Question

If $f\left(x\right)=\frac{\sin 3x+A\sin 2x+B\sin x}{x^5};x\ne 0$ is continuous at $x=0$ , then

• A. $A+B=2$
• B. $A+B=1$
• C. $A+B=0$
• D. $A-B=1$

Answer 1

Answer: (B)

$A+B=1$

$f\left(x\right)=\frac{sin3x+Asin2x+Bsinx}{x^5};x\ne 0$
$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{sin3x+Asin2x+Bsinx}{x^5}$
It is of the form $\frac{0}{0}$, so applying L-Hospital's rule
$=\underset{x\to 0}{lim}\frac{3\cos 3x+2A\cos 2x+B\cos x}{5x^4}$
As $x\to 0,D^r\to 0\Rightarrow N^r\to 0$
$\underset{x\to 0}{lim}3\cos 3x+2A\cos 2x+B\cos x=0$
$\Rightarrow 3+2A+B=0$ .....(i)
Again $\underset{x\to 0}{lim}\frac{3\cos 3x+2A\cos 2x+B\cos x}{5x^4}$ is of the form $\frac{0}{0}$
$=\underset{x\to 0}{lim}\frac{-9\sin 3x-4A\sin 2x-B\sin x}{20x^3}$
Again of the form $\frac{0}{0}$
$=\underset{x\to 0}{lim}\frac{-27\cos 3x-8A\cos 2x-B\cos x}{60x^2}$
As $x\to 0,D^r\to 0\Rightarrow N^r\to 0$
$\underset{x\to 0}{lim}-27\cos 3x-8A\cos 2x-B\cos x=0$
$\Rightarrow 27+8A+B=0$ .....(ii)
Solving (i) and (ii), we get
$A=-4,B=5$
Thus, $A+B=1$