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JEE Main 2019 · 10 Apr · Shift II

JEE Main 2019 Mathematics question on Limits Continuity and Differentiability

If lim⁡x→1x2−ax+bx−1=5\lim_{x \to 1} \dfrac{x^2 - ax + b}{x - 1} = 5, then a+ba + b is equal to
  1. A.
    -4
  2. B.
    1
  3. C.
    -7
  4. D.
    5
Show answer and solution

Correct option: C

Solution

For the limit to exist and equal 5, the numerator must vanish at x = 1 (otherwise the limit would be infinite). So 1 - a + b = 0, giving b = a - 1.

Factor: x² - ax + b = (x - 1)(x - (a-1)), so the limit equals -(a-1) = 1-(a-1) ... more carefully: quotient is x - (a-1), evaluated at x=1 gives 1-(a-1) = 2-a = 5, so a = -3, b = a-1 = -4.

Therefore a + b = -3 + (-4) = -7.

[JEE Main 2019 · 10 Apr · Shift II]

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