Akshita Kaushik posted an Question
December 13, 2020 • 18:02 pm 30 points
  • IIT JAM
  • Mathematics (MA)

0. for me n such that n2 m, the function f (r) 2n+ -sin x-1 vxe [1,2] withf(1) j) has 2 +1 (a) no zero in [1, 2] (b) exactly zeros in [1, 2] (c) exactly two zer

0. For me N such that n2 m, the function f (r) 2n+ -sin x-1 Vxe [1,2] withf(1) J) has 2 +1 (a) no zero in [1, 2] (b) exactly zeros in [1, 2] (C) exactly two zeros in 1, 2 d) countable zeros in [1, 2]

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  • Akshita kaushik

    it can also be true by IVT bcz the value at end points is of opposite sign

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    Satpal singh best-answer

    u can easily check this with the help of graphs the intersection point of y= x^2n -1 y= sinx will be root of given function. there is one zero in this interval

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