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JEE Main & Advanced Mathematics Limits, Continuity and Differentiability Question Bank Self Evaluation Test - Continuity and Differentiability

  • question_answer
    Suppose\[f(x)={{e}^{ax}}+{{e}^{bx}}\], where\[a\ne b\], and that \[f{{\,}^{n}}(x)-2f'(x)-15f(x)=0\] for all x. Then the product ab is

    A) 25

    B) 9

    C) -15

    D) -9

    Correct Answer: C

    Solution :

    [c] \[({{a}^{2}}-2a-15){{e}^{ax}}+({{b}^{2}}-2b-15){{e}^{bx}}=0\] \[or({{a}^{2}}-2a-15)=0\,\,and\,{{b}^{2}}-2b-15=0\] \[or(a-5)(a+3)=0\,\,and\,\,(b-5)(b+3)=0\] i.e., \[a=5\] or \[-3\] and \[b=5\] or \[-3\] \[\therefore a\ne b\]. Hence, \[a=5\] and \[b=-3\] or \[a=-3\] and \[b=5\] or \[ab=-15\].


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