Let S be the set of all functions f: [0,1] - R, which are continuous on [0, 1] and differentiable o

Q: Wait, why do we reject the negative root here? Or why can't we just use energy conservation?

This question doesn't involve energy conservation because it's a pure calculus problem about function properties. We reject the value $c=1$ because the Mean Value Theorem specifically guarantees $c$ lies in the *open* interval $(0, 1)$. If our calculation forces $c=1$, that condition fails the requirement of the theorem.

Q: Is it really okay to just pick random functions like $f(x) = x^2$?

Absolutely! In fact, it is a brilliant strategy. When a question claims a property holds for *all* functions in a set, testing a simple, well-behaved function is the fastest way to check if the claim is valid. If it fails for $f(x) = x^2$, it is definitely not true for the entire set.

Q: The question asks for a condition that holds for *every* $f \in S$. Does this mean I have to check infinitely many functions?

Not at all. You only need to check if the statement is universally true. If you find one counter-example, you have proven the statement false. If you cannot find a counter-example, you might need to look closer at the logic, but in JEE Advanced, counter-examples are often the intended path to quickly eliminate incorrect options.

Question

Let $S$ be the set of all functions $f: [0,1] 	o R$, which are continuous on $[0, 1]$ and differentiable on $(0, 1)$. Then for every $f$ in $S$, there exists a $c \in (0,1)$, depending on $f$, such that:

Options:

(A) $\frac{f(1)-f(c)}{1-c} = f'(c)$

(B) $|f(c) - f(1)| < |f'(c)|$

(C) $|f(c) + f(1)| < (1 + c)|f'(c)|$

(D) $|f(c) - f(1)| < (1 - c)|f'(c)|$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the edge of a vast, abstract landscape of functions. You are given a set $S$ of functions that are continuous on $[0, 1]$ and differentiable on $(0, 1)$.

The question asks you to find a condition that holds true for *every* single function in this set. This is the heart of a JEE Advanced challenge: it is not just about calculation; it is about logical rigor.

When you see a 'for all' statement, do not panic. Instead, think like a detective. You are looking for a flaw; if you can find just one function that breaks the rule, you have solved the puzzle.

### The Mean Value Theorem: Your North Star

The options provided involve $f(1)$, $f(c)$, and $f'(c)$. This is a massive clue. Whenever you see these terms together, your mind should immediately jump to the **Mean Value Theorem (MVT)**.

The MVT states that for a function continuous on $[c, 1]$ and differentiable on $(c, 1)$, there exists a point $c$ such that the slope of the secant line equals the slope of the tangent line at that point. The expression for the slope of the secant line is:

$$ \frac{f(1) - f(c)}{1 - c} $$

The derivative $f'(c)$ represents the slope of the tangent. Option 1 suggests these are always equal. Let us test this hypothesis.

### The Power of Simple Functions

Let us test Option 1 with the simplest function we can think of: $f(x) = x^2$. This function is continuous and differentiable everywhere.

For $f(x) = x^2$, we have $f(1) = 1$ and $f(c) = c^2$. The slope of the secant line becomes:

$$ \frac{1 - c^2}{1 - c} $$

Using the algebraic identity $1 - c^2 = (1 - c)(1 + c)$, this simplifies beautifully to $1 + c$. Now, the derivative is $f'(x) = 2x$, so $f'(c) = 2c$.

If Option 1 were true, we would have $1 + c = 2c$, which implies $c = 1$. But wait! The problem states $c \in (0, 1)$. Since our test function forces $c = 1$, Option 1 cannot be true for all functions. We have eliminated it with a single, elegant counter-example.

### Testing the Inequalities

Now, what about the inequalities? Let us try a constant function, $f(x) = k$. This is the ultimate test for inequalities.

For $f(x) = k$, we find $f(1) = k$, $f(c) = k$, and $f'(c) = 0$. Let us plug these into Option 4:

$$ |f(c) - f(1)| < (1 - c)|f'(c)| $$

This becomes $|k - k| < (1 - c)|0|$, which simplifies to $0 < 0$. This is a contradiction, as zero is not strictly less than zero. Therefore, Option 4 is also incorrect.

### The Bonus Revelation

This brings us to a fascinating moment in JEE history. In the 2014 JEE Advanced paper, this question was awarded as a **'Bonus'** because none of the options were universally true.

This is a powerful lesson: sometimes, the most important part of the problem is realizing that the standard path is blocked. Trust your counter-examples, trust your logic, and never be afraid to challenge the options.

You have navigated the landscape of functions and emerged victorious. Keep this mindset, and no calculus problem will ever be able to trap you again.

Animated Solution for Mathematics - Differentiation: Let $S$ be the set of all functions $f : [0, 1] \to R$ , which are continuous on $[0, 1]$ and differentiable on $(0, 1)$ . Then for every $f$ in $S$ , there exists a $c \in (0, 1)$ , depending on $f$ , such that:

Select Answer:

Visualized Solution

Defining the Set $S$

Analyzing the Options

Introducing Point $c$

Geometric Meaning of Option 1

The Tangent at $c$

Testing Option 1 with $f (x) = x^{2}$

Evaluating the Secant Slope

Evaluating the Tangent Slope

Equating LHS and RHS

Rejecting Option 1

Testing Inequalities with a Constant Function

Values for Constant Function

Testing Option 4

Testing Options 2 and 3

Final Conclusion

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Mean Value Theorem

The Power of Simple Functions

Testing the Inequalities

The Bonus Revelation

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