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NIMCET 2021 Mathematics PYQ — The area of the region bounded by the X-axis and the curves and i… | Mathem Solvex

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NIMCET 2021 Mathematics PYQ — The area of the region bounded by the X-axis and the curves and i… | Mathem Solvex | Mathem Solvex

Explanation

Area Calculation
The area of the region bounded by the X-axis and the curves $y = tan x, - \frac{π}{3} \leq x \leq \frac{π}{3}$ and $y = cot x, \frac{π}{6} \leq x \leq \frac{3 π}{2}$ is to be determined.
Step 1: Analyze the functions and intervals
The function $y = tan x$ is considered in the interval $[- \frac{π}{3}, \frac{π}{3}]$ . In this interval, $tan x$ is negative for $x \in [- \frac{π}{3}, 0)$ and positive for $x \in (0, \frac{π}{3}]$ . The function $y = cot x$ is considered in the interval $[\frac{π}{6}, \frac{3 π}{2}]$ . In this interval, $cot x$ is positive for $x \in [\frac{π}{6}, \frac{π}{2})$ and negative for $x \in (\frac{π}{2}, \frac{3 π}{2}]$ .
Step 2: Calculate the area under $y = tan x$
The area $A_{1}$ under $y = tan x$ from $x = - \frac{π}{3}$ to $x = \frac{π}{3}$ is given by the integral: $A_{1} = \int_{- \frac{π}{3}}^{\frac{π}{3}} ∣ tan x ∣ d x$ . Due to the symmetry of $∣ tan x ∣$ about the y-axis, this can be written as: $A_{1} = 2 \int_{0}^{\frac{π}{3}} tan x d x$ . The integral of $tan x$ is $- lo g ∣ cos x ∣$ . $A_{1} = 2 [- lo g ∣ cos x ∣]_{0}^{\frac{π}{3}} = 2 (- lo g (cos (\frac{π}{3})) - (- lo g (cos (0))))$ . $A_{1} = 2 (- lo g (\frac{1}{2}) + lo g (1)) = 2 (- lo g (\frac{1}{2}) + 0) = - 2 lo g (\frac{1}{2})$ . Using the property $lo g (\frac{a}{b}) = lo g a - lo g b$ and $lo g (a^{b}) = b lo g a$ : $A_{1} = - 2 (lo g 1 - lo g 2) = - 2 (0 - lo g 2) = 2 lo g 2$ .
Step 3: Calculate the area under $y = cot x$
The area $A_{2}$ under $y = cot x$ from $x = \frac{π}{6}$ to $x = \frac{3 π}{2}$ is given by the integral: $A_{2} = \int_{\frac{π}{6}}^{\frac{3 π}{2}} ∣ cot x ∣ d x$ . This integral needs to be split due to the change in sign of $cot x$ at $x = \frac{π}{2}$ : $A_{2} = \int_{\frac{π}{6}}^{\frac{π}{2}} cot x d x + \int_{\frac{π}{2}}^{\frac{3 π}{2}} ∣ cot x ∣ d x$ . The integral of $cot x$ is $lo g ∣ sin x ∣$ . For the first part: $\int_{\frac{π}{6}}^{\frac{π}{2}} cot x d x = [lo g ∣ sin x ∣]_{\frac{π}{6}}^{\frac{π}{2}} = lo g (sin (\frac{π}{2})) - lo g (sin (\frac{π}{6})) = lo g (1) - lo g (\frac{1}{2}) = 0 - lo g (\frac{1}{2}) = lo g 2$ . For the second part, $cot x$ is negative in $(\frac{π}{2}, \frac{3 π}{2})$ , so $∣ cot x ∣ = - cot x$ : $\int_{\frac{π}{2}}^{\frac{3 π}{2}} - cot x d x = - [lo g ∣ sin x ∣]_{\frac{π}{2}}^{\frac{3 π}{2}} = - (lo g ∣ sin (\frac{3 π}{2}) ∣ - lo g ∣ sin (\frac{π}{2}) ∣) = - (lo g ∣ - 1∣ - lo g ∣1∣) = - (lo g 1 - lo g 1) = 0$ . Therefore, $A_{2} = lo g 2 + 0 = lo g 2$ .
Step 4: Combine the areas
The total area is the sum of $A_{1}$ and $A_{2}$ : Total Area $= A_{1} + A_{2} = 2 lo g 2 + lo g 2 = 3 lo g 2$ .
Final Answer
The final answer is $None of these$ .

NIMCET 2021 — Mathematics PYQ

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