Question

The determinant $\begin{vmatrix}1&\left(x-3\right)&\left(x-3\right)^{2}\\ 1&\left(x-4\right)&\left(x-4\right)^{2}\\ 1&\left(x-5\right)&\left(x-5\right)^{2}\end{vmatrix}$ vanishes for

• A. 3 values of x
• B. 2 values of x
• C. 1 values of x
• D. No value of x

Answer 1

Answer: (D)

The given determinant vanishes, i.e.,
$\begin{vmatrix}1&\left(x-3\right)&\left(x-3\right)^{2}\\ 1&\left(x-4\right)&\left(x-4\right)^{2}\\ 1&\left(x-5\right)&\left(x-5\right)^{2}\end{vmatrix} = 0$
Expanding along $C_1$, we get
$(x-4)(x-5)^{2}-(x-5)(x-4)^{2}$
$-\left\{(x-3)(x-5)^{2}-(x-5)(x-3)^{2}\right\}$
$+(x-3)(x-4)^{2}-(x-4)(x-3)^{2}=0$
$\Rightarrow (x-4)(x-5)(x-5-x+4)$
$-(x-3)(x-5)(x-5-x+3)+(x-3)(x-4)(x-4-x+3)=0$
$\Rightarrow -(x-4)(x-5)+2(x-3)(x-5)-(x-3)(x-4)$
$=0$
$\Rightarrow -x 2+9 x-20+2 x 2-16 x+30-x 2+7 x-12$
$=0$
$\Rightarrow -32+30=0$
$\Rightarrow -2=0$
Which is not possible, hence no value of $x$ satisfies the given condition.