Question

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

• A. $(-2,-4)$
• B. $(-2,-1)$
• C. $(-1,\sqrt{3})$
• D. $(-2,-\sqrt{3})$

Answer 1

The correct option is A $(-2,-4)$

As denominator $\to 0$ as $x\to 0$ in $\frac{a+2\cos 2x}{x^2}$, Numerator should also $\to 0$ when $x\to 0$

$\Rightarrow a+2=0$

$\Rightarrow a=-2$

$\Rightarrow \underset{x\to 0}{lim}\frac{-2+2\cos 2x}{x^2}=\frac{-2(1-\cos 2x)}{x^2}$

$=\underset{x\to 0}{lim}\frac{-2\left(2{\sin }^{2}x\right)}{x^2}=-4\underset{x\to 0}{lim}{\left(\frac{\sin x}{x}\right)}^2$

$=-4$

R.H.L.

When $x\to 0^{+}$, $[x+4]\to 4$

$\Rightarrow b\tan \left(\frac{\pi }{[x+4]}\right)=b\tan \frac{\pi }{4}=b$

For f to be continuous at $x=0$, $b=-4$.