Question

Let $f\left(x\right)=\frac{x(1+a\cos x)-b\sin x}{x^3},x\ne 0$ and $f\left(0\right)=1$. The value of $a$ and $b$ so that $f$ is a continuous function are-

• A. $\frac{5}{2},\frac{3}{2}$
• B. $\frac{5}{2},-\frac{3}{2}$
• C. $-\frac{5}{2},-\frac{3}{2}$
• D. None of these

Answer 1

The correct option is C $-\frac{5}{2},-\frac{3}{2}$

$f\left(x\right)=\underset{x\to 0}{lim}\frac{x(1+a(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}......))-b(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!})}{x^3}$
$\underset{x\to 0}{lim}\frac{(1+a-b)+x^2(\frac{-a}{2!}+\frac{b}{3!})}{x^2}$
$\Rightarrow 1+a-b=0.....\left(i\right)$
and $\frac{-a}{2}+\frac{b}{6}=1.....\left(ii\right)$
solving $\left(i\right)$ and $\left(ii\right)$ we get
$a=\frac{-5}{2};b=\frac{-3}{2}$