KVPY Sample Paper KVPY Stream-SX Model Paper-13

  • question_answer
    There are three coplanar parallel lines. If any point p taken on each the lines, the maximum number of triangles with vertices at these points is

    A) \[3{{p}^{2}}(p-1)+1\]

    B) \[3{{p}^{2}}(p-1)\]

    C) \[{{p}^{2}}(4p-3)\]

    D) \[{{p}^{2}}(2p-3)\]

    Correct Answer: C

    Solution :

    Maximum number of triangles \[{}^{3P}{{C}_{3}}-3({}^{P}{{C}_{3}})\]
    \[=\frac{3p\,(3p-1)(3p-2)}{6}-\frac{3p\,(p-1)(p-2)}{6}\]
    \[=\frac{p}{2}[(3p-1)(3p-2)-(p-1)(p-2)]\]
    \[=\frac{p}{2}[9{{p}^{2}}-9p+2-{{p}^{2}}+3p-2]\]
    \[=\frac{p}{2}[8{{p}^{2}}-6p]={{p}^{2}}(4p-3)\]


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