Question

The equation $\begin{vmatrix}(1+x)^2 & (1-x)^2 & -\left(2+x^2\right) \\ 2 x+1 & 3 x & 1-5 x \\ x+1 & 2 x & 2-3 x\end{vmatrix}+\begin{vmatrix}(1+x)^2 & 2 x+1 & x+1 \\ (1-x)^2 & 3 x & 2 x \\ 1-2 x & 3 x-2 & 2 x-3\end{vmatrix}=0$

• A. has no real solution
• B. has 4 real solutions
• C. has two real and two non-real solutions
• D. has infinite number of solutions, real or non-real

Answer 1

Answer: (D)

$1^{\text {st }}$ two columns of $1^{\text {st }}$ determinant are same as $1^{\text {st }}$ two rows of $2^{\text {nd }}$.
Hence transpose the $2^{\text {nd }}$. Add the two determinants and use $C _1 \rightarrow C _1+ C _3 \Rightarrow D =0$