Question

Let $x+y\propto z+\frac{1}{z}$, $x-y\propto z-\frac{1}{z}$ and $z=2$, when $y=1,x=3$. Then

• A. $x=\frac{2}{15}(11z+\frac{1}{z})$
• B. $x=\frac{22}{15}z-\frac{2}{15}.\frac{1}{z}$
• C. $y=\frac{2}{15}z-\frac{22}{15}.\frac{1}{z}$
• D. $y=\frac{2}{15}(z+\frac{11}{z})$

Answer 1

Answer: (A), (D)

The correct options are
A $x=\frac{2}{15}(11z+\frac{1}{z})$
D $y=\frac{2}{15}(z+\frac{11}{z})$

Given:- $x+y$$\propto z+\frac{1}{z},x-y\propto z-\frac{1}{z}$

$z=2,y=1,x=3$

now after or removing propotional sign

$x+y=k(z+\frac{1}{z})$

by putting values of x,y and z

$3+1=k(2+\frac{1}{2})$ ...........(i)

simillary

$x-y=m(z-\frac{1}{z})$ ...........(ii)

from equation(1)

$3-1=k(2-\frac{1}{2})$

$4=k\left(\frac{5}{2}\right)$

$k=\frac{5}{2}$

from equation (2)

$2=m\left(\frac{4}{3}\right)$

$m=\frac{4}{3}$

Therefore,

$x+y=\frac{8}{5}(z+\frac{1}{z})$

$x-y=\frac{4}{3}(z-\frac{1}{z})$

adding both

$2x=\frac{8}{5}z+\frac{4}{3}z+\frac{8}{5z}-\frac{4}{3z}$

$2x=\frac{44z}{15}+\frac{4}{15z}$

$x=\frac{1}{2}(\frac{44z}{15}+\frac{4}{15z})$

$x=\frac{2}{15}(11z+\frac{1}{z})$

Put the value of x in eq.(2)

$y=\frac{2}{15}(z+\frac{11}{z})$