Continuity

Continuity at the origin demands that the assigned value k equal the limit of the expression as x \to 0, so the task is to evaluate that limit. Rationalize by multiplying numerator and denominator by \sqrt{1+x} + \sqrt{1-x}: the numerator becomes (1+x) - (1-x) = 2x, giving f(x) = \frac{2x}{x(\sqrt{1+x} + \sqrt{1-x})} = \frac{2}{\sqrt{1+x} + \sqrt{1-x}}. As x \to 0 the denominator tends to 1 + 1 = 2, so the limit is 2/2 = 1. Hence k = 1. Option 0 wrongly assumes the leading behaviour cancels to zero, but the cancellation is between equal-order roots, not a vanishing one. Option 1/2 stops after halving without completing the rationalization. Option 2 forgets to divide by the denominator sum of the two roots. The governing JEE Advanced pattern is conjugate rationalization to remove a 0/0 surd indeterminacy. Plausibility check: at x = 0.01, f \approx (1.00499 - 0.99499)/0.01 = 1.00003, confirming the limit and the continuity value k = 1.