If limx-1x2-ax-bx-1-5- then a-b is equal to -

Given lim_x→1x^2 ax+bx 1=5 In order to limit exist, numerator must be equal to 0 at x=1 ⇒ (1)^2 a(1)+b=0 1 a+b=0 a b=1 ...(1) Now, applying L'hospital Rule ...

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Byju's Answer

Standard XIII

Mathematics

L'Hospital Rule to Remove Indeterminate Form

If limx→1x2-a...

Question

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

1

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Solution

The correct option is D $- 7$
Given
$lim x \to 1 \frac{x^{2} - a x + b}{x - 1} = 5$
In order to limit exist, numerator must be equal to $0$ at $x = 1$
$\Rightarrow (1)^{2} - a (1) + b = 0$
$1 - a + b = 0$
$a - b = 1 . . . (1)$

Now, applying L'hospital Rule
$lim x \to 1 \frac{2 x - a}{1} = 5$
$\Rightarrow 2 - a = 5$
$\Rightarrow a = - 3 . . . (2)$
putting the value of $a$ in $(1)$
$∴ b = - 4$
$\Rightarrow a + b = - 7$

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

The correct option is D −7Given limx→1x2−ax+bx−1=5 In order to limit exist, numerator must be equal to 0 at x=1 ⇒(1)2−a(1)+b=0 1−a+b=0 a−b=1 ...(1) Now, applying L'hospital Rule limx→12x−a1=5 ⇒2−a=5 ⇒a=−3 ...(2) putting the value of a in (1) ∴b=−4 ⇒a+b=−7

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