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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

62.
The solution of the equation ${(3 + 2 \sqrt{2})}^{x^{2} - 8} + {(3 + 2 \sqrt{2})}^{8 - x^{2}}$ = 6 are

$3 \pm 2 \sqrt{2}$

$\pm 1$

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

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

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

$Given, {(3 + 2 \sqrt{2})}^{x^{2} - 8} + {(3 + 2 \sqrt{2})}^{8 - x^{2}} = 6 Let {(3 + 2 \sqrt{2})}^{x^{2} - 8} = y ∴ y + y^{- 1} = 6 \Rightarrow y^{2} - 6 y + 1 = 0 \Rightarrow y = \frac{6 \pm \sqrt{36 - 4}}{2 \times 1} = \frac{6 \pm 4 \sqrt{2}}{2} = 3 \pm 2 \sqrt{2} For' +' sign {(3 + 2 \sqrt{2})}^{x^{2} - 8} = 3 + 2 \sqrt{2} \Rightarrow x^{2} - 8 = 1 \Rightarrow x = \pm 3 For' -' sign {[{(3 + 2 \sqrt{2})}^{- 1}]}^{8 - x^{2}} = 3 - 2 \sqrt{2}$

$\Rightarrow {(3 - 2 \sqrt{2})}^{8 - x^{2}} = 3 - 2 \sqrt{2} \Rightarrow 8 - x^{2} = 1 \Rightarrow x^{2} = 7$

$\Rightarrow x = \pm \sqrt{7}$

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