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

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

Answers (1)

Answer: 

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}.

Solution:

\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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