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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Afternoon)

  • question_answer
    Let \[{{a}_{1}},{{a}_{2}},{{a}_{3}},\]......be an A. P. with \[{{a}_{6}}=2.\]Then the common difference of this A. P., which maximizes the produce \[{{a}_{1}}{{a}_{4}}{{a}_{5}},\]is : [JEE Main 10-4-2019 Afternoon]

    A) \[\frac{6}{5}\]                        

    B) \[\frac{8}{5}\]

    C) \[\frac{2}{3}\]                        

    D) \[\frac{3}{2}\]

    Correct Answer: B

    Solution :

    Let a is first term and d is common difference then, \[a+5d=2\] (given) ...(1) \[f(d)=(2-5d)(2-2d)(2-d)\] \[f'(d)=0\Rightarrow d=\frac{2}{3},\frac{8}{5}\] \[f''(d)<0\,\text{at}\,\,d=8/5\]\[\Rightarrow d=\frac{8}{5}\]


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