Question

If $f\left(x\right)=x+\frac{1}{2}$. Then, the number of real values of $x$ for which the three unequal terms $f\left(x\right)$, $f\left(2x\right)$ and $f\left(4x\right)$ are in HP is

• A. $1$
• B. $0$
• C. $3$
• D. $2$

Answer 1

The correct option is A

$1$

Explanation for the correct option:

Step 1: Finding $f\left(2x\right)$ and $f\left(4x\right)$:

Given that,

$f\left(x\right)=\frac{2x+1}{2}$

$\begin{array}{rcl}f\left(2x\right)& =& \frac{2\left(2x\right)+1}{2}\\ & =& \frac{4x+1}{2}\\ f\left(4x\right)& =& \frac{2\left(4x\right)+1}{2}\\ & =& \frac{8x+1}{2}\end{array}$

Step 2: Find the number of real values of $x$:

Since, $f\left(x\right)$, $f\left(2x\right)$ and $f\left(4x\right)$ are in HP then $\frac{1}{f\left(x\right)},\frac{1}{f\left(2x\right)},\frac{1}{f\left(4x\right)}$ are in AP.

By using formula $2b=a+c$

$\begin{array}{rcl}\frac{2}{{\displaystyle \frac{4x+1}{2}}}& =& \frac{2}{2x+1}+\frac{2}{8x+1}\\ & \Rightarrow & \frac{2}{4x+1}=\frac{8x+1+2x+1}{\left(2x+1\right)\left(8x+1\right)}\\ & \Rightarrow & 2\left(2x+1\right)\left(8x+1\right)=2\left(5x+1\right)\left(4x+1\right)\\ & \Rightarrow & 16x^2+2x+8x+1=20x^2+5x+4x+1\\ & \Rightarrow & 4x^2-x=0\\ & \Rightarrow & x\left(4x-1\right)=0\\ & \Rightarrow & x=0or\frac{1}{4}\end{array}$

For $x=0$, $f\left(x\right)$, $f\left(2x\right)$, $f\left(4x\right)$ becomes equal and it's given the terms are unequal then only one is the real value of $x$ i.e $\frac{1}{4}$.

Hence, the correct option is A.