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