JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let the solution curve $x = x(y), 0 < y < \frac{\pi}{2}$ , of the differential equation $(lo g_{e} (cos y))^{2} cos y d x - (1 + 3 x lo g_{e} (cos y)) sin y d y = 0$ satisfy $x (\frac{π}{3}) = \frac{1}{2 l o g _{e} 2}$ . $\text{If } x\left( \frac{\pi}{6} \right) = \frac{1}{\log_e m - \log_e n}, \text{ where } m \text{ and } n \text{ are co-prime, then } mn \text{ is equal to}$

Choose the correct answer:

Explanation

Given Equation:

(\cos y) \cdot (\ln (\cos y))^2 \, dx = (1 + 3x \ln \cos y) \sin y \, dy

Step 1: Rearranging to standard linear form

\frac{dx}{dy} = \frac{(1 + 3x \ln \cos y) \sin y}{(\ln \cos y)^2 \cos y}

= \tan y \left[ \frac{1}{(\ln \cos y)^2} + \frac{3x}{\ln \cos y} \right]

\frac{dx}{dy} - \left( \frac{3 \tan y}{\ln \cos y} \right) x = \frac{tan y}{(\ln \cos y)^2}

Step 2: Finding the Integrating Factor (I.F.)

This is a linear differential equation of the form $\frac{d x}{d y} + P x = Q$ .

\text{I.F.} = e^{\int P \, dy} = e^{-\int \frac{3 \tan y}{\ln \cos y} \, dy}

Let $u = ln cos y ⟹ d u = - tan y d y$ .

\text{I.F.} = e^{3 \int \frac{1}{u} \, du} = e^{3 \ln u} = u^3 = (\ln \cos y)^3

Step 3: General Solution

x \times (\ln \cos y)^3 = \int \left[ (\ln \cos y)^3 \times \frac{\tan y}{(\ln \cos y)^2} \right] dy

x \times (\ln \cos y)^3 = \int (\ln \cos y) \tan y \, dy

Solving the integral (using substitution $u = ln cos y$ ):

x \times (\ln \cos y)^3 = \frac{-(\ln \cos y)^2}{2} + C

Step 4: Boundary Condition

At $y = \frac{π}{3}$ ,

Step 5: Finding the constant $C$

\frac{1}{2 \ln 2} \times \left( \ln \left( \frac{1}{2} \right) \right)^3 = -\frac{\left( \ln \left( \frac{1}{2} \right) \right)^2}{2} + C

\Rightarrow C = 0

Step 6: General Equation with $C = 0$

\text{So, } x \times \ln^3 \cos y = \frac{-\ln^2 \cos y}{2}

Step 7: Calculating $x$ at $y = \frac{π}{6}$

\text{At } y = \frac{\pi}{6}, x \times \left( \ln \left( \frac{\sqrt{3}}{2} \right) \right)^3 = -\frac{1}{2} \left( \ln \left( \frac{\sqrt{3}}{2} \right) \right)^2

\Rightarrow x = -\frac{1}{2 \ln \left( \frac{\sqrt{3}}{2} \right)}

Step 8: Simplifying the expression for $x$

= -\frac{1}{2 [\ln \sqrt{3} - \ln 2]} = \frac{-1}{2 \left[ \frac{1}{2} \ln 3 - \ln 2 \right]}

= \frac{-1}{2 \left[ \frac{\ln 3 - \ln 4}{2} \right]} = \frac{1}{\ln 4 - \ln 3}

Final Result:

\Rightarrow m = 4, n = 3

\Rightarrow mn = 12

Choose the correct answer:

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