JEE Main 2021 (Online) 17th March Evening ShiftQuestion 1
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 $\begin{vmatrix}2\left(x-\frac{1}{2}\right)&x-1&x^{2}\\1&0&x\\x&1&0\end{vmatrix}$ is equal to $\_\_\_\_\_\_$ .
Determinant
Answer$2$
Solution
AP: $2\log_{10}(4^x-2)=1+\log_{10}(4^x+\tfrac{18}{5})$, so with $u=4^x$: $(u-2)^2=10u+36\Rightarrow u^2-14u-32=0\Rightarrow u=16$ ($u=-2$ rejected). So $x=2$. Determinant $=\begin{vmatrix}3&1&4\\1&0&2\\2&1&0\end{vmatrix}=3(0-2)-1(0-4)+4(1-0)=-6+4+4=2$.