JEE MAIN - Mathematics (2019 - 10th April Evening Slot - No. 4)
If $$\mathop {\lim }\limits_{x \to 1} {{{x^2} - ax + b} \over {x - 1}} = 5$$, then a + b is equal to :
1
- 4
- 7
5
Explanation
$$\mathop {\lim }\limits_{x \to 1} {{{x^2} - ax + b} \over {x - 1}} = 5$$
$$ \Rightarrow $$ $${(1)^2} - a(1) + b = 0$$
$$ \Rightarrow $$$$1 - a + b = 0$$
$$ \Rightarrow $$$$a - b = 1\,\,......(1)$$
Now 'L' hospital rule
2x - a = 5
$$ \Rightarrow $$2 - a = 5 ($$ \because $$ x = 1)
$$ \Rightarrow $$ a = - 3
By putting a = -3 in (1)
$$ \Rightarrow $$ b = -4
$$ \therefore $$ a + b = -7
$$ \Rightarrow $$ $${(1)^2} - a(1) + b = 0$$
$$ \Rightarrow $$$$1 - a + b = 0$$
$$ \Rightarrow $$$$a - b = 1\,\,......(1)$$
Now 'L' hospital rule
2x - a = 5
$$ \Rightarrow $$2 - a = 5 ($$ \because $$ x = 1)
$$ \Rightarrow $$ a = - 3
By putting a = -3 in (1)
$$ \Rightarrow $$ b = -4
$$ \therefore $$ a + b = -7
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