Question

The area (in sq. units) of the region bounded by \[ x=0,\quad x=\frac{\pi}{2}, \] \[ y=0,\quad y=\cos x,\quad \text{and}\quad y=\tan x \] is

• A. \[ \frac{\sqrt5-1}{2} + \frac12 \log\left(\frac{\sqrt5-1}{2}\right) \]
• B. \[ \frac{3-\sqrt5}{2} + \log\left(\frac{\sqrt5-1}{2}\right) \]
• C. \[ \frac{\sqrt5-1}{2} - \log\left(\frac{\sqrt5-1}{2}\right) \]
• D. \[ \frac{3-\sqrt5}{2} + \frac12 \log\left(\frac{\sqrt5+1}{2}\right) \]

Hint:

When multiple curves bound a region, first find their intersection points. Then split the area into intervals where the upper boundary changes and integrate accordingly.

Answer 1

The Correct Option is D

Concept: The required area is obtained by splitting the region at the point of intersection of \[ y=\cos x \] and \[ y=\tan x. \]

Step 1: Find the point of intersection. \[ \cos x=\tan x. \] \[ \cos x=\frac{\sin x}{\cos x}. \] \[ \cos^2x=\sin x. \] Using \[ \cos^2x=1-\sin^2x, \] \[ 1-\sin^2x=\sin x. \] \[ \sin^2x+\sin x-1=0. \] Let \[ \sin x=t. \] Then \[ t^2+t-1=0. \] \[ t=\frac{\sqrt5-1}{2}. \] Hence \[ \sin\alpha=\frac{\sqrt5-1}{2}, \] where \(\alpha\) is the point of intersection.

Step 2: Write the area integral. For \[ 0\le x\le \alpha, \] \[ \tan x\le \cos x. \] For \[ \alpha\le x\le \frac{\pi}{2}, \] \[ \cos x\le \tan x. \] Hence, \[ A = \int_0^\alpha \tan x\,dx + \int_\alpha^{\pi/2}\cos x\,dx. \]

Step 3: Evaluate the integrals. \[ A = \left[-\log(\cos x)\right]_0^\alpha + \left[\sin x\right]_\alpha^{\pi/2}. \] \[ = -\log(\cos\alpha) + 1-\sin\alpha. \] Since \[ \sin\alpha=\frac{\sqrt5-1}{2}, \] \[ 1-\sin\alpha = 1-\frac{\sqrt5-1}{2} = \frac{3-\sqrt5}{2}. \] Also, \[ \cos^2\alpha = 1-\sin^2\alpha. \] Using \[ \sin\alpha=\frac{\sqrt5-1}{2}, \] \[ \cos^2\alpha = \frac{\sqrt5-1}{2}. \] Therefore, \[ \cos\alpha = \sqrt{\frac{\sqrt5-1}{2}}. \] Hence, \[ -\log(\cos\alpha) = -\frac12 \log\left(\frac{\sqrt5-1}{2}\right). \] Using \[ \frac{2}{\sqrt5-1} = \frac{\sqrt5+1}{2}, \] \[ -\frac12 \log\left(\frac{\sqrt5-1}{2}\right) = \frac12 \log\left(\frac{\sqrt5+1}{2}\right). \] Thus, \[ A = \frac{3-\sqrt5}{2} + \frac12 \log\left(\frac{\sqrt5+1}{2}\right). \] Therefore, \[ \boxed{ A= \frac{3-\sqrt5}{2} + \frac12 \log\left(\frac{\sqrt5+1}{2}\right) } \] \[ \boxed{\text{Answer = (D)}} \]