Answers for NSSH exam paper 2 type of questions, based on the syllabus part 2 (includes 16)

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1 Answers for NSSH eam paper type of questions, based on the syllabus part (includes 6) Section Integration dy 6 6. (a) Integrate with respect to : d y c ( )d or d The curve passes through P(,) so = 6/ + + c this means c = 6 so that 6 the required equation is: y 6 dy 6 (b) Substitute = in: + the derivative is then 6 + = / d So the gradient of the tangent is /, the equation becomes y = / + c. P(, ) is on the line so = / + c this means c =. The equation of the tangent is: y = / +. d = d = [ ln + ] + c = ln + + c dp. Use the substitution p = then d p + c = p + c = + c. so dp= d so that d = dp = p dy dy. Integrate : with respect to : d d ( ) d or y = ½ + + c d (, ) is on the curve so = + + c so that c =. The equation of the curve is then: y = ½ +. dp 5. (a) Use the substitution p = then so that dp = d this means d d d dp ln p c substitute for p and you get: ½ ln ( ) + c as an answer. p (b) e d = 6. (b) (i) (ii) [ e e ] e du d use a substitution u = so that d = du or d = ½ du d d = e u u e du e e du d use a substitution u = + so that d = du or d = ½ du d d du } u ln u {ln ln.5 =

2 7 (b) Find d d ln( ) 8. d d d c ln. (i) d = ½ ln ( + ) + c. This can be solved as well solved with a substitution: dp Let p = + so so that ½dp = d. Then ln ln( ). d d dp pc p c = (ii) e d e e e [This is an eact answer so it can be left like this; the approimate value is 8.] Section Differentiation and finding gradients. First find the coordinate when y = ln 8: ln 8 = 5 ln ln or ln 8 + ln = 5 ln using the first law of logs we get: ln = 5 ln or ln = ln 5 [Third law of logs] so that 5 =. This means =. Differentiate the curve with respect to : d 5 ln ln 5. Substitute =, gives a gradient.5 d. d e d e. (a) f () = + stationary point f () = so + = (multiply with ) + = or = ½ =. 5.7 the y coordinate of the stationary point is f(. 5 ) = [(. 5 )] So the coordinates of the stationary point is (.7,.8) f () = > for < so the stationary point is a minimum. (b) see (a). (c) f () = + both terms are positive for > so (d) all gradients of the curve are positive in the first and fourth quadrant. y f(). 5 =.8. (a) d d 5 d d = 5 (b) d d d d 5. If y = d 6 + *) then [ 6 +] = 6. This equation gives the value of the gradients of d the tangents to the curve. The tangent has gradient so solve 6 = or = substitute this in *):

3 + = 6 so the point of touch is (, 6). This is as well the point the tangent is passing through. So substitute (, 6) in the tangent: 6 = + c so c =. d 6. (a) ln = d d (b) ln = d 7. (a) f () = + = ( + ) (b) f () = + = or ( + ) = so = or = so the stationary points are (, ) and (, 7) (c) Solution I A sign diagram of f () looks like: So (, 7) is a minimum and (,) is a point of inflection. Solution II First determine f () = + = ( + ) with f () > so (, 7) is a minimum. As well: f () = so (, ) is a point of inflection. 8. (a) d ln d. (a) let y = d (b) let y = 5 then dy. (a) Let y = 5 e then y = e. dy 5 d. (a) (b) dy ln( + ) = d d.5 d C.5ln( ) f()=^+^++. (a) f () = Try to solve = * the discriminant is: 6 = so the equation * has no solution. (b) Point of infleion f () = = so = so the point of inflection is (, f()) = (, ). (c) Use the table function on your calculator. Point of intersection with the -ais: (, ) and the y-ais (, ). (d) See graph.. d e d = e

