Question

If [x] denotes the greatest integer not exceeding $x$ and if the function $f$ defined by $f\left(x\right)=\{\begin{array}{cc}\frac{a+2\cos x}{x^2}& (x<0)\\ b\tan \frac{\pi }{[x+4]}& (x\ge 0)\end{array}$ is continuous at $x=0$, then the order pair (a, b) =

• A. $(-2,1)$
• B. $(-2,-1)$
• C. $(-1,\sqrt{3})$
• D. $(-2,-\sqrt{3})$

Answer 1

The correct option is B $(-2,-1)$

Since the function is continuous at $x=0,f\left(0^{-}\right)=f\left(0^{+}\right)$

$\Rightarrow \underset{x\to 0^{-}}{lim}\frac{a+2\cos x}{x^2}=\underset{x\to 0^{+}}{lim}b\tan \frac{\pi }{[x+4]}$

For a finite limit, $a$ has to be $-2$ since the numerator has to be

$0$ for the limit to be of the form $\frac{0}{0}$

Using L'Hospital rule, we get $\underset{x\to 0^{-}}{lim}\frac{-2\sin x}{2x}=-1$

$\Rightarrow b\tan \frac{\pi }{4}=-1$

$\Rightarrow b=-1$