Let y=y(x) be the solution of the differential equation \frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x, x \∊(- \frac{\\pi}{2}, \frac{\\pi}{2} ) such that if y(0)=1 ,then [10 Apr 2019 Shift 2]

Correct Answer is : y′ ( π4 )+y′ (− π4 )=−√2

y^{'} ( \frac{\\pi}{4} )-y^{'} (- \frac{\\pi}{4} )=\\pi-\sqrt{2}

y ( \frac{\\pi}{4} )-y (- \frac{\\pi}{4} )= \sqrt{2}

y ( \frac{\\pi}{4} )+y (- \frac{\\pi}{4} )=\\pi^{2}/2+\sqrt{2}

y^{'} ( \frac{\\pi}{4} )+y^{'} (- \frac{\\pi}{4} )= -\sqrt{2}

JEE Main 10 Apr 2019 Paper 2 Solved Paper

Section: Mathematics

Question : 72 of 90

Marks: +1, -0

Let y=y(x) be the solution of the differential equation

+y‌tan‌x=2x+x2‌tan‌x,x∊(−

) such that if y(0)=1 ,then

[10 Apr 2019 Shift 2]

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