Question

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

• A. $1$
• B. $0$
• C. $3$
• D. $2$

Answer 1

The correct option is A $1$

Given,

$f\left(x\right)=x+\frac{1}{2}=\frac{2x+1}{2}$
$\therefore f\left(2x\right)=2x+\frac{1}{2}$
$\Rightarrow f\left(2x\right)=\frac{4x+1}{2}$
and

$f\left(4x\right)=4x+\frac{1}{2}\Rightarrow f\left(4x\right)=\frac{8x+1}{2}$

Since, $f\left(x\right),f\left(2x\right)$ and $f\left(4x\right)$ are in HP.

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

$\Rightarrow \frac{1}{f\left(2x\right)}=\frac{\frac{1}{f\left(x\right)}+\frac{1}{f\left(4x\right)}}{2}$

$\Rightarrow \frac{2}{4x+1}=\frac{\frac{2}{2x+1}+\frac{2}{8x+1}}{2}$

$\Rightarrow \frac{2}{4x+1}=\frac{10x+2}{(2x+1)(8x+1)}$

$\Rightarrow (2x+1)(8x+1)=(5x+1)(4x+1)$
$\Rightarrow 16x^2+10x+1=20x^2+9x+1$
$\Rightarrow 4x^2-x=0$

$\Rightarrow x(4x-1)=0$
$\Rightarrow x=\frac{1}{4}$ $[\because x\ne 0]$
Hence, one real value of $x$ for which the three unequal terms are in HP.