Question

JEE MAINS MATHS QUESTION SEPTEMBER 2020

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

$\displaystyle \: \frac{dy}{2 + y} = \frac{ {e}^{x} }{5 + {e}^{x} }$

On integration

$\displaystyle \int\limits_{}^{}\frac{1}{2 + y} \, dy = \int\limits_{}^{}\frac{ {e}^{x} }{5 + {e}^{x} } \, dx$

Let

$u = 2 + y$

Differentiating both sides with respect to y

$\displaystyle \: \frac{du}{dy} = 1$

$\implies \: du = dy$

Let

$\displaystyle \: v = 5 + {e}^{x}$

Differentiating both sides with respect to x

$\displaystyle \: \frac{dv}{dx} = {e}^{x}$

$\implies \: dv = {e}^{x} dx$

So

$\displaystyle \int\limits_{}^{}\frac{1}{2 + y} \, dy = \int\limits_{}^{}\frac{ {e}^{x} }{5 + {e}^{x} } \, dy$

$\implies \: \displaystyle \int\limits_{}^{}\frac{du}{u} = \int\limits_{}^{}\frac{dv }{v }$

$\implies \: log \: u = log \: v + log \: c$

Where log c is a constant

$\implies \: log \: u = log \: (cv)$

$\implies \: u = cv$

$\implies \: 2 + y = c(5 + {e}^{x} )$

Now by the given condition x = 0, y = 4

So

$6 = c(5 + 1)$

$\therefore \: \: c = 1$

So

$2 + y = 5 + {e}^{x}$

$\implies \: y = 3 + {e}^{x}$

Now

$for \: \: x = log_{e}(13) \: \: we \: have$

$y = 3 + {e}^{ log_{e}(13) }$

$\implies \: y = 3 + 13 = 16$

RESULT

Hence the required value of y is 16