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Let f(x)=int1^xsqrt(2-t^2)dt Then the re...

Let `f(x)=int_1^xsqrt(2-t^2)dt` Then the real roots of the equation `x^2-f^(prime)(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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