Limits, Continuity and Differentiability – Continuity – JEE Main 23 Jan 2026 Shift 2

Solution:

For the function to be continuous at $x = 0$, we must have:
$$ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = b $$

First, we calculate the Left Hand Limit (LHL) at $x = 0$. Let $x = -h$ where $h \to 0$:
$$ \lim_{h \to 0} \frac{a|-h| + (-h)^2 – \sin(2|-h|)}{-h} $$

$$ = \lim_{h \to 0} \frac{ah + h^2 – \sin(2h)}{-h} $$

Using standard limits $\lim_{h \to 0} \frac{\sin 2h}{2h} = 1$:
$$ = \lim_{h \to 0} \left( -a – h + 2 \cdot \frac{\sin 2h}{2h} \right) = -a + 2 $$

Now, we calculate the Right Hand Limit (RHL) at $x = 0$. Let $x = h$ where $h \to 0$:
$$ \lim_{h \to 0} \frac{ah + h^2 – \sin(2h)}{h} $$

$$ = \lim_{h \to 0} \left( a + h – 2 \cdot \frac{\sin 2h}{2h} \right) = a – 2 $$

Equating LHL and RHL to $f(0) = b$:
$$ -a + 2 = a – 2 = b $$

From $-a + 2 = a – 2$, we get:
$$ 2a = 4 \Rightarrow a = 2 $$

Substituting $a = 2$ into the equation for $b$:
$$ b = 2 – 2 = 0 $$

Finally, the value of $a + b$:
$$ a + b = 2 + 0 = 2 $$

Ans. (2)