1.

Let `f : (-1, 1) -> R` be such that `f(cos4theta) = 2/(2-sec^2theta` for `theta in (0, pi/4) uu (pi/4, pi/2)`. Then the value(s) of `f(1/3)` is/areA. `1+-(sqrt3)/(2)`B. `1+-sqrt((2)/(3))`C. `1+-sqrt((1)/(3))`D. `1+-sqrt((1)/(2))`

Answer» Correct Answer - A
If `0in(0,(pi)/(4))uu((pi)/(4),(pi)/(2)), then 2thetain(0, (pi)/(2))uu((pi)/(2),pi)`
Therefore, `cos 2 theta` can be positive or negative.
Hence, `cos2theta=pmsqrt((1+cos4theta)/(2))`
`impliescos 2theta=pmsqrt((1+(1)/(3))/(2))=+-sqrt((2)/(3))" "["Taking"cos4theta=1/3]`
Now, `f(cos4theta)=(2)/(2-sec^(2)theta)=(2cos^(2)theta)/(2cos^(2)theta-1)=(1+cos2theta)/(cos2theta)`
`impliesf((1)/(3))=1+-sqrt((3)/(2)). [because cos4theta=1/3and cos2theta=+-sqrt((2)/(3))]`


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