Question

If $f\left(x\right)=\{\begin{array}{c}\frac{x(1+a\cos x)-b\sin x}{x^3},x\ne 0\\ 1,x=0\end{array}$ then $f$ is continuous for values of $a$ and $b$ given by-

• A. $\frac{3}{2},\frac{3}{2}$
• B. $\frac{3}{2},\frac{-3}{2}$
• C. $\frac{-5}{2},\frac{-3}{2}$
• D. $\frac{-3}{2},\frac{3}{2}$

Answer 1

Answer: (C)

$\frac{-5}{2},\frac{-3}{2}$

${lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}$

$=\frac{0(1+a)-b\left(0\right)}{0}\frac{0}{0}$form

Using L hospital

${lim}_{x\to 0}\frac{x\left(a(-\sin x)\right)+(a-b)-b\cos x}{3x^2}$

$=\frac{1-0+(a-b)}{0}$

But given it is continuous

$\Rightarrow 1+a-b=0\to 1$

Now $\frac{0}{0}$ form

Using L Hospital

$={lim}_{x\to 0}\frac{0-a(x\cos x+\sin x)+(a-b)(-\sin x)}{6x}$

$={lim}_{x\to 0}(-\frac{a\cos x}{6}-\frac{a\sin x}{6x}+\frac{(b-a)}{6}\cdot \frac{\sin x}{x})$

$=\frac{-a}{6}-\frac{a}{6}+\frac{(b-a)}{6}=\frac{b-3a}{6}$

${lim}_{x\to 0}f\left(x\right)=f\left(0\right)=1$ [continuous at $0$]

$\Rightarrow \frac{b-3a}{6}=1$

$b-3a=6\to 2$

From $1$ and $2$

$1-2a=6$

$a=-\frac{5}{2},b=-\frac{3}{2}$