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Let f(x) = int1^x sqrt(2 - t^2) dt. Then...

Let `f(x) = int_1^x sqrt(2 - t^2) dt`. Then the real roots of the equation `x^2 - f(x) = 0` are:

A

`+-1`

B

`+-1/(sqrt(2))`

C

`+-1/2`

D

0 and 1

Text Solution

Verified by Experts

We have `f(x)=int_(1)^(x)sqrt((2-t^(2)))dt`
`impliesf'(x)=sqrt((2-x^(2)))`
`:.x^(2)-f'(x)=0`
`impliesx^(2)=sqrt((2-x^(2)))=0impliesx^(4)+x^(2)-2=0`
`impliesx^(2)=1,-2`
`impliesx=+-1` [ only for real value of `x`]
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