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 2003

Multiple Choice Questions

71.

$\int \frac{1 + x + \sqrt{x + x^{2}}}{\sqrt{x} + \sqrt{1 + x}} dx is equal to$

$\frac{1}{2} \sqrt{1 + x} + C$
$\frac{2}{3} {(1 + x)}^{\frac{3}{2}} + C$
$\sqrt{1 + x} + C$
$2 {(1 + x)}^{\frac{3}{2}} + C$

72.

$\int (1 + x - x^{- 1}) e^{x + x^{- 1}} dx is equal to ˸$

$(1 + x) e^{x + x^{- 1}} + C$
$(x - 1) e^{x + x^{- 1}} + C$
$- {xe}^{x + x^{- 1}} + C$
${xe}^{x + x^{- 1}} + C$

73.

$\int_{- 2}^{2} |\begin{matrix} [x] \end{matrix}| dx is equal to$

74.

$\int_{0}^{1} \sin (2 \tan^{- 1} (\sqrt{\frac{1 + x}{1 - x}})) dx is equal to$

$\frac{π}{6}$
$\frac{π}{4}$
$\frac{π}{2}$
$π$

75.

$\int_{0}^{3} \frac{3 x + 1}{x^{2} + 9} dx$ is equal to :

$\log (2 \sqrt{2}) + \frac{π}{12}$
$\log (2 \sqrt{2}) + \frac{π}{2}$
$\log (2 \sqrt{2}) + \frac{π}{6}$
$\log (2 \sqrt{2}) + \frac{π}{3}$

76.

If [2, 6] is divided into four intervals of equal length, then the approximate value of $\int_{2}^{6} \frac{1}{x^{2} - x} dx$ using Simpson's rule, is

0.3222
0.2333
0.5222
0.2555

77.

The differential equation of the family of parabola with focus as the origin and the axis as X-axis, is

$y {(\frac{dy}{dx})}^{2} + 4 x \frac{dy}{dx} = 4 y$
$- y {(\frac{dy}{dx})}^{2} = 2 x \frac{dy}{dx} - y$
$y {(\frac{dy}{dx})}^{2} + y = 2 xy \frac{dy}{dx}$
$y {(\frac{dy}{dx})}^{2} + 2 xy \frac{dy}{dx} + y = 0$

78.

Soution of $\frac{dy}{dx} = \frac{xlog (x^{2}) + x}{\sin (y) + ycos (y)}$ is

$ysin (y) = x^{2} \log (x) + C$
$ysin (y) = x^{2} + C$
$ysin (y) = x^{2} + \log (x)$
$ysin (y) = xlog (x) + C$

79.
The general solution of $y^{2} dx + (x^{2} - xy + y^{2}) dy = 0$ is :

$\tan^{- 1} (\frac{y}{x}) = \log (y) + C$

$2 \tan^{- 1} (\frac{x}{y}) + \log (x) + C = 0$

$\log (y + \sqrt{x^{2} + y^{2}}) + \log (y) + C = 0$

$\sinh^{- 1} (\frac{x}{y}) + \log (y) + C = 0$

$\tan^{- 1} (\frac{y}{x}) = \log (y) + C$

$We have, y^{2} dx + (x^{2} - xy + y^{2}) dy = 0 \Rightarrow \frac{dy}{dx} = \frac{- y^{2}}{x^{2} - xy + y^{2}} It is homogneous linear differential equation Put y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} ∴ v + x \frac{dv}{dx} = \frac{- v^{2} x^{2}}{x^{2} - vx + x^{2} v^{2}} = \frac{- v^{2}}{v^{2} - v + 1} x \frac{dv}{dx} = \frac{- v^{2} - v^{3} + v^{2} - v}{v^{2} - v + 1} = \frac{- v^{3} - v}{v^{2} - v + 1} \Rightarrow \frac{(v^{2} - v + 1)}{- v^{3} - v} dv = \frac{1}{x} dx \Rightarrow \frac{- (v^{2} + 1) + v}{v (v^{2} + 1)} dv = \frac{1}{x} dx ∴ - \frac{1}{v} dv + \frac{1}{v^{2} + 1} dv = \frac{1}{x} dx On integarting both sides, we get - \log (v) + \tan^{- 1} (v) = \log (x) + C \Rightarrow \tan^{- 1} (v) = \log (xv) + C ∴ \tan^{- 1} (\frac{y}{x}) = \log (y) + C$