Continuity

Continuity at the origin requires m to equal the limit of the quotient as x \to 0, so we must evaluate that trigonometric limit precisely. Use the half-angle identity 1 - \cos 4x = 2\sin^2 2x, turning the expression into \frac{2\sin^2 2x}{x^2}. Rewrite as 2 \left(\frac{\sin 2x}{2x}\right)^2 \cdot \frac{(2x)^2}{x^2} = 2 \cdot 1 \cdot 4 = 8 in the limit, since \sin 2x / 2x \to 1 and (2x)^2/x^2 = 4. Hence m = 8. Option 2 keeps only the leading factor and forgets the 4 from the doubled angle squared. Option 4 captures the angle factor but drops the constant 2 from the identity. Option 16 double-counts a factor of 2. The governing JEE pattern is the half-angle identity combined with the fundamental sine limit, a pairing that converts an awkward cosine-difference into a perfect square whose limit is already known. The reason the angle multiplier enters squared is structural: the cosine term carries the angle 4x, the half-angle identity halves it to 2x inside a sine, and forming the difference quotient with x^2 forces that 2x to be compared against x, contributing a factor of four. Recognising this scaling lets a student predict the answer for any coefficient without redoing the algebra, which is exactly the kind of pattern transfer JEE Advanced rewards. Plausibility check: replacing 4x by a general 2ax gives the limit 2a^2; with a = 2 this is 2 \cdot 4 = 8, internally consistent and confirming the continuity value m = 8.