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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Using Lagrange's mean value theorem, prove that
$(\text{b}-\text{a})\sec^2\text{a}<\tan\text{b}-\tan\text{a}<(\text{b}-\text{a})\sec^2\text{b}$
where $0<\text{a}<\text{b}<\frac{\pi}{2}.$
✓

Answer

Consider the function as
$\text{f}(\text{x})=\tan\text{x},$ $\Big\{\text{x}\in[\text{a},\text{b}]\text{ such that }0<\text{a}<\text{b}<\frac{\pi}{2}\Big\}$
We know that $\tan\text{x}$ is continuous and differentiable in $\Big(0,\frac{\pi}{2}\Big),$ so, Lagrange's mean value theorem is applicable on (a, b), so there exist a point c such that,
$\text{f}'(\text{c})=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}$
$\Rightarrow\sec^2\text{c}=\frac{\tan\text{b}-\tan\text{a}}{\text{b}-\text{a}}\ ....(\text{i})$
Now,
$\text{c}\in(\text{a},\text{b})$
$\Rightarrow\text{a}<\text{c}<\text{b}$
$\Rightarrow\sec^2\text{a}<\sec^2\text{c}<\sec^2\text{b}$
$\Rightarrow\sec^2\text{a}<\Big(\frac{\tan\text{b}-\tan\text{a}}{\text{b}-\text{a}}\Big)<\sec^2\text{b}$
Using equation (i),
$(\text{b}-\text{a})\sec^2\text{a}<(\tan\text{b}-\tan\text{a})<(\text{b}-\text{a})\sec^2\text{b}$

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