The constant c of Lagrange's mean value theor | vedclass

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(B) The given function $f(x) = \cos x - \sin 2x$ is continuous on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and differentiable on $\left(-\frac{\pi

Vedclass · Accepted Answer

$\sin^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right)$

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(B) The given function $f(x) = \cos x - \sin 2x$ is continuous on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and differentiable on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.According to Lagrange's Mean Value Theorem,there exists $c \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $f'(c) = \frac{f(\pi/2) - f(-\pi/2)}{\pi/2 - (-\pi/2)}$.First,calculate $f(\pi/2) = \cos(\pi/2) - \sin(\pi) = 0 - 0 = 0$.Next,$f(-\pi/2) = \cos(-\pi/2) - \sin(-\pi) = 0 - 0 = 0$.Thus,$f'(c) = \frac{0 - 0}{\pi} = 0$.The derivative is $f'(x) = -\sin x - 2\cos 2x$.Setting $f'(c) = 0$,we get $-\sin c - 2\cos 2c = 0$.Using the identity $\cos 2c = 1 - 2\sin^2 c$,we have $-\sin c - 2(1 - 2\sin^2 c) = 0$.$-\sin c - 2 + 4\sin^2 c = 0$,which simplifies to $4\sin^2 c - \sin c - 2 = 0$.Using the quadratic formula for $\sin c$,we get $\sin c = \frac{1 \pm \sqrt{(-1)^2 - 4(4)(-2)}}{2(4)} = \frac{1 \pm \sqrt{1 + 32}}{8} = \frac{1 \pm \sqrt{33}}{8}$.Therefore,$c = \sin^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right)$.

The constant $c$ of Lagrange's mean value theorem for $f(x)=\cos x-\sin 2x$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is

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