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  • Verify mean value theorem for each of the functions given f(x) = sinx - sin2x in

Verify mean value theorem for each of the functions given
f(x) = sinx - sin2x in [0,\pi]

Answers (1)

Given:

f(x) = sinx - sin2x in [0,π]

Now, we have to show that f(x) verify the Mean Value Theorem

First of all, Conditions of Mean Value theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that

f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}

Condition 1:

f(x) = sinx - sin 2x

Since, f(x) is a trigonometric function and we know that, sine function are defined for all real values and are continuous for all x ∈ R

⇒ f(x) = sinx - sin 2x is continuous at x ∈ [0,π]

Hence, condition 1 is satisfied.

Condition 2:

f(x) = sinx - sin 2x

f’(x) = cosx - 2 cos2x 

\left[\because \frac{d}{d x} \sin x=\cos x\right]

⇒ f(x) is differentiable at [0,π]

Hence, condition 2 is satisfied.

Thus, Mean Value Theorem is applicable to the given function

Now,
\\ f(x)=\sin x-\sin 2 x x \in[0, \pi] \\ f(a)=f(0)=\sin (0)-\sin 2(0)=0\left[\because \sin \left(0^{\circ}\right)=0\right] \\ f(b)=f(\pi)=\sin (\pi)-\sin 2(\pi)=0-0=0 \\ {[\because \sin \pi=0 \& \sin 2 \pi=0]}

Now, let us show that there exist c ∈ (0,1) such that

\\f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\f(x)=\sin x-\sin 2 x
On differentiating above with respect to x, we get
f^{\prime}(x)=\cos x-2 \cos 2 x
Put x=c in above equation, we get

f’(c) = cos(c) - 2cos2c …(i)

By Mean Value Theorem,

\begin{aligned} &f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\\ &\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{\pi})-\mathrm{f}(0)}{\pi-0}\\ &\Rightarrow \cos c-2 \cos 2 c=\frac{0-0}{\pi-0}\\ &\Rightarrow \cos c-2 \cos 2 c=0\\ &\Rightarrow \cos c-2\left(2 \cos ^{2} c-1\right)=0\left[\because \cos 2 x=2 \cos ^{2} x-1\right]\\ &\Rightarrow \cos c-4 \cos ^{2} c+2=0\\ &\Rightarrow 4 \cos ^{2} c-\cos c-2=0\\ &\text { Now, let } \cos c=x\\ &\Rightarrow 4 x^{2}-x-2=0 \end{aligned}

Now, to find the factors of the above equation, we use

\\ x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ \Rightarrow x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4 \times 4 \times(-2)}}{2 \times 4} \\ \Rightarrow x=\frac{1 \pm \sqrt{1+32}}{8} \\ \Rightarrow x=\frac{1 \pm \sqrt{33}}{8}

\begin{aligned} &\Rightarrow \cos c=\frac{1 \pm \sqrt{33}}{8} \text { [above we let } \left.\cos c=x\right]\\ &\Rightarrow c=\cos ^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right)\\ &\text { So, value of }\\ &\mathrm{c}=\cos ^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right) \in(0, \pi) \end{aligned}

Thus, Mean Value Theorem is verified.

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