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

Solution

## 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) ⇒ [f(2x)]2 = f(x)f(4x) (2(2x)+1)2 = (2x + 1)(2 ×(4x) + 1)  (4x+1)2 = (2x+1)(8x+1) 16 x2 + 8x + 1 = 16 x2 + 10x + 1 ⇒ 2x = 0 x = 0 ⇒ Only one value

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