For every pair of continuous functions f, g: [0, 1] → R such that max {f(x): x ∈ [0, 1]} = max {g(x)

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

Great question! While you can compare them, defining $h(x)$ transforms the problem into finding the roots of a single function. It turns a comparison problem into a zero-crossing problem, which is exactly what the Intermediate Value Theorem is designed to solve. It simplifies the complexity of two functions into the behavior of one.

Q: Wait, why is $h(x_1) \ge 0$?

Remember that $M$ is the maximum of $g(x)$. So, $g(x_1)$ must be less than or equal to $M$. Since $h(x_1) = f(x_1) - g(x_1) = M - g(x_1)$, and $g(x_1) \le M$, the result $M - g(x_1)$ must be non-negative. It's all about the definition of the maximum!

Q: Why are options B and C wrong?

Options B and C imply that the intersection point $c$ must satisfy a specific algebraic relationship that isn't universally true for all continuous functions. Since we only proved $f(c) = g(c)$, any identity involving $f(c)$ and $g(c)$ that simplifies to $0=0$ is correct, but specific equations like $(f(c))^2 + f(c) = (g(c))^2 + 3g(c)$ are not guaranteed. We must stick to what is mathematically certain.

Question

For every pair of continuous functions $f, g: [0, 1] 	o R$ such that $\max \{f(x): x \in [0, 1]\} = \max \{g(x): x \in [0, 1]\}$, the correct statement(s) is (are):

Options:

(A) $(f(c))^2 + 3f(c) = (g(c))^2 + 3g(c)$ for some $c \in [0, 1]$

(B) $(f(c))^2 + f(c) = (g(c))^2 + 3g(c)$ for some $c \in [0, 1]$

(C) $(f(c))^2 + 3f(c) = (g(c))^2 + g(c)$ for some $c \in [0, 1]$

(D) $(f(c))^2 = (g(c))^2$ for some $c \in [0, 1]$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, rolling landscape. You have two paths, represented by two continuous functions, $f(x)$ and $g(x)$, both defined on the interval $[0, 1]$.

These are not just any paths; they are smooth, unbroken, and elegant. The problem states that they both reach the exact same maximum altitude, $M$.

This is our common ceiling. Think of it as two mountain climbers, each taking a different route, yet both reaching the same peak elevation. They might reach this peak at different times or locations, but the peak itself is shared.

### The Power of the Difference Function

When we face two functions, our instinct might be to compare them directly. In the world of JEE Advanced, we need a more surgical tool.

We define a new function:
$$h(x) = f(x) - g(x)$$

This function $h(x)$ captures the 'gap' between our two climbers. If $h(x) > 0$, then $f(x) > g(x)$. If $h(x) < 0$, then $g(x) > f(x)$.

If $h(x) = 0$, that is the moment of intersection, the moment our two paths cross. Because $f$ and $g$ are continuous, $h$ is also continuous. It cannot jump; it must flow.

### The Bridge of the Intermediate Value Theorem

Now, let's look at the peaks. Let $f$ reach its maximum $M$ at $x_1$, and $g$ reach its maximum $M$ at $x_2$.

At $x_1$, we evaluate the difference function:
$$h(x_1) = f(x_1) - g(x_1) = M - g(x_1)$$

Since $M$ is the maximum of $g$, $g(x_1)$ cannot exceed $M$. Thus, $h(x_1) \ge 0$.

At $x_2$, we evaluate the difference function:
$$h(x_2) = f(x_2) - g(x_2) = f(x_2) - M$$

Since $M$ is the maximum of $f$, $f(x_2)$ cannot exceed $M$. Thus, $h(x_2) \le 0$.

Our function $h(x)$ starts at a non-negative value and ends at a non-positive value. By the **Intermediate Value Theorem**, it must cross the zero line.

There exists some $c \in [0, 1]$ such that $h(c) = 0$, which implies:
$$f(c) = g(c)$$

### The Final Revelation

We have proven that $f(c) = g(c)$. This is the golden key to the problem.

Now, consider the options provided. For Option A:
$$(f(c))^2 + 3f(c) = (g(c))^2 + 3g(c)$$

Since $f(c) = g(c)$, let's call this common value $y$. The equation becomes $y^2 + 3y = y^2 + 3y$, which is an **identity**. It is always true.

Similarly, for Option D:
$$(f(c))^2 = (g(c))^2$$

This becomes $y^2 = y^2$, which is also an identity. Options B and C, however, are not identities; they are specific conditions that are not guaranteed to hold.

We have navigated the landscape, found the intersection, and verified the truth. This is the elegance of calculus.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: For every pair of continuous functions $f, g : [0, 1] \to R$ such that $max {f (x) : x \in [0, 1]} = max {g (x) : x \in [0, 1]}$ , the correct statement(s) is (are):

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Visualized Solution

The Setup: Continuous Functions

The Maximum Value Condition

Locating the Peaks

The Difference Function $h (x)$

Evaluating $h (x)$ at $x_{1}$

Evaluating $h (x)$ at $x_{2}$

Intermediate Value Theorem (IVT)

The Intersection Point $c$

Verifying the Options

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Power of the Difference Function

The Bridge of the Intermediate Value Theorem

The Final Revelation

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