2026 DHS Prelim P1 (Soln)

1 DHS 2026 Year 6 H2 Mathematics Preliminary Examination DHS 2026 H2 Math Preliminary Examination Paper 1 (Suggested Solutions) Qn Suggested Solutions 1 Let 32f( )x Ax Bx Cx D= + ++ Using vertical asymptote at 2x = − , when 2,"denominator = 0"x = − , i.e. f ( 2) 0−= ( ) ( ) ( ) 32 2 2 20 8 4 2 0.......(1) A B CD A B CD ⇒ − + − + −+ = ⇒− + − + = Using Remainder Theorem, f (2) 4⇒= ( ) ( ) ( ) 32 222 4A B CD⇒ + + += 8 4 2 4.......(2)A B CD+ + += Given 11,4− is a stationary point on the curve, ( )1, 4− is also a stationary point of f f '( 1) 0⇒ −= 2f '( ) 3 2x Ax Bx C⇒ = ++ ( ) ( ) 2 f '( 1) 3 1 2 1 0 3 2 0......(3) A BC A BC ⇒ −= − + −+ = ⇒ − += Given 11,4− is a point on the curve, when 11, . 4xy= −= 11 4 f ( 1)⇒= − f ( 1) 4⇒−= ( ) ( ) ( ) 32 1 1 14 4.......(4) A B CD ABCD ⇒ − + − + −+ = −+ − + = Using GC to solve (1)-(4), 1, 0, 3, 2AB C D== = −= Therefore, equation of curve is 3 1 32y xx= −+ Alternative Method 1: Since 2x = − is a vertical asymptote, ( ) 2x + is a factor of f( x) ( ) ( ) 2f( ) 2x x ax bx c⇒ =+ ++ Using Remainder Theorem, ( ) ( ) 2f (2) 4 2 2 (2) (2) 4 4 2 1.......(1) a bc a bc ⇒ = ⇒+ + += + += Given 11,4− is a stationary point on the curve, ( )1, 4− is also a stationary point of f f '( 1) 0⇒ −=

( )( ) ( ) ( ) 2 2 2 f '( ) 2 2 3 (4 2 ) 2 f '( 1) 3 1 (4 2 )( 1) 2 0 0......(2) x x ax b ax bx c ax a b x b c a a b bc ac ⇒ =+ ++ ++ = + + ++ ⇒ − = − + + −+ += ⇒− + = Given 11,4− is a point on the curve, when 11, . 4xy= −= ( ) ( ) 211 f ( 1) 4 1 2 ( 1) ( 1) 44 f ( 1) 4.......(3) a bc abc ⇒ = ⇒ − = ⇒−+ − + − + =− −+= Using GC to solve (1)-(3), 1, 2, 1ab c== −= Therefore, equation of curve is ( ) ( ) 32 11 .322 21y xxx xx= = −++ −+ Alternative Method 2: Using Remainder Factor Theorem, since dividing by ( ) 2x − gives a remainder of 4 ( ) ( ) 2f( ) 2 4xx x x α βγ⇒ =− + ++ Since 2x = − is a vertical asymptote, f ( 2) 0−= ( )( )f ( 2) 2 2 4 2 4 0 16 8 4 4.......(1) α βγ αβγ ⇒ − =−− − + + = − +−= − Given 11,4− is a stationary point on the curve, ( )1, 4− is also a stationary point of f f '( 1) 0⇒ −= ( )( ) ( ) ( ) ( ) ( ) ( ) ( )( ) 2 2 2 f '( ) 2 2 3 (2 4 ) 2 f '( 1) 1 2 2 1 1 1 0 7 4 0......(2) xx x x x xx αβ α βγ α βα γ β α βα β γ α βγ ⇒ = − ++ ++ = + − +− ⇒ −= − − −+ + − + −+ = ⇒ − += Given 11,4− is a point on the curve, when 11, . 4xy= −= ( )( ) 11 f ( 1) 44 f ( 1) f ( 1) 1 2 4 4 3 3 3 0.......(3) αβγ αβγ ⇒ = ⇒−=− ⇒ − =−− − + + = −+ −= Using GC to solve (1)-(3), 1, 2, 1αβ γ= = = Therefore, equation of curve is ( ) ( ) 32 11 .322 2 14y xxx xx= = −+− + ++

Qn Suggested Solutions 2(a) Let h and r be the height and radius of the water respectively. ( ) Using similar triangles, 8 836 Using similar triangles, 88 3 838 yy y h rhr += ⇒= + =⇒= + Volume of water, ( ) ( ) ( ) ( ) 22 2 3 11π 8 π(3 )(8)33 13π 8 8 24π38 3π 8 24 π (shown)64 V rh hh h = +− = + +− = +− 2(b) Differentiate with respect to t: ( ) ( ) 2 2 1 d9 πd 8d 64 d 9πd15 2 864 d d 16 d 15π 16 The height is falling at a rate of cm s.15π Vh htt h t h t − = +× −= + × −= ∴ 6 cm 3 cm 8 cm y cm h cm r cm

Qn Suggested Solutions 3(a) 1 for 0 2 for 0 nn nn n uu n u pu up n + >> +> −< > Since 1 for 0nnu un+− <− > , 11... ... 1nnuu u p+<− <− < <− =− < 1 (shown)p∴ >− 3(b)(i) 3 4 23 23 va vb = + = + 43 3 2 36 9 266 33 vv ba ba ba = += + = + = + 3(b)(ii) 6 2(2 3) 3 49 12 21 vb b a = ++ = + = +

