Let `f(x)=int_1^xsqrt(2-t^2)dtdot` Then the real roots of the equation , `x^2-f^(prime)(x)=0` are: `+-1` b. `+-1/(sqrt(2))` c.`+-1/2` d. `0&1`

Options

A. `+-1`
B. `+-1/(sqrt(2))`
C. `+-1/2`
D. 0 and 1

Explanation

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`]