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# JEE Mathematics Solved Question Paper 2009

#### Multiple Choice Questions

1.

The sum to infinity of the series • 2

• 3

• 4

• 6

B.

3 147 Views

2.

Let A and B denote the statements
A: cos α + cosβ + cosγ = 0
B : sinα + sinβ + sinγ = 0
If cos(β – γ) + cos(γ – α) + cos(α – β) = – 3/2, then

•  A is true and B is false

• A is false and B is true

• both A and B are true

• both A and B are false

C.

both A and B are true

cos(β–γ) + cos(γ – α) + cos(α – β) = –3/2
⇒ 2cos (β – γ) + 2cos(γ– α) + 2cos(α – β) = –3
⇒ Σ(cosβcosγ + 2Σsin α sinβ + 3 = 0
⇒ (cosα+cosβ + cosγ)2+ (sinα + sinβ +sinγ)2=0
⇒ cosα + cosβ + cosγ = 0
sinα + sinα + sinγ = 0

127 Views

3.

The ellipse x2+ 4y2= 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is

• x2+ 16y2= 16

• x2+ 12y2= 16

• 4x2+ 48y2= 48

• 4x2+ 64y2= 48

B.

x2+ 12y2= 16  371 Views

4.

The projections of a vector on the three coordinate axis are 6, - 3, 2 respectively. The direction cosines of the vector are

•  6, –3, 2

• 6/5, -3/5, 2/5

• 6/7, -3/7, 2/7

• -6/7, -3/7, 2/7

C.

6/7, -3/7, 2/7

Projection of a vector on coordinate axis are
x2-x1, y2-y1, z2-z1
x2-x1 = 6, y2-y1 = -3, z2-z1 = 2 199 Views

5.

Three distinct points A, B and C are given in the 2 – dimensional coordinate plane such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the point ( - 1, 0) is equal to 1/3 . Then the circumcentre of the triangle ABC is at the point

• (0,0)

• (5/4, 0)

• (5/2, 0)

• (5/3,0)

B.

(5/4, 0)

P = (1,0);Q (-1,0)
Let A = (x,y) ⇒ 3AP =AQ
⇒ 9AP2 = AQ2
⇒9 (x-1)2 +9y2
= (x+1)2 +y2
⇒ 9x2-18x +9 +9y2 = x2 +2x+ 1 +y2
⇒8x2-20x + 8y2 +8 = 0
⇒ x2+ y2-(5/2)+1 = 0 .. (2)
therefore A lies on the circle
Similarly B,C are also lies on the same circle
therefore, CIrcumcentre of ABC= Centre of circle (1) = (5/4, 0)

221 Views

6.

If the roots of the equation bx2+ cx + a = 0 be imaginary, then for all real values of x, the expression 3b2x2 + 6bcx + 2c2 is

• greater than 4ab

• less than 4ab

• greater than -4ab

• less than 4ab

C.

greater than -4ab

As, bx2 + cx + a = 0 has imaginary roots
So, c2< 4ab
Now, 3b2x2 + 6bcx + 2c2
= 3(bx + c)2– c2≥ – c2≥ – 4ab

134 Views

7.

The differential equation which represents the family of curves y=c1ec2xe, where c1 and c2 are arbitrary constants, is

• y' =y2

•  y″ = y′ y

• yy″ = y′

• yy″ = (y′)

D.

yy″ = (y′)

y c1ec2x = …..(i)
y' = c1c2ec2x
y' = c2y.....(from (i) ....(ii)
y" = c2y' ....... (iii)
from (ii) & (iii) 120 Views

8.

If the mean deviation of the numbers 1, 1 + d, 1+ 2d, ... , 1 + 100d from their mean is 255, then the d is equal to

• 10.0

• 20.0

• 10.1

• 20.2

C.

10.1 446 Views

9.

If A, B and C are three sets such that A ∩ B = A∩ C and A ∪ B = A ∪ C, then

• A = B

• A = C

• B = C

• A ∩ B = φ

C.

B = C

A ∪ B = A ∪ C
⇒ n (A ∪ B) = n(A ∪ C)
⇒ n(A) + n(B) – n(A ∩ B)
= n(A) + n(C) – n(A ∩C)
n(B) = n(C)

105 Views

10.

The remainder left out when 82n –(62)2n+1 is divided by 9 is

• 0

• 2

• 7

• 8

B.

2

82n – (62)2n + 1
⇒ (9 – 1)2n – (63 – 1)2n + 1
⇒ (2nC0 92n2nC1 92n – 1 + ….. + 2nC2n)
– (2n + 1C0 632n + 1–2n + 1C1 632n + ….
2n +1C2n + 1
Clearly remainder is ‘2’.

264 Views