If limx→1x2−ax+bx−1=5 , then a+b is equal to

limx→1x2−ax+bx−1=5→(1)limx→1(x2−ax+b)=01−a+b=0→(1)(1)⇒limx→12x−a1=5⇒2−a=5⇒a=−3(1)⇒1+3+b=0⇒b=−4a+b=−7

If limx→1x2−ax+bx−1=5 , then a+b is equal to

limx→1x2−ax+bx−1=5→(1)limx→1(x2−ax+b)=01−a+b=0→(1)(1)⇒limx→12x−a1=5⇒2−a=5⇒a=−3(1)⇒1+3+b=0⇒b=−4a+b=−7

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If $\lim_{x \to 1} \frac{x^{2} - a x + b}{x - 1} = 5,$ then $a + b$ is equal to

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$- 7$

$5$

$1$

$- 4$

answer is C.

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Detailed Solution

$\lim_{x \to 1} \frac{x^{2} - a x + b}{x - 1} = 5 \to (1)$
$\lim_{x \to 1} (x^{2} - a x + b) = 0$
$1 - a + b = 0 \to (1)$
$(1) \Rightarrow \lim_{x \to 1} \frac{2 x - a}{1} = 5 \Rightarrow 2 - a = 5$
$\Rightarrow a = - 3$
$(1) \Rightarrow 1 + 3 + b = 0 \Rightarrow b = - 4$
$a + b = - 7$