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Find the range of `f(x)=sec(pi/4cos^2x),`where `-oo |
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Answer» Correct Answer - `[1,sqrt(2)]` `f(x)="sec"((pi)/(4)"cos"^(2)x)` We know that `0 le cos^(x) le 1,` i.e., `0 le (pi)/(4)cos^(2)x le (pi)/(4)` For the above value of `theta =(pi)/(pi)/(4)cos^(2)x,sec x` is an increasing function. ` :. f(x) in [f(0),f((pi)/(4))] " or " f(x) in [1,sqrt(2)]` |
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