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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

$\frac{2 π}{3} + \frac{\sqrt{3}}{2}$

$\frac{π}{3} + \frac{\sqrt{3}}{2}$

$\frac{π}{3} - \frac{\sqrt{3}}{2}$

$\frac{2 π}{3} - \frac{\sqrt{3}}{2}$

$\frac{2 π}{3} - \frac{\sqrt{3}}{2}$

Intersection point of two circles

$x^{2} + y^{2} = 1 . . . (i) {(x - 1)}^{2} + y^{2} = 1 . . . (ii) is given by {(x - 1)}^{2} + {(1 - x)}^{2} = 1 \Rightarrow x^{2} + 1 - 2 x - x^{2} = 0 \Rightarrow x = \frac{1}{2} From eq . (i), \frac{1}{4} + y^{2} = 1 y^{2} = 1 - \frac{1}{4} \Rightarrow y = \pm \frac{\sqrt{3}}{2} Point A (\frac{1}{2}, \frac{\sqrt{3}}{2}) and C (\frac{1}{2}, \frac{- \sqrt{3}}{2}) So, Area of region OABDO = 2 \times Area of region OABDO . . . (iii) Area of OABDO = Area of OADO + Area of ABDA$

$= \int_{0}^{\frac{1}{2}} \sqrt{1 - {(x - 1)}^{2}} dx + \int_{\frac{1}{2}}^{1} \sqrt{1 - x^{2}} dx = {[\frac{1}{2} . (x - 1) \sqrt{1 - {(x - 1)}^{2}} + \frac{1}{2} \sin^{- 1} (\frac{x - 1}{1})]}^{\frac{1}{2}}_{0} + {[\frac{1}{2} x \sqrt{1 - x^{2}} + \frac{1}{2} \sin^{- 1} (\frac{x}{1})]}^{1}_{\frac{1}{2}} = [- \frac{1}{2} \frac{1}{2} \frac{\sqrt{3}}{2} + \frac{1}{2} \sin^{- 1} (\frac{- 1}{2}) - \frac{1}{2} 0 - \frac{1}{2} \sin^{- 1} (- 1)] + [\frac{1}{2} . 0 + \frac{1}{2} \sin^{- 1} (1) - \frac{1}{4} \frac{\sqrt{3}}{2} - \frac{1}{2} \sin^{- 1} (\frac{1}{2})]$

$= (- \frac{\sqrt{3}}{8}) - \frac{1}{2} \sin^{- 1} (\frac{1}{2}) + \frac{1}{2} \sin^{- 1} (1) + \frac{1}{2} \sin^{- 1} (1) - \frac{\sqrt{3}}{8} - \frac{1}{2} \sin^{- 1} (\frac{1}{2})$

$= \sin^{- 1} (1) - \sin^{- 1} (\frac{1}{2}) - \frac{\sqrt{3}}{4} = \frac{π}{2} - \frac{π}{6} - \frac{\sqrt{3}}{4} = (\frac{π}{3} - \frac{\sqrt{3}}{4}) From eq . (iii) Area of region OABCO = 2 \times (\frac{π}{3} - \frac{\sqrt{3}}{4}) = (\frac{2 π}{3} - \frac{\sqrt{3}}{2})$

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

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

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