If the function f(x) is differentiable at x=1, then what is the value of (a+b)?
Consider the following for the next two (02) items that follow: Let f(x)=\begin{cases}ax(x+1)+b,&x \lt 1\\ x-1,&1\le x\le2\end{cases}
- A. -\frac{1}{3} ✓
- B. -1
- C. 0
- D. 1
Correct Answer: A. -\frac{1}{3}
Explanation
If f(x) is differentiable at x=1, it is also continuous there. For continuity, \lim_{x \rightarrow 1^-} f(x) = f(1) \implies a(1)(2)+b = 1-1 \implies 2a+b = 0. For differentiability, LHD = RHD \implies \frac{d}{dx}(ax^2+ax+b) = \frac{d}{dx}(x-1) at x=1. Thus 2ax+a = 1 \implies 3a = 1 \implies a = \frac{1}{3}. Substituting a, 2(\frac{1}{3})+b = 0 \implies b = -\frac{2}{3}. Therefore, a+b = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}.
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