Question

If the value of determinant $\Delta =\left|\begin{array}{ccc}2+x^2& x& x^2\\ 2+x& \sqrt{x}& x\\ 2+x^4& x^2& x^4\end{array}\right|$ is $6$. Then which of the following statement is correct?

• A. $(\sqrt{x}-1)(x-1)(x\sqrt{x}-1)+3=0$
• B. $x(\sqrt{x}-1)(x-1)(x\sqrt{x}-1)-3=0$
• C. $x^2(\sqrt{x}-1)(x-1)(x\sqrt{x}-1)+3=0$
• D. $x^3(\sqrt{x}-1)(x-1)(x\sqrt{x}-1)-3=0$

Answer 1

The correct option is C $x^2(\sqrt{x}-1)(x-1)(x\sqrt{x}-1)+3=0$

Applying $C_1\to C_1-C_3,$
$\Delta =\left|\begin{array}{ccc}2& x& x^2\\ 2& \sqrt{x}& x\\ 2& x^2& x^4\end{array}\right|$
Taking $\left(2\right)$ common from $C_1,$ we get
$\Delta =\left(2\right)\left|\begin{array}{ccc}1& x& x^2\\ 1& \sqrt{x}& x\\ 1& x^2& x^4\end{array}\right|$
$\Rightarrow 2(x-\sqrt{x})(\sqrt{x}-x^2)(x^2-x)=6\because \left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=(a-b)(b-c)(c-a)\Rightarrow x^2(\sqrt{x}-1)(x-1)(x\sqrt{x}-1)+3=0$