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JEE Mathematics Solved Question Paper 2006

Multiple Choice Questions

31.

Seven balls are drawn simultaneously from a bag containing 5 white and 6 green balls. The probability of drawing 3 white and 4 green balls is :

$\frac{7}{{}^{11}C_{7}}$
$\frac{{}^{5}C_{3} + {}^{6}C_{4}}{{}^{11}C_{7}}$
$\frac{{}^{5}C_{2} {}^{6}C_{2}}{{}^{11}C_{7}}$
$\frac{{}^{6}C_{3} {}^{5}C_{4}}{{}^{11}C_{7}}$

32.

In a book of 500 pages, it is found that there are 250 typing errors. Assume that Poisson law holds for the number of errors per page. Then,the probability that a random sample of 2 pages will contain no error, is :

e ^{- 3}
e ^{- 5}
e ^{- 1}
e ^{- 2}

33.

The transformed equation of x² + 6xy + 8y² = 10 when the axes are rotated through an angle $\frac{π}{4}$ is :

15x² - 14xy + 3y²= 20
15x² + 14xy - 3y²= 20
15x² + 14xy + 3y²= 20
15x² - 14xy - 3y²= 20

34.

The lines x - y - 2 = 0, x + y - 4 = 0 and x + 3y = 6 meet in the common point :

(1, 2)
(2, 2)
(3, 1)
(1, 1)

35.

The equation of the line passing through the point of intersection of the lines x - 3y + 2 = 0 and 2x + 5y - 7 = 0 and perpendicular to the line 3x + 2y + 5 = 0 is :

2x - 3y + 1 = 0
6x - 9y + 11 = 0
2x - 3y + 5 = 0
3x - 2y + 1 = 0

36.

Let O be the origin and A be a point on the curve y = 4x. Then the locus of the mid point of OA is :

x² = 4y
x² = 2y
y² = 16x
y² = 2x

37.

The lines represented by the equation x² - y² - x + 3y - 2 = 0 are :

x + y - 1 = 0, x - y + 2 = 0
x - y - 2 = 0, x + y + 1 = 0
x + y + 2 = 0, x - y - 1 = 0
x + y - 1 = 0, x + y - 2 = 0

38.

The centroid of the triangle formed by the pair of straight lines 12x² - 20xy + 7y² = 0 and the line 2x - 3y + 4 = 0 is :

$(- \frac{7}{3}, \frac{7}{3})$
$(- \frac{8}{3}, \frac{8}{3})$
$(\frac{8}{3}, \frac{8}{3})$
$(\frac{4}{3}, \frac{4}{3})$

39.
If OA is equally inclined to OX, OY and OZ and if A is $\sqrt{3}$ units from the origin, then A is :

(3, 3, 3)

(- 1, 1, - 1)

(- 1, 1, 1)

(1, 1, 1)

(1, 1, 1)

$Since OA is equally inclined to OX, OY and OZ . ∴ Coordinates of A are (a, a, a) . Also, OA = \sqrt{3} ∴ \sqrt{{(a - 0)}^{2} + {(a - 0)}^{2} + {(a - 0)}^{2}} = \sqrt{3} \Rightarrow \sqrt{3 a^{2}} = \sqrt{3} \Rightarrow a^{2} = 1 \Rightarrow a = \pm 1 ∴ Co - ordinates of A are (1, 1, 1) or (- 1, - 1, - 1) .$