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

Q: Wait, why can't we just use the Mean Value Theorem (MVT) to prove Option (C)?

That is a fantastic question! The MVT states that for a function continuous on $[c, 1]$ and differentiable on $(c, 1)$, there exists some $\xi \in (c, 1)$ such that $\frac{f(1) - f(c)}{1 - c} = f'(\xi)$. Notice the difference: MVT guarantees the existence of a point $\xi$ *somewhere* in the interval, but Option (C) claims the derivative at the *specific* point $c$ must equal the secant slope. These are not the same thing!

Q: Why did we choose $f(x) = x^2$ as a counter-example?

When testing a claim, we want a function that is 'well-behaved' (continuous and differentiable) but has a non-constant derivative. $f(x) = x^2$ is the simplest non-linear function. If a property doesn't hold for a simple parabola, it certainly won't hold for all functions in the set $S$.

Q: Why does the constant function $f(x) = k$ break the inequalities in (A) and (B)?

The inequalities in (A) and (B) are *strict* inequalities (using the $<$ symbol). For a constant function, the derivative is $0$ everywhere, and the difference $f(c) - f(1)$ is also $0$. This leads to the statement $0 < 0$, which is mathematically false. Since the inequality must hold for *every* function in $S$, and it fails for this constant function, the options cannot be correct.

Question

Let $S$ be the set of all function $f : [0, 1] 	o \mathbb{R}$, 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) $|f (c) - f (1)| < (1 −c)| f' (c)|$

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

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

(D) None of these

Sigma Percentile · Accepted Answer

### The Detective's Guide to Calculus: Unmasking the Universal Truth

Welcome, future engineers! Today, we are not just solving a problem; we are embarking on a detective mission. We have a set $S$ of functions that are continuous on $[0, 1]$ and differentiable on $(0, 1)$.

We are looking for a property that holds true for *every* single function in this set. In the world of JEE Advanced, when you see the phrase 'for every function,' your internal alarm should ring. It is a call to arms for the most powerful tool in your arsenal: the **counter-example**.

### Phase 1: The Trap of Option (C)

Let's look at Option (C): 
$$\frac{f(1) - f(c)}{1 - c} = f'(c)$$

At first glance, this looks suspiciously like the Mean Value Theorem (MVT). But let's be precise. The MVT tells us that there exists some $\xi \in (c, 1)$ such that the slope of the secant line equals the derivative at $\xi$.

Option (C), however, makes a much bolder claim: it asserts that the derivative at the *exact* point $c$ must equal the secant slope. This is a much stronger condition. If we can find just one function where the tangent at $c$ is never parallel to the secant from $c$ to $1$, we can strike this option off our list.

### Phase 2: The Parabola Test

Let's test the simplest non-linear function we know: $f(x) = x^2$. This function is perfectly continuous and differentiable. Let's see if we can find a $c \in (0, 1)$ that satisfies the condition in Option (C).

First, the secant slope (the left-hand side):
$$\frac{f(1) - f(c)}{1 - c} = \frac{1^2 - c^2}{1 - c} = \frac{(1 - c)(1 + c)}{1 - c} = 1 + c$$

Next, the tangent slope (the right-hand side):
$$f'(x) = 2x \implies f'(c) = 2c$$

Now, we equate them: $1 + c = 2c$. Solving this gives us $c = 1$.

But wait! The problem explicitly requires $c \in (0, 1)$. Since $c = 1$ is not in the open interval $(0, 1)$, our parabola has failed the test. Option (C) is officially incorrect.

### Phase 3: The Constant Function Test

Now, let's turn our attention to Options (A) and (B). These involve strict inequalities. When dealing with inequalities, the most dangerous functions are those that make the expressions zero.

Let's test the constant function $f(x) = k$. For this function, $f(c) = k$ and $f(1) = k$. The derivative $f'(x) = 0$ for all $x$.

Substituting these into Option (A):
$$|f(c) - f(1)| < (1 - c)|f'(c)| \implies |k - k| < (1 - c)|0| \implies 0 < 0$$

And for Option (B):
$$|f(c) - f(1)| < |f'(c)| \implies |k - k| < |0| \implies 0 < 0$$

In both cases, we are left with the statement $0 < 0$. This is a blatant contradiction! Zero cannot be strictly less than itself. Because this inequality fails for our constant function, neither (A) nor (B) can be universally true.

### Conclusion: The Power of None

We have systematically dismantled every option. By choosing a parabola, we proved (C) false. By choosing a constant function, we proved (A) and (B) false.

This leaves us with only one logical conclusion: **Option (D), 'None of these.'**

Remember, in JEE Advanced, you don't always need to prove a theorem to solve a problem. Sometimes, you just need to be a good detective and find the one case that breaks the rule. Keep practicing, keep questioning, and keep falling in love with the logic behind the math!

Animated Solution for Mathematics - Differentiation: Let $S$ be the set of all function $f : [0, 1] \to R$ , 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

The Set of Functions $S$

Geometric Meaning of Option (C)

Counter-Example for Option (C)

Calculating the Secant Slope

Calculating the Tangent Slope

Solving for $c$

Counter-Example for (A) and (B)

Substituting into (A) and (B)

The Impossible Inequality

Final Answer

The Sigma Insight: Mean Value Theorems

The Detective's Guide to Calculus

Phase 1

Phase 2

Phase 3

Conclusion

Similar Questions

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:

Let $f$ be any function continuous on $[a, b]$ and twice differentiable on $(a, b)$ . If for all $x \in (a, b)$ , $f^{'} (x) > 0$ and $f^{''} (x) < 0$ , then for any $c \in (a, b)$ , $\frac{f ( c ) - f ( a )}{f ( b ) - f ( c )}$ is greater than:

If $f (x)$ is twice differentiable and continuous function in $x \in [a, b]$ also $f^{'} (x) > 0$ and $f^{''} (x) < 0$ and $c \in (a, b)$ then $\frac{f ( c ) - f ( a )}{f ( b ) - f ( c )}$ is greater than

For all twice differentiable functions $f : R \to R$ , with $f (0) = f (1) = f^{'} (0) = 0$

For all $x$ in $[0, 1]$ , let the second derivative $f^{''} (x)$ of a function $f (x)$ exist and satisfy $∣ f^{''} (x) ∣ < 1$ . If $f (0) = f (1)$ , then show that $∣ f^{'} (x) ∣ < 1$ for all $x$ in $[0, 1]$ .

If $f : R \to R$ is a twice differentiable function such that $f^{''} (x) > 0$ for all $x \in R$ , and $f (\frac{1}{2}) = f (1) = 1$ , then

If $f$ and $g$ are differentiable functions in $[0, 1]$ satisfying $f (0) = 2 = g (1), g (0) = 0$ and $f (1) = 6$ , then for some $c \in] 0, 1 [$

If $f (x)$ and $g (x)$ are differentiable function for $0 \leq x \leq 1$ such that $f (0) = 2, g (0) = 0, f (1) = 6, g (1) = 2$ , then show that there exist $c$ satisfying $0 < c < 1$ and $f^{'} (c) = 2 g^{'} (c)$ .

Let the function, $f : [- 7, 0] \to R$ be continuous on $[- 7, 0]$ and differentiable on $(- 7, 0)$ . If $f (- 7) = - 3$ and $f^{'} (x) \leq 2$ , for all $x \in (- 7, 0)$ , then for all such functions $f$ , $f (- 1) + f (0)$ lies in the interval:

Let $f$ be any continuous function on $[0, 2]$ and twice differentiable on $(0, 2)$ . If $f (0) = 0, f (1) = 1$ and $f (2) = 2$ , then