Question

If $\underset{x\to 1}{lim}\frac{a\sin (x-1)+b\cos (x-1)+4}{x^2-1}=-2$, then $|a+b|$ is equal to

• A. $8$
• B. $4$
• C. $2$
• D. None of these

Answer 1

Answer: (A)

$8$

$\underset{x\to 1}{lim}\frac{a\sin (x-1)+b\cos (x-1)+4}{x^2-1}=-2$

$\Rightarrow a\times \sin (1-1)+b\cos (1-1)+4=0$

$\Rightarrow 0+b+4=0$

$\Rightarrow b=-4$

$\underset{x\to 1}{lim}\frac{a\sin (x-1)-4\cos (x-1)+4}{(x+1)(x-1)}=-2$

$x=1+4$

$x\to 1,4\to 0$

$\underset{4\to 0}{lim}\frac{a\sin (1+4-1)-\left(4\cos \right(1+4-1)+4}{(1+4+1)(1+4-1)}$

$\underset{4\to 0}{lim}\frac{a\sin 4-4\cos 4+4}{(2+4)\times \left(4\right)}=-2$

$\underset{4\to 0}{lim}\frac{a\sin 4-4\cos 4+4}{(2+4)\times \left(4\right)}=-2$

$\underset{x\to 0}{lim}\frac{a\left(\frac{\sin 4}{4}\right)-4\left(\frac{1-\cos 4}{4}\right)}{2+4}=-2$

$\Rightarrow \underset{x\to 0}{lim}\frac{\sin x}{x}=1,\underset{x\to 0}{lim}\frac{1-\cos x}{x}$ [ L Hospital Rule]

$\Rightarrow \underset{x\to 0}{lim}\frac{0-c(-\sin x)}{1}=0$

$\Rightarrow \underset{4\to 0}{lim}\frac{a\times 1-4\times 0}{2+4}=-2$

$\Rightarrow \frac{a}{2}=-2$

$\Rightarrow a=-4$

$a=-4,b=-4$

$|a+b|=|-4-4|=|-8|$

$=8$