Question

Let $g\left(x\right)=\left|\begin{array}{ccc}f(x+c)& f(x+2c)& f(x+3c)\\ f\left(c\right)& f\left(2c\right)& f\left(3c\right)\\ f^{\prime }\left(c\right)& f^{\prime }\left(2c\right)& f^{\prime }\left(3c\right)\end{array}\right|$, where $c$ is constant, then find $\underset{x\to 0}{lim}\frac{g\left(x\right)}{x}$.

• A. $0$
• B. $1$
• C. $-1$
• D. $\frac{-1}{2}$

Answer 1

The correct option is A $0$

$g\left(x\right)=\left|\begin{array}{ccc}f(x+c)& f(x+2c)& f(x+3c)\\ f\left(c\right)& f\left(2c\right)& f\left(3c\right)\\ f^{\prime }\left(c\right)& f^{\prime }\left(2c\right)& f^{\prime }\left(3c\right)\end{array}\right|$
$\therefore g\left(0\right)=0$
Since first and second rows become identical.
$\therefore \underset{x\to 0}{lim}\frac{g\left(x\right)}{x};\left(\frac{0}{0}\right)$ form
$=\underset{x\to 0}{lim}\frac{g^{\prime }\left(x\right)}{1}$ (using L'Hospital's rule)
$=g^{\prime }\left(0\right)$
Now, $g^{\prime }\left(x\right)=\left|\begin{array}{ccc}{f}^{\prime }(x+c)& f^{\prime }(x+2c)& f^{\prime }(x+3c)\\ f\left(c\right)& f\left(2c\right)& f\left(3c\right)\\ f^{\prime }\left(c\right)& f^{\prime }\left(2c\right)& f^{\prime }\left(3c\right)\end{array}\right|$
$\therefore g^{\prime }\left(0\right)=\left|\begin{array}{ccc}{f}^{\prime }\left(c\right)& f^{\prime }\left(2c\right)& f^{\prime }\left(3c\right)\\ f\left(c\right)& f\left(2c\right)& f\left(3c\right)\\ f^{\prime }\left(c\right)& f^{\prime }\left(2c\right)& f^{\prime }\left(3c\right)\end{array}\right|=0$
Since first and third rows become identical.
$\therefore \underset{x\to 0}{lim}\frac{g\left(x\right)}{x}=0$