1.

Let `f:(-1,1)vecR`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 (are)(a)`1-sqrt(3/2)`(b) `1+sqrt(3/2)`(c) `1-sqrt(2/3)`(d) `1+sqrt(2/3)`

Answer» It is given that,
`f(cos4theta) = 2/(2-sec^2theta)`
We have to find `f(1/3)`.
So, we can consider,
`cos4theta = 1/3`
Now,
`cos4theta = 2cos^2 2theta -1`
`=>cos2theta = +-sqrt((1+cos4theta)/2 )`
`=>cos2theta = +-sqrt((1+(1/3))/2 )`
`=>cos2theta = +-sqrt(2/3)`
Now,
`f(1/3) = 2/(2-sec^2theta)`
`=(2cos^2theta)/(2cos^2theta - 1)`
`=(1+cos2theta)/(cos2theta)`
`=1/(cos2theta)+1`
`=+-sqrt(3/2) +1`
`=>f(1/3) = 1+-sqrt(3/2)`
So, option `a` and `b` are the correct options.


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