NEET Test Series from KOTA - 10 Papers In MS WORD
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Differential Equation
87551
Let \(y=y(x)\) be the solution of the differential equation \(\cos x(3 \sin x+\cos x+3) d y=[1+y \sin x\) \((3 \sin x+\cos x+3] d x, 0 \leq x \leq \frac{\pi}{2}, y(0)=0\) Then, \(y\left(\frac{\pi}{3}\right)\) is equal to
(B) : \(\cos x(3 \sin x+\cos x+3) d y=[1+y \sin x(3 \sin x\) \(+\cos \mathrm{x}+3)] \mathrm{dx}\) \(\cos x d y=\left(\frac{1}{3 \sin x+\cos x+3}+y \sin x\right) d x\) \(\frac{d y}{d x}=y \frac{\sin x}{\cos x}+\frac{1}{\cos x(3 \sin x+\cos x+3)}\) \(\frac{d y}{d x}-(\tan x) y=\frac{1}{\cos x(3 \sin x+\cos x+3)}\) Which is linear differential equation, \(\therefore\) Integrating factor (IF) \(=\mathrm{e}^{\int(-\tan x) \mathrm{dx}}\) \(=\mathrm{e}^{\log |\cos x|}=|\cos \mathrm{x}|\) \(\because \quad|\cos x|>0, \forall \mathrm{x} \in\left[0, \frac{\pi}{2}\right]\) \(\therefore \quad|\cos x|=\cos x\) Hence, solution of equation (i) is \(y(\cos x)=\int(\cos x) \cdot \frac{1}{\cos x(3 \sin x+\cos x+3)} d x\) \(y \cos x=\int \frac{1}{3 \sin x+\cos x+3} d x\) \(y \cos x=\int \frac{\sec ^{2} \frac{x}{2} d x}{2 \tan ^{2} \frac{x}{2}+6 \tan \frac{x}{2}+4} d x\) \(=\int \frac{\sec ^{2} \frac{x}{2} d x}{2\left(\tan ^{2} \frac{x}{2}+3 \tan \frac{x}{2}+2\right)} d x\) On putting \(\tan \frac{x}{2}=2\) \(\frac{1}{2} \sec ^{2} \frac{x}{2} d x=d z\) \(\therefore \mathrm{y} \cos \mathrm{x}=\int \frac{\mathrm{dz}}{\mathrm{z}^{2}+32+2}=\int \frac{\mathrm{dz}}{(\mathrm{z}+1)(\mathrm{z}+2)}\) \(=\int \frac{1}{(z+1)} d z-\int \frac{1}{z+2} d z=\log (z+1)-\log (z+2)+C\) \(y \cos x=\log \left|\frac{z+1}{z+2}\right|+C\) \(=\log \left|\frac{\tan \frac{\mathrm{x}}{2}+1}{\tan \frac{\mathrm{x}}{2}+2}\right|+\mathrm{C}\) \(\therefore \quad \mathrm{C}=-\log \left(\frac{1}{2}\right)=\log (2)\) From equation (i), \(\mathrm{y} \cos \mathrm{x}=\log \left|\frac{\tan \frac{\mathrm{x}}{2}+1}{\tan \frac{\mathrm{x}}{2}+2}\right|+\log 2\) \(\therefore y\left(\frac{\pi}{3}\right)=2\left[\log \left|\frac{\frac{1}{\sqrt{3}}+1}{\frac{1}{\sqrt{3}}+2}\right|+\log 2\right]\) \(=2 \log \left|2\left(\frac{\sqrt{3}+1}{2 \sqrt{3}+1}\right)\right|=2 \log \left|2\left(\frac{\sqrt{3}+1}{2 \sqrt{3}+1} \times \frac{2 \sqrt{3}-1}{2 \sqrt{3}-1}\right)\right|\) \(=2 \log \left|\frac{2 \sqrt{3}+10}{11}\right|\)
JEE Main 17.03.2021
Differential Equation
87552
Let \(y=y(x)\) be a solution curve of the differential equation \((y+1) \tan ^{2} x d x+\tan x\) \(d y+y d x=0, x \in\left(0, \frac{\pi}{2}\right) \cdot \lim _{x \rightarrow 0^{+}} x y(x)=1\), then value of \(y\left(\frac{\pi}{4}\right)\) is
1 \(-\frac{\pi}{4}\)
