NYGH 2014-S3EOY-IM2 with ans

3 1 If ( )( )( ) CBxxAxxxx ++−−≡+−+ 2311826176 23 , find the value of the constants A, B and C. [3] 2 A circular disc has a volume of 3cm 5 . If the radius of its circular cross-section is ( )cm 13+ , find the thickness of the disc in terms of π. Leave your answer as an exact value, in the form 3ba+ . [3] 3 Given that p−=°220cos , find in terms of p, (i) °40cos , [1] (ii) °140tan , [2] (iii) )320sin( °− . [1] 4 The quadratic equation 022 =++ kxx has roots α andβ . Find the quadratic equation with roots 2 2 +β α and 2 2 +α β . [5] 5 Sketch the graph of ( )xy −= 1ln , labelling the asymptote(s) and intercept(s) clearly. Find the equation of the straight line that needs to be added onto the same axes so as to solve the equation )1( ee2 xx −= . It is not necessary to draw the line on your sketch. [4] 6 The line 1+= mxy does not intersect the curve 522 ++= xxy . However, it intersects a second curve 32 −= mxy at two distinct points. Find the range of possible values of m. [6]

4 7 The curve bxay sin= has a maximum value of 4, a minimum value of − 4 and a period of 3 2π . (i) State the values of a and b. [2] (ii) Hence, sketch the graph of bxay sin= for π0 ≤≤ x , indicating clearly all intercepts. [2] 8 Answer the whole of this question on the INSERT provided. The graph of ( )xy f= is given in the diagram below. The graph passes through the points ( )8 ,5A , ( )0 ,3B and ( )1 ,2 −C . On the same axes, sketch the graph of (a) ( )3f −= xy , [2] (b) ( )xy f2 1 −= , [2] (c) ( )xy 1f−= , [2] showing clearly the coordinates of the image of each of the points A, B and C.

5 9 (a) The function f is defined by 33122:f 2 +− xxx : , for 61 ≤≤ x . (i) By using completing the square method, express 33122 2 +− xx in the form ( ) cbxa ++ 2 . [3] (ii) Find the range of f. [2] (iii) Does f have an inverse? Explain your answer. [1] (iv) The function g is defined by 33122:g 2 +− xxx : . Write down a possible domain for g so that g has an inverse. [1] (b) If a function h is defined as 32:h −x xx : for 2 3≠x , find, in similar form, an expression for -1h , stating the restriction on its domain. [3] 10 (a) Given that ax 2log= and by 2log= , express 2 3 2 8log a b in terms of x and y. [3] (b) Solve the equation ( ) 2232 12 +=+ xx . [4] (c) Solve the equation ( ) ( ) 12log15log 15 2 2 =−+ +xx . [5] 11 (a) The function ( ) 26f 23 −++= bxaxxx where a and b are constants, is exactly divisible by 1−x and leaves a remainder of − 36 when divided by 2+x . Find the values of a and b. [3] (b) (i) Solve the equation 01644 23 =+−− xxx . [4] (ii) Hence, solve the equation xxxx 48164 223 +=++ . [2] (c) A quadratic function ( )xf is such that ( ) 32f = , ( ) 10f = and ( ) 01f = . Find the expression for ( )xf . [3]

6 12 (a) Prove that 1cosec 2cotcosec 244 −≡− xxx . [3] (b) Solve the equation xx sin2cos2 2 += , for the interval π20 ≤≤ x , leaving your answer(s) as exact value(s). [4] (c) Solve the equation AA 2cot12tan6 =+ , for the interval 3600 ≤≤ A . [4] Bonus question 13 Prove that kk 1)]}[cos(tantan{sin 11 =−− , where k is a positive constant. [3] - End of Paper -

IM2 EOY Marking Scheme 1 𝐴 = 2 𝐵 = 4 𝐶 = 10 2 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑑𝑖𝑠𝑐 = 𝜋𝑟2ℎ 5 = 𝜋�√3 + 1� 2 ℎ ℎ = 5 𝜋�4 + 2√3� ℎ = 5 𝜋�4 + 2√3� × 4 − 2√3 4 − 2√3 ℎ = 20 − 10√3 4𝜋 ℎ = 5 2𝜋 �2 − √3� 𝑜𝑟 10 − 5√3 2𝜋 cm OR 5 = 𝜋�√3 + 1� 2 ℎ ℎ = 5 𝜋�√3 + 1� 2 ℎ = 5 𝜋�√3 + 1� 2 × �√3 − 1� 2 �√3 − 1� 2 ℎ = 5 𝜋 − 5 2𝜋 √3 cm 3i cos 40° = 𝑝 3ii tan 140° = − tan 40° = − �1 − 𝑝2 𝑝 3iii sin(−320°) = sin 40° = �1 − 𝑝2 4 𝛼 + 𝛽 = −2, 𝛼𝛽 = 𝑘

𝛼2 + 𝛽2 = (𝛼 + 𝛽)2 − 2𝛼𝛽 = 4 − 2𝑘 2𝛼 𝛽 + 2 + 2𝛽 𝛼 + 2 = 2(𝛼2 + 𝛽2) + 4(𝛼 + 𝛽) 2(𝛼 + 𝛽) + 4 + 𝛼𝛽 = 2(4 − 2𝑘) + 4(−2) 2(−2) + 4 + 𝑘 = −4 � 2𝛼 𝛽 + 2� � 2𝛽 𝛼 + 2� = 4𝛼𝛽 2(𝛼 + 𝛽) + 4 + 𝛼𝛽 = 4𝑘 2(−2) + 4 + 𝑘 = 4 Sum of roots: Product of roots: So the equation is 𝑥 2 + 4𝑥 + 4 = 0 5 𝑒2𝑥 = 𝑒(1 − 𝑥) 𝑒2𝑥−1 = 1 − 𝑥 2𝑥 − 1 = ln(1 − 𝑥) ∴ draw 𝑦 = 2𝑥 − 1

