The area (in square units) bounded by y=tan ^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The curves $y = 	an^{-1} x$ and $y = \cot^{-1} x$ intersect where $	an^{-1} x = \cot^{-1} x$. Since $\cot^{-1} x = \frac{\pi}{2} - 	an^{-1} x$,

Vedclass · Accepted Answer

$\log _e 2$

Vedclass · Answer

(B) The curves $y = 	an^{-1} x$ and $y = \cot^{-1} x$ intersect where $	an^{-1} x = \cot^{-1} x$. Since $\cot^{-1} x = \frac{\pi}{2} - 	an^{-1} x$,we have $2 	an^{-1} x = \frac{\pi}{2}$,which implies $	an^{-1} x = \frac{\pi}{4}$,so $x = 1$.The area bounded by the curves and the $Y$-axis $(x=0)$ from $x=0$ to $x=1$ is given by the integral of the upper curve minus the lower curve.For $x \in [0, 1]$,$\cot^{-1} x \ge 	an^{-1} x$.Area $= \int_0^1 (\cot^{-1} x - 	an^{-1} x) dx$$= \int_0^1 (\frac{\pi}{2} - 	an^{-1} x - 	an^{-1} x) dx = \int_0^1 (\frac{\pi}{2} - 2 	an^{-1} x) dx$$= \frac{\pi}{2} [x]_0^1 - 2 \int_0^1 	an^{-1} x dx$$= \frac{\pi}{2} - 2 [x 	an^{-1} x - \frac{1}{2} \log_e(1+x^2)]_0^1$$= \frac{\pi}{2} - 2 [ (1 \cdot \frac{\pi}{4} - \frac{1}{2} \log_e 2) - (0 - 0) ]$$= \frac{\pi}{2} - 2 [ \frac{\pi}{4} - \frac{1}{2} \log_e 2 ]$$= \frac{\pi}{2} - \frac{\pi}{2} + \log_e 2 = \log_e 2$.Thus,the correct option is $(b)$.

The area (in square units) bounded by $y=\tan ^{-1} x$,$y=\cot ^{-1} x$ and the $Y$-axis is:

Similar Questions

The area of the region given by $\{(x, y): xy \leq 8, 1 \leq y \leq x^2\}$ is :

The area of the region enclosed by the parabola $(y-2)^2=x-1$,the line $x-2y+4=0$,and the positive coordinate axes is

The area included between the parabolas $y^{2} = 5x$ and $x^{2} = 5y$ is

The area (in sq. units) of the region $\{(x, y) \in R^{2} | 4 x^{2} \leq y \leq 8 x+12\}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module