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

Continuity at a Point

Imagine drawing the graph of a function and putting your pen down at $x = a$. If the function is continuous there, you can draw straight through that point without lifting your pen — no jump, no hole, no break. That is the intuition; here is the precision.

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The Three-Condition Test

For $f(x)$ to be continuous at $x = a$, all three must hold. If even one fails, $f$ is discontinuous there.

Condition 1 says $a$ is in the domain — the pen must have somewhere to land. Condition 2 says the curve approaches a single value from both sides — no jump. Condition 3 says that common approach value actually matches the function's value at $a$ — no misplaced point.

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Why All Three Are Needed

$f(x) = \dfrac{x^2 - 1}{x - 1}$ has $\lim_{x \to 1} f(x) = 2$, yet $f(1)$ is undefined (zero denominator). Condition 1 fails, leaving a hole at $(1, 2)$.

A piecewise function shows the opposite can be fine:

Here $f(2) = 3$, both one-sided limits equal $3$, and they match $f(2)$ — so all three hold and $f$ is continuous at $x = 2$.

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Common Pitfalls

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A Quick Check …