Let S be the set of all twice differentiable functions f from ℝ to ℝ such that d²f/dx²(x) 0 for al

Q: Why do we define $g(x) = f(x) - x$? Can't we just solve $f(x) = x$ directly?

Great question! While you could try to solve $f(x) = x$ directly, it is like trying to untangle a knot in the dark. By defining $g(x) = f(x) - x$, we shift the problem to finding the roots of $g(x) = 0$. This simplifies the geometry significantly because we are now comparing a convex curve to the flat, horizontal x-axis, rather than a slanted line $y=x$. It turns a dynamic intersection problem into a static root-finding problem.

Q: Wait, why can't a convex function have 3 roots? Can't it just wiggle?

Ah, that is the beauty of the second derivative! If $f''(x) > 0$, the slope $f'(x)$ is strictly increasing. If it were to cross the x-axis three times, it would have to turn back down to cross the second time, and then turn back up to cross the third time. That would require the slope to decrease, which contradicts $f''(x) > 0$. A strictly convex function is \"always turning upwards\"—it can never turn back down.

Q: Does the interval $(-1, 1)$ really matter? Why not $(-\infty, \infty)$?

It matters immensely! If the interval were the entire real line, the function could potentially have no roots or two roots, but the \"one root\" case might be tricky depending on the function's behavior at infinity. By restricting us to $(-1, 1)$, the problem forces us to ensure our constructed examples actually cross the axis within that specific window. It is a constraint that keeps our feet on the ground and ensures we are testing the function's behavior in a controlled environment.

Question

Let $S$ be the set of all twice differentiable functions $f$ from $\mathbb{R}$ to $\mathbb{R}$ such that $\frac{d^2f}{dx^2}(x) > 0$ 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?

Options:

(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 $X_f = 1$

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Smile

Imagine you are standing on a vast, flat plain. You are asked to draw a curve that is always "smiling"—that is, it always bends upwards. This is the essence of a strictly convex function.

In the language of calculus, we say $f''(x) > 0$. Today, we are going to explore how this simple geometric property dictates the number of times a function can intersect a line. This is not just a math problem; it is a lesson in how constraints define reality.

### Phase 1: The Transformation

We begin with the equation $f(x) = x$. It looks innocent enough, but it is actually a bit of a nuisance. We are looking for the intersection of a curve $y = f(x)$ and a line $y = x$.

To make this manageable, we perform a classic mathematical maneuver: we define a new function $g(x) = f(x) - x$. Finding where $f(x) = x$ is identical to finding the roots of $g(x) = 0$.

We have effectively flattened the line $y = x$ into the x-axis. Now, instead of tracking a curve against a diagonal line, we are simply watching a curve cross the horizontal axis.

### Phase 2: The Power of the Second Derivative

Now, let us look at the soul of our function $g(x)$. We know that $f''(x) > 0$.

When we differentiate $g(x)$ twice, we get:

$$g''(x) = f''(x) - 0 = f''(x)$$

Since $f''(x) > 0$, it follows that $g''(x) > 0$. This is the "Aha!" moment. A function with a positive second derivative is strictly convex.

Visually, this means the slope of the function is always increasing. It starts steep, flattens out, and then gets steep again, but it never, ever turns back down.

### Phase 3: The Maximum Root Constraint

Think about this: if a curve is always bending upwards, how many times can it cross the x-axis? If it crosses once, it is heading downwards.

To cross again, it must turn back up. But if it turns back up, it can never cross a third time because it is now heading towards infinity. It cannot turn back down to cross the axis again.

Therefore, a strictly convex function can intersect the x-axis at most twice. This immediately proves that $X_f \leq 2$. Statement (B) is locked in as true.

### Phase 4: Constructing Reality

Now, we must verify if 0, 1, or 2 roots are actually possible. We need to build functions that satisfy $f''(x) > 0$ and test them.

1. **Zero Roots:** Let $f(x) = x^2 + x + 0.4$. Here, $f''(x) = 2 > 0$. The function $g(x) = x^2 + 0.4$ is always positive and never touches the x-axis. Thus, $X_f = 0$ is possible.

2. **Two Roots:** Let $f(x) = x^2 + x - 0.4$. Again, $f''(x) = 2 > 0$. Here, $g(x) = x^2 - 0.4$. Setting $g(x) = 0$ gives:

$$x = \pm \sqrt{0.4} \approx \pm 0.632$$

Both roots lie within our interval $(-1, 1)$. Thus, $X_f = 2$ is possible.

3. **One Root:** Let $f(x) = x^2 + x$. Then $g(x) = x^2$. The only root is $x = 0$, which is in our interval. Thus, $X_f = 1$ is possible.

### Conclusion: The Elegance of the Result

By constructing these examples, we have dismantled the problem piece by piece. We found that $X_f$ can be 0, 1, or 2.

Statement (D), which claimed that $X_f$ could not be 1, is proven false. We have navigated the geometry of convexity and emerged with a clear understanding of the limits of these functions.

Remember, in JEE Advanced, the math is often just a language describing a deeper geometric truth. Once you see the "smile" of the convex function, the algebra follows naturally.

Select Answer:

Visualized Solution

Define $g (x)$

Meaning of $X_{f}$

Convexity of $g (x)$

Shape of $g (x)$

Maximum Roots (Option B)

Zero Roots (Option A)

Two Roots (Option C)

One Root (Option D)

Final Conclusion

The Sigma Insight: Higher Order Derivatives

Analyzing the Geometry of the Smile

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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