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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer
    If \[{{\cos }^{-1}}p+{{\cos }^{-1}}q+{{\cos }^{-1}}r=\pi \] then\[{{p}^{2}}+{{q}^{2}}+{{r}^{2}}+2pqr\] is equal to :

    A)  3                                            

    B)  1

    C)  2                                            

    D)  - 1

    Correct Answer: B

    Solution :

    We have,\[{{\cos }^{-1}}p+{{\cos }^{-1}}q+{{\cos }^{-1}}r=\pi \]we know that, if \[y={{\cos }^{-1}}x\], then\[-1\le x\le 1\] and \[0\le y\le \pi \], Hence the given equation will hold only when each is k \[\therefore \]  \[p=q=r=\cos \pi =-1\] \[\therefore \]  \[{{p}^{2}}+{{q}^{2}}+{{r}^{2}}+2pqr\] \[={{(-1)}^{2}}+{{(-1)}^{2}}+{{(-1)}^{2}}+2(-1)\,\,(-1)\,(-1)\] \[=1+1+1-2=3-2=1\]


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