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AIEEE 2011

JEE Main 2011 Mathematics question on Limits Continuity and Differentiability

If lim⁡x→0[1+xlog⁡(1+b2)]1/x=2bsin⁡2θ\lim_{x \to 0} [1 + x \log(1 + b^2)]^{1/x} = 2b\sin^2\theta, b>0b > 0 and θ∈(−π,π]\theta \in (-\pi, \pi], then the value of θ\theta is
  1. A.
    ±π4\pm \dfrac{\pi}{4}
  2. B.
    ±π3\pm \dfrac{\pi}{3}
  3. C.
    ±π6\pm \dfrac{\pi}{6}
  4. D.
    ±π2\pm \dfrac{\pi}{2}
Show answer and solution

Correct option: D

Solution

The limit is of the form 1^∞. Using lim[1 + f(x)]^(1/x) = e^(lim f(x)/x), we get:

lim[1 + x·log(1+b²)]^(1/x) = e^(log(1+b²)) = 1 + b²

So the equation becomes 1 + b² = 2b·sin²θ.

Rearranging: 2b·sin²θ = 1 + b² ≥ 2b (by AM-GM, since b > 0), so sin²θ ≥ 1. But sin²θ ≤ 1 always, so sin²θ = 1, which forces equality in AM-GM, meaning b = 1.

With sin²θ = 1, we get sinθ = ±1, so θ = ±π/2.

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