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A function $f(x)$ is defined as $f(x) = \begin{cases}e^x & < 1 \\ \ln x + ax^2 + bx & \ge 1 \end{cases}$ where $x \in \mathbb R$ which one of the following statements is TRUE?

  1. $f(x)$ is NOT differentiable at $x=1$ for any values of $a$ and $b$.
  2. $f(x)$ is differentiable at $x=1$ for the unique values of $a$ and $b$.
  3. $f(x)$ is differentiable at $x = 1$ for all the values of $a$ and $b$ such that $a + b = e$
  4. $f(x)$ is differentiable at $x=1$ for all values of $a$ and $b$.

I'm checking the continuity of the function at $x=1$. Left hand limit at $x=1^{-1}$ gives $e$. $$\lim_{x \to 1^{-1}} f(x) = e$$ Similarly taking the right hand limit $$\lim_{x \to 1^{1^{+}}} f(x) = a + b$$ So, for the function to be continuous $a + b = e$. So, according to me the right answer is option (3). Am I right ?

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    $\begingroup$ What does #2 even mean? "... the unique values"? Generally continuity is necessary but not sufficient. Since you have multiple choice, however, you might find the answer based on a sufficient condition of none of the others are sufficient. $\endgroup$ – Brick Feb 16 '17 at 16:14
  • $\begingroup$ @Brick So, option (3) must be right ? $\endgroup$ – Ansh Kumar Feb 16 '17 at 16:15
  • $\begingroup$ Since I don't understand what #2 means, I cannot quickly be sure if it's right or not. I didn't check #3 beyond the work that you showed. $\endgroup$ – Brick Feb 16 '17 at 16:16
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    $\begingroup$ @AnshKumar, notice that continuity doesn't imply differentiability. $\endgroup$ – Galc127 Feb 16 '17 at 16:22
  • $\begingroup$ Your definition of $f(x)$ needs correction. $\endgroup$ – zhw. Feb 16 '17 at 17:28
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As you have calculated, for the function to be continuous at $x=1$, we must have $a+b=e$.

Now, if $a+b=e$, observe that $$f'(x)=\begin{cases}e^x &, x<1 \\ \frac1x +2ax +b &,x \ge 1\end{cases}$$

Then we have $$f'(1^+)=e$$ $$f'(1^-)=1+2a+b$$

So for differentiability of the function at $x=1$, we must have both $$a+b=e\tag1$$ $$1+2a+b=e\tag2$$ Solving this, we have $a=-1$ and $b=e+1$.

So the function will be differentiable only for $a=-1$ and $b=e+1$.

Hence, the option $(2.)$ is correct.

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  • $\begingroup$ Shouldn't it be $$f'(x)=\begin{cases}e^x &, x\color{red}{<}1 \\ \frac1x +2ax +b &,x \color{red}{>} 1\end{cases}$$ $\endgroup$ – Paracosmiste Feb 16 '17 at 22:29
  • $\begingroup$ Yeah, @whatever. $\endgroup$ – SchrodingersCat Feb 17 '17 at 4:00
  • $\begingroup$ I meant $>$ not $\geq$ (like @zipirovich answer). $\endgroup$ – Paracosmiste Feb 17 '17 at 11:26
  • $\begingroup$ @whatever Why $>$? The question writes $\ge$. $\endgroup$ – SchrodingersCat Feb 17 '17 at 13:47
  • $\begingroup$ Doesn't the equality disappear when we differentiate? $\endgroup$ – Paracosmiste Feb 17 '17 at 14:18
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Hint:

1) the function have to be continuous at $x=1$. So, as you have found, $a+b=e$

But we want also that it has a derivative at $x=1$, so the ''right'' and ''left'' derivative have to be the same at $x=1$ and this gives the other equation:

2) $ e=1+2a+b$.

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No, you're not right. The fact that continuity implies that $a+b=e$ certainly rules out option (4) — you already determined that $a$ and $b$ can't be arbitrary. But any of the first three options still could be the answer. Remember that continuity is only half of what you need to verify — you also need to check whether the derivatives from the left and from the right agree, so there will be a second condition. Maybe that second condition will contradict what you found from continuity, and then (1) will be the answer. Maybe that second condition together with the first one will give you a unique solution for $a$ and $b$, and then (2) will be the answer. Or maybe that second condition will show that any $a$ and $b$ satisfying $a+b=e$ guarantee differentiability — and only in that case will you conclude that (3) is the answer.

Back to the specific function. If you take the derivatives, you'll get that

$$f'(x)=\begin{cases} e^x & \text{for }x<1 \\ \frac{1}{x}+2ax+b & \text{for }x>1 \end{cases}$$

Comparing $\lim\limits_{x\to1^{-}}f'(x)$ and $\lim\limits_{x\to1^{+}}f'(x)$, you'll get another condition on $a$ and $b$. So you end up with a system of two equations with respect to $a$ and $b$. Solving that system, you'll see that there are how many options for $a$ and $b$?

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