If f(x) = (sin^4 x + cos^4 x),0

Question

(B) Given $f(x) = \sin^4 x + \cos^4 x$.We can rewrite this as:$f(x) = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x$Since $\sin^2 x + \cos^2 x = 1$,we

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$\frac{1}{2}, \frac{\pi}{4}$

Vedclass · Answer

(B) Given $f(x) = \sin^4 x + \cos^4 x$.We can rewrite this as:$f(x) = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x$Since $\sin^2 x + \cos^2 x = 1$,we have:$f(x) = 1 - 2 \sin^2 x \cos^2 x$Using the identity $\sin 2x = 2 \sin x \cos x$,we get $\sin^2 x \cos^2 x = \frac{\sin^2 2x}{4}$.Substituting this into the expression:$f(x) = 1 - 2 \left(\frac{\sin^2 2x}{4}\right) = 1 - \frac{1}{2} \sin^2 2x$.For $0 When $\sin 2x = 1$,$f(x) = 1 - \frac{1}{2}(1)^2 = 1 - \frac{1}{2} = \frac{1}{2}$.Thus,the minimum value is $\frac{1}{2}$ at $x = \frac{\pi}{4}$.

If $f(x) = (\sin^4 x + \cos^4 x)$,$0 < x < \frac{\pi}{2}$,then the function has a minimum value of $ . . . . . . $ at $x = . . . . . . $.

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