Question

Let f(x) = 2x + 1. Then the number of real values of x for which the three numbers f(x), f(2x), f(4x) are in G.P. is

Answer 1

f(x), f(2x), f(4x) are in G.P. We will replace x by 2x and 4x while finding f(2x) and f(4x)

$\Rightarrow$ ${\left[f\right(2x\left)\right]}^2$ = f(x)f(4x)

${\left(2\right(2x)+1)}^2$ = (2x + 1)(2 ×(4x) + 1)

${(4x+1)}^2$ = (2x+1)(8x+1)

16 $x^2$ + 8x + 1 = 16 $x^2$ + 10x + 1

$\Rightarrow$ 2x = 0
x = 0

$\Rightarrow$ Only one value