NIELIT 2019 Feb Scientist D - Section D: 19

Question

If $x= \frac{\sqrt{p^{2}+q^{2}}+\sqrt{p^{2}-q^{2}}}{{\sqrt{p^{2}+q^{2}}-\sqrt{p^{2}-q^{2}}}}$ then $q^{2}x^{2}-2p^{2}x+q^{2}$ equals to :

• A. $3$
• B. $-1$
• C. $-2$
• D. $0$

Answer 1

$ \begin{align} &\frac{x}{1}= \frac{\sqrt{p^{2}+q^{2}}+\sqrt{p^{2}-q^{2}}}{{\sqrt{p^{2}+q^{2}}-\sqrt{p^{2}-q^{2}}}} \\ \implies &\frac{x+1}{x-1} = \frac{\sqrt{p^2 + q^2} }{\sqrt{p^2-q^2}} \qquad \qquad \rightarrow \text{ apply Componendo and Dividendo} \\ \implies & \frac{(x+1)^2}{(x-1)^2} = \frac{p^2+q^2}{p^2-q^2} \qquad \qquad \rightarrow \text{take square on both side}  \\ \implies & \frac{(x+1)^2 + (x-1)^2}{(x+1)^2 - (x-1)^2} = \frac{p^2}{q^2}  \qquad \rightarrow \text{ apply Componendo and Dividendo} \\ \implies & \frac{x^2 + 1}{2x} = \frac{p^2}{q^2}  \\ \implies& q^2(x^2+1) = 2xp^2 \\ \implies & q^2x^2 -2p^2x +q^2 = 0 \end{align}$

Option D.