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JEE Advanced 2007Paper 2
Exam
JEE Advanced
Year
2007
Paper
Paper 2
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Advanced 2007 · Paper 2 · Mathematics · Inverse Trigonometric Functions

Let (x, y) be such that ^ -1 (a x)+ ^ -1 (y)+ ^ -1 (b x y)=(π)/(2)...

Let (x,y)(x, y) be such that sin⁡−1(ax)+cos⁡−1(y)+cos⁡−1(bxy)=π2\sin ^{-1}(a x)+\cos ^{-1}(y)+\cos ^{-1}(b x y)=\frac{\pi}{2}. Match the statements in Column I with the values in Column II. Let (x, y) be such that ^ -1 (a x)+ ^ -1 (y)+ ^ -1 (b x y)=(π)/(2) . Match the statements in Column I with the values

Options

  1. AA-p; B-q; C-q; D-p
  2. BA-p; B-q; C-p; D-s
  3. CA-s; B-p; C-q; D-p
  4. DA-r; B-q; C-p; D-r

Correct answer

B. A-p; B-q; C-p; D-s

Step-by-step solution

(A)Ifa=1andb=0If a=1 and b=0, then sin⁡−1x+cos⁡−1y=0\sin ^{-1} x+\cos ^{-1} y=0 \Rightarrow \quad \sin ^{-1} x=-\cos ^{-1} y \Rightarrow x^2+y^2=1 (B)Ifa=1andb=1If a=1 and b=1, then sin⁡−1x+cos⁡−1y+cos⁡−1xy=π2⇒cos⁡−1x−cos⁡−1y=cos⁡−1xy⇒xy+1−x21−y2=xy [taking sine on both the  sides] \begin{array}{rlrl} & & \sin ^{-1} x+\cos ^{-1} y+\cos ^{-1} x y & =\frac{\pi}{2} \\ \Rightarrow & & \cos ^{-1} x-\cos ^{-1} y & =\cos ^{-1} x y \\ \Rightarrow & x y+\sqrt{1-x^2} \sqrt{1-y^2} & =x y \quad \text{ [taking sine on both\text{ the } sides] } \end{array} (C)Ifa=1andb=2If a=1 and b=2, then sin⁡−1x+cos⁡−1y+cos⁡−1(2xy)=π2⇒cos⁡−1x−cos⁡−1y=cos⁡−1(2xy)⇒xy+1−x21−y2=2xy⇒x2+y2=1\begin{aligned} & & \sin ^{-1} x+\cos ^{-1} y+\cos ^{-1}(2 x y) & =\frac{\pi}{2} \\ \Rightarrow & & \cos ^{-1} x-\cos ^{-1} y & =\cos ^{-1}(2 x y) \\ \Rightarrow & & x y+\sqrt{1-x^2} \sqrt{1-y^2} & =2 x y \\ \Rightarrow & & x^2+y^2 & =1 \end{aligned} [on squaring] (D)Ifa=2andb=2If a=2 and b=2, then sin⁡−1(2x)+cos⁡−1(y)+cos⁡−1(2xy)=π2⇒2xy+1+4x21−y2=2xy⇒(4x2−1)(y2−1)=0.\begin{aligned} & & \sin ^{-1}(2 x)+\cos ^{-1}(y)+\cos ^{-1}(2 x y) & =\frac{\pi}{2} \\ \Rightarrow & & 2 x y+\sqrt{1+4 x^2} \sqrt{1-y^2} & =2 x y \\ \Rightarrow & & \left(4 x^2-1\right) \left(y^2-1\right) & =0 . \end{aligned}
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