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JEE Mathematics Solved Question Paper 2009

Multiple Choice Questions

61.
The solution of the differential equation $\frac{dy}{dx} = \frac{1}{x + y^{2}}$ is

y = - x² - 2x - 2 + ce^x

y = x² + 2x + 2 - ce^x

x = - y² - 2y + 2 - ce^y

x = - y² - 2y - 2 + ce^y

x = - y² - 2y - 2 + ce^y

Given differential equation is

$\frac{dy}{dx} = \frac{1}{x + y^{2}} \Rightarrow \frac{dx}{dy} - x = y^{2} Here, P = - 1, Q = y^{2} IF = e^{\int - 1 dy} = e^{- y} ∴ Solution is {xe}^{- y} = \int e^{- y} y^{2} dy = - e^{- y} y^{2} + 2 [- e^{- y} y + \int e^{- y} dy] + c = - e^{- y} y^{2} + 2 [- e^{- y} - e^{- y}] + c \Rightarrow {xe}^{- y} = e^{- y} (- y^{2} - 2 y - 2) + c \Rightarrow x = - y^{2} - 2 y - 2 + {ce}^{y}$

62.

The solution of the equation ${(3 + 2 \sqrt{2})}^{x^{2} - 8} + {(3 + 2 \sqrt{2})}^{8 - x^{2}}$ = 6 are

$3 \pm 2 \sqrt{2}$
$\pm 1$
$\pm 3 \sqrt{3}, \pm 2 \sqrt{2}$
$\pm 3, \pm \sqrt{7}$

63.

Let T_n denote the number of triangles which can equal to be formed by using the vertices of a regular polygon of n sides. If T_{n + 1} - T_n = 28, then n equals