JEE PYQ: Sequence And Series Question 4
Question 4 - 2021 (17 Mar Shift 2)
If $1$, $\log_{10}(4^x - 2)$ and $\log_{10}\left(4^x + \frac{18}{5}\right)$ are in arithmetic progression for a real number $x$, then the value of the determinant $\begin{vmatrix} 2\left(x - \frac{1}{2}\right) & x - 1 & x^2 \ 1 & 0 & x \ x & 1 & 0 \end{vmatrix}$ is equal to:
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Answer: (2)
Solution
$2\log_{10}(4^x - 2) = 1 + \log_{10}\left(4^x + \frac{18}{5}\right)$
$(4^x - 2)^2 = 10\left(4^x + \frac{18}{5}\right)$
$(4^x)^2 + 4 - 4(4^x) - 32 = 0$
$(4^x - 16)(4^x + 2) = 0$
$4^x = 16$
$x = 2$
$\begin{vmatrix} 3 & 1 & 4 \ 1 & 0 & 2 \ 2 & 1 & 0 \end{vmatrix} = 3(-2) - 1(0-4) + 4(1) = -6 + 4 + 4 = 2$
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