SATHEE: Area Question 47

Area Question 47

Question 47

Find the area of the region bounded by the curve $C : y = \tan x$ , tangent drawn to $C$ at $x = π / 4$ and the $X$ -axis.

$(1988, 5 M)$

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Solution:

Given, $y = \tan x \Rightarrow \frac{d y}{d x} = \sec^{2} x$

$∴ \frac{d y}{d x}_{x = \frac{π}{4}} = 2$

Hence, equation of tangent at $A \frac{π}{4}, 1$ is

$\frac{y - 1}{x - π / 4} = 2 \Rightarrow y - 1 = 2 x - \frac{π}{2}$

$\Rightarrow (2 x - y) = \frac{π}{2} - 1$

$∴$ Required area is $O A B O$

$\begin{aligned} = \int_{0}^{π / 4} (\tan x) d x - area of △ A L B & = [\log | \sec x |]_{0}^{π / 4} - \frac{1}{2} \cdot B L \cdot A L & = \log \sqrt{2} - \frac{1}{2} \frac{π}{4} - \frac{π - 2}{4} \cdot 1 & = \log \sqrt{2} - \frac{1}{4} sq unit \end{aligned}$

Area Question 47

Question 47

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Area Question 47

Question 47

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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