Question

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

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

Answer 1

The correct option is B $(1,\pm 3)$

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

Given, $t\left(0\right)=4$ (continuous at $x=2$ )

Applying $L,HospitalRULE$

$\left(\frac{-a\sin x+b\sin bx}{2x}\right)$

$-a\sin x+b\sin bx=0$ (at $x=0$)

Again, $L^{\prime }HOSPITALRULE$

$\frac{-a\cos x+b^2\cos bx}{2}=4(atx=0)$

$-a+b^2=8$

In equation $\left(i\right)$, numerator $=0$

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

$a-1=0$

$a=1$

$b=\pm 3$

$(1+,\pm 3)$