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JEE Mathematics Solved Question Paper 2010

Multiple Choice Questions

71.

$If \int (1 - \cos (x)) \csc^{2} (x) dx = f (x) + c, then f (x) is equal to$

$\tan (\frac{x}{2})$
$\cot (\frac{x}{2})$
$2 \tan (\frac{x}{2})$
$\frac{1}{2} \tan (\frac{x}{2})$

72.

$If I_{n} = \int_{0}^{\frac{n^{} π}{4}} \tan (x) dx, then I_{2} + I_{4}, I_{3} + I_{5}, I_{4} + I_{6}, . . ., are in$

arithmatic progression
geometric progression
harmonic progression
arithmetic-geometric progression

73.

The area (in square units) of the region enclosed by the two circles x² + y² = 1 and (x - 1)² + y² = 1 is

$\frac{2 π}{3} + \frac{\sqrt{3}}{2}$
$\frac{π}{3} + \frac{\sqrt{3}}{2}$
$\frac{π}{3} - \frac{\sqrt{3}}{2}$
$\frac{2 π}{3} - \frac{\sqrt{3}}{2}$

74.

The values of a function f(x) at different values of x are as follows

x 0 1 2 3 4 5

f(x) 2 3 6 11 18 27

Then, the approximate area (in square units) bounded by the curve y = f(x) and x-axis between x = 0 and 5, using the Trapezoidal rule, is

50
75
52.5
62.5

75.

$The solution of \tan (y) \frac{dy}{dx} = \sin (x + y) + \sin (x - y) is$

sec(y) = 2cos(x) + c
sec(y) = - 2cos(x) + c
tan(y) = - 2cos(x) + c
sec²(y) = - 2cos(x) + c

76.
A family of curves has the differential equation $xy \frac{dy}{dx} = 2 y^{2} - x^{2}$ . Then, the family of curves is

$y^{2} = {cx}^{2} + x^{3}$

$y^{2} = {cx}^{4} + x^{3}$

$y^{2} = x + {cx}^{4}$

$y^{2} = x^{2} + {cx}^{4}$

$y^{2} = x^{2} + {cx}^{4}$

$xy \frac{dy}{dx} = 2 y^{2} - x^{2} \frac{dy}{dx} = \frac{2 y}{x} - \frac{x}{y} \frac{dy}{dx} - \frac{2 y}{x} = - \frac{x}{y} \Rightarrow y \frac{dy}{dx} - \frac{2 y^{2}}{x} = - x . . . (i) Put v = y^{2} . . . (ii) \Rightarrow (\frac{dv}{dx} = 2 y \frac{dy}{dx}) \Rightarrow (\frac{1}{2} \frac{dv}{dx} = y \frac{dy}{dx}) From eq . (i) \frac{1}{2} \frac{dv}{dx} - \frac{2 v}{x} = - x \frac{dv}{dx} - \frac{4 v}{x} = - 2 x IF = e^{\int pdx} = e^{- \frac{4}{x} dx} = e^{- 4 \log (x)} = e^{\log (x^{- 4})} = x^{- 4} Complete solution is x^{- 4} v = \int (- 2 x) x^{- 4} dx + c \frac{v}{x^{4}} = - 2 \int \frac{dx}{x^{3}} + c \Rightarrow \frac{v}{x^{4}} = \frac{1}{x^{2}} + c v = x^{2} + {cx}^{4} y^{2} = {cx}^{4} + x^{2} (from eq (ii)) which is the required family of curves .$

77.

The length of the common chord of the circles of radii 15 and 20, whose centres are 25 unit of distance apart, is

78.

Let M be the foot of the perpendicular from a point P on the parabola y = 8(x - 3) onto its directrix and let S be the focus ofthe parabola. If $∆$ SPM is an equilateral triangle, then P is equal to