Akshita kaushik Asked a Question
December 13, 2020 12:32 pmpts 30 pts
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 thankyou
    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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