If $\mathrm{f}(\mathrm{x})+2 \mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=3 \mathrm{x}, \mathrm{x} \neq 0$, and $\mathrm{S}=\{\mathrm{x} \in \mathrm{R}: \mathrm{f}(\mathrm{x})=\mathrm{f}(-\mathrm{x})\}$; then S :

(B) contains exactly two elements.

(A) contains exactly one element.

(C) contains more than two elements.

JEE Main · 2016 · 03 Apr 2016 · Offline · Sets Relations and Functions

f (x) + 2 f (\frac{1}{x}) = 3 x, x \neq = 0

, and

S = {x \in R : f (x) = f (- x)}

; then S :

Answer: (B)

Write the relation once as

f (x) + 2 f (\frac{1}{x}) = 3 x

and again with

x

replaced by

\frac{1}{x}

f (\frac{1}{x}) + 2 f (x) = \frac{3}{x}

, then treat them as two equations in

f (x)

and

f (1/ x)

. Eliminating

f (1/ x)

gives

3 f (x) = \frac{6}{x} - 3 x

, so

f (x) = \frac{2}{x} - x

. The condition

f (x) = f (- x)

becomes

\frac{2}{x} - x = - \frac{2}{x} + x

, i.e.

\frac{4}{x} = 2 x

, so

x^{2} = 2

and

x = \pm 2

. The set S has exactly two elements.

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option