Question

If $y=\frac{1}{t^2-t-6}$ and $t=\frac{1}{x-2}$, then the values of $x$ for which make the function y discontinuous, are:

• A. $2,\frac{2}{3},\frac{7}{3}$
• B. $2,\frac{3}{2},\frac{7}{3}$
• C. $2,\frac{3}{2},\frac{3}{7}$
• D. None of the above

Answer 1

The correct option is B $2,\frac{3}{2},\frac{7}{3}$

Clearly, the function $t=\frac{1}{x-2}$ is discontinuous at x = 2 and the function $y=\frac{1}{t^2-t-6}$ discontinuous at the points, where

$t^2-t-6=0$

$\Rightarrow (t+2)(t-3)=0$

$\Rightarrow t=-2,3$

But $t=-2$

$\Rightarrow \frac{1}{x-2}=-2$

$\Rightarrow x-2=\frac{1}{2}$

$\Rightarrow x=\frac{3}{2}$

and $t=3$

$\Rightarrow \frac{1}{x-2}=3$

$\Rightarrow x-2=\frac{1}{3}$

$\Rightarrow x=\frac{7}{3}$

Therefore, the value of x which make the function y discontinuous are $x=2,\frac{3}{2}$ and $\frac{7}{3}$.