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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

BITSAT 2013 · Mathematics · Inverse Trigonometric Functions

The greatest and least values of ( ^ -1 x)^ 2 +( ^ -1 x)^ 2 are...

The greatest and least values of $\left(\sin ^{-1} x\right)^{2}+\left(\cos ^{-1} x\right)^{2} are respectively

Options

  1. Aπ24\frac{\pi^{2}}{4} and 0
  2. Bπ2\frac{\pi}{2} and −π2\frac{-\pi}{2}
  3. C5π24\frac{5 \pi^{2}}{4} and π28\frac{\pi^{2}}{8}
  4. Dπ24\frac{\pi^{2}}{4} and −π24\frac{-\pi^{2}}{4}

Correct answer

C. 5π24\frac{5 \pi^{2}}{4} and π28\frac{\pi^{2}}{8}

Step-by-step solution

We have, (sin⁡−1x)2+(cos⁡−1x)2=(sin⁡−1x+cos⁡−1x)2−2sin⁡−1x⋅cos⁡−1x=π24−2sin⁡−1x(π2−sin⁡−1x)=π24−πsin⁡−1x+2(sin⁡−1x)2=2[(sin⁡−1x)2−π2sin⁡−1x+π28]=2[(sin⁡−1x−π4)2+π216]Thus,theleastvalueis2(π216)\left(\sin ^{-1} x\right)^{2}+\left(\cos ^{-1} x\right)^{2}=\left(\sin ^{-1} x+\cos ^{-1} x\right)^{2}-2 \sin ^{-1} x \cdot \cos ^{-1} x=\frac{\pi^{2}}{4}-2 \sin ^{-1} x\left(\frac{\pi}{2}-\sin ^{-1} x\right)=\frac{\pi^{2}}{4}-\pi \sin ^{-1} x+2\left(\sin ^{-1} x\right)^{2}=2\left[\left(\sin ^{-1} x\right)^{2}-\frac{\pi}{2} \sin ^{-1} x+\frac{\pi^{2}}{8}\right]=2 \left[\left(\sin ^{-1}x-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{16}\right] Thus, the least value is 2 \left(\frac{\pi^{2}}{16}\right) i.e. π28\frac{\pi^{2}}{8} and the greatest value is 22\left[ \left(\frac{-\pi}{2}-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{16}\right]i.e.i.e.\frac{5 \pi^{2}}{4} .
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