SSC CGL 2023 PYQ — Question 74

The correct answer is option 4 i.e. $\sqrt{13}$.

Given that, $x=\frac{1}{x-3}, (x > 0)$

⇒ x² - 3x - 1 = 0

Using discriminant formula, the value of x,

⇒ $x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

⇒ $x = {3 \pm \sqrt{9 + 4} \over 2}$

⇒ $x = {3 \pm \sqrt{13} \over 2}$

Since x > 0,

⇒ $x = {\sqrt{13} + 3 \over 2}$

Now, ${1 \over x} = {2 \over {\sqrt{13} + 3}}$

On doing rationalisation,

${1 \over x} = {2 \over {\sqrt{13} + 3}} \times {{ \sqrt{13} - 3} \over {\sqrt{13} - 3}}$

⇒ ${1 \over x} = {{2 \times ({\sqrt{13} - 3})} \over {13 - 9}}$

⇒ ${1 \over x} = {{\sqrt{13} - 3} \over 2}$

Now, according to the given question,

$x+\frac{1}{x}$ = ${\sqrt{13} + 3 \over 2}$ + ${{\sqrt{13} - 3} \over 2}$

⇒ ${\sqrt{13} + 3 + \sqrt{13} - 3} \over 2$

⇒ ${2\sqrt{13}} \over 2$ = $\sqrt {13}$

Hence the value of $x+\frac{1}{x}$ is $\sqrt {13}$.