Question

The function $f\left(x\right)=\{\begin{array}{c}\frac{\cos 3x-\cos 4x}{x^2},forx\ne 0\\ \frac{7}{2},forx=0\end{array}$ at $x=0$ is

• A. right continuous only
• B. discontinuous
• C. left continuous only
• D. continuous

Answer 1

The correct option is A right continuous only

Now,

Checking for continuity,

$\begin{array}{c}{lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{\cos 3x-{\cos }^{2}x}{x^2}\\ ={lim}_{x\to 0}\frac{1-{\frac{\left(3x\right)}{2!}}^2+{\frac{\left(3x\right)}{4!}}^4.........\left(1-\frac{{\left(4x\right)}^2}{2!}+\frac{{\left(4x\right)}^4}{4!}....\infty \right)}{x^2}\\ ={lim}_{x\to 0}\frac{\left(\frac{-9}{2!}{x}^2....+\frac{16x^2}{2!}......\right)}{x^2}\\ Puttingonthelimit=\frac{16-9}{2!}=\frac{7}{2}\\ Checkingleftcontinuity;\\ ={lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{\cos 3x-{\cos }^{2x}}{x^2}\\ ={lim}_{x\to 0}\frac{1-{\frac{\left(3x\right)}{2!}}^2+{\frac{\left(3x\right)}{4!}}^4.........\left(1-\frac{{\left(4x\right)}^2}{2!}+\frac{{\left(4x\right)}^4}{4!}......\infty \right)}{x^2}\\ =\frac{7}{2}\\ Hence,f\left(x\right)iscontinuousatx=0\end{array}$