Qn Suggested Solutions 4(a) 2 2 ln d 1 11ln d ln 1 d ln 1 , where is an arbitrary constant x xx xx x xx x xxx A Axxx    = −−−       = −+ = − −+ ⌠⌡ ⌠⌡ ⌠⌡ 4(b) 4 2 3 4 3 ln12 d ln 112 (fro m part ) x xx x xx = −− ⌠⌡ (a) ln 4 1 ln 3 112 44 33 = − − −− −   1 3ln 4 4 ln 3 1 6 ln 2 4 ln 3 = −+ = −+ 4 3 2 d ( 1)( 2)x xxx−−⌠⌡ 4 3 212 d 21 xxx= −  −−⌠⌡ 4 3 2 2 ln 2 ln 1xx=  − − − ( )2 2 ln 2 ln 3 2 ln1 ln 2= −− − 6 ln 2 2 ln 3= − OR 4 2 3 2 d32x xxx −+⌠⌡ ( ) ( ) 44 22 23 13 3 22 23 3 d d32x xxxx x−= +−+ −−⌠⌠ ⌡ ⌡ 43 142 22 3113 2 22 3 1ln 3 2 3 ln 2( ) xxx x −−= −+ +  −+ [ ] ( ) 21ln 6 ln 2 3 ln ln 32 6ln 3 ln 2 ln 3 ln 22 6 ln 2 2 ln 3 =−+ −  = + −+ = − 2 2 1 d1ln d d1 d 1 uux xx v xxx vx x − − = ⇒= = = = −= −

OR 4 2 3 2 d32x xxx −+⌠⌡ 44 2 3 3 23 3 d d ( 1)( 2)32x xx xxxx −= + −−−+ ⌠⌠ ⌡ ⌡ 442 3 3 4 42 33 11ln 3 2 3 d21 ln 3 2 3 ln 2 ln 1 ln 6 ln 2 3(ln 2 ln 3 ln1 ln 2) 6 ln 2 2 ln 3 xx xxx xx x x = −+ + − −− = − + + − − −  = − + − −+ = − ⌠⌡ ( ) ( ) 4 22 3 12 ln 2+ d 32 1 6 ln 2 4 ln 3 6 ln 2 2 ln 3 1 2 ln 3 1 ln 9 where 1 and 9 xx xx xx CD ∴  −+ = −+ + − = + = += = ⌠⌡

Qn Suggested Solutions 5(a) 10 bk = − The function would be undefined at .10bxk= = − 5(b) 1 Let 10 10 10 10 1 f () 10 1 axy xb xy by a x xy x a by a byx y a bxx x− −= + += − +=− −= + −= + 1fR\ 10b− = −  (or equivalent) 5(c) Given that 1f () f ()xx −= for all x in the domain of f 10 10 1 a x a bx xb x −−⇒= ++ Hence, by observation, 1.b = For the inverse function to exist, note that f must be a one-one function. Notice that: f( ) 110 1 10 10ax axx x x−−= = + − −− 1 11 10 if , f ( ) 110 1010 10 xax x−−⇒= − = = − − −− which is a constant function, for which 1f − does not exist 1 10a∴ ≠− 1\ 10a ∈−  or equivalent 5(d) 2( ) 11 ab−

Qn Suggested Solutions 6(a) 1. Translate 1 unit in the negative x-direction 2. Scale by a factor of 3 parallel to the x-axis OR 1. Scale by a factor of 3 parallel to the x-axis 2. Translate 3 units in the negative x-direction 6(b)(i) 6(b)(ii) y x O y x O 1( , )a b 1( , )c d 0x = ( ,0)a ( ,0)c 0y =

Qn Suggested Solutions 7(a) 22 24 14 16xykk += 7(b) ( ) ( ) 22 24 22 2 4 2 4 22 2 2 22 2 22 22 14 16 4 16 16 4 44 44 24 xy kk kx y k y k kx y kkx y kkx y kk x += += = − = − = − = − ( ) 22 22 24 24 A xy xk k x kx k x = = − = − ( )2 22 2 244A kx k x= − 4 23d2 32 16dAA kx kxx = − Let d 0dAx = ( )2 2216 2 0kx k x⇒ −= ( ) ( ) 216 2 2 0kx k x k x⇒ − += 0x = (reject) or 2xk= or 2xk= − (reject) By first derivative test, 2xk −= 2xk= 2xk += 2kx− +ve 0 –ve d d A x +ve 0 –ve Slope Area is maximum when 2xk= Maximum area of the rectangular patio OMPN 2 2 32224 2 4 m .A kkk k k= −= Alternative method By second derivative test, 2 2 4 22 2 dd2 2 32 48d dAA A k kxx x += − when 2xk= ,

( ) ( ) ( ) 2 2 3 4 22 2 d2 0 2 4 32 48 2 d Ak k kkx+= − 2 2 d 8d A kx = − < 0 Area is maximum when 2xk= Maximum area of the rectangular patio OMPN 2 2 32224 2 4 m .A kkk k k= −= 7(c) Total area of the garden plot OXY = ( ) ( ) 2 32 1 π2 44 2πm kk k = = Maximum Ratio = 3 3 42 π2 k kπ = which is a constant value, independent of k.