If lim _{x rightarrow 1} frac{x^2-ax+b}{x-1}= | vedclass

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(C) Given $\lim _{x \rightarrow 1} \frac{x^2-ax+b}{x-1}=7$. Since the limit exists and the denominator approaches $0$ as $x \rightarrow 1$,the numerat

Vedclass · Accepted Answer

-$11$

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(C) Given $\lim _{x \rightarrow 1} \frac{x^2-ax+b}{x-1}=7$. Since the limit exists and the denominator approaches $0$ as $x \rightarrow 1$,the numerator must also approach $0$ at $x=1$. Therefore,$(1)^2 - a(1) + b = 0$,which implies $1 - a + b = 0$,or $a - b = 1 \dots (i)$. Using $L$'$H$ôpital's rule or factoring,we have $\lim _{x \rightarrow 1} \frac{d}{dx}(x^2 - ax + b) = 7$. $\Rightarrow \lim _{x}$ ${\rightarrow 1} (2x - a) = 7$. Substituting $x=1$,we get $2(1) - a = 7$,so $2 - a = 7$,which gives $a = -5$. Substituting $a = -5$ into equation $(i)$,we get $-5 - b = 1$,so $b = -6$. Thus,$a + b = -5 + (-6) = -11$.

If $\lim _{x \rightarrow 1} \frac{x^2-ax+b}{x-1}=7$,then $a+b$ is equal to

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