Question

Let $f(x) = x + 1/2$. then the number of real values of x for which the three unequal terms $f(x), f(2x), f(4x)$ are in $H.P.$ is

• A. 1
• B. 0
• C. 3
• D. 2

Answer 1

Answer: (A)

Given, $f(x)=x+\frac{1}{2}=\frac{2 x+1}{2}$
$\therefore f(2 x)=2 x+\frac{1}{2}$
$\Rightarrow f(2 x)=\frac{4 x+1}{2}$
and $\quad f(4 x)=4 x+\frac{1}{2} $
$\Rightarrow f(4 x)=\frac{8 x+1}{2}$
Since, $f(x), f(2 x)$ and $f(4 x)$ are in $HP$.
$\therefore \frac{1}{f(x)}, \frac{1}{f(2 x)}$ and $\frac{1}{f(4 x)}$ are in AP.
$\Rightarrow \frac{1}{f(2 x)}=\frac{\frac{1}{f(x)}+\frac{1}{f(4 x)}}{2}$
$\Rightarrow \frac{2}{4 x+1}=\frac{\frac{2}{2 x+1}+\frac{2}{8 x+1}}{2}$
$\Rightarrow \frac{2}{4 x+1}=\frac{10 x+2}{(2 x+1)(8 x+1)}$
$\Rightarrow (2 x+1)(8 x+1)=(5 x+1)(4 x+1)$
$\Rightarrow 16 x^{2}+10 x+1=20 x^{2}+9 x+1$
$\Rightarrow 4 x^{2}-x=0 $
$\Rightarrow x(4 x-1)=0$
$\Rightarrow x=\frac{1}{4} \,\,\,[\because x \neq 0]$
Hence, one real value of $x$ for which the three unequal terms are in $HP$.