Let $S$ be the set of all twice differentiable functions $f$ from $\mathbb{R}$ to $\mathbb{R}$ such that $\frac{d^2 f}{dx^2}(x) > 0$ for all $x \in (-1, 1)$. For $f \in S$, let $X_f$ be the number of points $x \in (-1, 1)$ for which $f(x) = x$. Then, which of the following statements is (are) true? (a) There exists a function $f \in S$ such that $X_f = 0$ (b) For every function $f \in S$, we have $X_f \leq 2$ (c) There exists a function $f \in S$ such that $X_f = 2$ (d) There does not exist any function $f \in S$ such that $X_f = 1$

1. Turning the problem into a standard form

Take $g(x)=f(x)-x.$ Because $g''(x)=f''(x)>0\quad\text{for all }x\in(-1,1),$ the function $g$ is strictly convex in $(-1,1)$.

2. Geometry of a strictly convex curve

For any strictly convex $g$ on an interval:

1. The graph lies below the straight line joining any two points on it.
2. Therefore, if $g$ has two distinct zeros, say at $x=a$ and $x=b$ with $a<b$, then
   $g(x)<0 \;\text{for all }x\in(a,b).$
3. A strictly convex function cannot have more than two real zeros in a connected interval; three would contradict the property in (2).

3. Consequences for the number of fixed points $X_f$

Because the fixed points of $f$ are precisely the zeros of $g$:

• Upper bound: $X_f\le 2.$
• Possibilities: $X_f$ can be $0,1,2$.

4. Constructing explicit examples

1. $X_f=0$:
   $f(x)=x + 2 + x^2.$
   Then $g(x)=2+x^2>0\;\forall x,$ so no fixed points.
2. $X_f=1$:
   $f(x)=x^2.$
   Here $g(x)=x^2-x=x(x-1),$ which vanishes only at $x=0$ inside $(-1,1)$.
3. $X_f=2$:
   $f(x)=x + x^2 - \tfrac14.$
   Then $g(x)=x^2-\tfrac14=(x-\tfrac12)(x+\tfrac12),$ giving zeros at $x=\pm\tfrac12$. All examples satisfy $f''(x)=2>0$ within $(-1,1)$.

5. Verifying the options

• (a) True – example 1.
• (b) True – proven upper bound $X_f\le2$.
• (c) True – example 3.
• (d) False – example 2 contradicts it.

What is the question?

We are looking at very smooth curves $f(x)$ on the number line such that the curve always bends upwards between $x=-1$ and $x=1$.

What do we count?

We count how many times that curve meets the diagonal line $y = x$ inside the window $(-1,1)$. Every meeting point is called a "fixed point" because there $f(x)=x$.

Why does the curve always bend up?

Because $f''(x)>0$ in $(-1,1)$. A positive second derivative means the curve is strictly convex – picture a bowl that always opens upward.

Important facts of a bowl-shaped curve minus the diagonal

Define another curve $g(x)=f(x)-x$. Because $f''(x)>0$ and $d^2(x)/dx^2=0$, we have
$g''(x)=f''(x)>0.$
So $g(x)$ is also strictly convex.

A strictly convex curve can:

• cross the $x$-axis zero, one, or two times (never three or more). Think of a smile that may stay above the axis, touch it once at its lowest dip, or cut it twice.

What the choices really ask

• (a) Is a bowl that never touches the diagonal possible? (Yes.)
• (b) Can we ever get 3 or more touch points? (No.)
• (c) Can we design a curve that touches exactly twice? (Yes.)
• (d) Is it impossible to touch exactly once? (No, it is possible.)

So the correct statements are (a), (b), (c).

Step 1: Reformulate with $g(x)=f(x)-x$

$g(x)=f(x)-x, \qquad g''(x)=f''(x)>0 \text{ on }(-1,1).$ So $g$ is strictly convex.

Step 2: Prove an upper bound on zeros of a strictly convex function

Assume, for contradiction, that $g$ has three distinct zeros $a<b<c$ in $(-1,1)$. By convexity, $g\bigl(\tfrac{a+c}{2}\bigr)<\tfrac12\bigl(g(a)+g(c)\bigr)=0,$ which contradicts $g(b)=0$ because the curve would cross the axis more than twice. Hence $X_f=\#\{x\in(-1,1):g(x)=0\}\le2.$ This proves statement (b).

Step 3: Produce explicit examples

1. No fixed point $(X_f=0)$
   Choose $f(x)=x+2+x^2.$
   Then $g(x)=2+x^2>0,\;\forall x\in(-1,1)\Rightarrow X_f=0.$
   Hence (a) is true.

2. Exactly two fixed points $(X_f=2)$
   Choose $f(x)=x+x^2-\tfrac14.$
   Here $g(x)=x^2-\tfrac14=(x-\tfrac12)(x+\tfrac12),$ so zeros at $x=\pm\tfrac12$. Therefore $X_f=2$, proving (c) is true.

3. Exactly one fixed point exists, so statement (d) is false
   Take $f(x)=x^2.$
   Then $g(x)=x(x-1)$ vanishes only at $x=0$ in $(-1,1)$, giving $X_f=1$. Hence (d) is false.

Step 4: Final verdict

True statements: (a), (b), (c) False statement: (d)