4 Section Equations (including trig equations and identities).(a) Divide sin + cos = by cos sin, it becomes: or tan + = sec cos cos Substitute this in the given equation: (tan + ) + tan 5 = so the equation is: tan + tan = (b) Let tan = p then the quadratic equation will become: p + p = find two numbers product 6; sum. These are and so the eq. changes to p + p p = or p(p + ) (p + ) = or (p )(p + ) = so p = or tan = which means =.5 rad =.785 rad or.5rad =. rad or p =.5 or tan =.5 tan - (.5) =.87 so = (.87) rad =.6 rad or =.87 = 5. rad sin cot cos. (a) (i) cos ec sin LHS = sin cos sin sin cos sin cos sin Using sin + cos = we get: LHS = (ii) For = º (the denominator is zero) sin after multiplying numerator and denominator with sin. which is indeed equal to the RHS. Q.e.d. (b) (i) Divide sin + cos = by cos sin, it becomes: or tan + = sec cos cos Make tan subject: tan = sec and substitute in: tan θ sec θ = ; you get: (sec ) sec θ = or sec sec =. (ii) Let sec = p the equation changes to p p = factorize: p p + p = p(p ) + (p ) = or (p )(p + ) =. This gives solutions p = or sec = or /cos = this means cos = ½ so = or =. As well: p = ½ or sec = ½ or /cos = ½ this means cos = so no further solution here. (c) (i) -8º -º º (ii) Etend the graphs a bit, so there is one solution of the equation f() = g() in the interval º º. sin cot. (a) cosec cos LHS = sin ( cos ) cos cos sin sin ( cos ) sin cos sin cos sin cos sin sin cos ec q.e.d.

5 (b) cot ½ θ =.87 tan ½ θ = /.87 =.78 so that ½ θ = º k; k This means θ = ºk = 7º+ 6ºk so θ = º [Taking k = ] (c) (i) 5 7 sin cos = *) substitute cos = sin so *) becomes: 5 7 sin ( sin ) = Rewrite: sin 7 sin + = **) (ii) Let sin = p then **) becomes: p 7p + = or p 6p p + = or p(p ) (p ) = So that (p )(p ) = this gives sin = ½ so that = º or = 5º in radians: = 6 radians or = 6 5 radians; p = or sin = does not give any solution. [Remark: one can epress 6 radians and 6 5 radians as decimals. This is not wrong but at the same not asked for. If you round off wrongly, marks can/will be subtracted.]. (a) sin = ½ = sin - (½) = k rad with k Z.. or = k rad k Z (b) gives: = k that means =. rad (k = ) gives: = k that means =. rad (k = ) cos tan cot. sin cos LHS = sin cos cos sin cos ( cos ) cos cos q.e.d. sin sin cos sin cos sin cos cos 5. (a) sec tan sin cos ( sin ) cos ( sin ) LHS = cos sin cos cos (b) cosec = sin =.5 sin = k rad. or = k rad. gives =.6 + k so =.6 radians or =. rad gives =.88 + k radians or =. + k this give the solutions =. radians (k = ) or =. =.7 radians (k = ) sin cos tan f()=sin (c) (i) 5cos sin = *) substitute sin = cos *) becomes: 5cos ( cos ) = Rewrite: cos 5cos + =. (ii) Substitute cos = p p 5p + = or p p p + = or p(p ) (p ) = so that (p )(p ) = this gives p = [No sol.] and p = ½ or cos = ½ this means = 6º or = º 6. (a) cos general solution is: = + k with k an integer or = leads to = + k which will give solutions = 5 ; and 7 k Z] + k 5

6 (b) leads to = 5 + k which will give solutions = 7 ; cos ec tan tan cot 7. (a) LHS = sin LHS = sin sin sin sin cos sin cos cos sin cos cos sin sin cos sin cos cos sin cos sin cos sin cos tan tan cos sec cosec cos sin cos sin RHS = sec cosec cos sin cos sin So LHS = RHS. and 5 sin cos sin cos sin sin cos sin cos sin cos sin cos sin = RHS. (b) cot = or tan = then = tan - ( ) = 8.º + k 8º with kz, divide by : =.7º + k º. Take k = then = 7º Take k = then = º Take k = then = º. (c) Solve for : cos sin =..* for radians Use cos = sin then * ( sin ) sin = rewrite: sin sin =. Rewrite: (sin + )(sin ) =. The only acceptable solution is sin = ½. = 6 radians. So the solutions are: = 6 5 radians or = 6 radians. 8. [Paper 5 Q ] cos cos cos ( cos ) cos ( cos ) cos cos (a) (i) LHS = cot cos cos cos sin sin sin (ii) The identity is not defined when cos θ = for θ = (b) (i) tan sec =.*. Use the identity tan + = sec in eq * you get: sec sec = or sec sec = (ii) tan sec = or sec sec = which is (sec )(sec + ) = so sec = ➀ or sec =.➁ ➀ can be rewritten as cos = ½ or = 6 and ➁ as cos = which has no solution in [, ] (c) (i) (ii) No solutios. y cosec y = cos y = sin - 6

