Q.If $f(x) = \begin{cases} mx + 1, & x \le \dfrac{\pi}{2} \\ \sin x + n, & x > \dfrac{\pi}{2} \end{cases}$ is continuous at $x = \dfrac{\pi}{2}$, then
(A) $m = 1,\ n = 0$
(B) $m = \dfrac{n\pi}{2} + 1$
(C) $n = \dfrac{m\pi}{2}$
(D) $m = n = \dfrac{\pi}{2}$

Concept: Continuity Condition — For $f$ to be continuous at $x = \frac{\pi}{2}$, the left-hand limit must equal the right-hand limit, and both must equal $f(\frac{\pi}{2})$.

Step 1: Left-hand limit (using $x \le \frac{\pi}{2}$):

Step 2: Right-hand limit (using $x > \frac{\pi}{2}$):

$\lim_{x \to \frac{\pi}{2}^+} f(x) = \sin\left(\frac{\pi}{2}\right) + n = 1 + n$ …

The core idea is the Continuity Condition: a function is continuous at a point if the limit from the left equals the limit from the right, and both equal the function's value at that point. For a piecewise function that changes its rule at a boundary, this condition forces a relationship between the parameters on either side.

Here, the boundary is $x = \frac{\pi}{2}$. The function is defined by $mx + 1$ for $x \le \frac{\pi}{2}$ and by $\sin x + n$ for $x > \frac{\pi}{2}$. At $x = \frac{\pi}{2}$ itself, the definition uses the first piece (since $x \le \frac{\pi}{2}$ includes the equality). So $f(\frac{\pi}{2}) = m \cdot \frac{\pi}{2} + 1$.

Now, for continuity, the limit as $x$ approaches $\frac{\pi}{2}$ from the left must equal the limit from the right, and both must equal that function value.

1. Left-hand limit ($x \to \frac{\pi}{2}^-$): For $x$ just less than $\frac{\pi}{2}$, the function is $mx + 1$. Since this is a polynomial (hence continuous everywhere), the limit is simply the value at $x = \frac{\pi}{2}$:

2. Right-hand limit ($x \to \frac{\pi}{2}^+$): For $x$ just greater than $\frac{\pi}{2}$, the function is $\sin x + n$. The sine function is continuous everywhere, so the limit is:

3. Continuity condition: We require:

The left-hand limit already equals $f(\frac{\pi}{2})$ by definition, so the key equation is:

4. Simplify: Subtract $1$ from both sides:

So $n = \frac{m\pi}{2}$. …

Method: Finding a Parameter Relation from Continuity of a Piecewise Function

Use this whenever a piecewise function involves unknown constants and you're told it is continuous at the point where the pieces join.

Steps

Step 1: Identify exactly which piece defines the function AT the boundary point, based on which inequality includes equality ($\le$ or $\ge$).

That piece gives you $f(a)$ directly — do not use the other piece for this value.

Step 2: Compute the left-hand limit using the piece that applies for $x$ approaching $a$ from below.

Step 3: Compute the right-hand limit using the piece that applies for $x$ approaching $a$ from above.

Step 4: Apply the continuity condition — set left-hand limit $=$ right-hand limit $= f(a)$.

$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$ …

Common Mistakes

Mistake 1: Using the wrong piece of the definition to compute $f\left(\frac{\pi}{2}\right)$ itself.

Why it's wrong: since the first piece is defined for $x \le \frac{\pi}{2}$ (which includes equality), $f\left(\frac{\pi}{2}\right)$ must come from $mx+1$, not from $\sin x + n$ — using the second piece here changes the whole equation and leads to a wrong relation between $m$ and $n$.

Correct approach: always check carefully which inequality includes the boundary point (the $\le$ or $\ge$ side) before deciding which formula gives $f(a)$.

Mistake 2: Expecting the continuity condition to pin down unique numeric values for both $m$ and $n$. …