2007 TJC H2 MA 9740 PRELIM P1_Qns n sol

2007 TJC H2 MA 9740 PRELIM P1_Qns n sol - 2 1 Expand x 4 3...

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TJC/MA9740/P1/PRELIM2007 2 1 Expand   3 2 4 x as a series in ascending powers of x , up to and including the term in 3 x . State the values of x for which this expansion is valid. [5] Solution:   3 3 3 2 2 2 4 4 1 4 x x    23 3 5 3 5 7 13 2 2 2 2 2 1 8 2 4 2! 4 3! 4 x x x                           1 3 15 35 1 8 8 128 1024 x x x Expansion valid for 4 x . 2 (a) Find the number of ways four couples can be seated in a row such that each couple is seated together? [2] (b) Karen tries to recall the 4-digit PIN code for her safe but cannot remember it exactly. She knows that it consists of 4 different digits excluding 0 and 9, and that it starts and ends with an even number. For each attempt, she is allowed to key in 6 wrong codes before the safe shuts down for half an hour as a security precaution. If she wants to make sure that she gets the right PIN code within one week(7 days), and assuming she attempts an equal number of codes per day, at least how many attempts of 6 possible codes must she make each day? [4] Solution: (a) No of arrangements = 4! 2 4 = 384 (b) Number of possible PIN codes = 4 6 5 3 = 360 7 6 360 8.57 She must make at least 9 attempts each day to make sure that she gets the right PIN code within a week.
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TJC/MA9740/P1/PRELIM2007 3 3 Given that 22 2 25 (1 )(3 ) Lx x , L , x > 0. By implicit differentiation or otherwise, find the exact value of x for which L is minimised. [6] Solution: 2 25 )(3 ) x Diff wrt x , 32 2 25 2 )2(3 ) ( 50 )(3 ) dL L x x x dx x   Put 0 dL dx , 23 50 50 (2 ) (3 ) ( 3) 0 xx    2 ( 3)[(2 50) 50(3 )] 0 x x x x 3 3 ( 3)(2 150) 0 75 x x x    (since x > 0) X 3 ( 75) 3 75 3 ( 75) dL dx -ve Zero +ve Therefore, L is minimum when 3 75 x 4 Relative to an origin O , position vectors of A , B and C are a , b and c respectively where a , b and c are non-parallel vectors. M is the mid-point of AC and P is on the line AB produced such that AB : BP = 2:3. The line PM meets line BC at a point S . Show that the position vector of S is 1 8 (5 b + 3 c ). [6] Solution: By Ratio theorem, 1 2 OM  ( a + c ) and 1 2 OP  (5 b – 3 a )   1 OS OP OM  2 (5 b – 3 a ) + 1 2 ( a + c ) = 1 2 2    a + 5 2 b + 1 2 c C, S and B are collinear OS  c +   1 b Since a , b and c are non-parallel vectors, 1 5 1 2 0; 1 ; 2 2 2   13 , 48  1 8 OS  (5 b + 3 c ) (shown) B P M C S 3 1 1 A 2
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TJC/MA9740/P1/PRELIM2007 4 5 In an Argand diagram, the point P represents the complex number z such that 33 zi    and 3 arg( 2 ) 44    .
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2007 TJC H2 MA 9740 PRELIM P1_Qns n sol - 2 1 Expand x 4 3...

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