JEE Main 2019MathematicsLimitsEasyMCQ

JEE Main 2019 — Limits Question with Solution

JEE Main 2019 (10 Apr Shift 2)

Question

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

Choose an option

Show full solutionCorrect option: D
Correct answer
D–7

Step-by-step explanation

Given L=limx→1⁡x2-ax+bx-1=5

As x→1, the denominator becomes zero, hence for finite limit, the numerator must also approach to zero as x→1

⇒1-a+b=0

⇒b=a-1   …(1)

Now L=limx→1x2-ax+bx-1  00 form

By using L'Hospital Rule,

⇒L=limx→12x-a1

⇒L=limx→12x-a1=5

⇒2-a=5 ⇒a=-3

From the equation 1, we get

⇒b=a-1=-4

⇒a+b=-7.

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About this question

This is a previous-year question from JEE Main 2019, covering the Limits chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.