Question

If $f\left(x\right)=\frac{\sin 3x+A\sin 2x+B\sin x}{x^5},x\ne 0$ then the value of $\left(f\right(0)+A+B)$ equal to $\left(Consideringf\right(x)tobecontinuousatx=0)$

• A. $0$
• B. $2$
• C. $-2$
• D. None of these

Answer 1

The correct option is B $2$

$\because f\left(x\right)iscontinuousatx=0$

$\underset{x\to 0}{lim}=f\left(0\right)$

$considerL=\underset{x\to 0}{lim}\left[\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}\right]$

using L - H rule

$L=\underset{x\rightarrow 0}{\lim}\left [\dfrac{3\cos 3x+2A\cos 2x+B\cos x}{5x^4} \right ]

$

for existence of limit

$0=3+2A+B....\left(i\right)$

further from L - H rule

$L=\underset{x\to 0}{lim}\left[\frac{9\sin 3x+4A\sin 2x+B\sin x}{20x^3}\right]$

$=\underset{x\to 0}{lim}\left[\frac{27\cos 3x+8A\cos 2x+B\cos x}{60x^2}\right]$

Now, 8A+B+27=O ....( ii)

$\Rightarrow A=-4,B=5$

$L=\underset{x\to 0}{lim}\left[\frac{81\sin 3x+16A\sin 2x+B\sin x}{120x}\right]$

$=1=f\left(0\right)$

hence $f\left(0\right)+A+B=2$