Let S = (0,2\pi ) - \ \pi \over 2,3\pi \over 4,3\pi \over 2,7\pi \over 4 \ . Let y = y(x) , x \in S, be the solution curve of the differential equation dy \over dx = 1 \over 1 + \sin 2x,\,y( \pi \over 4 ) = 1 \over 2 . If the sum of abscissas of all the points of intersection of the curve y = y(x) with the curve y = \sqrt 2 \sin x is k\pi \over 12 , then k is equal to \underline\hspace2cm .

Differential Equations PYQ — JEE Main 2022 · 26 Jun · Shift 1 · Q41 with Solution

View written solutionFree

Correct answer: 42

Given differential equation

We need to solve

\frac{dy}{dx}=\frac{1}{1+\sin 2x}, \qquad y\left(\frac\pi4\right)=\frac12.

Also,

1+\sin 2x=1+2\sin x\cos x=(\sin x+\cos x)^2.

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

Using

(\sec x+\tan x)'=\sec x\tan x+\sec^2 x=\frac{1+\sin x}{\cos^2 x}=\frac{1}{(\sin x+\cos x)^2}?

That is not the most direct form, so instead simplify as

1+\sin 2x=\sin^2x+\cos^2x+2\sin x\cos x=(\sin x+\cos x)^2.

Hence

\frac{1}{1+\sin 2x}=\frac{1}{(\sin x+\cos x)^2}.

Now note that

\frac{d}{dx}\left(\frac{\sin x-\cos x}{2(\sin x+\cos x)}\right)=\frac{1}{(\sin x+\cos x)^2}.

So integrating,

y=\frac{\sin x-\cos x}{2(\sin x+\cos x)}+C.

Use the initial condition

At $x=\frac\pi4$ ,

\sin\frac\pi4=\cos\frac\pi4=\frac{1}{\sqrt2}.

Therefore,

\frac{\sin x-\cos x}{2(\sin x+\cos x)}\Bigg|_{x=\pi/4}=0.

Since $y\left(\frac\pi4\right)=\frac12$ , we get

C=\frac12.

Thus

y=\frac12+\frac{\sin x-\cos x}{2(\sin x+\cos x)} =\frac{\sin x}{\sin x+\cos x}.

So the solution curve is

y(x)=\frac{\sin x}{\sin x+\cos x}, \qquad x\in S.

Intersection with $y=\sqrt2\sin x$

Set

\frac{\sin x}{\sin x+\cos x}=\sqrt2\sin x.

Rearrange:

\sin x=\sqrt2\sin x(\sin x+\cos x).

\sin x\Big(1-\sqrt2(\sin x+\cos x)\Big)=0.

Hence intersections occur when either:

Case 1: $\sin x=0$

In $(0,2\pi)$ , this gives

x=\pi.

(Neither $0$ nor $2\pi$ belong to the interval.)

Case 2: $1-\sqrt2(\sin x+\cos x)=0$

\sin x+\cos x=\frac{1}{\sqrt2}.

Using

\sin x+\cos x=\sqrt2\sin\left(x+\frac\pi4\right),

we get

\sqrt2\sin\left(x+\frac\pi4\right)=\frac{1}{\sqrt2}

\sin\left(x+\frac\pi4\right)=\frac12.

Thus

x+\frac\pi4=\frac\pi6+2n\pi \quad \text{or} \quad \frac{5\pi}6+2n\pi.

Therefore

x=-\frac\pi{12}+2n\pi \quad \text{or} \quad \frac{7\pi}{12}+2n\pi.

Now restrict to $(0,2\pi)$ :

From $x=-\frac\pi{12}+2n\pi$ , we get $x=\frac{23\pi}{12}$ .
From $x=\frac{7\pi}{12}+2n\pi$ , we get $x=\frac{7\pi}{12}$ .

Both lie in $S$ (they are not excluded points).

So all intersection abscissas are

\pi,\quad \frac{7\pi}{12},\quad \frac{23\pi}{12}.

