H2 Math: Promo Exam Functions

Transcription

1 H Math: Promo Eam Functions S/No Topic AJC k = [,0) (iii) ( a) < ( b), R \{0} ACJC, :, (, ) Answers(includes comments and graph) 3 CJC g : ln ( ), (,0 ) - : a, R, > a, R (, ) = g y a a - 4 DHS - The graph o is the relection o the graph o in the line y =. (iii) g : e, R, > 0 R g = (, ) g : 4, 5 HCI y (, b + ) e + b y = b O Page

2 6 IJC ( b b ] R =, +. The value o k is b = ln R = 0, (iii) ( ) (iv) [ ] g ( ) =, D = (0,) (iii) g :, < < 0, ( ) Dg =,0, R g = (0,) 7 JJC 4 ( ) = ( ) = 0 ( ) =. Since Rg D, g eists. cos + (iii) g :, 0 < < π. cos (iv) Rg =, 8 MI/PU/ Since any horizontal line H/Promo is one-one and its inverse eists. cuts the graph at most once, y y=() y= y= - () They are relections o each other about the line. MI/PÚ/ h ( ) = + e, Dh = R; = ; < m < H/Promo 9 NYJC R = 3, ; ( ) 6 D =,6 ; (iii) 0 NJC (0,) [ ) g O (,0) α = ( ] = 3 69 α 0 α Greatest k = α. Page

3 PJC a = 6 ; ( ) = α, ( 0, α (iii) = 6 y = ; - : 4 4 = RI(JC) 3 (a) 3 < <. (b) k = 3 3 < 3. 3 RVHS 4 SAJC 5 SRJC + +, (iii) least value o b = (iv) g :, [, ) ; π R = [ + a, ) R - = a, π a g (iii) : or R, 4 7 Largest value o k = 3 ; : +, > 0 4 (iii) y y = ( ) y = = (iv) R (,3) = 3 0 g 3 y = ( ) 6 TPJC For to be a one-one unction, every horizontal line y = a, where a 9, cuts the graph o at eactly one point. This is possible or 3 or 3. Since k, the least value o k is 3. : , 9 Domain o = [ 9, ) (iii) (0, e 3 ] 7 TJC ln ( ) =, R, λ ; λ = Page 3

4 8 VJC ( ) ( ) Or = =, so is not one-one and thereore has no inverse unction. (iii) b < 0. (iv) R b = 0,. 4 b gh 3 The horizontal line y = 3 cuts the graph y =() more than once, thus is not oneone and has no inverse. (Draw the correct graph rom GC, many did not draw asymptotes.) g ( ) = +. D = R =, ( 0, ) g g 4 3 Page 4

5 AJC/0/H/Promo/5 The unctions and g are deined by Page 5 ( ) : + ln, R, k < < 0, g : State the least value o k such that Using the value o k in, Show that the composite unction, R. eists. [] g eist and ind the range o g. [4] (iii) Given that a > b where a and b are elements in the domain o, write down an inequality relating (a) and (b). Hence, solve the inequality g( ) >. [3] ACJC/0/H/Promo/6 Functions and g are deined as ollows: :, + + < a ( ) g : ln, < < 0 Write down the largest value o a such that the inverse unction eists, and deine in a similar orm. [4] Using the value o a ound in, show that the composite unction g eists. Deine 3 CJC/0/H/Promo/9 The unctions and g are deined by g in similar orm and ind its range eactly. [3] : + a, R, > 0 a g : e, R It is given that a is a positive constant. Show that is one-one and deine Sketch, on a single diagram, the graphs o and - - in a similar orm. [4] -. State the geometrical relationship between and. [3] (iii) Eplain why the composite unction g eists. Deine g in a similar orm and state its range. [4] 4 DHS/0/H/Promo/ Functions and g are deined by :, 4 R,,, g :,. Show that the composite unction g eists. []

6 Page 6 Deine g in a similar orm. [] (iii) Solve the inequality g( ) <. []

7 5 HCI/0/H/Promo/9 The unction is deined as ( + ) : e, + b R, where b is a positive constant. Sketch the graph o y ( ) =. Hence state the range o in terms o b. [3] Eplain why does not eist. The domain o is restricted to (, k] such that eists and the range o remains unchanged. State the value o k. [] Use the given restricted domain and the value o k ound in part or the rest o the question. (iii) Find the rule o in terms o b. [3] The unction g is deined as ( ) g : b, R. (iv) Find the range o g. [] 6 IJC/0/H/Promo/9 The unction is deined by :, >. By using dierentiation, show that () increases as increases. [3] Find ( ) and its domain. [4] The unction g is deined by g :, < < 0. + (iii) Find g and state the domain and range o g. [3] 7 JJC/0/H/Promo/5 + The unction is deined by :, or R,. Find ( ) and 0 ( ).[3] The unction g is deined by g : cos, or 0 < < π. Page 7

