In [0, 1] Lagranges Mean Value theorem is NOT applicable to (A) f(x) = {1/2 - x, x

Q: Wait, the function is continuous at $x = \frac{1}{2}$ because the limits match. If it's continuous, why doesn't LMVT apply?

That is a fantastic observation! You are absolutely right that continuity is satisfied. However, continuity is only the first hurdle. Think of continuity as the function being 'connected'—it doesn't break. But differentiability is about the function being 'smooth.' A function can be connected but still have a sharp, jagged corner (like the absolute value function $|x|$ at $x=0$). LMVT requires that 'smoothness' so that a tangent line can exist at every point. If there is a sharp corner, you cannot define a unique tangent line, and the theorem breaks down.

Q: Do I always have to check the junction point for piecewise functions?

Yes, absolutely! In the context of JEE Advanced, the junction point is almost always where the 'trap' is hidden. When a function changes its definition, it is prone to having a discontinuity or a sharp corner. Always treat the junction point as a crime scene—investigate it first for continuity, and then for differentiability. It is the most efficient way to solve these problems.

Q: What if the LHD and RHD were equal? Would that mean LMVT applies?

If the LHD and RHD were equal, it would mean the function is differentiable at that point. If the function is differentiable at the junction and continuous everywhere else, then yes, the conditions for LMVT would be satisfied! In this specific problem, however, the slopes are $-1$ and $0$, which are clearly different, confirming the 'kink' in the graph.

Question

In $[0, 1]$ Lagranges Mean Value theorem is NOT applicable to

Options:

(A) $f(x) = \begin{cases} \frac{1}{2} - x, & x < \frac{1}{2} \ (\frac{1}{2} - x)^2, & x \ge \frac{1}{2} \end{cases}$

(B) $f(x) = \begin{cases} \frac{\sin x}{x}, & x 
eq 0 \ 1, & x = 0 \end{cases}$

(C) $f(x) = x |x|$

(D) $f(x) = |x|$

Sigma Percentile · Accepted Answer

### The Philosophy of LMVT

Imagine you are driving your car from point $A$ to point $B$ over a time interval $[0, 1]$. If your average speed is $60$ km/h, LMVT guarantees that at some specific moment, your speedometer read exactly $60$ km/h.

This intuition relies on one crucial assumption: your journey was smooth. You didn't teleport, and you didn't undergo infinite acceleration (a sharp, instantaneous change in velocity).

In mathematical terms, the function must be continuous on $[a, b]$ and differentiable on $(a, b)$. If your path has a sharp corner—a point where you instantly change direction—the theorem fails. That is exactly what we are hunting for today.

### The Investigation: Analyzing the Piecewise Function

Let us examine the function in Option A:

$$f(x) = \begin{cases} \frac{1}{2} - x, & x < \frac{1}{2} \ (\frac{1}{2} - x)^2, & x \ge \frac{1}{2} \end{cases}$$

This function is a hybrid. On the left side of $x = \frac{1}{2}$, it is a linear function with a negative slope. On the right side, it is a parabola.

Our mission is to check if this function is 'smooth' enough for LMVT on the interval $[0, 1]$.

### Step 1: The Continuity Check

Before we worry about derivatives, we must ensure the graph is connected. We check the junction at $x = \frac{1}{2}$.

For the Left Hand Limit (LHL):

$$\lim_{x 	o \frac{1}{2}^-} f(x) = \lim_{x 	o \frac{1}{2}^-} \left( \frac{1}{2} - x ight) = 0$$

For the Right Hand Limit (RHL):

$$\lim_{x 	o \frac{1}{2}^+} f(x) = \lim_{x 	o \frac{1}{2}^+} \left( \frac{1}{2} - x ight)^2 = 0$$

Since the LHL equals the RHL, the function is continuous. There is no jump, and the graph is connected. Continuity is merely the entry ticket; we must now verify differentiability.

### Step 2: The Differentiability Check (The Trap)

Now, we test for the 'smoothness'—the differentiability. We need to see if the slope coming from the left matches the slope coming from the right at $x = \frac{1}{2}$.

Let us calculate the Left Hand Derivative (LHD):

$$f'(x) = \frac{d}{dx} \left( \frac{1}{2} - x ight) = -1$$

So, the LHD at $x = \frac{1}{2}$ is $-1$. This means the graph is approaching the junction with a downward slope of $-1$.

Now, let us calculate the Right Hand Derivative (RHD) using the chain rule:

$$f'(x) = \frac{d}{dx} \left( \frac{1}{2} - x ight)^2 = 2 \left( \frac{1}{2} - x ight)(-1) = 2x - 1$$

Evaluating this at $x = \frac{1}{2}$:

$$f' \left( \frac{1}{2} ight) = 2 \left( \frac{1}{2} ight) - 1 = 0$$

### The Verdict: A Sharp Corner

Look at the results: the LHD is $-1$, but the RHD is $0$. Because $-1 
eq 0$, the slopes do not match.

Geometrically, this means that at $x = \frac{1}{2}$, the graph makes a sharp turn. It is not differentiable at this point.

Because this point lies within our interval $(0, 1)$, the second condition of LMVT is violated. **The theorem cannot be applied.**

### Final Thoughts

This problem teaches us a vital lesson: never assume a function is 'well-behaved' just because it is continuous. In the world of JEE Advanced, the devil is in the details—specifically, in the derivatives at the junction points.

By rigorously checking the LHD and RHD, you have successfully navigated the trap. Keep this analytical mindset, and you will find that even the most complex calculus problems begin to reveal their secrets.

Animated Solution for Mathematics - Differentiation: In $[0, 1]$ Lagranges Mean Value theorem is NOT applicable to

Select Answer:

Visualized Solution

Interval of Interest

LMVT Conditions

Analyzing Option A

Graphing the Left Piece

Graphing the Right Piece

The Critical Junction

Checking Continuity: LHL

Checking Continuity: RHL

Continuity Established

Checking Differentiability: LHD

Checking Differentiability: RHD

The Sharp Corner

Conclusion

The Sigma Insight: Mean Value Theorems

The Philosophy of LMVT

The Investigation

Step 1

Step 2

The Verdict

Final Thoughts

Similar Questions

The value of $c$ in the Lagrange's mean value theorem for the function $f (x) = x^{3} - 4 x^{2} + 8 x + 11$ when $x \in [0, 1]$ is :

The value of $c$ in Lagrange’s mean value theorem for the function $f (x) = x^{3} - 4 x^{2} + 8 x + 11$ , where $x \in [0, 1]$ is :

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 :

Let $f (x)$ be a non-constant twice differentiable function defined on $(- \infty, \infty)$ such that $f (x) = f (1 - x)$ and $f^{'} (\frac{1}{4}) = 0$ . Then,

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 $[0, 1]$ .

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:

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

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 $g : R \to R$ be a non constant twice differentiable such that $g^{'} (\frac{1}{2}) = g^{'} (\frac{3}{2})$ . If a real valued function $f$ is defined as $f (x) = \frac{1}{2} [g (x) + g (2 - x)]$ , then

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

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In [0,1] Lagranges Mean Value theorem is NOT applicable to