JEE Main & Advanced Sample Paper JEE Main Sample Paper-38

  • question_answer
    Statement-1: If\[1/2\le x\le 1\], then \[{{\cos }^{-1}}x+{{\cos }^{-1}}\left[ \frac{x}{2}+\frac{\sqrt{3-3{{x}^{2}}}}{2} \right]\]is equal to\[\pi /3\]
    Statement 2:\[{{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)=2{{\sin }^{-1}}x,\]if\[x\in \left[ -1/\sqrt{2},\,\,1/\sqrt{2} \right]\]

    A)  Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

    B)  Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

    C)  Statement-1 is true, Statement-2 is false.

    D)  Statement-1 is false, Statement-2 is true.

    Correct Answer: B

    Solution :

     In statement \[1\] put \[x=\cos \theta \] then\[0\le \theta \le \pi /3\] \[L.H.S={{\cos }^{-1}}(\cos \theta )+{{\cos }^{-1}}\left[ \frac{1}{2}\cos \theta +\frac{\sqrt{3}}{2}\sin \theta  \right]\]\[=\theta +{{\cos }^{-1}}(\cos (\pi /3-\theta ))=\theta +\pi /3-\theta =\frac{\pi }{3}\]so statement \[1\] is true. In statement \[2\], put \[x=\sin \theta \] then\[-\pi /4\le \theta \le \pi /4\] so\[{{\sin }^{-1}}(2x\sqrt{1-{{x}^{2}}})={{\sin }^{-1}}(2\sin \theta \cos \theta )\] \[={{\sin }^{-1}}(\sin 2\theta )=2\theta =2{{\sin }^{-1}}x\]

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