8 Eplain why the composite unction g eists. [] (iii) Deine g, giving its domain. [] (iv) Find the range o g. [] 8 MI/0/PU/H/Promo/H/0 The unction is deined as Show that eists. [] Deine in a similar orm and state its range. [3] (iii) Find the eact coordinates o the intersection point between and. [3] (iv) Sketch the graphs o and on the same diagram, labelling each graph clearly, and describe the relationship between them. [3] MI/0/PU/H/Promo II/H/6 Functions g and h are deined by g : m, + + R where m is a constant, ( ) h : ln +, >. Sketch the graph o y = h( ), stating the eact coordinates o the point o intersection with the ais and the equation o the asymptote. [] Find h ( ) and write down the domain o h. [3] (iii)on the same diagram as in part, sketch the graph o y = h ( ). Find the solution o the equation h( ) = h ( ). [3] (iv) State the condition or the composite unction hg to eist. Find the maimum range o values o m such that hg eists. [3] 9 NYJC/0/H/Promo/9 Functions and g are deined by ( )( ) : 3 + 5, or R, ; g : 5, or R, < 5. Show that the composite unction g eists and ind its range. [3] Find ( ) and state the domain o. [4] (iii) Find the eact solution o ( ) ( ) =. [] Page 8

9 0 NJC/0/H/Promo/7 The unctions and g are deined by α R, 0 < k, :, g:, R, <, 5 or some positive constant α. Sketch the graph o y = α, labelling clearly the aial intercepts. State the greatest value o k or the inverse unction to eist. [3] Hence, ind ( ) and state its domain. [] For α = 5, solve the equation PJC/0/H/Promo/8 The unction is deined by g( ) =. [], or R, 0. : 8 + has an inverse i its domain is restricted to a. State the least value o a. Find, in similar orm, corresponding to the restricted domain. [5] Write down the equation o the line in which the graph o y ( ) in order to obtain the graph o y ( ) equation ( ) ( ) RI(JC)/0/H/Promo/0 The unctions and g are deined by = must be relected =, and hence ind the eact solution o the =. [3] ( ) : 50 4, k k, 3 3 g :, R, where k is a positive real number. (a) Without using a calculator, solve the inequality g( ) >. [3] (b) It is given that the range o is [ 50,50]. Find the value o k. [] Without the use o a calculator, solve the inequality g( ) > 0. [4] Page 9

10 3 RVHS/0/H/Promo/9 The unction is deined by ( ) Eplain why which : ln , R. does not eist, and ind the largest domain in the orm (, a] on eists. [] For the domain ound in, ind The unction g is deined by g : ln 3, [ b, ) in a similar orm. [3]. (iii) (iv) State the least value o b such that the composition For the value o b ound in part (iii), ind g g eists. [] and state its range. [3] 4 SAJC/0/H/Promo/6 Functions and g are deined by : + a or R, where a is an unknown constant π π g : tan or R, < <. Find the range o in terms o a. [] - - Show that the composite unction g eists. Find the range o g in terms o a. [3] (iii) It is given that a = 3. The domain o is now restricted to{ R : }. For the new domain, give a deinition (including the domain) o 5 SRJC/0/H/Promo/3 The unctions and g are deined by + R : 7, -. [3] g :, R, > 4 4 Eplain why does not eist. [] (iii) (iv) The domain o is now restricted to the set A where A = { : < k} R such that eists. Determine the largest possible value o k and deine in a similar orm using this value o k. [4] Sketch and on the same graph and hence solve the equation ( ) = ( ). [4] Eplain why g eists and ind its range. [3] Page 0

11 6 TPJC/0/H/Promo/5 Functions and g are deined by : 6 or R, g: e - or R. I the domain o is urther restricted to k, state with a reason the least value o k or which is a one-to-one unction. [] In the rest o the question, use the domain deined in part. Find the rule or, stating its domain. [3] (iii) Find the range o the composite unction g. [] 7 TJC/0/H/Promo/8 The unctions and g are deined as ollows: λ : e, R, 0 and λ >, g : ln, R, > 0. Find (), stating its domain. [3] On the same aes, sketch the graphs o, and. [3] (iii) Show that the composite unction g eists. Given that g(λ) = 8, ind the value o λ. [3] Page

12 8 VJC/0/H/Promo/ The unction is deined as ollows :,,, R. Eplain why does not have an inverse. [] The unction g is deined as ollows g :, <,. g and state the domain o g. [4] Find ( ) The graph o unction y h ( ) = is shown below. y y = b O Given that b is negative and gh eists, ind = (iii) the range o values o b. [] (iv) the range o gh, leaving your answer in terms o b. [3] Page