7 sin. (a) sec tan cos LHS sec tan cos sin cos cos ( sin ) sin RHS qed cos cos cos cos sin cos sin sin cos cos cos ( sin ) cos ( sin ) ( sin ) sin (b) cos = ½ so = cos ( ½ ) + kπ..➀ or = π cos ( ½ ) + kπ.➁ with k Z. Divide ➀ by gives: =. + kπ which gives: =. rad (k = ) or =.8 rad (k = ) Divide ➁ by gives: = π. + kπ which gives: =.8 (k = ) or =. (k = ). This solution is acceptable but not very accurate. Some calculators give eact solutions in terms of π, which are preferable. Even without that calculator more eact solutions can be found using the special and the graph of the cosine function. f()=cos From the cos = ½ or cos 6 π = ½ Then cos 6 π = ½..➀ or cos 6 π = ½ ➁ 5 5 ➀ gives: = 6 π + kπ = π + kπ. 5 7 With solutions: = π (k = ) or = π (k = ) 7 7 ➁ gives: = 6 π + kπ = π + kπ. 7 With solutions: = 5 π (k = ) or = π (k = ). Section Equations and graphs of absolute values. (a) (b) 5 for = ½ Using the definition of absolute values we get: for ½ 5 becomes 5 + = or for < ½ 5 = or =. becomes = or = or =. (; 5) 5 (; ) (.5; ) (; ). (a) + = + = for =. (b) For the equation becomes: + = or = 6. For < the equation becomes: = = or = so no solution here because it is not in line with <. y. (a) (i) (ii) y.5 (6, ) (,) (, ) (, ) (, ) 7

8 (b) Solve the inequality < 8 *) For < *) becomes: + < 8 < or >. For *) becomes < 8 < or < 6 Taking and together we get: < < 6.. (a) gf() =g( ) = g() = () 5 () = 5 = (b) See graph on the right (c) 6 = + *) For > *) changes to 6 = + or = 7 =.5 For *) changes to + 6 = + or = 5 =.5 y (,5) (, ) 5. Solve: *) For < *) changes to 6 so 8. This results to 8 < For *) changes to so. This results to Taking and together we get (a) = 6 can be rewritten as: = 6 so = 6 or = this means = or = or = this means = y (b) (, ) (c) < this means < < or < < 5 7 (a) The dotted lines are not the solution but are assisting the drawing of the final graph. In the final eam it is better to erase them. (, ) y y = + 5 y = + + (b) > = for = y = + For then > changes into: > or > 6 or > For < the inequality ( ) > or + > so that > or < So > for > or <. 8. (a) (i) Range of f is: y (ii) Range of g is y (b) For or for (c) The line passing through (, ) and (, ) has gradient so a = then b = (d) Both f and g do not eist. [Use the horizontal line test in the graph to see that there always originals with more than one image] (e) The zeros of f are and so f() = p( + )( ). (, ) is on the graph so = p() () so p = 8

