The area of the region (in square units) above the x-axis bounded by the curve $$y = \tan x$$, $$0 \leq x \leq \frac{\pi}{2}$$ and the tangent to the curve at $$x = \frac{\pi}{4}$$ is:

The problem requires finding the area above the x-axis bounded by the curve $$y = \tan x$$ for $$0 \leq x \leq \frac{\pi}{2}$$ and the tangent to this curve at $$x = \frac{\pi}{4}$$. First, find the equation of the tangent at $$x = \frac{\pi}{4}$$. The derivative of $$y = \tan x$$ is $$\frac{dy}{dx} = \sec^2 x$$. At $$x = \frac{\pi}{4}$$, $$\sec \frac{\pi}{4} = \sqrt{2}$$, so $$\sec^2 \frac{\pi}{4} = (\sqrt{2})^2 = 2$$. Thus, the slope of the tangent is 2. At $$x = \frac{\pi}{4}$$, $$y = \tan \frac{\pi}{4} = 1$$, so the point of tangency is $$\left( \frac{\pi}{4}, 1 \right)$$. Using the point-slope form, the equation of the tangent is: $$$ y - 1 = 2 \left( x - \frac{\pi}{4} \right) $$$ Simplifying: $$$ y - 1 = 2x - \frac{\pi}{2} $$$ $$$ y = 2x - \frac{\pi}{2} + 1 $$$ Next, find where this tangent intersects the x-axis by setting $$y = 0$$: $$$ 0 = 2x - \frac{\pi}{2} + 1 $$$ $$$ 2x = \frac{\pi}{2} - 1 $$$ $$$ x = \frac{\frac{\pi}{2} - 1}{2} = \frac{\pi}{4} - \frac{1}{2} $$$ So, the tangent intersects the x-axis at $$\left( \frac{\pi}{4} - \frac{1}{2}, 0 \right)$$. Denote $$a = \frac{\pi}{4} - \frac{1}{2}$$ and $$b = \frac{\pi}{4}$$. The bounded region above the x-axis is enclosed by: - The x-axis from $$(0, 0)$$ to $$(a, 0)$$ - The tangent line from $$(a, 0)$$ to $$\left( \frac{\pi}{4}, 1 \right)$$ - The curve $$y = \tan x$$ from $$\left( \frac{\pi}{4}, 1 \right)$$ back to $$(0, 0)$$ To find the area, split the region at $$x = a$$: - From $$x = 0$$ to $$x = a$$, the upper boundary is $$y = \tan x$$ and the lower boundary is the x-axis ($$y = 0$$). - From $$x = a$$ to $$x = b$$, the upper boundary is $$y = \tan x$$ and the lower boundary is the tangent line $$y = 2x - \frac{\pi}{2} + 1$$. The total area $$A$$ is: $$$ A = \int_{0}^{a} \left[ \tan x - 0 \right] dx + \int_{a}^{b} \left[ \tan x - \left( 2x - \frac{\pi}{2} + 1 \right) \right] dx $$$ Compute each integral separately. **First integral:** $$\int_{0}^{a} \tan x dx$$ The antiderivative of $$\tan x$$ is $$-\ln |\cos x|$$. Since $$\cos x \gt 0$$ for $$x \in [0, \frac{\pi}{2})$$, this simplifies to $$-\ln (\cos x)$$. Evaluate from 0 to $$a$$: $$$ \left[ -\ln (\cos x) \right]_{0}^{a} = -\ln (\cos a) - \left( -\ln (\cos 0) \right) = -\ln (\cos a) + \ln (1) = -\ln (\cos a) $$$ since $$\cos 0 = 1$$ and $$\ln 1 = 0$$. Now, $$a = \frac{\pi}{4} - \frac{1}{2}$$, so: $$$ \cos a = \cos \left( \frac{\pi}{4} - \frac{1}{2} \right) = \cos \frac{\pi}{4} \cos \frac{1}{2} + \sin \frac{\pi}{4} \sin \frac{1}{2} = \frac{\sqrt{2}}{2} \cos \frac{1}{2} + \frac{\sqrt{2}}{2} \sin \frac{1}{2} = \frac{\sqrt{2}}{2} \left( \cos \frac{1}{2} + \sin \frac{1}{2} \right) $$$ Thus: $$$ \int_{0}^{a} \tan x dx = -\ln \left( \frac{\sqrt{2}}{2} \left( \cos \frac{1}{2} + \sin \frac{1}{2} \right) \right) = -\left[ \ln \left( \frac{\sqrt{2}}{2} \right) + \ln \left( \cos \frac{1}{2} + \sin \frac{1}{2} \right) \right] $$$ $$$ \ln \left( \frac{\sqrt{2}}{2} \right) = \ln (\sqrt{2}) - \ln 2 = \frac{1}{2} \ln 2 - \ln 2 = -\frac{1}{2} \ln 2 $$$ So: $$$ \int_{0}^{a} \tan x dx = -\left[ -\frac{1}{2} \ln 2 + \ln \left( \cos \frac{1}{2} + \sin \frac{1}{2} \right) \right] = \frac{1}{2} \ln 2 - \ln \left( \cos \frac{1}{2} + \sin \frac{1}{2} \right) $$$ **Second integral:** $$\int_{a}^{b} \left[ \tan x - \left( 2x - \frac{\pi}{2} + 1 \right) \right] dx$$ Split into three parts: $$$ \int_{a}^{b} \tan x dx - \int_{a}^{b} \left( 2x - \frac{\pi}{2} + 1 \right) dx = \int_{a}^{b} \tan x dx - 2 \int_{a}^{b} x dx + \left( \frac{\pi}{2} - 1 \right) \int_{a}^{b} dx $$$ Compute each: - $$\int_{a}^{b} \tan x dx = \left[ -\ln (\cos x) \right]_{a}^{b} = -\ln (\cos b) + \ln (\cos a) = \ln \left( \frac{\cos a}{\cos b} \right)$$ Since $$b = \frac{\pi}{4}$$, $$\cos b = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$, and $$\cos a = \frac{\sqrt{2}}{2} \left( \cos \frac{1}{2} + \sin \frac{1}{2} \right)$$, so: $$$ \frac{\cos a}{\cos b} = \frac{\frac{\sqrt{2}}{2} \left( \cos \frac{1}{2} + \sin \frac{1}{2} \right)}{\frac{\sqrt{2}}{2}} = \cos \frac{1}{2} + \sin \frac{1}{2} $$$ Thus: $$$ \int_{a}^{b} \tan x dx = \ln \left( \cos \frac{1}{2} + \sin \frac{1}{2} \right) $$$ - $$\int_{a}^{b} x dx = \left[ \frac{x^2}{2} \right]_{a}^{b} = \frac{b^2 - a^2}{2}$$ $$$ b = \frac{\pi}{4}, \quad a = \frac{\pi}{4} - \frac{1}{2} $$$ $$$ b^2 = \left( \frac{\pi}{4} \right)^2 = \frac{\pi^2}{16} $$$ $$$ a^2 = \left( \frac{\pi}{4} - \frac{1}{2} \right)^2 = \left( \frac{\pi}{4} \right)^2 - 2 \cdot \frac{\pi}{4} \cdot \frac{1}{2} + \left( \frac{1}{2} \right)^2 = \frac{\pi^2}{16} - \frac{\pi}{4} + \frac{1}{4} $$$ $$$ b^2 - a^2 = \frac{\pi^2}{16} - \left( \frac{\pi^2}{16} - \frac{\pi}{4} + \frac{1}{4} \right) = \frac{\pi}{4} - \frac{1}{4} $$$ So: $$$ \int_{a}^{b} x dx = \frac{1}{2} \left( \frac{\pi}{4} - \frac{1}{4} \right) = \frac{\pi - 1}{8} $$$ - $$\int_{a}^{b} dx = \left[ x \right]_{a}^{b} = b - a = \frac{\pi}{4} - \left( \frac{\pi}{4} - \frac{1}{2} \right) = \frac{1}{2}$$ Now combine: $$$ \int_{a}^{b} \left[ \tan x - \left( 2x - \frac{\pi}{2} + 1 \right) \right] dx = \ln \left( \cos \frac{1}{2} + \sin \frac{1}{2} \right) - 2 \cdot \frac{\pi - 1}{8} + \left( \frac{\pi}{2} - 1 \right) \cdot \frac{1}{2} $$$ Simplify: $$$ -2 \cdot \frac{\pi - 1}{8} = -\frac{\pi - 1}{4} $$$ $$$ \left( \frac{\pi}{2} - 1 \right) \cdot \frac{1}{2} = \frac{\pi}{4} - \frac{1}{2} $$$ So: $$$ \int_{a}^{b} \left[ \tan x - \left( 2x - \frac{\pi}{2} + 1 \right) \right] dx = \ln \left( \cos \frac{1}{2} + \sin \frac{1}{2} \right) - \frac{\pi - 1}{4} + \frac{\pi}{4} - \frac{1}{2} $$$ $$$ -\frac{\pi - 1}{4} + \frac{\pi}{4} = -\frac{\pi}{4} + \frac{1}{4} + \frac{\pi}{4} = \frac{1}{4} $$$ Thus: $$$ \int_{a}^{b} \left[ \tan x - \left( 2x - \frac{\pi}{2} + 1 \right) \right] dx = \ln \left( \cos \frac{1}{2} + \sin \frac{1}{2} \right) + \frac{1}{4} - \frac{1}{2} = \ln \left( \cos \frac{1}{2} + \sin \frac{1}{2} \right) - \frac{1}{4} $$$ Now, the total area $$A$$ is: $$$ A = \left[ \frac{1}{2} \ln 2 - \ln \left( \cos \frac{1}{2} + \sin \frac{1}{2} \right) \right] + \left[ \ln \left( \cos \frac{1}{2} + \sin \frac{1}{2} \right) - \frac{1}{4} \right] $$$ The logarithmic terms cancel: $$$ A = \frac{1}{2} \ln 2 - \frac{1}{4} $$$ Factor out $$\frac{1}{2}$$: $$$ A = \frac{1}{2} \left( \ln 2 - \frac{1}{2} \right) $$$ Comparing with the options: - A: $$\frac{1}{2}(\log 2 - \frac{1}{2})$$ - B: $$\frac{1}{2}(1 + \log 2)$$ - C: $$\frac{1}{2}(1 - \log 2)$$ - D: $$\frac{1}{2}(\log 2 + \frac{1}{2})$$ Here, $$\log$$ denotes the natural logarithm (common in JEE contexts). Thus, option A matches the result. Hence, the correct answer is Option A.