Question

If $f^{\prime }\left(x\right)=\left|\begin{array}{ccc}mx& mx-p& mx+p\\ n& n+p& n-p\\ mx+2n& mx+2n+p& mx+2n-p\end{array}\right|$, then $y=f\left(x\right)$ represents

• A. a straight line parallel to x-axis
• B. a straight line parallel to y-axis
• C. a parabola
• D. a straight line with negative slope

Answer 1

The correct option is A a straight line parallel to x-axis

Let $y=f\left(x\right)=a_0+a_{1}x+a_2{x}^2+....$
Given, $f^{\prime }\left(x\right)=\left|\begin{array}{ccc}mx& mx-p& mx+p\\ n& n+p& n-p\\ mx+2n& mx+2n+p& mx+2n-p\end{array}\right|$
$\Rightarrow f\left(x\right)=\left|\begin{array}{ccc}\frac{m{x}^2}{2}& \frac{m{x}^2}{2}-px& \frac{m{x}^2}{2}+px\\ nx& nx+px& nx-px\\ \frac{m{x}^2}{2}+2nx& \frac{m{x}^2}{2}+2nx+px& \frac{m{x}^2}{2}+2nx-px\end{array}\right|$
$C_2\to C_2+C_3$
$a_0+a_{1}x+a_2{x}^2+....=\left|\begin{array}{ccc}\frac{m{x}^2}{2}& 2\frac{m{x}^2}{2}& \frac{m{x}^2}{2}+px\\ nx& 2nx& nx-px\\ \frac{m{x}^2}{2}+2nx& 2(\frac{m{x}^2}{2}+2nx)& \frac{m{x}^2}{2}+2nx-px\end{array}\right|$
$\Rightarrow a_0=0$
Substitute $x=0$ in $f^{\prime }\left(x\right)$
$a_1=\left|\begin{array}{ccc}0& -p& p\\ n& n+p& n-p\\ 2n& 2n+p& 2n-p\end{array}\right|$
$\Rightarrow a_1=0$
$\Rightarrow y=f\left(x\right)=0$
Hence,a straight line parallel to x-axis