Question

If $f(x)$ = $\left|\begin{array}{ccc}x+1& x& 1\\ x\left(x+1\right)& x\left(x-1\right)& 2x\\ x\left(x+1\right)\left(x-1\right)& x\left(x-1\right)\left(x-2\right)& 3x\left(x-1\right)\end{array}\right|$ ,
then $f(1000)$ is equal to

• A. 1
• B. 1000
• C. -1000
• D. 0

Answer 1

Answer: (D)

Given,
$f(x)=\left|\begin{array}{ccc}x+1& x& 1\\ x(x+1)& x(x-1)& 2x\\ x(x+1)(x-1)& x(x-1)(x-2)& 3x(x-1)\end{array}\right|$
Taking $x$ common from $R_2$ and $x(x-1)$ common from $R_3$,
$=x\times x(x-1)\left|\begin{array}{ccc}x+1& x& 1\\ x+1& x-1& 2\\ x+1& x-2& 3<br/>\end{array}\right|$
$=x^2(x-1)(x+1)\left|\begin{array}{ccc}1& x& 1\\ 1& x-1& 2\\ 1& x-2& 3\end{array}\right|$
Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$,
$=x^2(x-1)(x+1)\left|\begin{array}{ccc}1& x& 1\\ 0& -1& 1\\ 0& -2& 2\end{array}\right|$
$\Rightarrow =x^2\left(x^2-1\right)[1(-2+2)]=0$
$\therefore f(x)=0$
$\therefore f(1000)=0$