Question

If $f\left(x\right)=\frac{a\cos x-\cos bx}{x^2},x\ne 0$ and $f\left(0\right)=4$ continuous at $x=0$, then the ordered pair $(a,b)$ is

• A. $(\ne 1,3)$
• B. $(1,\ne 3)$
• C. $(-1,-3)$
• D. $(1,\pm 3)$

Answer 1

The correct option is A $(\ne 1,3)$

Solution:-

$f\left(x\right)=\frac{a\cos x-\cos bx}{x^2},x\ne 0$

$f\left(0\right)=4$

Given that the function is continuous at $x=0$.

$\therefore {lim}_{x\to 0}f\left(x\right)=f\left(0\right)=4$

The function does not tend to infinity as $x\to 0$ even though the denominator tends to zero.

$\therefore$ at $x=0$,

$a\cos x-\cos bx=0$

$\Rightarrow a-1=0[\because \cos \left(0\right)=1]$

$\Rightarrow a=1$

Applying L'Hopital's rule, we get

${lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{a\cos x-\cos bx}{x^2}={lim}_{x\to 0}\frac{-a\sin ax+b\sin bx}{2x}={lim}_{x\to 0}\frac{-a\cos x+b^2\cos bx}{2}=4$

$\Rightarrow \frac{-a+b^2}{2}=4$

$\Rightarrow -1+b^2=8[\because a=1]$

$\Rightarrow b^2=9$

$\Rightarrow b=\pm 3$

Hence the ordered pair $(a,b)=(1,\pm 3)$.

According to the given option, $\left(D\right)$ will be the required answer.