Q.If a function defined by 𝑓(𝑥) = { 𝑘𝑥 + 1, 𝑥 ≤ 𝜋 cos 𝑥 , 𝑥 > 𝜋 is continuous at 𝑥 = 𝜋, then the value of 𝑘 is
(A) 𝜋
(B) −1 𝜋
(C) 0
(D) −2 𝜋

Concept: Continuity At A Point — A function is continuous at $x = \pi$ if the left-hand limit, right-hand limit, and $f(\pi)$ are all equal.

Step 1: For $x \le \pi$, $f(x) = kx + 1$, so

$f(\pi) = k\pi + 1$ and $\lim_{x \to \pi^-} f(x) = k\pi + 1$.

Step 2: For $x > \pi$, $f(x) = \cos x$, so

$\lim_{x \to \pi^+} f(x) = \cos \pi = -1$.

Step 3: Continuity at $x = \pi$ requires …

The idea of continuity at a point is beautifully simple: a function is continuous at $x = a$ if you can draw its graph through that point without lifting your pen. More formally, three things must match — the function's value at $a$, the limit as you approach from the left, and the limit as you approach from the right. If any one of these is different, there's a break, a jump, or a hole.

Here, the function is defined in two pieces, meeting at $x = \pi$. The left piece is $kx + 1$ (a straight line), and the right piece is $\cos x$ (a wavy curve). For continuity at the seam, the line must exactly meet the curve at $x = \pi$.

Let's work through it step by step.

1. Find the left-hand limit as $x \to \pi^-$. For $x \leq \pi$, the function is $f(x) = kx + 1$. So as we approach $\pi$ from values slightly less than $\pi$, we use this expression:

2. Find the right-hand limit as $x \to \pi^+$. For $x > \pi$, the function is $f(x) = \cos x$. Approaching $\pi$ from the right, we get:

And $\cos \pi = -1$. So the right-hand limit is $-1$.

3. Find the function's value at $x = \pi$. Since the definition says $f(x) = kx + 1$ for $x \leq \pi$, the point $x = \pi$ itself belongs to the left piece. So:

4. Apply the continuity condition. For continuity at $x = \pi$, we need: $\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^+} f(x) = f(\pi).$ …

Method: Solving for an Unknown in a Piecewise Function Using the Continuity Condition

This method applies to any problem where a piecewise function has an unknown constant, and you're told the function is continuous at the point where the pieces meet — you then find the constant.

Steps

Step 1: Identify the pieces and the junction point

Write down which formula applies just left of the junction and which applies just right of it, and note carefully which side includes the equality (e.g. $x \le a$ vs. $x < a$) — that tells you which piece actually defines $f(a)$.

Step 2: Compute the one-sided limits at the junction

For a piece built from standard continuous functions (polynomials, trig functions, exponentials), the one-sided limit equals direct substitution of the junction value into that piece:

Step 3: Apply the continuity condition

Continuity at $a$ requires all three of $f(a)$, the left-hand limit, and the right-hand limit to agree. Since the piece containing the equality already supplies $f(a)$ and matches its own one-sided limit automatically, the real content of the condition is usually just: …

Common Mistakes

Mistake 1: Evaluating $\cos \pi$ incorrectly

Some students recall $\cos 0 = 1$ and mistakenly carry that value over, writing $\cos \pi = 1$ instead of $\cos \pi = -1$. Why it's wrong: $\cos \pi$ is a distinct standard value (the angle $\pi$ radians points in the opposite direction on the unit circle). Correct approach: memorize the standard values $\cos 0=1, \cos(\pi/2)=0, \cos\pi=-1$ precisely, or sketch the cosine curve to check the sign at $\pi$.

Mistake 2: Using the wrong piece to evaluate $f(a)$ at the junction …