Trigonometry · JEE Main previous year questions

2017 3 questions

1. Q212017 · 9 Apr 2017 · Shift 1

   A value of x satisfying the equation $\sin [{\cot }^{-1}(1+x)]=\cos [{\tan }^{-1}x]$, is :

   1. A.$-\frac{1}{2}$
   2. B.$-1$
   3. C.$0$
   4. D.$\frac{1}{2}$
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   Answer: A

2. Q232017 · 8 Apr 2017 · Shift 1

   If $$S=\left\{x\in [0,2\pi ]:\left|\begin{array}{ccc}0& \cos x& -\sin x\\ \sin x& 0& \cos x\\ \cos x& \sin x& 0\end{array}\right|=0\right\},$$ then ${\sum }_{x\in S}\tan \left(\frac{\pi }{3}+x\right)$ is equal to :

   1. A.$4+2\sqrt{3}$
   2. B.$-2+\sqrt{3}$
   3. C.$-2-\sqrt{3}$
   4. D.$4-2\sqrt{3}$
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   Answer: D

3. Q242017 · 8 Apr 2017 · Shift 1

   The value of ${\tan }^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right]$, $|x|<\frac{1}{\sqrt{2}}$, $x\ne 0$, is equal to :

   1. A.$\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}{x}^2$
   2. B.$\frac{\pi }{4}+{\cos }^{-1}{x}^2$
   3. C.$\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}{x}^2$
   4. D.$\frac{\pi }{4}-{\cos }^{-1}{x}^2$
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   Answer: A