2 \(\frac{\pi}{4}-1\)
3 \(\frac{\pi}{4}+1\)
4 \(\frac{\pi}{4}\)
Explanation:
(D) : \((y+1) \tan ^{2} x d x+\tan x d y+y d x=0\) \(\left[(y+1) \tan ^{2} x+y\right] d x+\tan x d y=0\) \(\frac{d y}{d x}+(y+1) \tan x+\frac{y}{\tan x}=0\) \(\frac{d y}{d x}+\frac{y \tan ^{2} x+\tan ^{2} x+y}{\tan x}=0\) \(\frac{d y}{d x}+\frac{y \sec ^{2} x}{\tan x}+\tan ^{2} x=0\) \(\frac{d y}{d x}+\left(\frac{\sec ^{2} x}{\tan x}\right) y=-\tan ^{2} x\) This is a linear differential equation, \(\therefore \quad I F=\mathrm{e}^{\int \frac{\sec ^{2} x}{\tan x}}=\mathrm{e}^{\ln (\tan x)}=\tan x\) So, solution is given by- \((y \tan x)=\int-\tan ^{2} x d x=\int\left(1-\sec ^{2} x\right) d x=x-\tan x+C\) \(y=x \cot x-1+C \cot x\) Now, \(\quad \lim _{x \rightarrow 0} x \cdot y=1\) \(\lim _{x \rightarrow 0^{+}}\left|x^{2} \cot x-x+C x \cot x\right|=1\) \(\lim _{x \rightarrow 0^{+}}\left|x \cdot \frac{x}{\tan x}-x+\frac{C x}{\tan x}\right|=1\) \(0-0+\mathrm{C}=1\) \(\mathrm{C}=1\) \(\therefore \quad \mathrm{y}=\mathrm{x} \cot \mathrm{x}-1+\cot \mathrm{x}\) Now, \(\quad \mathrm{x}=\frac{\pi}{4}\) \(\mathrm{y}=\frac{\pi}{4}-1+1=\frac{\pi}{4}\)
JEE Main 26.08.2021
Differential Equation
87556
Let \(y=y(x)\) be the solution of the differential equation, \(\frac{2+\sin x}{y+1} \cdot \frac{d y}{d x}=-\cos x, y>0, y(0)=1\). If \(y(\pi)=a\) and \(\frac{d y}{d x}\) at \(x=\pi\) is \(b\), then the ordered pair \((a, b)\) is equal to
87551
Let \(y=y(x)\) be the solution of the differential equation \(\cos x(3 \sin x+\cos x+3) d y=[1+y \sin x\) \((3 \sin x+\cos x+3] d x, 0 \leq x \leq \frac{\pi}{2}, y(0)=0\) Then, \(y\left(\frac{\pi}{3}\right)\) is equal to
(B) : \(\cos x(3 \sin x+\cos x+3) d y=[1+y \sin x(3 \sin x\) \(+\cos \mathrm{x}+3)] \mathrm{dx}\) \(\cos x d y=\left(\frac{1}{3 \sin x+\cos x+3}+y \sin x\right) d x\) \(\frac{d y}{d x}=y \frac{\sin x}{\cos x}+\frac{1}{\cos x(3 \sin x+\cos x+3)}\) \(\frac{d y}{d x}-(\tan x) y=\frac{1}{\cos x(3 \sin x+\cos x+3)}\) Which is linear differential equation, \(\therefore\) Integrating factor (IF) \(=\mathrm{e}^{\int(-\tan x) \mathrm{dx}}\) \(=\mathrm{e}^{\log |\cos x|}=|\cos \mathrm{x}|\) \(\because \quad|\cos x|>0, \forall \mathrm{x} \in\left[0, \frac{\pi}{2}\right]\) \(\therefore \quad|\cos x|=\cos x\) Hence, solution of equation (i) is \(y(\cos x)=\int(\cos x) \cdot \frac{1}{\cos x(3 \sin x+\cos x+3)} d x\) \(y \cos x=\int \frac{1}{3 \sin x+\cos x+3} d x\) \(y \cos x=\int \frac{\sec ^{2} \frac{x}{2} d x}{2 \tan ^{2} \frac{x}{2}+6 \tan \frac{x}{2}+4} d x\) \(=\int \frac{\sec ^{2} \frac{x}{2} d x}{2\left(\tan ^{2} \frac{x}{2}+3 \tan \frac{x}{2}+2\right)} d x\) On putting \(\tan \frac{x}{2}=2\) \(\frac{1}{2} \sec ^{2} \frac{x}{2} d x=d z\) \(\therefore \mathrm{y} \cos \mathrm{x}=\int \frac{\mathrm{dz}}{\mathrm{z}^{2}+32+2}=\int \frac{\mathrm{dz}}{(\mathrm{z}+1)(\mathrm{z}+2)}\) \(=\int \frac{1}{(z+1)} d z-\int \frac{1}{z+2} d z=\log (z+1)-\log (z+2)+C\) \(y \cos x=\log \left|\frac{z+1}{z+2}\right|+C\) \(=\log \left|\frac{\tan \frac{\mathrm{x}}{2}+1}{\tan \frac{\mathrm{x}}{2}+2}\right|+\mathrm{C}\) \(\therefore \quad \mathrm{C}=-\log \left(\frac{1}{2}\right)=\log (2)\) From equation (i), \(\mathrm{y} \cos \mathrm{x}=\log \left|\frac{\tan \frac{\mathrm{x}}{2}+1}{\tan \frac{\mathrm{x}}{2}+2}\right|+\log 2\) \(\therefore y\left(\frac{\pi}{3}\right)=2\left[\log \left|\frac{\frac{1}{\sqrt{3}}+1}{\frac{1}{\sqrt{3}}+2}\right|+\log 2\right]\) \(=2 \log \left|2\left(\frac{\sqrt{3}+1}{2 \sqrt{3}+1}\right)\right|=2 \log \left|2\left(\frac{\sqrt{3}+1}{2 \sqrt{3}+1} \times \frac{2 \sqrt{3}-1}{2 \sqrt{3}-1}\right)\right|\) \(=2 \log \left|\frac{2 \sqrt{3}+10}{11}\right|\)
JEE Main 17.03.2021
Differential Equation
87552
Let \(y=y(x)\) be a solution curve of the differential equation \((y+1) \tan ^{2} x d x+\tan x\) \(d y+y d x=0, x \in\left(0, \frac{\pi}{2}\right) \cdot \lim _{x \rightarrow 0^{+}} x y(x)=1\), then value of \(y\left(\frac{\pi}{4}\right)\) is
1 \(-\frac{\pi}{4}\)
2 \(\frac{\pi}{4}-1\)
3 \(\frac{\pi}{4}+1\)
4 \(\frac{\pi}{4}\)
Explanation:
(D) : \((y+1) \tan ^{2} x d x+\tan x d y+y d x=0\) \(\left[(y+1) \tan ^{2} x+y\right] d x+\tan x d y=0\) \(\frac{d y}{d x}+(y+1) \tan x+\frac{y}{\tan x}=0\) \(\frac{d y}{d x}+\frac{y \tan ^{2} x+\tan ^{2} x+y}{\tan x}=0\) \(\frac{d y}{d x}+\frac{y \sec ^{2} x}{\tan x}+\tan ^{2} x=0\) \(\frac{d y}{d x}+\left(\frac{\sec ^{2} x}{\tan x}\right) y=-\tan ^{2} x\) This is a linear differential equation, \(\therefore \quad I F=\mathrm{e}^{\int \frac{\sec ^{2} x}{\tan x}}=\mathrm{e}^{\ln (\tan x)}=\tan x\) So, solution is given by- \((y \tan x)=\int-\tan ^{2} x d x=\int\left(1-\sec ^{2} x\right) d x=x-\tan x+C\) \(y=x \cot x-1+C \cot x\) Now, \(\quad \lim _{x \rightarrow 0} x \cdot y=1\) \(\lim _{x \rightarrow 0^{+}}\left|x^{2} \cot x-x+C x \cot x\right|=1\) \(\lim _{x \rightarrow 0^{+}}\left|x \cdot \frac{x}{\tan x}-x+\frac{C x}{\tan x}\right|=1\) \(0-0+\mathrm{C}=1\) \(\mathrm{C}=1\) \(\therefore \quad \mathrm{y}=\mathrm{x} \cot \mathrm{x}-1+\cot \mathrm{x}\) Now, \(\quad \mathrm{x}=\frac{\pi}{4}\) \(\mathrm{y}=\frac{\pi}{4}-1+1=\frac{\pi}{4}\)
JEE Main 26.08.2021
Differential Equation
87556
Let \(y=y(x)\) be the solution of the differential equation, \(\frac{2+\sin x}{y+1} \cdot \frac{d y}{d x}=-\cos x, y>0, y(0)=1\). If \(y(\pi)=a\) and \(\frac{d y}{d x}\) at \(x=\pi\) is \(b\), then the ordered pair \((a, b)\) is equal to
