Question

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

• A. 10
• B. 4
• C. 8
• D. 16

Answer 1

The Correct Option is A

Given that the terms \(4^{1+x} + 4^{1-x}\), \(\frac{K}{2}\), and \(16^x + 16^{-x}\) are three consecutive terms of an arithmetic progression, we can set up the condition for an arithmetic progression:

\[ 2 \times \left(\frac{K}{2}\right) = \left(4^{1+x} + 4^{1-x}\right) + \left(16^x + 16^{-x}\right). \]

Simplifying:

\[ K = 4^{1+x} + 4^{1-x} + 16^x + 16^{-x}. \]

Step 1: Express Each Term in Simplified Form

Recall that:

\[ 4^{1+x} = 4 \cdot 4^x, \quad 4^{1-x} = 4 \cdot 4^{-x}, \quad 16^x = (4^x)^2, \quad 16^{-x} = (4^{-x})^2. \]

Thus:

\[ 4^{1+x} + 4^{1-x} = 4 \left(4^x + 4^{-x}\right), \quad 16^x + 16^{-x} = (4^x)^2 + (4^{-x})^2. \]

So:

\[ K = 4 \left(4^x + 4^{-x}\right) + \left((4^x)^2 + (4^{-x})^2\right). \]

Step 2: Substitute and Simplify

Let \(t = 4^x + 4^{-x}\). Then:

\[ (4^x)^2 + (4^{-x})^2 = t^2 - 2 \quad \text{(by the identity } (a+b)^2 = a^2 + b^2 + 2ab\text{)}. \]

So:

\[ K = 4t + (t^2 - 2). \]

Step 3: Minimize \(K\)

To find the least value of \(K\), we need to minimize \(t\) subject to \(t \geq 2\) (since \(4^x + 4^{-x} \geq 2\) for \(x \geq 0\)):

\[ K = 4t + t^2 - 2. \]

The minimum value of \(t\) is 2, so substituting \(t = 2\):

\[ K = 4 \cdot 2 + 2^2 - 2 = 8 + 4 - 2 = 10. \]

Therefore, the least value of \(K\) is 10.