The minimum value of the function f(x) = 2cos | vedclass

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Question

(D) Let $f(x) = 2\cos 2x - \cos 4x$. Using the identity $\cos 4x = 2\cos^2 2x - 1$,we can rewrite the function as: $f(x) = 2\cos 2x - (2\cos^2 2x - 1)

Vedclass · Accepted Answer

$-3$

Vedclass · Answer

(D) Let $f(x) = 2\cos 2x - \cos 4x$. Using the identity $\cos 4x = 2\cos^2 2x - 1$,we can rewrite the function as: $f(x) = 2\cos 2x - (2\cos^2 2x - 1) = -2\cos^2 2x + 2\cos 2x + 1$. Let $t = \cos 2x$. Since $0 \le x \le \pi$,the range of $2x$ is $0 \le 2x \le 2\pi$,so the range of $t = \cos 2x$ is $[-1, 1]$. Now,we have a quadratic function $g(t) = -2t^2 + 2t + 1$ for $t \in [-1, 1]$. To find the minimum value,we check the endpoints of the interval $[-1, 1]$ because the parabola opens downwards. For $t = 1$: $g(1) = -2(1)^2 + 2(1) + 1 = -2 + 2 + 1 = 1$. For $t = -1$: $g(-1) = -2(-1)^2 + 2(-1) + 1 = -2 - 2 + 1 = -3$. Comparing these values,the minimum value is $-3$.

The minimum value of the function $f(x) = 2\cos 2x - \cos 4x$ in the interval $0 \le x \le \pi$ is:

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