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Continuity & Differentiability · Question 112 of 125

Question 112

HardOriginal question

For which aa and bb is f(x)={ax2+1,x≤1x+b,x>1f(x) = \begin{cases} ax^{2} + 1, & x \le 1 \\ x + b, & x > 1 \end{cases} differentiable at x=1x = 1?

Options for question 112

Show the answer and explanation
CorrectNot quiteAnswer A

Correct answer: A

Explanation

  1. Values must match at x=1x = 1: a+1=b+1a + 1 = b + 1.

  2. Slopes must match: 2a=12 a = 1, so a=12a = \frac{1}{2}, b=12b = \frac{1}{2}.

Worth noting

a=1, b=1a = 1,\ b = 1 joins the values but leaves slopes 2 and 1.

Concept
Constants for differentiability
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Ten questions from Continuity & Differentiability, at your own pace.