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JEE Main 201412 AprilOnline
Exam
JEE Main
Year
2014
Date
12 April
Mode
Online
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2014 · 12 April · Online · Mathematics · Inverse Trigonometric Functions

Statement I: The equation ( ^ -1 x)^3+ ( ^ -1 x)^3-a π^3=0 has a...

Statement I: The equation (sin⁡−1x)3+(cos⁡−1x)3−aπ3=0\left(\sin ^{-1} \mathrm{x}\right)^3+ \left(\cos ^{-1} \mathrm{x}\right)^3-\mathrm{a} \pi^3=0 has a solution for all a≥132\mathrm{a} \geq \frac{1}{32}. Statement II: For any x∈R\mathrm{x} \in \mathrm{R}, sin⁡−1x+cos⁡−1x=π2\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} and 0≤(sin⁡−1x−π4)2≤9π2160 \leq\left(\sin ^{-1} x-\frac{\pi}{4}\right)^2 \leq \frac{9 \pi^2}{16}

Options

  1. ABoth statements I and II are true.
  2. BBoth statements I and II are false.
  3. CStatement I\mathrm{I} is true and statement II\mathrm{II} is false.
  4. DStatement I\mathrm{I} is false and statement II is true.

Correct answer

A. Both statements I and II are true.

Step-by-step solution

sin⁡−1x∈[−π2,π2]⇒−3π4≤(sin⁡−1x−π4)≤π40≤(sin⁡−1x−π4)2≤916π2\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}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.\\ &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}Putting this value in equation (1) 0 \leq π248(\frac{\pi^2}{48}(32 a-1) \leq \frac{9}{16} \pi^2 ⇒0≤32a−1≤27132\leqa≤78\begin{aligned}&\Rightarrow 0 \leq 32 a-1 \leq 27 \\ &\frac{1}{32} \leqa \leq \frac{7}{8} \end{aligned}Statement-I is also true.
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