Now that we've seen some of the tricky questions in JEE Physics 2009, let's move on to Mathematics. The level of difficulty for JEE Math 2009 paper was moderate.

How many of the following can you solve? The answer key is provided at the end of post. Go on, challenge yourself!

How many of the following can you solve? The answer key is provided at the end of post. Go on, challenge yourself!

**Question #1**

Tangents drawn from the point P(1,8) to the circle x^2 + y^2 - 6x - 4y - 11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is

- x^2 + y^2 + 4x - 6y + 19 = 0
- x^2 + y^2 - 4x - 10y + 19 = 0
- x^2 + y^2 - 2x + 6y - 29 = 0
- x^2 + y^2 + 6x - 4y +19 = 0

**Question #2**

Let

*z = cosΘ + i sinΘ*. Then the value of

*Σ(m=1 to 15) lm(z^(2m-1))*at

*Θ=2 degree*is

- 1/sin 2
- 1/3sin 2
- 1/2sin 2
- 1/4sin2

**Question #3**

The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2, and 3 only is

- 55
- 66
- 77
- 88

**Question #4**

In a triangle ABC with fixed base BC, the vertex A moves such that

*cos B + cos C = 4 sin^2 (A/2).*

If

*a*,

*b*and

*c*denote the lengths of the sides of the triangle opposite to the angles

*A*,

*B*and

*C*, respectively, then

*b*+*c*= 4*a**b*+*c*= 2*a*- locus of point
*A*is an ellipse - locus of point
*A*is a pair of straight lines

**Question #5**

If ((sin^4 x) / 2) + ((cos^4 x) / 3) = 1/5, then

- tan^2 x = 2/3
- ((sin^8 x) / 8)+((cos^8 x) / 27) = 1/125
- tan^2 x = 1/3
- ((sin^8 x) / 8)+((cos^8 x) / 27) = 2/125

**Question #6**

If the sum of the first n terms of an A.P. is cn^2, then the sum of the squares of these n terms is

- [n(4n^2-1)c^2]/6
- [n(4n^2+1)c^2]/3
- n(4n^2-1)c^2]/3
- [n(4n^2+1)c^2]/6

**Question #7**

The normal at a point P on the ellipse X^2+4Y^2=16 meets the x-axis at Q. If M is the mid point of the line segment PQ, then the locus of M intersects the latus rectums of the given ellipse at the points

- [3 sqrt(5)/2, 2/7)
- [3 sqrt(5)/2, sqrt(19)/4]
- [2 sqrt(3), 1/7]
- [2 sqrt(3), 4 sqrt(3)/7]

**Question #8**

If ⌠(-π to +π)

**[sin nx/(1+**

**π^x) sinx]**dx, n=0,1,2,.... then

- In=In+2
- Σ (from m=1 to 10) I2n+1=10π
- Σ (from m=1 to 10) I2m=0
- In=In+1

**Answer key:**

Question #1: A

Question #2: D

Question #3: C

Question #4: B,C (Multiple answer correct)

Question #5: A,B (Multiple answer correct)

Question #6: C

Question #7: C

Question #8: A, B, C (Multiple answer correct)

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