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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
Question
If $1, \log _{10}\left(4^{x}-2\right)$ 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 $\left|\begin{array}{ccc}2\left(x-\frac{1}{2}\right) & x-1 & x^{2} \\ 1 & 0 & x \\ x & 1 & 0\end{array}\right|$ is equal to ...... .
✓

Answer

d
$2 \log _{10}\left(4^{ x }-2\right)=1+\log _{10}\left(4^{ x }+\frac{18}{5}\right)$

$\left(4^{ x }-2\right)^{2}=10\left(4^{ x }+\frac{18}{5}\right)$

$\left(4^{ x }\right)^{2}+4-4\left(4^{ x }\right)-32=0$

$\left(4^{ x }-16\right)\left(4^{ x }+2\right)=0$

$4^{ x }=16$

$x =2$

$\left|\begin{array}{lll}3 & 1 & 4 \\ 1 & 0 & 2 \\ 2 & 1 & 0\end{array}\right|=3(-2)-1(0-4)+4(1)$

$=-6+4+4=2$

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