Let f : R → R be a function defined as : f(x) = { 5, if x ≤ 1; a + bx, if 1

Q: Why do we check continuity at $x=1, 3,$ and $5$?

These are the 'junction points' where the definition of the function changes. A function can only be continuous if it is continuous at every point in its domain. If it breaks at any of these transition points, the entire function is considered discontinuous.

Q: What if the limits match but the function value is different?

That would be a 'removable discontinuity.' For a function to be continuous, the limit must exist AND equal the function value. If $LHL = RHL \neq f(c)$, the function is still discontinuous at that point.

Q: Why did we get a contradiction when solving for $a$ and $b$?

This happens because the system is 'overdetermined.' We have three constraints (one for each junction) but only two variables ($a$ and $b$). The math is telling us that there is no pair of $(a, b)$ that can satisfy all three conditions simultaneously. It's a classic JEE trap to see if you can identify an inconsistent system.

Question

Let $f : R 	o R$ be a function defined as : $f(x) = \begin{cases} 5, & 	ext{if } x \le 1 \ a + bx, & 	ext{if } 1 < x < 3 \ b + 5x, & 	ext{if } 3 \le x < 5 \ 30, & 	ext{if } x \ge 5 \end{cases}$. Then, f is :

Options:

(A) continuous if a = 5 and b = 5

(B) continuous if a = -5 and b = 10

(C) continuous if a = 0 and b = 5

(D) not continuous for any values of a and b

Sigma Percentile · Accepted Answer

### The Illusion of a Smooth Path

Imagine you are walking along a path defined by a piecewise function. This path isn't just one long, straight road; it is constructed from four distinct segments, each governed by its own rule.

As you travel from $x = -\infty$ to $x = +\infty$, you encounter three major checkpoints: $x = 1$, $x = 3$, and $x = 5$. For this path to be truly continuous, you must be able to walk from one segment to the next without ever having to jump or teleport.

In the language of calculus, this means the path must be connected at every single junction.

### The Mathematical Blueprint

To ensure our path is seamless, we rely on the fundamental definition of continuity at a point $c$. We require that the limit of the function as we approach $c$ from the left (the Left Hand Limit, or LHL) must equal the limit as we approach from the right (the Right Hand Limit, or RHL), and both must equal the actual value of the function at that point, $f(c)$.

Mathematically, we write this as:

$$\lim_{x 	o c^-} f(x) = \lim_{x 	o c^+} f(x) = f(c)$$

### Analyzing the Junctions

At the first checkpoint, $x = 1$, the path transitions from a constant value of $5$ to the linear expression $a + bx$. For continuity, the value from the left must match the value from the right:

$$LHL = f(1) = 5$$

$$RHL = \lim_{x 	o 1^+} (a + bx) = a + b$$

Equating these, we get our first constraint: $a + b = 5$. This is our first piece of the puzzle.

Moving to the second checkpoint, $x = 3$, the path shifts from $a + bx$ to $b + 5x$. Again, we equate the limits:

$$LHL = \lim_{x 	o 3^-} (a + bx) = a + 3b$$

$$RHL = f(3) = b + 5(3) = b + 15$$

Setting these equal, we have $a + 3b = b + 15$, which simplifies beautifully to $a + 2b = 15$.

Finally, at the third checkpoint, $x = 5$, the path transitions from $b + 5x$ to a constant $30$:

$$LHL = \lim_{x 	o 5^-} (b + 5x) = b + 25$$

$$RHL = f(5) = 30$$

Equating these gives us $b + 25 = 30$, which immediately reveals that $b = 5$.

### The Reality Check

Now, let's see if these values hold up. If $b = 5$, we can substitute this into our first equation, $a + b = 5$, to find that $a = 0$.

So, we have a candidate solution: $a = 0$ and $b = 5$. But wait! We have a third equation, $a + 2b = 15$, that we haven't fully tested yet.

Let's plug our values into this equation:

$$LHS = a + 2b = 0 + 2(5) = 10$$

$$RHS = 15$$

Since $10 
eq 15$, our system is inconsistent. The math is telling us that it is physically impossible to connect all these segments at the same time.

No matter how you choose $a$ and $b$, you will always find a gap at one of the junctions. Therefore, the function is **not continuous for any values of $a$ and $b$**.

This is the beauty of such problems—they teach us that sometimes, the answer isn't a number, but the realization that a condition cannot be met.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : R \to R$ be a function defined as : $f (x) = ⎩ ⎨ ⎧ 5, a + b x, b + 5 x, 30, if x \leq 1 if 1 < x < 3 if 3 \leq x < 5 if x \geq 5$ . Then, f is :

Select Answer:

Visualized Solution

The Piecewise Function

Condition for Continuity

Checking at $x = 1$

First Equation

Checking at $x = 3$

Second Equation

Checking at $x = 5$

Third Equation

Finding the value of $a$

The Consistency Check

Verifying Equation 2

Final Conclusion

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

The Illusion of a Smooth Path

The Mathematical Blueprint

Analyzing the Junctions

The Reality Check

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