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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

For the least possible value of n ∈ Z the solution (x, y) of the...

For the least possible value of n∈Zn \in \mathbf{Z} the solution (x,y)(x, y) of the equations cos⁡−1x+(sin⁡−1y)2=nπ24\cos ^{-1} x+\left(\sin ^{-1} y\right)^2=\frac{n \pi^2}{4} and cos⁡−1x(sin⁡−1y)2=π416\cos ^{-1} x\left(\sin ^{-1} y\right)^2=\frac{\pi^4}{16}, is

Options

  1. A(π24,±1)\left(\frac{\pi^2}{4}, \pm 1\right)
  2. B(π24,sin⁡π216)\left(\frac{\pi^2}{4}, \sin \frac{\pi^2}{16}\right)
  3. C(cos⁡(π24),±1)\left(\cos \left(\frac{\pi^2}{4}\right), \pm1\right)
  4. D(sin⁡(π24),cos⁡π4)\left(\sin \left(\frac{\pi^2}{4}\right), \cos \frac{\pi}{4}\right)

Correct answer

C. (cos⁡(π24),±1)\left(\cos \left(\frac{\pi^2}{4}\right), \pm1\right)

Step-by-step solution

\left(\sin ^{-1} y\right)^2+\cos ^{-1} x=\frac{n \pi^2}{4}and(cos⁡−1x)(sin⁡−1y)2=π416and \left(\cos ^{-1} x\right)\left(\sin ^{-1} y\right)^2=\frac{\pi^4}{16} Let cos⁡−1x=a\cos ^{-1} x=a and sin⁡−1y=b⇒a+b2=nπ24…(i)a⋅b2=π416…(ii)∴a−b=(a+b2)2−4ab=n2π416−4π416\sin ^{-1} y=b \Rightarrow a+b^2=\frac{n \pi^2}{4} \ldots(i)a \cdot b^2=\frac{\pi^4}{16} \ldots(ii)\therefore \quad a-b=\sqrt{\left(a+b^2\right)^2-4 a b}=\sqrt{\frac{n^2 \pi^4}{16}-\frac{4 \pi^4}{16}} Now, least value of nn can be 2 a−b2=4π416−4π416⇒a−b2=0…(iii)a-b^2=\sqrt{\frac{4 \pi^4}{16}-\frac{4 \pi^4}{16}} \Rightarrow a-b^2=0 \ldots(iii) Adding Eqs. (i) and (iii), we get 2a=nπ24 or 2a=2π24(∵n=2)2 a=\frac{n \pi^2}{4} \text { or } 2 a=\frac{2 \pi^2}{4} \quad(\because n=2) or a=π24⇒cos⁡−1x=π24\Rightarrowx=cos⁡\quad a=\frac{\pi^2}{4} \Rightarrow \cos ^{-1} x=\frac{\pi^2}{4} \Rightarrowx=\cos\left(\frac{\pi^2}{4}\right) Now, from Eq. (iii), a=b\left(\sin ^{-1}y\right)^2=π24 or sin⁡−1\frac{\pi^2}{4} \text { or } \sin ^{-1}y= ±π2or\pm \frac{\pi}{2} or\quad y=\sin \left( \pm \frac{\pi}{2}\right)orory= \pm 1Therefore,orderedpairis(cos⁡π24,±1)Therefore, ordered pair is \left(\cos \frac{\pi^2}{4}, \pm 1\right).
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