Class CMI: Test-B
[Q1] [Total: 8 marks]
[Sol]
(a) Prove, by mathematical induction, that 2 + 4 + 6 + + 2n = n (n + 1) for all positive integers n. [5] (b) By using the identity (k + 2)3 - k 3 = 6k 2 + 12k + 8, or otherwise, find 22 + 42 + 62 + + 1002 . [3] [Sol] (a) For n = 1, L.H.S. = 1 = R.H.S. The statement is true for n = 1. Assume that the statement is true for some integer k 1, i.e., 2 + 4 + 6 + + 2k ) = k (k + 1) For n = k + 1,
k (k +1)
2 + 4 + 6 + + 2(k + 1) = 2 + 4 + 6 + + 2k +(2k + 2) = k (k + 1) + 2(k + 1) = (k + 1)[(k + 1) + 1] The statement is true for n = k + 1. Hence, by the principle of M.I., the statement is true for all positive integers n . 43 - 23 = 6(2)2 + 12(2) + 8 63 - 43 = 6(4)2 + 12(4) + 8 83 - 63 = 6(6)2 + 12(6) + 8 . (b) For k = 2, 4, 6, . . . , 100, we have . . . 3 3 102 - 100 = 6(100)2 + 12(100) + 8 Adding up these 50 equations, we have 1023 - 23 = 6(22 + 42 + 62 + + 1002 ) + 12(2 + 4 + 6 + + 100) + 8 50 22 + 42 + 62 + + 1002 = 1061200 - 12 50(51) - 400 = 171700. 6
[ALT]
22 + 42 + 62 + + 1002 = 4(12 + 22 + 32 + + 502 ) 50(50 + 1)[2(50) + 1] 6 = 171700 =4
AD: College Mathematics I [2011-12]
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CMI: Class Test-B
[Q2] [Total: 12 marks] (a) 1 - x
9
3
[Sol]
x
is expanded in descending powers of x . [2] [2] [2]
(i) Write down the general term of the expansion. (ii) Find the coefficient of the x -1/3 term in the expansion. (iii) Find the 7th term in the expansion.
(b) Expand (8 -x )-1/3 in ascending powers of x up to and including the term in x 2 and state the set of values of x for which the expansion is valid. [4] (c) Hence, find the decimal value of [Sol]
1 3
16 correct to 2 decimal places.
1 18-5r
[2]
9 9 (a) (i) The general term = C r (-x 3 )9-r (x - 2 )r = (-1)9-r C r x 6 . 18 - 5r 1 (ii) Set = - , we have r = 4. 6 3 9 The coefficient of the x -1/3 term is (-1)9-4C 4 = -126. 9 (iii) The 7th term, put r = 6, is (-1)9-6C 6 x
18-30 6
= -84x -2 .
(b)
x -1/3 (8 - x )-1/3 = 8-1/3 1 - 8 1 (- 3 )(- 1 - 1) x 2 1 x 1 3 = 1 + (- )(- ) + (- ) + 2 3 8 2 8 =
1 1 2 1 + x+ x + 2 48 576 x The expansion is valid when | | < 1, i.e., -8 < x < 8. 8 3 (c) Let x = , we have 16 1
3 3
3 8 - ( 16 )
=
3
16 1 1 3 1 3 2 = + ( )+ ( ) + 5 2 48 16 576 16 16 = 5 1 1 3 1 3 2 + ( )+ ( ) + 2 48 16 576 16 = 2.52 (to 2 d.p.)
AD: College Mathematics I [2011-12]
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CMI: Class Test-B
[Q3] [Total: 15 marks]
[Sol]
- - - - - - (a) Let OA = i + 4 j and O B = 5 i + 2 j . C is a point on A B such that AC : C B = 1 : r . - (i) Find A B . - - - (ii) Express OC in terms of r, i and j . (iii) If OC A B , find r . (b) The line L meets the plane at the point A, where L: x -4 y +1 z -3 = = 3 -1 2 and - - - - : r ( i - j - k )=0
[1] [2] [3]
(i) Find the coordinates of the point A. (ii) Find the general Cartesian equation of the plane 1 which containing L and perpendicular to . (iii) Find the acute angle in degrees, to correct 2 decimal places, between L and . [Sol]
[3] [4] [2]
AD: College Mathematics I [2011-12]
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CMI: Class Test-B
[Sol] - - - - - (a) (i) A B = O B - OA = 4 i - 2 j . - - - - - - - OA + r O B r( i +4 j )+5 i +2 j (ii) OC = = = 1+r 1+r - - (iii) OC A B. OC A B = 0, i.e.,
[Sol]
5+r 1+r
- 2 + 4r i + 1+r
- j .
