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Directions: This section contains integer type questions. The answer to each of the question is a single digit integer, ranging from 0 to 9. Choose the correct option.If sin-1(6x/1+9x2) =2 tan-1(ax) then a =(a) 3 (b) 8 (c) 6 (d) 9 |
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Answer» (a) \(sin^{-1}\left(\frac{6x}{1+9x^2}\right)=sin^{-1}\left(\frac{2(3x)}{1+(3x)^2}\right)\) Let 3x \(=tan\theta\) \(\Rightarrow\theta=tan^{-1}(3x)\) Then, \(sin^{-1}\left(\frac{2\times3x}{1+(3x)^2}\right)=sin^{-1}\left(\frac{2tan\theta}{1+tan^2\theta}\right)\) \(=sin^{-1}(sin\,2\theta)\) \((\because sin\,2\theta=\frac{2tan\theta}{1+tan^2\theta})\) \(=2\theta\) \(=2\,tan^{-1}3x\,\,(\because\theta=tan^{-1}3x)\) \(=2tan^{-1}3x=2tan^{-1}(ax)\) \(\therefore\) a = 3 |
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