JEE Main & Advanced Sample Paper JEE Main - Mock Test - 25

  • question_answer
    If\[3{{p}^{2}}=5p+2\]and\[3{{q}^{2}}=5q+2\]where \[p\ne q\], then the equation whose roots are 3p - 2q and 3q - 2p is

    A) \[3{{x}^{2}}-5x-100=0\]

    B) \[5{{x}^{2}}+3x+100=0\]

    C) \[3{{x}^{2}}-Sx+100=0\]                     

    D) \[5{{x}^{2}}-3x-100=0\]

    Correct Answer: A

    Solution :

    [a] : According to question, p and q are roots of \[3{{x}^{2}}-5x-2=0\] \[\therefore \]\[p+q=\frac{5}{3}\]and\[pq=\frac{-2}{3}.\] We have to find the equation whose roots are 3p - 2q and 3q - 2p. Clearly, sum of roots \[=(3p-2q)+\left( 3q-2p \right)=(p+q)=\frac{5}{3}\] and product of roots \[=(3p-2q)(3q-2p)\] \[=9pq-6{{q}^{2}}-6{{p}^{2}}+4pq=13pq-2(3{{p}^{2}}+3{{q}^{2}})\] \[=13\left( \frac{-2}{3} \right)-2(5p+2+5q+2)\] \[=13\left( \frac{-2}{3} \right)-2\left[ 5\left( \frac{5}{3} \right)+4 \right]\] \[=\frac{-26}{3}-2\left[ \frac{5}{3}+4 \right]=\frac{-100}{3}\] Hence, required equation is \[3{{x}^{2}}-5x-100=0\].


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