Aspire Faculty ID #13843 · Topic: JAMIA MILLIA ISLAMIA MCA 2021 · Just now
JAMIA MILLIA ISLAMIA MCA 2021

The minimum value of $4^x + 4^{1-x},\ x \in \mathbb{R}$ is:

Solution

Let $4^x = t,\ t > 0$. Then expression $= t + \dfrac{4}{t}$. By AM ≥ GM, $t + \dfrac{4}{t} \ge 2\sqrt{t \cdot \dfrac{4}{t}} = 4.$ Equality when $t = 2$, i.e., $4^x = 2 \Rightarrow x = \dfrac{1}{2}$. $\boxed{\text{Answer: (B) 4}}$
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