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: Check statement \(P\).
For \(x>0\),
\[ f(x)=xg(x)+x^2h(x) \]
For \(x\leq0\),
\[ f(x)=cx+dx^2 \]
At \(x=0\),
\[ f(0)=0 \]
The right-hand derivative at \(0\) is
\[ \lim_{x\to0^+}\frac{f(x)-f(0)}{x} = \lim_{x\to0^+}\frac{xg(x)+x^2h(x)}{x} \]
\[ = \lim_{x\to0^+}\left(g(x)+xh(x)\right) \]
Since \(h\) is bounded,
\[ xh(x)\to0 \]
But \(g\) is only bounded, not necessarily continuous at \(0\).
So,
\[ \lim_{x\to0^+}g(x) \] may not exist or may not be equal to \(g(0)\).
Therefore, even if
\[ c=g(0)=h(0), \] \(f\) need not be differentiable at \(x=0\).
Hence, statement \(P\) is NOT correct.

Step 2: Check statement \(Q\).
Now suppose \(f,g,h\) are thrice differentiable and
\[ g(0)=0 \] with
\[ g'(0)+h(0)>0 \]
Since \(f\) is differentiable at \(0\), the left and right derivatives must match.
For \(x\leq0\),
\[ f(x)=cx+dx^2 \] so
\[ f'(0^-)=c \]
For \(x>0\), since \(g(0)=0\), the right derivative at \(0\) is
\[ f'(0^+)=g(0)=0 \]
Hence,
\[ c=0 \]

Step 3: Use second derivative test.
For \(x>0\),
\[ f(x)=xg(x)+x^2h(x) \]
Near \(x=0\), since \(g\) and \(h\) are differentiable,
\[ g(x)=g(0)+g'(0)x+\cdots \] and
\[ h(x)=h(0)+\cdots \]
Since \(g(0)=0\),
\[ xg(x)\approx g'(0)x^2 \] and
\[ x^2h(x)\approx h(0)x^2 \]
Therefore, near \(0\),
\[ f(x)\approx \left(g'(0)+h(0)\right)x^2 \]
Since
\[ g'(0)+h(0)>0, \] we get
\[ f''(0)>0 \]
Thus, \(f\) has a local minimum at \(x=0\).
Hence, statement \(Q\) is correct.

Step 4: Final conclusion.
Statement \(P\) is not correct because boundedness of \(g\) does not ensure differentiability at \(0\).
Statement \(Q\) is correct because the given condition gives a positive second derivative at \(0\).
Therefore, the correct answer is
\[ \boxed{(B)} \]