Question

Let $f\left(x\right)=x+1/2$, 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 H.P is

• A. $1$
• B. $0$
• C. $3$
• D. $2$

Answer 1

The correct option is A $1$

Given f(x), f(2x), f(4x) are in H.P.

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

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

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

$\therefore 2x+\frac{1}{2}=\frac{2(x+\frac{1}{2})(4x+\frac{1}{2})}{(x+\frac{1}{2})+(4x+\frac{1}{2})}$

$\Rightarrow \frac{4x+1}{2}=\frac{(2x+1)(8x+1)}{(5x+1)\left(2\right)}$

$\Rightarrow 20x^2+9x+1=16x^2+10x+1$

$\Rightarrow 4x^2-x=0$

$x=0$ or $x=\frac{1}{4}$.

For $x=0$, we get each of $f\left(x\right),f\left(2x\right),$ and $f\left(4x\right)$ are all equal.

Hence, $x=\frac{1}{4}$ is the only real value of $x$ that meets the criteria.