9 So f() = ( + )( ) or f() = +..(a) + 8 = 6 divide by : + = or + = so + = ± so = or = 7 (b) + <. First solve: + = that is for = then solve ( + ) = for = so + < for < <. (c) See diagram. y -5.5 O.5-5 Section 5 Particle mechanics (-.5, -8) t ds t. (a) s e t so v ( t) e so v()=.5 =.5 m/s dt (b) v(t) = e t =.5 so e t =.5 take natural log from both sides: t = ln.5 so t = 5.8 s In three significant figures: t = 5. s t (c) v ( t) e dv t so the acceleration is: a( t) e so a(5.) =.5 =.5 m/s dt dv d. (a) (i) a(t) = (t t ) t dt dt. (a) (ii) s(t) = v t) dt (t t ) dt.5t t ( c if s(t) epresses the distance from flower F then s() = so c = the correct formula for the displacement is then s(t) =.5t t. (b) At B the speed will be, this is when t t = or t( t) = so v = for t = sec (c) s() =.5 7 =.5 =.5 m (d) a(t) = t so a(t) = for t =.5 sec. and v(.5) =.5.5 =.5 m/s (b) a( t) dv dt v( t d dt 5e t 7e t t t s( t) ) dt 5e dt 7.5e c since s() =, c = 7.5 m so s(t) = 7.5e t s() = e = 7. m. (a) v( t) a( t) dt (t ) dt t t c since v() = m/s, c = so the velocity is v(t) = t t +. (b) v(t) = so t t + = or t 7t t + = or t(t 7) (t 7) = so that (t 7)(t ) = So v(t) = for t = seconds or t =.5 seconds. (c) s( t) v( t) dt (t t ) dt t 5.5t t c since s() = 5; c = 5. So s(t) = t 5.5t t 5 (d) s() = = 5 6 m. So the distance travelled from O after s is 5 6 m; the distance travelled in the first second is = 6 m [s(t) is the distance to a fied point O and at t = the particle was already 5 m from O]

10 5. (a) (i) v(t) = t t = t( t) so for t = 6 there is a maimum velocity of 6 6 = 6m/s. [Eplanation: the graph of v(t) is a parabola with zero s at t = and t =. The line of symmetry is at t = 6 so v(6) is the maimum velocity.] (ii) s( t) v( t) dt (t t ) dt 6t t c. With c = because s() = so s(t) = 6t t OB = s(6) = 6 7 = m. (b) (i) s( t) v( t) dt (6 t ) dt 6t.5t c with c =. So s(t) = 6t.5t. Then s(t) = 6T.5T = divide by.5 you get: T T + 8 =. (ii) T T + 8 = (T )(T ) = so T = s or T = s. 6. (a) v =e t t then the acceleration is: dv 6e. dt t t (b) s(t) = v( t) dt e dt e + c with s() = so that c = s() = 6.5 m 6. So s(t) = t e 6 7. (a) Integrate: dv then v(t) = dv dt dt t c dt dt Since v() = it means c = so v(t) = t (b) s(t) = ( t) dt t 5t c with c = because at moment t = the stone is still at the ground. So s(t) = t 5t. (c) (i) v = (ii) v = so t = so t =. sec. (iii) s(.) =. 5. = 7. m. Section 6 Area and volume calculation through integration. (a) Shaded area is A = d dp y d. Use a substitution: p = so dp= d d When = then p = and when = then p = so A = p dp p = ( ) = (b) V = y d ( ) d. (a) First make y the subject of the formula: y =, required volume is: V = y d ( ) d = = {(½ ) (½ )} = ½ (b) y dy = y ( ) dy or so the gradient at B is d d.5.5 ( ) () The equation is y =.5 + c. B(, ) on the line so =.5 + c so c =. So the equation is y =.5. (a) Point A : y coordinate is zero so = so = A(, ) Point B : coordinate is zero so y = so y =

