)^{2}}dx$

  Put  $\tan^{2}x=1\Rightarrow 3\tan^{3}x \sec^{2}x dx=dt$

$\therefore $      $I=\frac{1}{3}\int_{}^{} \frac{dt}{( 1+t)^{2}}$

$\Rightarrow$     I=  $\frac{-1}{3( 1+t)}+C$

$\Rightarrow$    $I= \frac{-1}{3( 1+\tan^{3}x)}+C$

The length of the projection of the line segment joining the points (5,-1,4) and (4,-1,3) on the plane , x+y+z=7 is

Answer:

Explanation:

Key Idea  length ofv projection of the line segment joining a₁ and a₂ on the plane r .n

= d is $\mid\frac{(a_{2}-a_{1})\times n}{\mid n\mid}\mid$

 Length of projection the line segment joining the points (5, -1, 4 ) and (4,-1,3)

   on the plane x+y+z = 7 is

$AC= \mid\frac{(a_{2}-a_{1})\times n}{\mid n\mid}\mid$

 $AC=  \frac{\mid (-\hat{i}-\hat{k})\times (\hat{i}+\hat{j}+\hat{k)} \mid }{ \mid i+j+k\mid}$

$AC= \frac{\mid \widetilde{i}-\widetilde{k}\mid}{\sqrt{3}}$

$AC= \frac{\sqrt{2}}{\sqrt{3}}$

$AC= \sqrt{\frac{2}{3}}$

Alternative method

  Clearly , DR's of AB are 4-5, -1+1, 3-4, 

i.e. -1,0,-1

 and DR's of normal to plane are 1,1,1. Now, let θ be the angle between the line and plane , then θ is given by

  $ \sin\theta=\frac{\mid -1+0-1\mid}{\sqrt{(-1)^{2}+(-1)^{2}}\sqrt{1^{2}+1^{2}+1^{2}}}$

$=\frac{2}{\sqrt{2}\sqrt{3}}=\sqrt{\frac{2}{3}}$

$\Rightarrow\sin\theta=\sqrt{\frac{2}{3}}\Rightarrow \cos\theta=\sqrt{1-\sin^{2}\theta}$

= $\sqrt{1-\frac{2}{3}}=\frac{1}{\sqrt{3}}$

Clearly, length of projection

= $AB \cos\theta=\sqrt{2}\frac{1}{\sqrt{3}}$     [ $\therefore AB=\sqrt{2}$]

= $=\sqrt{\frac{2}{3}}$

• Mathematics (2018) - Solved questions (2018)