Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{A+3\cos x}{x^2},& ifx<0\\ B\tan \left(\frac{\pi }{[x+3]}\right),& ifx\ge 0\end{array}$ Where $[.]$ represents the greatest integer function, is continuous at $x=0$ Then.

• A. $A=-3,B=-\sqrt{3}$
• B. $A=3,B=-\frac{\sqrt{3}}{2}$
• C. $A=-3,B=-\frac{\sqrt{3}}{2}$
• D. $A=-\frac{\sqrt{3}}{2},B=-3$

Answer 1

The correct option is C $A=-3,B=-\frac{\sqrt{3}}{2}$

For given function $f\left(x\right)$ to be continuous at $x=0$

Condition 1-${lim}_{x\to 0^{-}}f\left(x\right)and{lim}_{x\to 0^{+}}f\left(x\right)$ must be finite and

Condition 2-${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{+}}f\left(x\right)$

For Condition 1 to satisfy

${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{-}}\frac{A+3\cos x}{x^2}$ must be finite

We can see that its denominator will tends to 0 as x tends to 0

So, in order for this to exist, the numerator must also tend to 0 as x tends to 0

$\Rightarrow {lim}_{x\to 0^{-}}(A+3\cos x)=0$

$\Rightarrow A=-3$

Now applying L'Hospital's formula to evaluate ${lim}_{x\to 0^{-}}f\left(x\right)[\because \frac{0}{0}form]$

$\Rightarrow {lim}_{x\to 0^{-}}\frac{-3\sin x}{2x}$

$\Rightarrow {lim}_{x\to 0^{-}}\frac{-3\cos x}{2}$

$\Rightarrow \frac{-3}{2}$

Now, for condition 2 to satisfy

$\Rightarrow {lim}_{x\to 0^{+}}f\left(x\right)=\frac{-3}{2}$

$\Rightarrow B\tan \left(\frac{\pi }{3}\right)=\frac{-3}{2}$

$\Rightarrow B=\frac{-\sqrt{3}}{2}$

Hence, answer is option $C$