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NDA 2019 Maths · Mathematics · Continuity and Differentiability · easy

The function f(x)=begincases2x+a,&x<1\\5,&x=1\\3x-b,&x>1endcases is continuous at x=1. The value of a+b is

Single correct option

The function f(x)={2x+a,x<15,x=13x−b,x>1f(x)=\begin{cases}2x+a,&x<1\\5,&x=1\\3x-b,&x>1\end{cases} is continuous at x=1x=1. The value of a+ba+b is
  1. A. 55
  2. B. −1-1
  3. C. 11
  4. D. 88
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Answer: (C) 11

Continuity needs lim⁡x→1−f=f(1)=lim⁡x→1+f\lim_{x\to1^-}f=f(1)=\lim_{x\to1^+}f. The left limit gives 2+a=52+a=5, so a=3a=3. The right limit gives 3−b=53-b=5, so b=−2b=-2. Therefore a+b=1a+b=1. The value 88 results from taking b=5b=5 and forgetting the sign in 3−b3-b.
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Original question written in the NDA 2019 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake