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provide solution for RD Sharma maths class 12 chapter Mean Value Theoram exercise 14.2 question 11

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Hint: Use Lagrange’s mean value theorem formula.

Given:Use Lagrange’s mean value theorem on (b-a) \sec ^{2} a<\tan b-\tan a<(b-a) \sec ^{2} b , where 0<a<b<\frac{\pi}{2}.


\text { Let } f(x)=\tan x \text { in }(a, b) \text { where } 0<a<b<\frac{\pi}{2}

\therefore f (x) is continuous in [a, b] and differentiable in (a, b)

Now, f^{\prime}(x)=\sec ^{2} x

\Rightarrow f^{\prime}(c)=\sec ^{2} c

By Mean Value Theorem,

\begin{aligned} f^{\prime}(c) &=\frac{f(b)-f(a)}{b-a} \\ f^{\prime}(c) &=\frac{\tan b-\tan a}{b-a} \end{aligned}                                    c \in(a, b)

\begin{aligned} &\Rightarrow a<c<b \\ &\Rightarrow \sec ^{2} a<\sec ^{2} c<\sec ^{2} b \\ &\Rightarrow \sec ^{2} a<\frac{\tan b-\tan a}{b-a}<\sec ^{2} b \end{aligned}

Hence, proved.


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