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Mathematics · 2022

JEE Main · 26 June 2022, Shift 1 · Q30

Let S =(0,2 π)-{(π)/2, (3 π)/4, (3 π)/2, (7 π)/4}. Let y=y(x), x ∈ S, be the solution curve of the differential equation (d y)/(d x)=1/(1+ sin 2 x),…

Let $\displaystyle \mathrm{S}=(0,2 \pi)-\left\{\frac{\pi}{2}, \frac{3 \pi}{4}, \frac{3 \pi}{2}, \frac{7 \pi}{4}\right\}$. Let $\displaystyle y=y(x), x \in \mathrm{~S}$, be the solution curve of the differential equation $\displaystyle \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{1+\sin 2 x}, y\left(\frac{\pi}{4}\right)=\frac{1}{2}$. If the sum of abscissas of all the points of intersection of the curve $\displaystyle y=y(x)$ with the curve $\displaystyle y=\sqrt{2} \sin x$ is $\displaystyle \frac{\mathrm{k} \pi}{12}$, then k is equal to \_\_\_\_.
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.