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Please solve RD Sharma class 12 chapter Mean Value Theoram exercise 14.2 question 1 sub question (xv) maths textbook solution

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Answer:   \cos ^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right)

Hint: You must know the formula of Lagrange’s Mean Value Theorem.

Given:f(x)=\sin x-\sin 2 x-x \text { on }[0, \pi]


f(x)=\sin x-\sin 2 x-x

f(x)  is a polynomial function.

It is continuous in \left [ 0,\pi \right ]

f^{\prime}(x)=\cos x-2 \cos 2 x-1

(Which is defined in \left [ 0,\pi \right ])

\therefore f(x) is differentiable in \left [ 0,\pi \right ]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

\begin{aligned} &c \in[0, \pi] \\ &f^{\prime}(c)=\frac{f(\pi)-f(0)}{\pi-0} \end{aligned}

\begin{aligned} &\cos c-2 \cos 2 c-1=\frac{(\sin \pi-\sin 2 \pi-\pi)-(\sin 0-\sin 0-0)}{\pi} \\ &\cos c-2 \cos 2 c-1=\frac{(0-0-\pi)-(0-0-0)}{\pi} \end{aligned}

\begin{aligned} &\cos c-2 \cos 2 c-1=\frac{-\pi}{\pi} \\ &\cos c-2 \cos 2 c-1=-1 \\ &\cos c-2 \cos 2 c=-1+1 \end{aligned}

\begin{aligned} &\cos c-2 \cos 2 c=0 \\ &\cos c-2\left(2 \cos ^{2} c-1\right)=0 \\ &\cos c-4 \cos ^{2} c+2=0 \\ &4 \cos ^{2} c-\cos c-2=0 \end{aligned}

\begin{aligned} &\cos c=\frac{-(-1) \pm \sqrt{(-1)^{2}-4 \times 4 \times(-2)}}{2 \times 4} \\ &\cos c=\frac{1 \pm \sqrt{1+32}}{8} \end{aligned}

\begin{aligned} &\cos c=\frac{1 \pm \sqrt{33}}{8} \\ &c=\cos ^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right) \end{aligned}


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