JEE Main 2025 Jan 23 Shift 1, Mathematics Q3: Continuity of Function

JEE Main2025Jan 23, Shift 1Mathematics

If the function

$f (x) = ⎩ ⎨ ⎧ \frac{2}{x} {sin (k_{1} + 1) x + sin (k_{2} - 1) x}, x < 0 4, x = 0 \frac{2}{x} lo g_{e} (\frac{2 + k _{1} x}{2 + k _{2} x}), x > 0$

is continuous at $x = 0$ , then $k_{1}^{2} + k_{2}^{2}$ is equal to

A
$8$
B
$20$
C
$5$
D
$10$

Solution

For continuity at $0$ , both one-sided limits must equal $f (0) = 4$ .

Left limit: $lim_{x \to 0^{-}} \frac{2}{x} [sin (k_{1} + 1) x + sin (k_{2} - 1) x] = 2 (k_{1} + 1) + 2 (k_{2} - 1) = 2 (k_{1} + k_{2})$ . Setting this to $4$ gives $k_{1} + k_{2} = 2$ .

Right limit: $\frac{2}{x} ln (1 + \frac{( k _{1} - k _{2} ) x}{2 + k _{2} x}) \to \frac{2}{x} \cdot \frac{( k _{1} - k _{2} ) x}{2} = k_{1} - k_{2}$ as $x \to 0^{+}$ .

Setting this to $4$ gives $k_{1} - k_{2} = 4$ .

Solving: $k_{1} = 3$ , $k_{2} = - 1$ .

$∴ k_{1}^{2} + k_{2}^{2} = 9 + 1 = 10$ .

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JEE Main 2025 Jan 23 Shift 1, Mathematics Q3: Continuity of Function

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