Question

The values of x where the function $f\left(x\right)=\frac{tanxlog(x-2)}{x^2-4x+3}$ is discontinuous are given by:

• A. $(-\infty ,2]\cup \left\{3\right\}$
• B. $(-\infty ,3)$
• C. $(-\infty ,2]\cup \{3,n\pi +\frac{\pi }{2}:n\ge 1\}$
• D. R-{0,1,3}

Answer 1

The correct option is C $(-\infty ,2]\cup \{3,n\pi +\frac{\pi }{2}:n\ge 1\}$

$f\left(x\right)=\frac{tanxlog(x-2)}{x^2-4x+3}$, being product and quotient of functions tan x,
log(x-2) and polynomial $(x^2-4x+3)$ must be continuous in its domain of definition. Tan x is discontinuous in $\left\{\right(2n+1)\frac{\pi }{2}:nϵZ\}$,
log(x-2) is discontinuous for $x\le 2$ and $x^2-4x+3=0$
for x = 1 and 3.
Hence f(x) is discontinuous in $(-\infty ,2]\cup \{3,n\pi +\frac{\pi }{2}:nϵZ\}$
If $n\le 0,n\pi +\frac{\pi }{2}<2$ so the set of points of discontinuities of f(x) can be defined as $(-\infty ,2]\cup \{3,n\pi +\frac{\pi }{2}:n\ge 1\}$