JEE Main 2022 · 26 Jun · Shift 1 · Mathematics Q30

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MathematicsInteger+4 / -1Question 30 of 90

Let $S=(0,2\pi)-\left\{\dfrac\pi2,\dfrac{3\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4\right\}$ . Let $y=y(x)$ , $x\in S$ , be the solution curve of the differential equation $\dfrac{dy}{dx}=\dfrac1{1+\sin2x}$ , $y\left(\dfrac\pi4\right)=\dfrac12$ . If the sum of abscissas of all the points of intersection of the curve $y=y(x)$ with the curve $y=\sqrt2\sin x$ is $\dfrac{k\pi}{12}$ , then $k$ is equal to ______.

Your answer

Answer and worked solution

Official answer: 42

Solve. $\dfrac{dy}{dx}=\dfrac1{(\sin x+\cos x)^2}=\dfrac{\sec^2x}{(1+\tan x)^2}$ , so $y=-\dfrac1{1+\tan x}+C$ ; $y\left(\frac\pi4\right)=\frac12$ gives $C=1$ and $y=\dfrac{\sin x}{\sin x+\cos x}$ .

Intersections with $\sqrt2\sin x$ . Either $\sin x=0$ : $x=\pi$ ; or $\sin x+\cos x=\frac1{\sqrt2}$ , i.e. $\sin\left(x+\frac\pi4\right)=\frac12$ : $x=\frac{7\pi}{12}$ , $\frac{23\pi}{12}$ .

Sum. $\dfrac{(12+7+23)\pi}{12}$ , so $k=42$ (taking the same constant on every interval of S, as the key does).

Answer: 42

Topics:Variable separable Trigonometric equations

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JEE Main 2022 · 26 Jun · Shift 1, Mathematics question 30