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 slips. • Some students erroneously assumed that chain rule involves addition of derivatives instead of multiplication of derivatives.

9758/01/J1PROMO/2025 3 Without using a calculator, solve the inequality ( )( ) 3 2 5 22 x xx x <− +− . [3] Hence, solve the following inequalities. (a) ( )( ) e3 e2 5e 2 2 e x x xx<− +− [2] (b) ( )( ) 12 4 5 44 x xx x >− +− [2] Solution ( )( ) 3 2 5 22 x xx x <− +− ( )( ) 3 02 5 22 x xx x −<− +− ( )( ) 3 0252 2 x x xx +<− +− ( ) ( )( ) 523 052 2 xx xx ++ <+− ( )( ) 25 23 052 2 xx xx ++ <+− ( )( ) 2 235 55 052 2 xx xx  ++ <+− ( )( ) 22 1 315 5 55 052 2 x xx   + +−     <+− ( )( ) 2 1 145 5 25 052 2 x xx  ++ <+− Since numerator 2 1 145 55x ++ is always positive for all real values of x, So only need to consider ( )( ) 1 052 2xx <+− Hence, 2 25 x−<< Alternative Method

9758/01/J1PROMO/2025 (for showing numerator is always positive) ( )( ) 25 23 052 2 xx xx ++ <+− Consider 25 2 30xx+ += ( )( ) 22 4 2 45 3 0 b ac−= − < Additionally, 50a= > . Hence 25 23xx++ is always positive for all values of x. (a) Replace x with ex 2 e25 x−< < 0e 2 x<< (since ex > 0 for all real values of x) ln 2x< (b) Substitute x by 2 x into ( )( ) 3 2 5 22 x xx x >− +− we get 32 2 5 222 22 x x xx>    − +−      Multiplying by 2 2 to both sides of inequality, ( ) 6 4 5 42 2 x xx x >− +−  ( )( ) 12 4 5 44 x xx x >− +− Hence, using above answer, 2 or 225 2 xx<− > 4 or 45xx<− >

9758/01/J1PROMO/2025 Examiners’ Report Generally, students brought all the terms to one side before making the expression a single fraction. However, students had trouble explaining why the expression 25 23xx++ is always positive. Students either • totally ignore 25 23xx++ and proceed to solve the remaining inequality • attempting to use discriminant to show that 25 23xx++ is always positive without stating that the coefficient of 2x is 5 (which is positive) • attempting to complete the square without explaining that it is always positive In the above situations, students will never be awarded full credit due to omission of essential working. A number of students also made errors in completing the square. This is an assumed knowledge that H2 Math students are expected to have. (a) Students were able to identify the correct replacement. However, a number of them struggled to arrive at the final answer due to errors made in the earlier part or in solving inequalities involving exponential function. (b) Many students struggled to identify the correct replacement. This was a hence question, therefore, no marks were awarded if students solved the inequality from scratch.

9758/01/J1PROMO/2025 4 Aenicillin, a type of bacteria, is grown in a petri dish. Its population doubles every six hours. Denoting the amount of Aenicillin at the beginning of the first day as 0x , (a) write down a recurrence relation for nx , n ≥ 1, the amount of Aenicillin in the petri dish at the end of nth day, [1] (b) find nx , expressing your answer in the form of f ( ), 1nx nn= ≥ . [1] The growth rate of another bacteria, Benicillin, is known to follow the recurrence relation 1 ,1nnu u nn−=+≥ where nu is the amount of Benicillin at the end of nth day. The amount of Benicillin in the petri dish at the beginning of the first day, 0 ,u is 1 unit. (c) Write down 10uu− , 21uu− and 32uu− . Hence find a quadratic expression for ( ) ( ) ( ) ( )10 21 1 2 1... n n nnuu uu u u uu −− −− + − ++ − + − in terms of n. [3] (d) Show that ( )10 1 n rr n r uu uu − = −= −∑ . Using your results in part (c), find nu expressing your answer in the form of f ( ), 1nu nn= ≥ . [3] Solution (a) 4 121nnx xn −= ≥ Or, 116 1nnx xn −= ≥ (b) ( ) 4 0 2 n nxx= Or, ( )0 16 n nxx= (c) When n =1, 10 1uu−= When n =2, 21 2uu−= When n =3, 32 3uu−= Hence ( ) ( ) ( ) ( ) ( ) ( ) 10 21 32 1 2 1... 1 2 3 ... ( 1) 12 n n nnuu uu uu u u uu nn n n −− −− + − + − ++ − + − = ++++−+ = + OR Hence ( ) ( ) ( ) ( )10 21 1 2 1... n n nnuu uu u u uu −− −− + − ++ − + − = 2an bn c++ When n = 1, 10 1u u abc− ==++ When n = 2, ( ) ( )22 21 12 4 2uu uu a b c− + − =+= + + When n = 3, ( ) ( ) ( )32 21 10 123 9 3u u u u u u a bc− + − + − =++= + + From GC, 11, ,022abc= = = ( ) ( ) ( ) ( )10 21 1 2 1... n n nnuu uu u u uu −− −− + − ++ − + − = 211 22nn+

9758/01/J1PROMO/2025 (d) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 10 21 32 1 2 1 0 ... n rr r n n nn n uu uu uu uu u u uu uu − = −− − − = − + − + − ++ − + − = −+ ∑ OR ( ) ( ) 11 1 11 123 1 012 2 1 0 ... ... n nn rr r r r rr nn n n n uu u u uuu u u uuu u u uu −− = = = − −− −=− =++++ +− ++++ + = − ∑ ∑∑ 1+2+3+4+…+n 0 nuu= −+