87551
Let \(y=y(x)\) be the solution of the differential equation \(\cos x(3 \sin x+\cos x+3) d y=[1+y \sin x\) \((3 \sin x+\cos x+3] d x, 0 \leq x \leq \frac{\pi}{2}, y(0)=0\) Then, \(y\left(\frac{\pi}{3}\right)\) is equal to
(B) : \(\cos x(3 \sin x+\cos x+3) d y=[1+y \sin x(3 \sin x\) \(+\cos \mathrm{x}+3)] \mathrm{dx}\) \(\cos x d y=\left(\frac{1}{3 \sin x+\cos x+3}+y \sin x\right) d x\) \(\frac{d y}{d x}=y \frac{\sin x}{\cos x}+\frac{1}{\cos x(3 \sin x+\cos x+3)}\) \(\frac{d y}{d x}-(\tan x) y=\frac{1}{\cos x(3 \sin x+\cos x+3)}\) Which is linear differential equation, \(\therefore\) Integrating factor (IF) \(=\mathrm{e}^{\int(-\tan x) \mathrm{dx}}\) \(=\mathrm{e}^{\log |\cos x|}=|\cos \mathrm{x}|\) \(\because \quad|\cos x|>0, \forall \mathrm{x} \in\left[0, \frac{\pi}{2}\right]\) \(\therefore \quad|\cos x|=\cos x\) Hence, solution of equation (i) is \(y(\cos x)=\int(\cos x) \cdot \frac{1}{\cos x(3 \sin x+\cos x+3)} d x\) \(y \cos x=\int \frac{1}{3 \sin x+\cos x+3} d x\) \(y \cos x=\int \frac{\sec ^{2} \frac{x}{2} d x}{2 \tan ^{2} \frac{x}{2}+6 \tan \frac{x}{2}+4} d x\) \(=\int \frac{\sec ^{2} \frac{x}{2} d x}{2\left(\tan ^{2} \frac{x}{2}+3 \tan \frac{x}{2}+2\right)} d x\) On putting \(\tan \frac{x}{2}=2\) \(\frac{1}{2} \sec ^{2} \frac{x}{2} d x=d z\) \(\therefore \mathrm{y} \cos \mathrm{x}=\int \frac{\mathrm{dz}}{\mathrm{z}^{2}+32+2}=\int \frac{\mathrm{dz}}{(\mathrm{z}+1)(\mathrm{z}+2)}\) \(=\int \frac{1}{(z+1)} d z-\int \frac{1}{z+2} d z=\log (z+1)-\log (z+2)+C\) \(y \cos x=\log \left|\frac{z+1}{z+2}\right|+C\) \(=\log \left|\frac{\tan \frac{\mathrm{x}}{2}+1}{\tan \frac{\mathrm{x}}{2}+2}\right|+\mathrm{C}\) \(\therefore \quad \mathrm{C}=-\log \left(\frac{1}{2}\right)=\log (2)\) From equation (i), \(\mathrm{y} \cos \mathrm{x}=\log \left|\frac{\tan \frac{\mathrm{x}}{2}+1}{\tan \frac{\mathrm{x}}{2}+2}\right|+\log 2\) \(\therefore y\left(\frac{\pi}{3}\right)=2\left[\log \left|\frac{\frac{1}{\sqrt{3}}+1}{\frac{1}{\sqrt{3}}+2}\right|+\log 2\right]\) \(=2 \log \left|2\left(\frac{\sqrt{3}+1}{2 \sqrt{3}+1}\right)\right|=2 \log \left|2\left(\frac{\sqrt{3}+1}{2 \sqrt{3}+1} \times \frac{2 \sqrt{3}-1}{2 \sqrt{3}-1}\right)\right|\) \(=2 \log \left|\frac{2 \sqrt{3}+10}{11}\right|\)
JEE Main 17.03.2021
Differential Equation
87552
Let \(y=y(x)\) be a solution curve of the differential equation \((y+1) \tan ^{2} x d x+\tan x\) \(d y+y d x=0, x \in\left(0, \frac{\pi}{2}\right) \cdot \lim _{x \rightarrow 0^{+}} x y(x)=1\), then value of \(y\left(\frac{\pi}{4}\right)\) is
1 \(-\frac{\pi}{4}\)
2 \(\frac{\pi}{4}-1\)
3 \(\frac{\pi}{4}+1\)
4 \(\frac{\pi}{4}\)