Sum of abscissas

\pi+\frac{7\pi}{12}+\frac{23\pi}{12} =\pi+\frac{30\pi}{12} =\pi+\frac{5\pi}{2} =\frac{7\pi}{2}.

Write this as

\frac{7\pi}{2}=\frac{42\pi}{12}.

Hence

k=42.

Comparison with stored answer

Stored correct answer = $42$ , which matches our result.

More from Differential Equations

View written solutionFree

Correct answer: 42

Given differential equation

We need to solve

\frac{dy}{dx}=\frac{1}{1+\sin 2x}, \qquad y\left(\frac\pi4\right)=\frac12.

Also,

1+\sin 2x=1+2\sin x\cos x=(\sin x+\cos x)^2.

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

Using

(\sec x+\tan x)'=\sec x\tan x+\sec^2 x=\frac{1+\sin x}{\cos^2 x}=\frac{1}{(\sin x+\cos x)^2}?

That is not the most direct form, so instead simplify as

1+\sin 2x=\sin^2x+\cos^2x+2\sin x\cos x=(\sin x+\cos x)^2.

Hence

\frac{1}{1+\sin 2x}=\frac{1}{(\sin x+\cos x)^2}.

Now note that

\frac{d}{dx}\left(\frac{\sin x-\cos x}{2(\sin x+\cos x)}\right)=\frac{1}{(\sin x+\cos x)^2}.

So integrating,

y=\frac{\sin x-\cos x}{2(\sin x+\cos x)}+C.

Use the initial condition

At $x=\frac\pi4$ ,

\sin\frac\pi4=\cos\frac\pi4=\frac{1}{\sqrt2}.

Therefore,

\frac{\sin x-\cos x}{2(\sin x+\cos x)}\Bigg|_{x=\pi/4}=0.

Since $y\left(\frac\pi4\right)=\frac12$ , we get

C=\frac12.

Thus

y=\frac12+\frac{\sin x-\cos x}{2(\sin x+\cos x)} =\frac{\sin x}{\sin x+\cos x}.

So the solution curve is

y(x)=\frac{\sin x}{\sin x+\cos x}, \qquad x\in S.

Intersection with $y=\sqrt2\sin x$

Set

\frac{\sin x}{\sin x+\cos x}=\sqrt2\sin x.

Rearrange:

\sin x=\sqrt2\sin x(\sin x+\cos x).

\sin x\Big(1-\sqrt2(\sin x+\cos x)\Big)=0.

Hence intersections occur when either:

Case 1: $\sin x=0$

In $(0,2\pi)$ , this gives

x=\pi.

(Neither $0$ nor $2\pi$ belong to the interval.)

Case 2: $1-\sqrt2(\sin x+\cos x)=0$

\sin x+\cos x=\frac{1}{\sqrt2}.

Using

\sin x+\cos x=\sqrt2\sin\left(x+\frac\pi4\right),

we get

\sqrt2\sin\left(x+\frac\pi4\right)=\frac{1}{\sqrt2}

\sin\left(x+\frac\pi4\right)=\frac12.

Thus

x+\frac\pi4=\frac\pi6+2n\pi \quad \text{or} \quad \frac{5\pi}6+2n\pi.

Therefore

x=-\frac\pi{12}+2n\pi \quad \text{or} \quad \frac{7\pi}{12}+2n\pi.

Now restrict to $(0,2\pi)$ :

From $x=-\frac\pi{12}+2n\pi$ , we get $x=\frac{23\pi}{12}$ .
From $x=\frac{7\pi}{12}+2n\pi$ , we get $x=\frac{7\pi}{12}$ .

Both lie in $S$ (they are not excluded points).

So all intersection abscissas are

\pi,\quad \frac{7\pi}{12},\quad \frac{23\pi}{12}.

Sum of abscissas

\pi+\frac{7\pi}{12}+\frac{23\pi}{12} =\pi+\frac{30\pi}{12} =\pi+\frac{5\pi}{2} =\frac{7\pi}{2}.

Write this as

\frac{7\pi}{2}=\frac{42\pi}{12}.

Hence

k=42.

Comparison with stored answer

Stored correct answer = $42$ , which matches our result.

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