ExamHabit
Practise free

JEE Main 2019 · 9 April · Shift II

JEE Main 2019 Mathematics question on Limits Continuity and Differentiability

If the function f(x)={a∣π−x∣+1,x≤5b∣x−π∣+3,x>5f(x) = \begin{cases} a|\pi - x| + 1, x \leq 5 \\ b|x - \pi| + 3, x > 5 \end{cases} is continuous at x=5x = 5, then the value of a−ba - b is
  1. A.
    −2π+5\dfrac{-2}{\pi + 5}
  2. B.
    2π+5\dfrac{2}{\pi + 5}
  3. C.
    2π−5\dfrac{2}{\pi - 5}
  4. D.
    25−π\dfrac{2}{5 - \pi}
Show answer and solution

Correct option: D

Solution

For continuity at x = 5, the left and right limits must be equal.

Left limit (x ≤ 5): f(5) = a|π - 5| + 1 = a(5 - π) + 1, since π < 5 so |π - 5| = 5 - π.

Right limit (x > 5): lim → a(5-π) + 1 = b|5 - π| + 3 = b(5 - π) + 3.

Setting equal: a(5 - π) + 1 = b(5 - π) + 3, so (a - b)(5 - π) = 2, giving a - b = 2/(5 - π).

[JEE Main 2019 · 9 April · Shift II]

Practise Limits Continuity and Differentiability the way it is asked

A timed set of real JEE Main Limits Continuity and Differentiability questions, graded as the exam grades. No sign-up.

Start a Limits Continuity and Differentiability test

More: all JEE Main Limits Continuity and Differentiability questions · JEE Main 2019 · Mathematics chapter list