InterviewSolution
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Let I_(n)=int_(0)^(pi//4) tan^(n)x dx. Statement-1: (1)/(n+1)lt 2I_(n) lt (1)/(n-1) for all n=2,3,4,….. Statement-2: I_(n)+I_(n-2)=(1)/(n-1),n=3,4,5,…... |
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Answer» Statement-1 is true, Statement-2 is True,Statement-2 is a correct explanation for Statement-1. `I_(n)+I_(n-2)=UNDERSET(0)overset(PI//4)int tan^(n)x dx+underset(0)overset(pi//4)int tan^(n-2)x dx` `rArr I_(n)-I_(n-2)=underset(0)overset(pi//4)int tan^(n-2)x sec^(2)x dx=[(tan^(n-1)x)/(n-1)]_(0)^(pi//4)=(1)/(n-1)` So, statement-2 is true. We know that `0 lt tan xlt1` for ll `x in (0,(pi)/(4))` `rArr tan^(n+2)xlt tan^(n-2)x` for all `x in (0,(pi)/(4))` `rArr underset(0)overset(pi//4)int tan^(n+2)x dx lt underset(0)overset(pi//4)int tan^(n-2)x dx` `rArr I_(n+2)lt I_(n)lt I_(n-2)` `rArr I_(n)+I_(n+2)lt 2I_(n)+I_(n-2)` `rArr (1)/(n+1)ltI_(n)lt(1)/(n-1)""`[Using statement-2] So, statement-1 is true and statement-2 is a correct explanation for statement-1. |
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