JEE Main & Advanced Sample Paper JEE Main Sample Paper-5

  • question_answer
    The greater of the two angles \[A=2{{\tan }^{-1}}(2\sqrt{2}-1)\]and \[B=3{{\sin }^{-1}}\left( \frac{1}{3} \right)+\]\[{{\sin }^{-1}}\left( \frac{3}{5} \right)\] is

    A)  B                                           

    B)  A

    C)  C                                           

    D)  None of these

    Correct Answer: B

    Solution :

     Idea \[\because \]\[x=\sin \theta ,\theta ={{\sin }^{-1}}x,-1\le x\le 1\]and \[-\frac{\pi }{2}\le \theta \le \frac{\pi }{2},x=\tan \theta ,\theta ={{\tan }^{-1}}x,-\infty <x<\infty \]and\[\frac{-\pi }{2}<\theta <\frac{\pi }{2}\] Given angles are A = 2 \[{{\tan }^{-1}}(2\sqrt{2}-1)\] and    \[B=3{{\sin }^{-1}}\left( \frac{1}{3} \right)+{{\sin }^{-1}}\left( \frac{3}{5} \right)\] Consider \[A=2{{\tan }^{-1}}(2\sqrt{2}-1)=2{{\tan }^{-1}}(1.828)\] \[\therefore \]\[A>2{{\tan }^{-1}}\sqrt{3}\]       \[(\because \sqrt{3}=1.732<1.828)\] \[\Rightarrow \]               \[A>\frac{2\pi }{3}\]                                       ?(i) Also       \[{{\sin }^{-1}}\left( \frac{1}{3} \right)<{{\sin }^{-1}}\left( \frac{1}{2} \right)=\frac{\pi }{6}\] \[\Rightarrow \]               \[3{{\sin }^{-1}}\left( \frac{1}{3} \right)<\frac{\pi }{6}\] So, using the identity \[\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta ,\]we have \[{{\sin }^{-1}}\left( \frac{1}{3} \right)={{\sin }^{-1}}\left( 3\times \frac{1}{3}-4{{\left( \frac{1}{3} \right)}^{3}} \right)={{\sin }^{-1}}\left( \frac{23}{27} \right)\] \[={{\sin }^{-1}}(0.852)\] \[\therefore \]\[3{{\sin }^{-1}}\left( \frac{1}{3} \right)<{{\sin }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\left\{ \because \frac{\sqrt{3}}{2}=0.868>0.852 \right\}\]i.e., \[3{{\sin }^{-1}}\left( \frac{1}{3} \right)<\frac{\pi }{3}\] also\[{{\sin }^{-1}}\left( \frac{3}{5} \right)={{\sin }^{-1}}(0.6)<{{\sin }^{-1}}\left( \frac{\sqrt{3}}{2} \right)=\frac{\pi }{3}\] \[\therefore \]  \[{{\sin }^{-1}}\left( \frac{3}{5} \right)=\frac{\pi }{3}\] \[\therefore \]  \[B=3{{\sin }^{-1}}\left( \frac{3}{5} \right)+\sin \left( \frac{3}{5} \right)\] \[<\frac{\pi }{3}+\frac{\pi }{3}=\frac{2\pi }{3}\]                  \[\therefore \]\[B<\frac{2\pi }{3}\] Thus, from Eqs. (i) and (iii), we have A > B \[\therefore \] A is the greater angle. TEST Edge Application of inverse trigonometric function based questions are asked. To solve such type of question, students are advised to understand the concept of these functions.

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