Explanation:
(D) : \((y+1) \tan ^{2} x d x+\tan x d y+y d x=0\) \(\left[(y+1) \tan ^{2} x+y\right] d x+\tan x d y=0\) \(\frac{d y}{d x}+(y+1) \tan x+\frac{y}{\tan x}=0\) \(\frac{d y}{d x}+\frac{y \tan ^{2} x+\tan ^{2} x+y}{\tan x}=0\) \(\frac{d y}{d x}+\frac{y \sec ^{2} x}{\tan x}+\tan ^{2} x=0\) \(\frac{d y}{d x}+\left(\frac{\sec ^{2} x}{\tan x}\right) y=-\tan ^{2} x\) This is a linear differential equation, \(\therefore \quad I F=\mathrm{e}^{\int \frac{\sec ^{2} x}{\tan x}}=\mathrm{e}^{\ln (\tan x)}=\tan x\) So, solution is given by- \((y \tan x)=\int-\tan ^{2} x d x=\int\left(1-\sec ^{2} x\right) d x=x-\tan x+C\) \(y=x \cot x-1+C \cot x\) Now, \(\quad \lim _{x \rightarrow 0} x \cdot y=1\) \(\lim _{x \rightarrow 0^{+}}\left|x^{2} \cot x-x+C x \cot x\right|=1\) \(\lim _{x \rightarrow 0^{+}}\left|x \cdot \frac{x}{\tan x}-x+\frac{C x}{\tan x}\right|=1\) \(0-0+\mathrm{C}=1\) \(\mathrm{C}=1\) \(\therefore \quad \mathrm{y}=\mathrm{x} \cot \mathrm{x}-1+\cot \mathrm{x}\) Now, \(\quad \mathrm{x}=\frac{\pi}{4}\) \(\mathrm{y}=\frac{\pi}{4}-1+1=\frac{\pi}{4}\)
JEE Main 26.08.2021
Differential Equation
87556
Let \(y=y(x)\) be the solution of the differential equation, \(\frac{2+\sin x}{y+1} \cdot \frac{d y}{d x}=-\cos x, y>0, y(0)=1\). If \(y(\pi)=a\) and \(\frac{d y}{d x}\) at \(x=\pi\) is \(b\), then the ordered pair \((a, b)\) is equal to
87551
Let \(y=y(x)\) be the solution of the differential equation \(\cos x(3 \sin x+\cos x+3) d y=[1+y \sin x\) \((3 \sin x+\cos x+3] d x, 0 \leq x \leq \frac{\pi}{2}, y(0)=0\) Then, \(y\left(\frac{\pi}{3}\right)\) is equal to
(B) : \(\cos x(3 \sin x+\cos x+3) d y=[1+y \sin x(3 \sin x\) \(+\cos \mathrm{x}+3)] \mathrm{dx}\) \(\cos x d y=\left(\frac{1}{3 \sin x+\cos x+3}+y \sin x\right) d x\) \(\frac{d y}{d x}=y \frac{\sin x}{\cos x}+\frac{1}{\cos x(3 \sin x+\cos x+3)}\) \(\frac{d y}{d x}-(\tan x) y=\frac{1}{\cos x(3 \sin x+\cos x+3)}\) Which is linear differential equation, \(\therefore\) Integrating factor (IF) \(=\mathrm{e}^{\int(-\tan x) \mathrm{dx}}\) \(=\mathrm{e}^{\log |\cos x|}=|\cos \mathrm{x}|\) \(\because \quad|\cos x|>0, \forall \mathrm{x} \in\left[0, \frac{\pi}{2}\right]\) \(\therefore \quad|\cos x|=\cos x\) Hence, solution of equation (i) is \(y(\cos x)=\int(\cos x) \cdot \frac{1}{\cos x(3 \sin x+\cos x+3)} d x\) \(y \cos x=\int \frac{1}{3 \sin x+\cos x+3} d x\) \(y \cos x=\int \frac{\sec ^{2} \frac{x}{2} d x}{2 \tan ^{2} \frac{x}{2}+6 \tan \frac{x}{2}+4} d x\) \(=\int \frac{\sec ^{2} \frac{x}{2} d x}{2\left(\tan ^{2} \frac{x}{2}+3 \tan \frac{x}{2}+2\right)} d x\) On putting \(\tan \frac{x}{2}=2\) \(\frac{1}{2} \sec ^{2} \frac{x}{2} d x=d z\) \(\therefore \mathrm{y} \cos \mathrm{x}=\int \frac{\mathrm{dz}}{\mathrm{z}^{2}+32+2}=\int \frac{\mathrm{dz}}{(\mathrm{z}+1)(\mathrm{z}+2)}\) \(=\int \frac{1}{(z+1)} d z-\int \frac{1}{z+2} d z=\log (z+1)-\log (z+2)+C\) \(y \cos x=\log \left|\frac{z+1}{z+2}\right|+C\) \(=\log \left|\frac{\tan \frac{\mathrm{x}}{2}+1}{\tan \frac{\mathrm{x}}{2}+2}\right|+\mathrm{C}\) \(\therefore \quad \mathrm{C}=-\log \left(\frac{1}{2}\right)=\log (2)\) From equation (i), \(\mathrm{y} \cos \mathrm{x}=\log \left|\frac{\tan \frac{\mathrm{x}}{2}+1}{\tan \frac{\mathrm{x}}{2}+2}\right|+\log 2\) \(\therefore y\left(\frac{\pi}{3}\right)=2\left[\log \left|\frac{\frac{1}{\sqrt{3}}+1}{\frac{1}{\sqrt{3}}+2}\right|+\log 2\right]\) \(=2 \log \left|2\left(\frac{\sqrt{3}+1}{2 \sqrt{3}+1}\right)\right|=2 \log \left|2\left(\frac{\sqrt{3}+1}{2 \sqrt{3}+1} \times \frac{2 \sqrt{3}-1}{2 \sqrt{3}-1}\right)\right|\) \(=2 \log \left|\frac{2 \sqrt{3}+10}{11}\right|\)
JEE Main 17.03.2021
Differential Equation
87552
Let \(y=y(x)\) be a solution curve of the differential equation \((y+1) \tan ^{2} x d x+\tan x\) \(d y+y d x=0, x \in\left(0, \frac{\pi}{2}\right) \cdot \lim _{x \rightarrow 0^{+}} x y(x)=1\), then value of \(y\left(\frac{\pi}{4}\right)\) is
1 \(-\frac{\pi}{4}\)
2 \(\frac{\pi}{4}-1\)
3 \(\frac{\pi}{4}+1\)
4 \(\frac{\pi}{4}\)
Explanation:
(D) : \((y+1) \tan ^{2} x d x+\tan x d y+y d x=0\) \(\left[(y+1) \tan ^{2} x+y\right] d x+\tan x d y=0\) \(\frac{d y}{d x}+(y+1) \tan x+\frac{y}{\tan x}=0\) \(\frac{d y}{d x}+\frac{y \tan ^{2} x+\tan ^{2} x+y}{\tan x}=0\) \(\frac{d y}{d x}+\frac{y \sec ^{2} x}{\tan x}+\tan ^{2} x=0\) \(\frac{d y}{d x}+\left(\frac{\sec ^{2} x}{\tan x}\right) y=-\tan ^{2} x\) This is a linear differential equation, \(\therefore \quad I F=\mathrm{e}^{\int \frac{\sec ^{2} x}{\tan x}}=\mathrm{e}^{\ln (\tan x)}=\tan x\) So, solution is given by- \((y \tan x)=\int-\tan ^{2} x d x=\int\left(1-\sec ^{2} x\right) d x=x-\tan x+C\) \(y=x \cot x-1+C \cot x\) Now, \(\quad \lim _{x \rightarrow 0} x \cdot y=1\) \(\lim _{x \rightarrow 0^{+}}\left|x^{2} \cot x-x+C x \cot x\right|=1\) \(\lim _{x \rightarrow 0^{+}}\left|x \cdot \frac{x}{\tan x}-x+\frac{C x}{\tan x}\right|=1\) \(0-0+\mathrm{C}=1\) \(\mathrm{C}=1\) \(\therefore \quad \mathrm{y}=\mathrm{x} \cot \mathrm{x}-1+\cot \mathrm{x}\) Now, \(\quad \mathrm{x}=\frac{\pi}{4}\) \(\mathrm{y}=\frac{\pi}{4}-1+1=\frac{\pi}{4}\)
JEE Main 26.08.2021
Differential Equation
87556
Let \(y=y(x)\) be the solution of the differential equation, \(\frac{2+\sin x}{y+1} \cdot \frac{d y}{d x}=-\cos x, y>0, y(0)=1\). If \(y(\pi)=a\) and \(\frac{d y}{d x}\) at \(x=\pi\) is \(b\), then the ordered pair \((a, b)\) is equal to
