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Limits: Roots

JEE Maths question with a full step-by-step solution.

Question
If α,β\alpha,\beta are roots of x2−1−cos⁡2θ x+θ=0x^{2}-\sqrt{1-\cos 2\theta}\,x+\theta=0 where 0<θ<π20<\theta<\dfrac{\pi}{2}, then
lim⁡θ→0−(1α+1β)= ?\displaystyle\lim_{\theta\to 0^{-}}\left(\dfrac{1}{\alpha}+\dfrac{1}{\beta}\right)=\,?
A12\dfrac{1}{\sqrt{2}}
B−2-\sqrt{2}correct
C2\sqrt{2}
DDoesn't exist
Solution
Step 1: By Vieta's formulas: α+β=1−cos⁡2θ\alpha+\beta=\sqrt{1-\cos 2\theta} and αβ=θ\alpha\beta=\theta. Step 2: Simplify using 1−cos⁡2θ=2sin⁡2θ1-\cos 2\theta=2\sin^{2}\theta:
1−cos⁡2θ=2sin⁡2θ=2 ∣sin⁡θ∣\sqrt{1-\cos 2\theta}=\sqrt{2\sin^{2}\theta}=\sqrt{2}\,|\sin\theta|
 Note: The square root gives ∣sin⁡θ∣|\sin\theta|, not just sin⁡θ\sin\theta, because x2=∣x∣\sqrt{x^{2}}=|x|. Step 3: Express the required quantity:
1α+1β=α+βαβ=2 ∣sin⁡θ∣θ\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{\alpha+\beta}{\alpha\beta}=\dfrac{\sqrt{2}\,|\sin\theta|}{\theta}
 Step 4: As θ→0−\theta\to 0^{-}, sin⁡θ<0\sin\theta<0, so ∣sin⁡θ∣=−sin⁡θ|\sin\theta|=-\sin\theta:
lim⁡θ→0−2 (−sin⁡θ)θ=−2⋅1=−2\lim_{\theta\to 0^{-}}\dfrac{\sqrt{2}\,(-\sin\theta)}{\theta}=-\sqrt{2}\cdot 1=-\sqrt{2}
 Correct answer: (2) −2-\sqrt{2}
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