Name: Example of Mean Value Theorem
Uploaded: 2019-07-19
Duration: 2 min 17 s
Description: You've already learned about the Mean Value Theorem. Let’s explore more about the theorem with some examples in this video. Watch and learn!

Question

Using Lagrange's mean value theorem, prove that

(b - a) \sec^{2} a < \tan b - \tan a < (b - a) \sec^{2} b, where 0 < a < b < \frac{π}{2}

.

Accepted Answer

Let $f (x) = \tan x$ on $[a, b] . . . . . . . . . . . . . (1)$

Since, tangent function is continuous and derivable for all values of $x > 0$ , Therefore,

$1)$ $f (x)$ is continuous in the closed interval $[a, b]$ .

$2)$ $f (x)$ is differentiable in the open interval $(a, b)$ .

Thus, $f (x)$ satisfies both the condition of Lagrange's Mean Value Theorem.

$∴$ There exists at least one $c \in (a, b)$ such that

$f' (c) = \frac{f (b) - f (a)}{b - a} . . . . . . . . . . (2)$

On differentiating both sides equation $(1)$ with respect to $x$ , we get

$\frac{d}{d x} f (x) = \frac{d}{d x} \tan x$

$f' (x) = \sec^{2} x$

Put $x = c$

$\Rightarrow f' (c) = \sec^{2} c . . . . . . . . . . (3)$

$∴$ from equation $(2)$ and $(3)$ , we have

$\sec^{2} c = \frac{f (b) - f (a)}{b - a}$

$\sec^{2} c = \frac{\tan b - \tan a}{b - a} . . . . . . . . . . (4)$

Since, $c \in (a, b)$

$\Rightarrow a < c < b$

$\Rightarrow \sec^{2} a < \sec^{2} c < \sec^{2} b$

$\Rightarrow \sec^{2} a < \frac{\tan b - \tan a}{b - a} < \sec^{2} b$ $[$ $∵$ using equation $(4)$ $]$

$\Rightarrow (b - a) s e c^{2} a < \tan b - \tan a < (b - a) s e c^{2} b$

Hence, proved.

Using Lagrange's mean value theorem, prove that $(b - a) \sec^{2} a < \tan b - \tan a < (b - a) \sec^{2} b, where 0 < a < b < \frac{π}{2}$ .

Important Questions on Mean Value Theorems

Learn and Prepare for any exam you want !

Using Lagrange's mean value theorem, prove that (b-a)sec2a<tanb-tana<(b-a)sec2b, where 0<a<b<π2.

Important Questions on Mean Value Theorems

Learn and Prepare for any exam you want !

Using Lagrange's mean value theorem, prove that $(b - a) \sec^{2} a < \tan b - \tan a < (b - a) \sec^{2} b, where 0 < a < b < \frac{π}{2}$ .