For x geqslant 0 , the least value of K , for which 4^1+x+4^1-x, K/2, 16^x+16^-x are three consecutive terms of an A.P., is equal to :

Correct answer: (C) 10. k=4 (4^x+frac14^x )+ (4^2 x+frac14^2 x ) ≥ 2 ≥ 2 k ≥ 10

Q33 / 90 · JEE Main 2024 (05 Apr Shift 2) · Mathematics

For $x \geqslant 0$, the least value of $\mathrm{K}$, for which $4^{1+x}+4^{1-x}, \frac{\mathrm{K}}{2}, 16^x+16^{-x}$ are three consecutive terms of an A.P., is equal to :

A
8
B
4
Correct answer
C
10
D
16

Solution

$\begin{gathered}\mathrm{k}=4\left(4^{\mathrm{x}}+\frac{1}{4^{\mathrm{x}}}\right)+\left(4^{2 \mathrm{x}}+\frac{1}{4^{2 \mathrm{x}}}\right) \\ \qquad \geq 2 \quad\qquad\qquad \geq 2\end{gathered}$ $\mathrm{k} \geq 10$

Hence, the correct answer is option: 'C'.

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