6 If 𝑦 = 𝑚𝑥 + 1 does not intersect 𝑦 = 𝑥2 + 2𝑥 + 5, then 𝑥2 + 𝑥(2 − 𝑚) + 4 = 0 has no real roots, and hence (2 − 𝑚)2 − 4(1)(4) < 0 𝑚2 − 4𝑚 − 12 < 0 (𝑚 + 2)(𝑚 − 6) < 0 −2 < 𝑚 < 6 If 𝑦 = 𝑚𝑥 + 1 intersects 𝑦 = 𝑚𝑥 2 − 3 at two distinct points, then 𝑚𝑥2 − 𝑚𝑥 − 4 = 0 has 2 roots, and hence, 𝑚2 − 4(𝑚)(−4) > 0 𝑚(𝑚 + 16) > 0 𝑚 < −16 or 𝑚 > 0 Combining the two inequalities, obtain the range of possible values of m: 0 < 𝑚 < 6 7i 𝑎 = 4 𝑏 = 3 7ii

9ai 2𝑥2 − 12𝑥 + 33 = 2(𝑥2 − 6𝑥) + 33 𝑜𝑟 = 2 �𝑥2 − 6𝑥 + 33 2 � = 2(𝑥 − 3)2 − 18 + 33 = 2(𝑥 − 3)2 + 15 9aii 𝑓(1) = 23 𝑓(6) = 33 15 ≤ f(𝑥) ≤ 33 9aiii f does not have an inverse. It is not a one-to-one function. 9aiv Any acceptable answer Eg. 𝑥 ≤ 3 𝑜𝑟 𝑥 ≥ 3 9b 𝐿𝑒𝑡 𝑦 = 𝑥 2𝑥 − 3 2𝑥𝑦 − 3𝑦 = 𝑥 𝑥(2𝑦 − 1) = 3𝑦 𝑥 = 3𝑦 2𝑦 − 1 Hence, h−1: 𝑥 ↦ 3𝑥 2𝑥 − 1 , 𝑥 ≠ 1 2

10a log2 √𝑏3 8𝑎2 = 3 2 [log2 𝑏] − [log2 8 + 2 log2 𝑎] = 3 2 (𝑦) − (3 + 2𝑥) = −2𝑥 + 3 2 𝑦 − 3 OR log2 √𝑏3 8𝑎2 = log2 𝑏 3 2 − [log2 8 + log2 𝑎2] = 3 2 log2 𝑏 − [log2 23 + 2 log2 𝑎] = 3 2 𝑦 − 3 − 2𝑥 10b 22𝑥+1 = 3(2𝑥) + 2 2(2𝑥)2 − 3(2𝑥) − 2 = 0 Let 2 𝑥 = 𝑦, hence 2𝑦2 − 3𝑦 − 2 = 0 (2𝑦 + 1)(𝑦 − 2) = 0 2 𝑥 = 2 ⇒ 𝑥 = 1 2(2𝑥) = −1 (𝑁𝐴) 10c log2(5𝑥 + 1)2 − log(5𝑥+1) 2 = 1 log2(5𝑥 + 1)2 − log2 2 log2(5𝑥 + 1) = 1 2[log2(5𝑥 + 1)]2 − 1 = log2(5𝑥 + 1) Let 𝑦 = log 2(5𝑥 + 1), hence (2𝑦 + 1)(𝑦 − 1) = 0 log2(5𝑥 + 1) = − 1 2 ⇒ 𝑥 = 1 5 �√2 2 − 1� 𝑜𝑟 − 0.0586 or log2(5𝑥 + 1) = 1 ⇒ 𝑥 = 1 5 𝑜𝑟 0.2 11a f(1) = 0, hence, 6 + 𝑎 + 𝑏 − 2 = 0 𝑎 + 𝑏 = −4 f(−2) = −36, hence, −48 + 4𝑎 − 2𝑏 − 2 = −36 2𝑎 − 𝑏 = 7 Solving simultaneously, obtain

𝑎 = 1 𝑏 = −5 11bi f(2) = 0, hence (𝑥 − 2) is a factor. 𝑥 3 − 4𝑥2 − 4𝑥 + 16 = 0 (𝑥 − 2)(𝑥2 − 2𝑥 − 8) = 0 (𝑥 − 2)(𝑥 − 4)(𝑥 + 2) = 0 Hence, 𝑥 = 2, −2, or 4 11bii 01644 23 =+−− xxx |𝑥| = 2, −2 (𝑟𝑒𝑗), or 4 𝑥 = 2, −2, 4 or − 4 11c Consider the fact that f(1) = 0, and that f(𝑥) is a quadratic. This allows us to deduce that f(𝑥) = (𝑎𝑥 + 𝑏)(𝑥 − 1) Since f(2) = 3, 2𝑎 + 𝑏 = 3 Since f(0) = 1, −𝑏 = 1 Solving simultaneously, we have 𝑎 = 2, 𝑏 = −1 Hence, 𝑓(𝑥) = (2𝑥 − 1)(𝑥 − 1), OR Let f(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 f(0) = 1 ⇒ c = 1 Since f(2) = 3, 4𝑎 + 2𝑏 + 1 = 3 Since f(1) = 0, 𝑎 + 𝑏 + 1 = 0 Solving simultaneously, we have 𝑎 = 2, 𝑏 = −3 Hence, 𝑓(𝑥) = 2𝑥2 − 3𝑥 + 1.