2025_CJC_JC1_H2_Math_Promo_Solution_for_sharing_with_students

9758/01/J1PROMO/2025 1 A sequence nT is given by ( ) 2ln a nT n bn c= ++ for n +∈  . The first three terms of the sequence are 9, 17.306 and 31.901. (a) Find the values of a, b and c, correct to the nearest integers. [3] (b) Using the integer values found in part (a), find the sum of the first hundred terms of nT . [1] Solution (a) ( ) ( ) ( ) 2 1 ln 1 1 9 ---------- 1 aT bc bc = ++ = + ( ) ( ) ( ) 2 2 ln 2 2 17.306 ln 2 4 ---------- 2 aT bc a bc = ++ = ++ ( ) ( ) ( ) 2 3 ln 3 3 31.901 ln 3 9 ---------- 3 aT bc a bc = ++ = ++ Solving, 1, 3 and 6ab c= −= = (nearest integers) (b) Required sum = ( )( ) 100 100 12 11 ln 3 6 1015286.261 1020000 (3 s.f.) n nn T nn − = = = ++ = = ∑∑ Examiners’ Report (a) This part question was generally well done. The majority of the students were able to write down the 3 equations involving 3 unknowns and used G.C. to solve directly. A small percentage of students solved the 3 questions algebraically at the expense of their time which could have been used for other questions. A small percentage of students did not read the question carefully that the final answers should be rounded off to the nearest integer or mindlessly wrote rounded off to the nearest integers, but the final answers were still rounded off to 3 significant figures. A small percentage of students did not know how to formulate the 3 equations (by substituting the values of n and values of terms) or had wrongly substituted the values of the terms (T 1, T2, T3) as n. (b) This part question was not well done. Many students applied the sum formulae of A.P./G.P. even though there was no common difference/common ratio. Some students did not recognize that the question asked for sum of first 100 terms, not the 100 th term. A small percentage of students spilt up the summation and used G.C. to evaluate the 3 summations separately, which was unnecessary. A small percentage of students keyed in the summation wrongly (e.g. the wrong placement of brackets) which resulted in the wrong value.

9758/01/J1PROMO/2025 2 Differentiate 1 2 tan 1 x x −   − with respect to x. You should simplify your answer as a single fraction in its simplest form. [4] Solution [~2.5m] 1 2 2 2 2 d tand 1 1d d 11 1 x x x x xx x x −   − =  −+ − ( ) 1 22 2 2 22 2 11 (1) 1 ( 2 ) 21 11 xx x x x xx x − −− − −  − = −+ −  2 1 1 x = − Examiners’ Report • Many students erroneously differentiated ( )( ) 1tan f x− into ( )2 1 f1 xx ′ + instead of ( ) ( )2 1 f. 1f x x ′ + • A notable number of students curiously assumed that ( ) ( ) 1 1 12 2 d ddtan tan tan 1 ,d dd 1 x xxx xx x − −− =−− − a confusion that is probably spawned by the quotient law of logarithm. • Most students who struggle with this question bungled up the application of quotient rule in one way or another. But mistakes appear to be mostly careless sli