Question

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

• A. $A=-4$
• B. $B=5$
• C. $f\left(0\right)=1$
• D. $A=-3$

Answer 1

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

The correct options are
A $A=-4$
B $B=5$
C $f\left(0\right)=1$

We have $\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}$

$=\underset{x\to 0}{lim}\frac{3\cos 3x+2A\cos 2x+B\cos x}{{5x}^4}$
For a finite limit to exists the numerator must be $0$ at $x=0$
Since the denominator is $0$ at $x=0$
i.e $3+2A+B=0$ ...(1)
Now we have,
$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{2\cos 3x+2A\cos 2x+B\cos x}{{5x}^4}$
$=\underset{x\to 0}{lim}\frac{-9\sin 3x-4A\sin 2x-B\sin x}{{20x}^3}$

$=\underset{x\to 0}{lim}\frac{-27\cos 3x-8A\cos 2x-6\cos x}{60x^2}$
For a finite limit to exists the numerator must be $0$ at $x=0$
Since the denominator is $0$ at $x=0$
i.e $27+8A+B=0$ ...(2)
Now we have,
$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{81\sin 3x+16A\sin 2x+B\sin x}{120x}$

$=\underset{x\to 0}{lim}(\frac{81}{40}\left(\frac{\sin 3x}{3x}\right)+\left(\frac{\sin 2x}{2x}\right)+\frac{B}{120}\left(\frac{\sin x}{x}\right))$

$=\frac{81}{40}+\frac{16A}{60}+\frac{B}{120}$
From (1) and (2)
$A=-4,B=5$
For $f$ to be continuous at $x=0$ we have
$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)=\frac{81}{40}-\frac{64}{60}+\frac{5}{120}=1$