PH 2022 AMath Prelim Paper 2 Anskey

1 2022 PHSS Prelim AMATH Paper 2 Solutions Question Solutions 1 Solve the equation 2 2 53 e 4 e x x  − =− , giving your answer(s) in exact form. ( ) 2 2 222 53 e 4 e 3 e 4e 15 0 x x xx  − =− + − = Let 2e xy= ( )( ) 23 4 15 0 3 5 3 0 5 33 yy yy y or y + − = − + = = =− 2 2 2 5e e 3 (reject as e 0)3 52 ln 3 15ln23 x x x or x x = =−  = = 2 The equation of a polynomial is given by 32f ( ) 2 8 21x x x x= + − + . (i) Show that 3x+ is a factor of f ( )x . 32f ( ) 2 8 21x x x x= + − + Subst 3x=− into f(x) 32f ( 3) 2( 3) ( 3) 8( 3) 21 54 9 24 21 0 − = − + − − − + =− + + + = Since f ( 3) 0−= , by Factor Theorem, 3x+ is a factor of f ( )x . 2 (ii) Hence, show that the equation f ( ) 0x = has only one real root. By inspection, ( )( ) 3 2 22 8 21 3 2 7x x x x x bx+ − + = + + +

2 By comparing Coeff. of x: 3 7 8 5 b b + =− =− ( )( ) 3 2 22 8 21 3 2 5 7x x x x x x + − + = + − + For ( )( ) 2 f ( ) 0 3 2 5 7 0 x x x x = + − + = 22 5 7 0 3x x or x− + = =− ( ) 2 5 4(2)(7) 31 0 D= − − =−  Since discriminant is less than 0, 22 5 7 0xx− + = has no real roots. Therefore, the equation f ( ) 0x = has only one real root which is 3x=− . 3 (i) Differentiate 3 3ln 2x x with respect to x. ( ) ( ) ( ) ( ) 32 36 22 6 4 3 3 3ln 2d 3ln 2 d 3 9 ln 2 3 9ln 2 x x xx x x x x x x x x x x − = − = −= 3 (ii) Hence show that ( ) 2 41 4ln 2 1d ln 218 x x a b x =+ , where a and b are integer values to be determined.

3 ( ) 2 41 2 41 22 431 1 2 33 1 4ln 2 d 4 9ln 2 d9 4 3 3ln 2 d9 4 1 3ln 2 9 4 1 3ln 4 3ln 2 19 8 8 1 4 7 3ln 2 3ln 29 8 4 4 7 9 ln 29 8 4 7 ln 218 1 7 18ln 2 (Shown)18 x x x x x x xx xx x xx =  =−   = − −  = − − − − −   = − + =+  =+ =+    7 18ab== 4 The fourth term in the binomial expansion of 2 2 n x x  − , where n is a positive integer, is a constant a. (a) Show that 9n= and hence find the value of a. (a) 2 2 n x x  − ( ) ( ) ( ) 1 2 3 2 2 r nr r r n r nTx r x n xr − + −  =−  =− Fourth term is when r = 3 3(3) 0 90 nxx n − = −= 9n= (Shown)

4 ( ) 39 23 672 a =− =− (b) Find the coefficient of 6x in the expansion of ( ) 9 4 2 2 13xx x  −+ . ( ) 9 4 2 2 13xx x  −+ ( ) ( ) ( ) ( )( ) 2 239 8 7 22 9 6 3 2 3 9 4 4229 672 ... 1 4 3 3 3 ...2 2 3 18 144 672 .. 1 12 54 108 ... x x x x x x xx x x x x x x          = + − + − − + + + + +                     = − + − + + + + + Term in 6x = ( )( ) 6 3 318 144 108x x x=− + Therefore, coefficient of 6x 18 15552 15534 =− + = 5 (a) The equation of a quadratic curve is 23 4 5y x x=− + − . The line 2y mx=− is a tangent to the curve at the point Q where m > 0. Find the value of the constant m and hence find the coordinates of Q. ( ) 2 2 3 4 5 2 3 4 3 0 x x mx x m x − + − = − + − + = Since