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

• Mathematics (2018) - Solved questions (2018)

Let P be a point in the first octant , whose image Q in the plane x+y= 3 (that is, the line segment PQ is perpendicular to the plane x+y=3 and the mid- point of PQ lies in the plane x+y=3) lies on the Z-axis. Let the distance of P from the X-axis be 5. If R is the image of P  in the XY-plane, then the length of PR is .....

Answer:

Explanation:

Let $P\left( \alpha, \beta, \gamma\right)$  and R  is the image of P in the XY - plane.

$\therefore$  $R\left( \alpha, \beta, \gamma\right)$ 

   Also, Q is the image of P in the plane  x+y=3

$\therefore$   $\frac{x-\alpha}{1}= \frac{y-\beta}{1}=\frac{z-\gamma}{0}$

                     = $\frac{-2(\alpha+\beta-3)}{2}$

           $x=3-\beta ,y=3-\alpha ,z=\gamma$

 Since , Q is lies on Z-axis

  $\therefore$  $\beta =3, \alpha =3 , z=\gamma$

   $\therefore$  $ P (3,3, \gamma )$

Given , distance of  P  from  X-axis  be 5

  $\therefore$            $5= \sqrt{3^{2}+\gamma^{2}}$

                       $25-9=\gamma^{2}$

$\Rightarrow$     $\gamma \pm 4$

Then  , PR = $\mid2\gamma \mid =\mid 2\times 4\mid=8$

Let  $f:R\rightarrow R$  be a differentiable function with f(0) =1 and satisfying the equation f(x+y) =f(x)  f ' (y)+ f ' (x) f(y)  for 

all x , $y \in R $, Then, the value of log_e (f(4)) is..............

Answer:

Explanation:

Given, $ f(x+y) = f  (x) f ' (y) +f '(x) f (y), \forall x,y \in R$  and f (0)=1

Put x=y=0, we get

f(0) = f(0) f '(0) +f '(0) f (0)

Let $f:R\rightarrow R$ be a differentiable function with f(0)=0,  y=f(x) satisfies the differential equation

 $\frac{\text{d}y}{\text{d}x}= (2+5y)(5y-2)$  , then the value of  $\lim_{x \rightarrow \infty} f(x) $  is .............

Answer:

Explanation:

We have , $\frac{\text{d}y}{\text{d}x}=(2+5y)(5y-2)$

$\Rightarrow $     $\frac{dy}{25y^{2}-4}=dx$

$\Rightarrow $     $\frac{1}{25}(\frac{dy}{y^{2}-\frac{4}{25}})=dx$

  On integrating both sides, we get ,$\frac{1}{25}\int_{}^{} \frac{dy}{y^{2}-(\frac{2}{5})^{2}}=\int_{}^{} dx$

$\Rightarrow $    $\frac{1}{25}\times\frac{1}{2\times 2/5}\log_{}{\mid\frac{y-2/5}{y+2/5}}\mid =x+C$

$\Rightarrow $  $log_{}{\mid\frac{5y-2}{5y+2}}\mid =20(x+C)$

$\Rightarrow $     ${\mid\frac{5y-2}{5y+2}}\mid =Ae^{20x}$    [ $\therefore$       $e^{20C}=A$ ]

when x=o    $\Rightarrow $  y =0 then A =1

$\therefore$           $\mid \frac{5y-2}{5y+2}\mid =e^{20x}$

 $\lim_{x \rightarrow -\infty} \mid \frac{5f(x)-2}{ 5f(x)+2}\mid =\lim_{x \rightarrow -\infty}e^{20x}$

$\Rightarrow $   $\Rightarrow \lim_{n \rightarrow -\infty} \mid \frac{5f(x)-2}{ 5f(x)+2}\mid =0$

$\Rightarrow  \lim_{n \rightarrow -\infty} 5 f(x) -2=0$

$\Rightarrow  \lim_{n \rightarrow -\infty} f(x) =\frac{2}{5}=0.4$

Let X be a set with exactly 5 elements and Y be a set  exactly  7 elements. If α is the numbers of one - one functions from X to Y  and β is the number of onto functions from Y to X  , then the value of $\frac{1}{5!}(\beta -\alpha) $ is ............

Answer:

Explanation:

Given , X has exactly 5 elements and Y has exactly  7 elements.

       

The value of integral $\int_{0}^{1/2} \frac{1+\sqrt{3}}{((x+1)^{2}(1-x)^{6})^{1/4}}dx$ is ............

Answer:

Explanation:

 Let  I=  $\int_{0}^{1/2} \frac{1+\sqrt{3}}{((x+1)^{2}(1-x)^{6})^{1/4}}dx$

     $\Rightarrow$      I = $\int_{0}^{1/2} \frac{1+\sqrt{3}}{[(1-x)^{2}[(\frac{1-x}{1+x})^{6}]^{1/4}}dx$

   Put  $\frac{I-x}{I+x}=t \Rightarrow \frac{-2dx}{(1+x)^{2}}\Rightarrow dt$

  when x=0 , t=1, x= $\frac{1}{2}$ , t= $\frac{1}{3}$

$\therefore$                I= $\int_{0}^{1/3} \frac{\left(1+\sqrt{3}\right)dt}{-2(t)^{\frac{6}{4}}}$

  $\Rightarrow$  $I= \frac{-(1+\sqrt{3})}{2}\begin{bmatrix}  \underline{-2} \\  \sqrt{t} \end{bmatrix}^{1/3}_{1}$

$\Rightarrow$       $I= (1+\sqrt{3}) (\sqrt{3}-1)\Rightarrow I=3-1=2$

• Mathematics (2018) - Solved questions (2018)