5+r 2 + 4r (4) + (-2) = 0 1+r 1+r Solving, we have r = 4. (b) (i) Let A = (4 + 3t , -1 - t , 3 + 2t ). Then < 4 + 3t , -1 - t , 3 + 2t > < 1, -1, -1 > = 0 4 + 3t + 1 + t - 3 - 2t = 0 Hence, A = (1, 0, 1). - (ii) Let n 1 be the normal vector of 1 . Then - i - - - n1 = d n = 3 1 (4, -1, 3) lies on L, which lies on 1 . - - j k - - - -1 2 = 3 i + 5 j - 2 k -1 -1 (4, -1, 3) lies on 1 . t = -1
Hence, the general Cartesian equation of 1 : (3)(x - 4) + (5)(y + 1) + (-2)(z - 3) = 0, (iii) = sin-1 | < 3, -1, 2 > < 1, -1, -1 > | 9+1+4 1+1+1 i.e., 3x + 5y - 2z - 1 = 0 2 14 3
= sin-1
= 17.98o (to 2 d.p.)
AD: College Mathematics I [2011-12]
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CMI: Class Test-B
[Q4] [Total: 15 marks] (a) (i) If arg(z + 1) = 5 and arg(z - 1) = , find z . 3 6 z - (1 - 2i ) = 2. z - (-1 + 2i )
[Sol]
[3] [3]
(ii) Find the Cartesian equation of the locus of (b) The complex number z P =
-1 + 3i is represented by the point P in the Argand diagram, 2-i origin O. The complex numbers z R and z Q are represented by the points R and Q such that OPQR is a parallelogram and OR is obtained by enlarging OP by a factor of 2 and then rotating about the origin through an angle 60o in clockwise direction. (i) Find the modulus and the principal argument in radians of z P . (ii) Find z R and z Q in the form of p + q i where p and q are real numbers. [2] [2] [2]
(iii) Hence, or otherwise, find the exact values of
cos 75o
and
sin 75o .
(c) Solve the equation z 3 - 3z 2 + 3z = 1 - i , giving your answers in the form of p + r cis , where p, r > 0 and - < . [3] [Sol] 1 3 (a) (i) From the Argand diagram, z = (-1 + 2 sin 30o cos 60o ) + (2 sin 30o sin 60o )i = - + i. 2 2
[ALT] Let z = x + i y . Then y 5 1 y = tan( ) = 3 and = tan =- . x +1 3 x -1 6 3 1 3 1 3 Solving, we have x = - , y = . z =- + i. 2 2 2 2 (ii) Let z = x + i y . Then (x - 1) + (y + 2)i =2 (x + 1) + (y - 2)i (x - 1)2 + (y + 2)2 = 4(x + 1)2 + 4(y - 2)2 x 2 - 2x + y 2 + 4y + 5 = 4x 2 + 8x + 4y 2 - 16y + 20 3x 2 + 3y 2 + 10x - 20y + 15 = 0
AD: College Mathematics I [2011-12]
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CMI: Class Test-B
[Sol] (b)
[Sol]
-1 + 3i 2 + i = -1 + i . 2-i 2+i 3 |z P | = 2 and arg z P = . 4 (ii) z R = z P 2 cis (-60o ) = (-1 + i )(1 - 3i ) = ( 3 - 1) + ( 3 + 1)i . (i) z P = z Q = z P + z R = (-1 + i ) + ( 3 - 1) + ( 3 + 1)i = ( 3 - 2) + ( 3 + 2)i . (iii) z R = 2 2(cos 75o + i sin 75o ) = ( 3 - 1) + ( 3 + 1)i . 3-1 3+1 cos 75o = and sin 75o = . 2 2 2 2 (c) z 3 - 3z 2 + 3z = 1 - i
(z - 1)3 = -i = cis (- ) = cis (2k - ), k 2 2 (4k - 1) z = 1 + cis , k = -1, 0, 1 6
AD: College Mathematics I [2011-12]
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