Let f: R → R be a function given by
f(x)= (1-cos 2 x)/x², x < 0 ; α, x=0, ; (β √(1-cos x))/x, x>0
where α, β ∈ R. If f is continuous at x=0, then α²+β² is equal to :

Solution

Maths sir · 4 steps

Use 1-cos θ = 2 sin² θ/2 on both sides, then match both limits to α.

1. Left-hand limit. For x < 0, write 1-cos 2x = 2 sin² x. Then (2 sin² x)/x² = 2((sin x)/x)² → 2 as x → 0⁻.
2. Right-hand limit. For x > 0, √(1-cos x) = √2 |sin x/2|, and sin x/2 > 0 for small positive x. So the limit becomes the one below.

   lim(x → 0⁺) (β√2 sin x/2)/x = lim(x → 0⁺) β√2/2·(sin x/2)/(x/2) = β/√2

3. Match the limits. Continuity needs both limits to equal f(0). So α = 2, and β/√2 = 2 gives β = 2√2, so β² = 8.
4. So. α² + β² = 4 + 8 = 12, which is option (d).

The idea to keep. For continuity at a point, equate the left limit, the right limit and f(a). When you see 1-cos θ, rewrite it as 2 sin² θ/2 and form the standard (sin t)/t limit.

Answer from the NTA final answer key. Solution by PrepWiser.