JEE Main 2014 — Mathematics Trigonometry

$$\begin{aligned} &\sin ^{-1} x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\ &\Rightarrow-\frac{3 \pi}{4} \leq\left(\sin ^{-1} x-\frac{\pi}{4}\right) \leq \frac{\pi}{4} \\ &0 \leq\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2 \leq \frac{9}{16} \pi^2 \end{aligned}$$​sin−1x∈[−2π​,2π​]⇒−43π​≤(sin−1x−4π​)≤4π​0≤(sin−1x−4π​)2≤169​π2​ Statement II is true $$\begin{aligned} &\left(\sin ^{-1} x\right)^3+\left(\cos ^{-1} x\right)^3=a \pi^3 \\ \Rightarrow &\left(\sin ^{-1} x+\cos ^{-1} x\right)\left[\left(\sin ^{-1} x+\cos ^{-1} x\right)^2-\right.\\ &\left.3 \sin ^{-1} x \cos ^{-1} x\right]=a \pi^3 \\ \Rightarrow & \frac{\pi^2}{4}-3 \sin ^{-1} x \cos ^{-1} x=2 a \pi^2 \\ \Rightarrow & \sin ^{-1} x\left(\frac{\pi}{2}-\sin ^{-1} x\right)=\frac{\pi^2}{12}(1-8 a) \\ \Rightarrow &\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2=\frac{\pi^2}{12}(8 a-1)+\frac{\pi^2}{16} \\ \Rightarrow &\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2=\frac{\pi^2}{48}(32 a-1) \end{aligned}$$⇒⇒⇒⇒⇒​(sin−1x)3+(cos−1x)3=aπ3(sin−1x+cos−1x)[(sin−1x+cos−1x)2−3sin−1xcos−1x]=aπ34π2​−3sin−1xcos−1x=2aπ2sin−1x(2π​−sin−1x)=12π2​(1−8a)(sin−1x−4π​)2=12π2​(8a−1)+16π2​(sin−1x−4π​)2=48π2​(32a−1)​ Putting this value in equation (1) $$0 \leq \frac{\pi^2}{48}(32 a-1) \leq \frac{9}{16} \pi^2$$0≤48π2​(32a−1)≤169​π2 $$\begin{aligned} &\Rightarrow 0 \leq 32 a-1 \leq 27 \\ &\frac{1}{32} \leq a \leq \frac{7}{8} \end{aligned}$$​⇒0≤32a−1≤27321​≤a≤87​​ Statement-I is also true.