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IIT JEE 2009 Mathematics Paper2 Code 1 Solutions

3.     A line with positive direction cosines passes through the point P (2, -1, 2) and makes equal angles with the coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals

        (A)    1

        (B)    √2

        (C)    √3

        (D)    2

Sol.   (C)

        Direction Cosines of the line are 1/√3 , 1/√3 , 1/√3.

        Any point on the line at a distance t from P(2, -1, 2) is

        (2 + t/√3 , -1 + t/√3 ,  2 + t/√3)

        Which clearly lies on 2x + y + z = 9

        Hence ,

        => t = √3

4.     If the sum of the first n terms of an A.P. is cn2, then the sum of squares of these n terms is

       (A) (n(4n2 - 1) c2)/6 

       (B) (n(4n2 + 1) c2)/3 

       (C) (n(4n2 - 1) c2)/3 

       (D) (n(4n2 + 1) c2)/6

Sol.   (C)

        tn = c {n2 - (n - 1)2}

        = c (2n - 1)

        => tn2 = c2 (4n2 - 4n + 1)

        => ∑n=1n  tn2 = c2 {4n(n + 1)(2n + 1)/6 - 4n(n + 1)/2 + n}

        = c2n/3 {4(n+1)(2n+1) - 12(n+1)+6}

        = c2n/3 {4n2 + 6n + 2 - 6n - 6 + 3} = c2/3  n(4n2 - 1)

MULTIPLE CORRECT CHOICE TYPE

5.     The tangent PT and the normal PN to the parabola y2 = 4ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whose

        (A)    vertex is (2a/3 , 0)

        (B)    directrix is x = 0

        (C)    latus rectum is 2a/3

        (D)    focus is (a, 0)

Sol.   (A, D)

         ellipse 
G ≡ (h, k) 

=> h = (2a + at2)/3, k = 2at/3 

=> ((3h-2a)/a) = (9k2)/4a2 

=> Hence , required parabola is 

9y2)/4a2 =((3x - 2a))/a = 3/a ( x - 2a/3) 

=> y2 = 4a/3 (x-2a/3) 

Vertex ≡ (2a/3,0); Focus ≡ (a, 0)

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