11 (b) Area shaded region = dp y d d use substitution = p so that or d= dp d.5.5 Area shaded region = p dp p p { } ( 7).5 8 sq units. 5 ( ) (8 8 ) 8 6. (c) Generated volume = dy y dy y y dy y y y. (a) dy d d d d { 6.}.6.5 dy the value of at = is.5 6 = d 8. This means the tangent has equation y = 8 + c. The tangent passes through T(,) so = 8 + c this means c = ½ so the equation is y = (b) Area shaded region is: Area trapezium d to integrate a substitution is used: p = + ; dp d d ( ) so dp = d d Area shaded region is = ½ h(a + b) p dp (.5 ) p {68} } (c) V = y d ( ) d.5 {6 5. (a) For A, y = so A(6, ) and for B, = so y = + 6 =. This means B(, ) (b) Area shaded region = y d 6 d for this integration a substitution can be used: p = p dp p {6 } 5 dy ( y 6) dy ( y y 56) dy.y y 56y dp = d, but the boundary points change: Area shaded region = (c) Generated volume = 6. (a) = [. 8 +] = (. + ) = 8 5 = 56 5 dy y..*differentiate with respect to : so the tangent at = to the curve will have gradient d =. [This is possible if you check the graph in the diagram.] The tangent passes through (, ) [substitute = in *] So the equation of the tangent is y = + c or = + c c = 6. So the equation of the tangent is y = + 6. [Again the diagram gives further assurance that this is the correct answer.] (b) Area shaded region = area under the curve area trapezium = ( ) d ( ) = [ ln ].5 8 ln (.5 ln ).5 ln ln.86 5

12 (c) Requested volume = d ( 6 ) d { 6.5 ( 8 )} ( 6 7. (a) For A,y = so = 6 y so = this means A(, ). For B, = so y = 6 this means B(, 6). (b) Make y subject: y = 6 ; the shaded area = 6 ( 6 ) d [6 ] 6 6 (c) Volume = 6 dy (6 y) dy [6y y ] (56 8) = 8 8. (a) y = ; then 6 dy d (b) Calculate: d ( ) d (c) The integration substitute = and you get for the gradient dp d can best be done through a substitution: let p = + then so ½d p = d d and the boundaries change from to. Requested area = d area trapezium =.5 dp 8 p { 7} p ½.( + ) =. (a) A(,) and B(.) dy (b) so the tangent has gradient. So the eq. is y = + c. (,) on the curve so c = 8. d The tangent has equation y = + 8. (c) Area shaded region = ( ) d (d) Volume = dy ( y) dy y y {(6 8) (.5)} (87.5). 5 Section 7 D and D vectors. (a) Given: OA i + k ; OB i j + 5k and OC BA BO + OA (i j + 5k ) + i + k = i + j k. (b) Use the scalar product: (c) i + j k then BA BC BA BC cos ABˆ C = product components BA = i + j k and BC = BO OC = (i j + 5k) + (i + j k ) = i + j 8k ( ) ( ) ( 8) Cos ABC = so ABC = 8.º ( ) ( ) ( 8) 56 OD 5 OD 5 AC so AC = AO OC = (i + k) i + j k = 5i + j 5k so AC = i + j k so D(,, )

13 . (a) If A is (,, ) then OA= i j + k as well OB OA AB = i j + k i + j k = i + k So B(,, ) (b) (i) AB and CD are parallel q AB = CD so q(i + j k.) = i + pj + k looking at the i component one can conclude q = this means p = 6. (ii) AB and CD are perpendicular the scalar product is zero so + p + () = or 7 +p = or p = 5. (iii) AB = i + j k and CD = i + j + k. a b ab ab Use the formula: cos a b The required angle is cos - ( ) 7 8 = 8.5º ( ). (a) A has coordinates (,, ) i + j + k thenob = OA + AB = i + j + k i + j 6k = 5i + 6j k so B has coordinates (5, 6, ). (b) (i) If AB and CD are parallel then pab = CD or p(i + j 6k) = 8i j + tk. Comparing the i component you get p = so t =. (ii) Perpendicular scalar product is zero so AB CD =() 8 + () + (6) t = or (c) 6 6 6t = or 6t = 5 t = 8 Use the formula: ab ab ab cos =.5555 a b 6 6 The requested angle is cos (.5555) = 7.º. (a) OA OB = OA OB cos AOB = a b + a b cos AOB = a b ab OA OB Cos - 5 = 5º 5 (b) AB = AO + OB = so AB = 8 + ) = 85 =. 5 p p (c) AC = AO + OC =. 5 p p 5 OC AC (p + p ) = [(p ) + (p 5) ] after squaring both sides you get: p = p p + + p p + 5 or 5p = 5 so that p =½ 8 angle AOB = cos - ab ab OA OB = 5. (a) ac ac ac 5 5 cos requested angle is cos - 5 =.8 o a c 6 (b) AB = AO + OB = p p 5 and AC = AO + OC = 6 8 p p

