For `x epsilonR`, and a continuous function `f` let `I_(1)=int_(sin^(2)t)^(1+cos^(2)t)xf{x(2-x)}dx` and `I_(2)=int_(sin^(2)t)^(1+cos^(2)t)f{x(2-x)}dx`

Question

For `x epsilonR`, and a continuous function `f` let `I_(1)=int_(sin^(2)t)^(1+cos^(2)t)xf{x(2-x)}dx` and `I_(2)=int_(sin^(2)t)^(1+cos^(2)t)f{x(2-x)}dx`

For `x epsilonR`, and a continuous function `f` let `I_(1)=int_(sin^(2)t)^(1+cos^(2)t)xf{x(2-x)}dx` and `I_(2)=int_(sin^(2)t)^(1+cos^(2)t)f{x(2-x)}dx`.
Then `(I_(1))/(I_(2))` is
A. `-1`
B. `1`
C. `2`
D. `3`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
`I_(1)=int_(sin^(2)t)^(1+cos^(2)t)xf(x(2-x))dx`
`=int_(sin^(2)t)^(1+cos^(2)t)(2-x)f(x(2-x))dx=2I_(2)-I_(1)`
or `2I_(1)=2I_(2)` or `(I_(1))/(I_(2))=1`

5 views Answered 2023-02-05 15:29