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← JEE Main 2014 · 12 Apr · Shift 1
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JEE Main 2014 · 12 Apr · Shift 1, Mathematics question 88

Statement I : The equation (sin⁡−1x)3+(cos⁡−1x)3−aπ3=0(\sin^{-1}x)^3+(\cos^{-1}x)^3-a\pi^3=0 has a solution for all a≥132a\ge\dfrac1{32}.

Statement II : For any x∈Rx\in\mathbb R, sin⁡−1x+cos⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\dfrac\pi2 and 0≤(sin⁡−1x−π4)2≤9π2160\le\left(\sin^{-1}x-\dfrac\pi4\right)^2\le\dfrac{9\pi^2}{16}.

Choose one option
Answer and worked solution

Official answer: D

Let u=sin⁡−1x∈[−π2,π2]u=\sin^{-1}x\in[-\frac\pi2,\frac\pi2]. Then
u3+(π2−u)3=π38−3π24u+3π2u2=π332+3π2(u−π4)2.u^3+\left(\tfrac\pi2-u\right)^3=\frac{\pi^3}8-\frac{3\pi^2}4u+\frac{3\pi}2u^2=\frac{\pi^3}{32}+\frac{3\pi}2\left(u-\frac\pi4\right)^2.
With 0≤(u−π4)2≤9π2160\le(u-\frac\pi4)^2\le\frac{9\pi^2}{16} (Statement II, true), aa ranges over [132,78]\left[\frac1{32},\frac78\right]. So it is not every a≥132a\ge\frac1{32}: Statement I is false.

Topics:Inverse trigonometric functions
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