CAT 2018

Slot 1

QA

Question & Solution

Given that $U^{2}+(U-2V-1)^{2}$= −$4V(U+V)$ 

$\Rightarrow$ $U^{2}+(U-2V-1)(U-2V-1)$= −$4V(U+V)$

$\Rightarrow$ $U^{2}+(U^2-2UV-U-2UV+4V^2+2V-U+2V+1)$ = −$4V(U+V)$ 

$\Rightarrow$ $U^{2}+(U^2-4UV-2U+4V^2+4V+1)$ = −$4V(U+V)$ 

$\Rightarrow$ $2U^2-4UV-2U+4V^2+4V+1=−4UV-4V^2$

$\Rightarrow$ $2U^2-2U+8V^2+4V+1=0$

$\Rightarrow$ $2[U^2-U+\dfrac{1}{4}]+8[V^2+\dfrac{V}{2}+\dfrac{1}{16}]=0$

$\Rightarrow$ $2(U-\dfrac{1}{2})^2+8(V+\dfrac{1}{4})^2=0$

Sum of two square terms is zero i.e. individual square term is equal to zero. 

$U-\dfrac{1}{2}$ = 0 and $V+\dfrac{1}{4}$ = 0

U = $\dfrac{1}{2}$ and V = $-\dfrac{1}{4}$

Therefore, $U+3V$ = $\dfrac{1}{2}$+$\dfrac{-1*3}{4}$ = $\dfrac{-1}{4}$.

Hence, option C is the correct answer.