Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{\sin 3x+a\sin 2x+b\sin x}{x^5}& x\ne 0\\ c,& x=0\end{array}$ is continuous at $x=0$, find the values of $a,b,c$.

• A. $a=-4$
• B. $b=5$
• C. $c=1$
• D. $c=-4$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $a=-4$
B $b=5$
C $c=1$

If the function is continuous at x=0,then $\underset{x\to 0}{lim}f\left(x\right)=value=c$
Using expansion $\underset{x\to 0}{lim}f\left(x\right)$ $=\underset{x\to 0}{lim}\frac{x(3+2a+b)+x^3(-\frac{9}{2}-\frac{4a}{3}-\frac{b}{6})+\frac{x^5}{5!}(3^5+2^{5}a+b)}{x^5}$

Above limit will be equal to value c if $3+2a+b=0,-\frac{9}{2}-\frac{4a}{3}-\frac{b}{6}=0$ and $\frac{1}{120}(243+32a+b)=c$

The first two relations give $a=-4,b=5.$ and hence from third $c=\frac{1}{120}(243-128+5)=\frac{120}{120}=1$
Ans: A,B,C