Question

Let the functions f and g be defined by $f\left(x\right)=(x-3)$ and $g\left(x\right)={\int }_{k,whenx=-3}^{\frac{x^2-9}{x+3},whenx\ne -3}$ find the value of k such that f(x) = g(x) for all $x\in R$

Answer 1

$f\left(x\right)=(x-3)$

$g\left(x\right)=\{\begin{array}{cc}\int \frac{x^2-9}{x+3}d,& whenx\ne -3\\ k& whenx=3\end{array}$

$\Rightarrow g\left(x\right)=\{\begin{array}{cc}\int \frac{(x+3)(x-3)}{(x+3)}d,& whenx\ne -3\\ k& whenx=3\end{array}$

$=\{\begin{array}{cc}\frac{x^2}{2}+3x& whenx\ne -3\\ k& whenx=-3\end{array}$

If $f\left(x\right)=g\left(x\right)$

when $x=-3$

$x-3=\frac{x^2}{2}+3x$

$\frac{x^2}{2}+2x+3=0$

$x^2+4x+6=0$

$(x+2{)}^2+2=0$

not possible

So, $f\left(x\right)=g\left(x\right)$ only when $x=-3$

$\Rightarrow (x-3)=k$

$\Rightarrow -3-3=k$

$\therefore k=-6$