Question

Sol. [Key Concept: Both numerator and denominator vanish at $x = 2$; use the standard limits $sin\ t/t$ and $ln(1 + t)/t$.]

$\because$ Denominator $= 0$ at $x = 2$

⇒ Numerator $= 0$ at $x = 2$

$$\Rightarrow 2^{3} - 5(2)^{2} + a(2) + b = 0$$

$$\begin{array}{r} \Rightarrow 2a + b = 12\#(1) \end{array}$$

$$m = \underset{Li \rightarrow 2}{Lim}\frac{\frac{\sin\left( x^{3} - 5x^{2} + ax + b \right)}{\left( x^{3} - 5x^{2} + ax + b \right)} \cdot \left( x^{3} - 5x^{2} + ax + b \right)}{\left( \frac{x - 1 - 1}{\sqrt{x - 1} + 1} \right) \cdot \frac{\log_{e}\left( 1 + (x - 2) \right)}{(x - 2)} \cdot (x - 2)} $$

$$\Rightarrow m = {Lim}_{x \rightarrow 2}2\left( \frac{x^{3} - 5x^{2} + ax + b}{(x - 2)^{2}} \right) $$

$$\Rightarrow m = {Lim}_{x \rightarrow 2}2\left( \frac{3x^{2} - 10x + a}{2(x - 2)} \right) $$∵ Denominator $= 0$ at $x = 2$

⇒ Numerator $= 0$ at $x = 2$

$$\Rightarrow 3(2)^{2} - 10(2) + a = 0 \Rightarrow a = 8 $$put $a = 8$ in (1), we get $b = - 4$

$${\therefore m = {Lim}_{x \rightarrow 2}2\left( \frac{6x - 10}{2} \right) \Rightarrow m = 2 }{\therefore a + b + m = 6 }$$