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

A
5
B
-4
C
-7
D
1

Solution:

Given, $lim_{x \to 1} \frac{x^{2} - a x + b}{x - 1} = 5$

For existence of limit, $1 - a + b = 0$

$\Rightarrow b = a - 1$ …(i)

$∴ lim_{x \to 1} \frac{x^{2} - a x + a - 1}{x - 1} = 5$ [Using (i)]

$\begin{array}{l} \Rightarrow lim_{x \to 1} \frac{x^{2} - 1 - a (x - 1)}{x - 1} = 5 \Rightarrow lim_{x \to 1} \frac{(x - 1) (x + 1) - a (x - 1)}{x - 1} = 5 \\ \Rightarrow lim_{x \to 1} \frac{(x - 1) (x + 1 - a)}{x - 1} = 5 \Rightarrow lim_{x \to 1} (x + 1 - a) = 5 \\ \Rightarrow 2 - a = 5 \Rightarrow a = - 3 and b = - 3 - 1 = - 4 [Using (i)] \\ ∴ a + b = - 3 - 4 = - 7 \end{array}$

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

Solution:

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