JEE MainMathematicsToughFunctions

Range of $f(x) = 4^{x} + 2^{x} + 1$ is

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f(x) = 4^{x} + 2^{x} + 1

Let

2^{x} = t > 0,\forall x \in R

\therefore

f(x) = g(t) = t^{2} + t + 1,t > 0

g(t) = \left( t + \frac{1}{2} \right)^{2} + \frac{3}{4}

\left( t + \frac{1}{2} \right) > \frac{1}{2} \Rightarrow

\left( t + \frac{1}{2} \right)^{2} > \frac{1}{4} \Rightarrow

\left( t + \frac{1}{2} \right)^{2} + \frac{3}{4} > 1

Range is

(1,\infty)

Source: Grass-G / M-19 Mathematics Paper 28 Nov Fourth.docx

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