14 AC AB = + p (c) BAC = º AC AB = or + p = so that p =. 6. (a) C(,, ) (b) BA = BO + OA = (j + k) + (i j + k) = i j k. BC = BO + OC = (j + k) + (i j) = i j k (c) Angle ABC = cos 6 6 cos 6 cos 6 ) ( ) ( ) ( ) ( ) ( =.7 rad [. rad. is correct as well since angles are to rounded off to one decimal place] 7. (a) use the scalar product: b a b a b a b a AOB ˆ cos = ( ) ( 5) so angle AOB = º (b) First find vector AB and AC. 5 OB AO AB and c c OC AO AC BAC = º so the scalar product is zero: + () ( + c) + 5 = work out brackets: + c + 5 = c = 8 so that c = (c) d d OD BO BD then ) ( d so that d = 8. (a) OA = +() + ( ) ] = [6 + + ] = 6 = 8. (b) Then OB = [i j + k] = 8i j + k so B(8,, ) (c) (i) OA.OC = so a b + a b + a b = OA OB cosaoc = a b + a b + a b = or + () p = or 8 = p p = (ii) cos 6º =.5 so + () p = 8 ( + p ).5 rewrite: 8 p = ( + p ) Divide by : p = ( + p ) square both sides: ( p) = + p Work out the brackets and write to standard form: p + p = + p p + p =. So p = or p =. (a) Use the scalar product: cos b a b a a b AOB OB OA OB OA so cos AOB = 5 this means angle AOB = cos - (/ 5) =. (b) OB AO AB and 5 c c OC AO AC The scalar product is then zero, so (c ) + = or (c ) = c = 7 so that c =. (a) (i) OX = 5j + k (ii) XB = 6i 5j k (iii) AC = 6i + 5j k (b) Then XO = 5j k so 7 5 ˆ cos b a b a b a a b XB O so angle OXB =.8 radians.

15 Section 8 Factor and remainder theorem. (a) Let f() = + 7 check f() = = indeed so ( + ) is a factor. + + Based on the long division we now know: 7 + f() = + 7 = ( + )( ) The quadratic part has to be solved: = 7 6 or 7 so the solutions are: ;.56 or.56 (b) f() = + a + b has a factor ( + ) f() = so + a b = rewritten: a b =. When f() is divided by ( + ), it has the same remainder as when it is divided by ( ). f() = f(). This will provide another relation between a and b: 8 + a b = 8 + a + b or 6 = b So that b = and using we get a =.. (a) Work out the long division: (b) f() = = 7 5 = so ( ) is a factor of f(). Conclusion: a = 6; b = ; c = and R = 5 (c) Work out the long division: Conclusion: 6 + = ( )(6 ) = + + ( )(6 + 8 ) = ( )[( ) + ( )] = + ( )( )( + ). (a) f() = = so ( + ) is a factor of f(). (b) Work out: This means = ( + )( + 6) so that a = ; b = and c = 6 (c) ( + )( + 6) = ( + )[ ( + ) ( + )] = ( + )( + )( ) = so = or = or = ½. 5

16 . (a) substitute = in p +, you will get: + p + = 5 + p = so p = (b) (i) + = = or ( 5)( + ) = so that = 5 or =. (ii) using (i) we can write down: a 5 or a = 6. [The root will not give a solution since a square root always gives a positive answer.] 5. (a) So f() = 6 or a 8 = 6 a = or a = 7 So f() = (b) (i) Work out f(): = indeed so ( + ) is factor. (ii) [One can work this out with a process of long division, but that takes time. Smarter is the following:] c = 8 so c = 8. For b you can say: b + c = 7 [Because ( + )( + b + c) = + 7 8]. So b 8 = 7, this means b =. So f() = ( + )( + 8) (iii) = is one solution; + 8 = is the other part of the equation. Solve with abc-formula: b b ac..( 8) ½ ½ =.7 or =.7 a (c) (i) f () = + 7 turning points have coordinates + 7 = or = so that ( + 7) ( + 7) = ( + 7)( ) = = or = 7 /. (ii) f () = 6 + but f ( 7 / ) < for = 7 / there is a maimum and f () > so for = there is a minimum. 6. (a) (i) > so a =. (ii) range is R (iii) f() = ln( ) or y = ln( ) this means e y = or e y + = swop and y: y = e + So f () = e + (iv) y f - () y= f() (b) (i) gf() = g(ln ) = ln + =.6 (ii) fg() = f() but f is only defined for > so fg() does not eist. 7. (a) When f() is divided by ( + ), the remainder is 5 f() = 5 or 5 = () + () + 5() + c 5 = c so that c = (b) (i) ( + ) is a factor then f() = substitute in f() = it becomes: = which is true. Conclusion: ( + ) of f(). 6

