QJEEM16Mac1ef: JEE Main 2016 Mathematics - Class 11 JEEM16Mac1ef Solution: If , and ; then

Question

Verified Solution • 2025–26 Session moderate

If

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

, and

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

; then

\mathrm{S}:

Accepted Answer

We have,

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

\ldots

(i)

On replacing

x

by

\frac{1}{x}

in the above equation, we get

\begin{aligned}&amp; f\left(\frac{1}{x}\right)+2 f(x) =\frac{3}{x} \\\\\Rightarrow &amp; \,\, 2 f(x)+f\left(\frac{1}{x}\right) =\frac{3}{x} \,\,\,\,\,...(ii)\end{aligned}

On multiplying Eq. (ii) by 2, we get

4 f(x)+2 f\left(\frac{1}{x}\right)=\frac{6}{x}\quad...(iii)

and subtracting Eq. (i) from Eq. (iii), we get

[4 f(x)+2 f\left(\frac{1}{x}\right)] - [f(x)+2 f\left(\frac{1}{x}\right)]=\frac{6}{x} - 3x

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

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

Now, consider

\quad f(x)=f(-x)

\begin{aligned}&amp;\Rightarrow \frac{2}{x}-x =-\frac{2}{x}+x \\\\&amp;\Rightarrow \frac{4}{x} =2 x \\\\&amp;\Rightarrow 2 x^2 =4 \\\\&amp;\Rightarrow x^2 =2 \\\\&amp;\Rightarrow x =\pm \sqrt{2}\end{aligned}

Hence,

S

contains exactly two elements.

Step-by-Step Verified Solution