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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

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

$\tan (y) \frac{dy}{dx} = \sin (x + y) + \sin (x - y) \tan (y) \frac{dy}{dx} = 2 \sin (\frac{2 x}{2}) \cos (\frac{2 y}{2}) [∵ \sin (C) + \sin (D) = 2 \sin (\frac{C + D}{2}) \cos (\frac{C - D}{2})] \Rightarrow \tan (y) \frac{dy}{dx} = 2 \sin (x) \cos (y) \Rightarrow \frac{\sin (y)}{\cos (y)} \frac{dy}{dx} = 2 \sin (x) \cos (y) On integrating \Rightarrow \int \frac{\sin (y)}{\cos^{2} (y)} dy = \int 2 \sin (x) dx [Let t = \cos (y) \frac{dt}{dy} = - \sin (y) - dt = \sin (y) dy] \Rightarrow \int \frac{dt}{t^{2}} = 2 (- \cos (x)) + c \Rightarrow - (- \frac{1}{t}) = - 2 \cos (x) + c \Rightarrow \frac{1}{\cos (y)} = - 2 \cos (x) + c \Rightarrow \sec (y) = - 2 \cos (x) + c$

76.

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

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

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