CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer If \[{{\sin }^{-1}}\frac{x}{5}+\cos e{{c}^{-1}}\frac{5}{4}=\frac{\pi }{2}\], then \[x\] is equal to :

    A)  1                                            

    B)  4

    C)  3                                            

    D)  5

    Correct Answer: C

    Solution :

    We have, \[{{\sin }^{-1}}\frac{x}{5}+\cos e{{c}^{-1}}\frac{5}{4}=\frac{\pi }{2}\] \[\Rightarrow \]               \[{{\sin }^{-1}}\frac{x}{5}+{{\sin }^{-1}}\frac{4}{5}=\frac{\pi }{2}\] \[\Rightarrow \,{{\sin }^{-1}}\left[ \frac{x}{5}\times \sqrt{1-{{(4/5)}^{2}}}+\frac{4}{5}\times \sqrt{1-{{(x/5)}^{2}}} \right]=\pi /2\]or \[\frac{x}{5}\times \frac{3}{5}+\frac{4}{5}\times \frac{\sqrt{25-{{x}^{2}}}}{5}=\sin \pi /2\] \[\Rightarrow \,\,3x+4\sqrt{25-{{x}^{2}}}=25\] \[\Rightarrow \]               \[4\sqrt{25-{{x}^{2}}}=25-3x\] Squaring both sides, we get                 \[16\,(25-{{x}^{2}})=625+9{{x}^{2}}-150x\] \[\Rightarrow \,\,\,4500-16{{x}^{2}}=625+9{{x}^{2}}-150x\] \[\Rightarrow \,\,\,\,16{{x}^{2}}+9{{x}^{2}}-150x+625-400=0\] \[\Rightarrow \]               \[25{{x}^{2}}-150x+225=0\] \[\Rightarrow \]                               \[{{x}^{2}}-6x+9=0\] or            \[{{x}^{2}}-3x-3x+9=0\]                 \[x\left( x-3 \right)-3\left( x-3 \right)=0\] \[\Rightarrow \]                               \[{{\left( x-3 \right)}^{2}}=0\] \[\therefore \]                                  \[x=3\]

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