If x 2lim (x^3-5x^2+ax+b)(x-1-1) _e(x-1)=m, then a+b+m is equal to :

Question

If $\underset{x	o 2}{lim}\frac{\sin ({x}^{3}-5{x}^{2}+ax+b)}{(\sqrt{x-1}-1){\log }_{e}(x-1)}=m$, then $a+b+m$ is equal to :

Accepted Answer

$	herefore $ Denominator $=0$ at $x=2$

$\Rightarrow$ Numerator $=0$ at $x=2$

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

$\Rightarrow 2a+b=12\, \ldots (1)$

$	herefore \, m=\underset{x	o 2}{lim}\frac{\frac{\sin ({x}^{3}-5{x}^{2}+ax+b)}{{x}^{3}-5{x}^{2}+ax+b}\cdot ({x}^{3}-5{x}^{2}+ax+b)}{((\frac{x-1-1}{\sqrt{x-1}+1}))\cdot \frac{{\log }_{e}(1+(x-2))}{x-2}\cdot (x-2)}$

$\Rightarrow m=\underset{x	o 2}{lim}2(\frac{{x}^{3}-5{x}^{2}+ax+b}{(x-2{)}^{2}})$

$\Rightarrow m=\underset{x	o 2}{lim}2(\frac{3{x}^{2}-10x+a}{2(x-2)})$

$	herefore $ Denominator $=0$ at $x=2$

$\Rightarrow$ Numerator $=0$ at $x=2$

$\Rightarrow 3(2{)}^{2}-10(2)+a=0	o a=8$

put $a=8$ in (1), we get $b=-4$

$	herefore \, m=\underset{x	o 2}{lim}2(\frac{6x-10}{2})	o m=2$

$	herefore \, a+b+m=6$

Solutions · 1