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VITEEE 2025 Paper

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Question : 94 of 125

Marks: +1, -0

5^{1+x}+5^{1-x},\; \frac{a}{2},\; 5^{2x}+5^{-2x}

are in A.P., then the value of a is:

$a < 12$
$a \leq 12$
a $\geq 42$
$a \geq \frac{17}{4}$

Solution:

Given :

5^{1+x}+5^{1-x},\; \frac{a}{2},\; 5^{2x}+5^{-2x}

are in A.P.

\; \therefore \;\; 2 \cdot \frac{a}{2}=5^{1+x}+5^{1-x}+5^{2x}+5^{-2x}

\; \Rightarrow a=5 \cdot 5^{x}+5 (5^{x})^{-1}+ (5^{x})^{2}+ (5^{x})^{-2}

\; \text{ Let }\; 5^{x}=t \therefore \;\; a=5 t+\frac{5}{t}+t^{2}+\frac{1}{t^{2}}

\; \Rightarrow a=t^{2}+\frac{1}{t^{2}}+5\left(t+\frac{1}{t}\right)

\; \Rightarrow a=\left(t+\frac{1}{t}\right)^{2}-2+5\left(t+\frac{1}{t}\right)

\; \text{ Put }\; t+\frac{1}{t}=A

\; \therefore a=A^{2}+5A-2 \;\; \left[\; \text{ add \& subtract }\; \left(\frac{b}{2a}\right)^{2}\right]

\; \Rightarrow a= \left[A^{2}+5A-\left(\frac{5}{2}\right)^{2}\right] + \left(\frac{5}{2}\right)^{2} - 2

\; \Rightarrow a= \left(A-\frac{5}{2}\right)^{2}+\frac{17}{4} \Rightarrow a \geq \frac{17}{4}

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