Question

Let \(h:\mathbb{R}\to\mathbb{R}\) and \(g:\mathbb{R}\to\mathbb{R}\) be two bounded functions. Define the function \(f:\mathbb{R}\to\mathbb{R}\) by

• A. \(P\) is correct and \(Q\) is NOT correct
• B. \(P\) is NOT correct and \(Q\) is correct
• C. Both \(P\) and \(Q\) are correct
• D. Neither \(P\) nor \(Q\) is correct

Hint:

Boundedness does not imply continuity or differentiability. For differentiability at a joining point, always check the left-hand and right-hand derivative limits.

Answer 1

The Correct Option is B

Step 1: Test differentiability for $P$.
For $x>0$, $f(x)=xg(x)+x^2h(x)$. The right hand slope at $0$ is $\lim_{x\to0^+}(g(x)+xh(x))$.

Step 2: Spot the gap.
The term $xh(x)\to0$ since $h$ is bounded, but $g$ is only bounded, so $\lim g(x)$ may not even exist. Matching $c=g(0)=h(0)$ does not save it, so $P$ is not correct.

Step 3: Set up $Q$.
Now $f,g,h$ are smooth, $g(0)=0$, and $g'(0)+h(0)>0$. Matching left and right slopes at $0$ forces $c=0$.

Step 4: Expand near zero.
With $g(0)=0$, $f(x)\approx (g'(0)+h(0))x^2$ for small $x$, so $f''(0)>0$ and $f$ has a local minimum at $0$. Thus $Q$ is correct.

Step 5: Conclude.
$P$ wrong, $Q$ right, option (B).
\[ \boxed{(B)} \]