JEE Main 2026 Jan 23 Shift 2, Mathematics Q1: Continuity of Function

JEE Main2026Jan 23, Shift 2Mathematics

If $f (x) = ⎩ ⎨ ⎧ \frac{a ∣ x ∣ + x ^{2} - 2 ( sin ∣ x ∣ ) ( cos ∣ x ∣ )}{x}, x \neq = 0 b, x = 0$ is continuous at $x = 0$ , then $a + b$ is equal to :

Solution

Note $2 sin ∣ x ∣ cos ∣ x ∣ = sin (2∣ x ∣)$ . For continuity at $0$ , the left-hand and right-hand limits must equal $f (0) = b$ .

Right-hand limit $(x \to 0^{+})$ : $∣ x ∣ = x$ , so

$lim_{x \to 0^{+}} \frac{a x + x ^{2} - sin 2 x}{x} = lim_{x \to 0^{+}} (a + x - \frac{sin 2 x}{x}) = a + 0 - 2 = a - 2$ .

Left-hand limit $(x \to 0^{-})$ : $∣ x ∣ = - x$ , so

$lim_{x \to 0^{-}} \frac{- a x + x ^{2} - sin ( - 2 x )}{x} = lim_{x \to 0^{-}} (- a + x + \frac{sin 2 x}{x}) = - a + 0 + 2 = 2 - a$ .

Setting LHL = RHL: $a - 2 = 2 - a \Rightarrow a = 2$ , and so $b = 0$ .

$∴ a + b = 2$ .

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JEE Main 2026 Jan 23 Shift 2, Mathematics Q1: Continuity of Function

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