CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer If \[0\le x\le \pi \] and \[{{81}^{{{\sin }^{2}}x}}+{{81}^{{{\cos }^{2}}x}}=30\], then \[x\] is equal to :

    A)  \[\frac{\pi }{6}\]                                             

    B)  \[\frac{\pi }{2}\]

    C)                  \[\frac{\pi }{4}\]                                             

    D)                  \[\frac{3\pi }{4}\]

    Correct Answer: A

    Solution :

    \[{{81}^{{{\sin }^{2}}x}}+{{81}^{{{\cos }^{2}}x}}=30\] \[\Rightarrow \]               \[{{81}^{{{\sin }^{2}}x}}+{{81}^{1-{{\sin }^{2}}x}}=30\] \[\Rightarrow \]               \[{{81}^{{{\sin }^{2}}x}}+\frac{81}{{{81}^{1-{{\sin }^{2}}x}}}=30\] Let \[{{81}^{{{\sin }^{2}}x}}=y\] \[\therefore \]                  \[y+\frac{81}{y}=30\] \[\Rightarrow \]               \[{{y}^{2}}-30y+81=0\] \[\Rightarrow \]               \[(y-27)\,(y-3)=0\] if,            \[y-27=0\]                                 \[y=27\]                 \[{{81}^{{{\sin }^{2}}x}}=27\]                                 \[{{3}^{4{{\sin }^{2}}x}}={{3}^{3}}\]                                 \[{{\sin }^{2}}x=\frac{3}{4}\]                                 \[\sin x=\frac{\sqrt{3}}{2}=\sin \frac{\pi }{3}\]                 \[\Rightarrow \]               \[x=\pi /3\]         Now if, \[y-3=0\]                 \[y=3\]                 \[{{81}^{{{\sin }^{2}}x}}=3\]                 \[{{3}^{4{{\sin }^{2}}x}}={{3}^{1}}\] \[\Rightarrow \]               \[4\,{{\sin }^{2}}x=1\] \[\Rightarrow \]                               \[\sin x=\frac{1}{2}=\sin \frac{\pi }{6}\] \[\Rightarrow \]                               \[x=\frac{\pi }{6}\]

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