Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

JEE Mathematics Solved Question Paper 2008

Multiple Choice Questions

51.

If xdy = y(dx + y dy), y(1) = 1 and y(x) > 0, then y(- 3) is equal to

52.

If $\int_{a}^{b} x^{3} dx = 0 and \int_{a}^{b} x^{2} dx = \frac{2}{3}$ , then the values of a and b are respectively

1, - 1
- 1, 1
1, 1
- 1, - 1

53.

The differential equation representing the family of curves y = 2c (x + c³), where c is a positive parameter, is of

order 1, degree 1
order 1, degree 2
order 1, degree 3
order 1, degree 4

54.

The differential equation representing the family of curves y = xe^cx (c is a constant) is

$\frac{dy}{dx} = \frac{y}{x} (1 - \log (\frac{y}{x}))$
$\frac{dy}{dx} = \frac{y}{x} \log (\frac{y}{x}) + 1$
$\frac{dy}{dx} = \frac{y}{x} (1 + \log (\frac{y}{x}))$
$\frac{dy}{dx} + 1 = \frac{y}{x} (\log (\frac{y}{x}))$

55.
The solution of $\frac{dy}{dx} = 1 + y + y^{2} + x + xy + {xy}^{2}$ is

$\tan^{- 1} (\frac{2 y + 1}{\sqrt{3}}) = x + xy + {xy}^{2}$

$4 \tan^{- 1} (\frac{2 y + 1}{\sqrt{3}}) = \frac{\sqrt{3}}{2} (2 x + x^{2})$

$\sqrt{3} \tan^{- 1} (\frac{3 y + 1}{\sqrt{3}}) = 4 (1 + x + x^{2}) + c$

$\tan^{- 1} (\frac{2 y + 1}{\sqrt{3}}) = \sqrt{3} (2 x + x^{2}) + c$

$\tan^{- 1} (\frac{2 y + 1}{\sqrt{3}}) = \sqrt{3} (2 x + x^{2}) + c$

$\frac{dy}{dx} = 1 + y + y^{2} + x (1 + y + y^{2}) \Rightarrow \frac{dy}{1 + y + y^{2}} = (1 + x) dx \Rightarrow \int \frac{dy}{1 + y + y^{2}} = \int (1 + x) dx \Rightarrow \frac{1}{\frac{\sqrt{3}}{2}} \tan^{- 1} (\frac{y + \frac{1}{2}}{\frac{\sqrt{3}}{2}}) = x + \frac{x^{2}}{2} + \frac{c}{2} 4 \tan^{- 1} (\frac{2 y + 1}{\sqrt{3}}) = \sqrt{3} (2 x + x^{2}) + c$