Question

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★ STANDARD QUESTION 15 ★

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Question is -

If a root the equation n²Sin²x - 2Sinx - ( 2n + 1 ) = 0 lies in [ 0 , π / 2 ] , then the minimum positive integral value of " n " is ...


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Answer 1

n²sin²x-2sinx-2n+1=0
sinx=1+√2n³+n²+1/n²

0≤ 1+√2n³+n²+1 ≤n²

√2n³+n²+1 ≤  n²-1

n^4-2n³-3n² ≥ 0
(n-3)(n+1) ≥ 0
n≥3

n=3,4,5........

minimum value of n = 3

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