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JEE Main 29-Jan-2025 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 29-Jan-2025 Paper - Shift 14 Marks
MCQ
Let $y=y(x)$ be the solution of the differential equation $\cos x\left(\log _{e}(\cos x)\right)^{2} d y+\left(\sin x-3 y \sin x \log _{e}(\cos x)\right) d x=0$, $\mathrm{x} \in\left(0, \frac{\pi}{2}\right)$. If $\mathrm{y}\left(\frac{\pi}{4}\right)=\frac{-1}{\log _{\mathrm{e}} 2}$, then $\mathrm{y}\left(\frac{\pi}{6}\right)$ is :
  • A
    $\frac{2}{\log _{e}(3)-\log _{e}(4)}$
  • B
    $\frac{1}{\log _{e}(4)-\log _{e}(3)}$
  • C
    $-\frac{1}{\log _{e}(4)}$
  • ✓
    $\frac{1}{\log _{e}(3)-\log _{e}(4)}$
✓

Answer

Correct option: D.
$\frac{1}{\log _{e}(3)-\log _{e}(4)}$
(D)
Sol. $\cos x(\ln (\cos x))^{2} d y+(\sin x-3 y(\sin x) \ln (\cos x)) d x=0$
$
\begin{aligned}
& \cos x(\ln (\cos x))^{2} \frac{d y}{d x}-3 \sin x \cdot \ln (\cos x) y=-\sin x \\
& \frac{d y}{d x}-\frac{3 \tan x}{\ln (\cos x)} y=\frac{-\tan x}{(\ln (\cos x))^{2}} \\
& \frac{d y}{d x}+\frac{3 \tan x}{\ln (\sec x)} y=\frac{-\tan x}{(\ln (\sec x))^{2}} \\
& \text { I.F. }=e^{\int \frac{3 \tan x}{\ln (\sec x)} d x}=(\ln (\sec x))^{3}
\end{aligned}
$
$
\mathrm{y} \times(\ln (\sec \mathrm{x}))^{3}=-\int \frac{\tan \mathrm{x}}{(\ln (\sec \mathrm{x}))^{2}}(\ln (\sec \mathrm{x}))^{3} \mathrm{dx}
$
$
y \times(\ln (\sec x))^{3}=-\frac{1}{2}(\ln (\sec x))^{2}+C
$
$
\text { Given : } \mathrm{x}=\frac{\pi}{4}, \mathrm{y}=-\frac{1}{\ln 2}
$
$
\frac{-1}{\ln 2} \times(\ln \sqrt{2})^{3}=-\frac{1}{2} \times(\ln \sqrt{2})^{2}+\mathrm{C}
$
$
\Rightarrow \frac{-1}{8 \ln 2} \times(\ln 2)^{3}=\frac{-1}{2} \times \frac{1}{4}(\ln 2)^{2}+\mathrm{C}
$
$
-\frac{1}{8}(\ln 2)^{2}=\frac{-1}{8}(\ln 2)^{2}+\mathrm{C}
$
$
\Rightarrow \mathrm{C}=0
$
$\therefore y(\ln (\sec x))^{3}=\frac{-1}{2}(\ln (\sec x))^{2}+0$
$y=\frac{-1}{2 \ln (\sec x)}$
$y=\frac{1}{2 \ln (\cos x)}$
$\therefore y\left(\frac{\pi}{6}\right)=\frac{1}{2 \ln \left(\cos \frac{\pi}{6}\right)}$
$=\frac{1}{2 \ln \left(\frac{\sqrt{3}}{2}\right)}$
$=\frac{1}{2\left(\frac{1}{2} \ln 3-\ln 2\right)}$
$=\frac{1}{\ln 3-\ln 4}$

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