Q66 / 125 · BITSAT 2013 · Mathematics
The determinant $\left|\begin{array}{ccc}1 & (x-3) & (x-3)^{2} \\ 1 & (x-4) & (x-4)^{2} \\ 1 & (x-5) & (x-5)^{2}\end{array}\right|$
- A3 values of $x$
- B2 values of $x$
- C1 values of $x$
- Correct answerDNo value of $x$
Solution
The given determinant vanishes, i.e.,
$$
\left|\begin{array}{ccc}
1 & x-3 & (x-3)^{2} \\
1 & x-4 & (x-4)^{2} \\
1 & x-5 & (x-5)^{2}
\end{array}\right|=0
$$
Expanding along $\mathrm{C}_{1},$ we get
$$
\begin{array}{l}
(x-4)(x-5)^{2}-(x-5)(x-4)^{2}-\left\{(x-3)(x-5)^{2}\right. \\
\left.-(x-5)(x-3)^{2}\right\}+(x-3)(x-4)^{2} \\
-(x-4)(x-3)^{2}=0 \\
\Rightarrow(x-4)(x-5)(x-5-x+4) \\
-(x-3)(x-5)(x-5-x+3) \\
+(x-3)(x-4)(x-4-x+3)=0
\end{array}
$$
$$
\Rightarrow-(x-4)(x-5)+2(x-3)(x-5)-(x-3)
$$
$(x-4)=0$
$$
\Rightarrow-x^{2}+9 x-20+2 x^{2}-16 x+30-x^{2}+7 x-12=0
$$
$\Rightarrow-32+30=0 \Rightarrow-2=0$
Which is not possible, hence no value of $\mathrm{x}$ satisfies the given condition.
Hence, the correct answer is option: 'D'.