17 (ii) Work out a long division: (a) Work out f() = = so ( + ) is a factor. (b) Divide: So a = ; b = and c = 6 Alternatively: The value of a, b and c can as well be obtained by inspection: f() = ( + )(a +b + c).= 7 6 a = can be seen immediately since a is the only product giving the. As well c = 6 delivers c = 6. The b is a bit more difficult since there are two products contributing to 7. (c) So f() = 7 6 = ( + )( 6) = 5 + 6) = ( + )[5( ) + ( )] = ( + )( )(5 + ) f() = for = or =.5 or = This means = ( + )( + + ) = ( + )( +)( + ) =( + ) ( +) (c) (i) Work out f () = ; f () = = or ( + 5)( + ) = this means there is a turning point for = and for = (ii) Work out f (); f () = 6 +, f () > so = is a minimum; f ( ) < so = is maimum. It is given that f() = p, where p is a constant. 6 (a) Then f()= or f() = p = or p = so that p = (b) (i) p =, so = 5 or 6 = or ( )( + ) = = or =.. (ii) a a 5 (a ) = or a = so a =. only will give a solution since a square root has a positive outcome.. (a) f() = (a + b + c)( + ) + R = a + (a + b) + (b + c) + c + R this is identical to f() = so a = 6; b = ; c = 5 and R = 5. (b) Find f(): f() = = so + is a factor. (c) One can write down immediately: f() = ( + ) (6 + p + ) = with p = So f() = ( + ) (6 + ) = ( + ) (6 5 + ) = ( + )[( 5) ( 5) = ( + )( 5)( ) so f() = ( + )( 5)( ) 7

18 . (a) So f() = or 8 + c = so that c = 6 indeed. (b) f() = 7 6. One can easily see that f( ) = so + is a factor. Then 7 6 = ( + )(a + b + d) * One can easily see that a = and d = 6 so * changes into: 7 6 = The 7 is generated by multiplying b and ( 6) so b 6 = 7 so b =. f() = ( + )( 6) = ( + )( )( + ). [The way shown here is a shortcut, one can use long division as well.] (c) (i) f() = 7 6 f '() = 7. Solve f '() = or 7 = or = 7 or = (ii) f ''() = 6 this means for = there is a maimum value and for = there is a minimum value. Section Trigonometric functions. (a) Amplitude is [Half the range]; period is 8º. [ varies from º to 6º when varies from º to 8º.] (b) f()=5-*cos Maimum points: (º, ) And (7º, ) (c) 5 cos = or cos = so that cos = ½ this means = 6º or = º.. (a) T(½ (b) period of radians. (c) y (d) D,. (a) (i) a = and b = (ii) range is y (b) (i) p = and q = (ii) period is º (c) f()g() < for º < < º or for º < < º or for 5º < < 8º.. (a) a = ; b = ; c = (b) g : tan 5. (a) (i) a = ; b = ; c = (ii) sin () + = sin = ½ so = º + k 6º with k Z this means = º + k or = 5º + k 6º with k Z this means = 5º + k.. gives that = º when you take k = but gives = 7º. After checking the graph = 7º 8

19 (b) (i) p = q sin cos = cos + sin or sin 7cos = divide by cos : tan = 7 so that =. rad. (ii) p + q = ( sin cos ) + ( cos + sin ) = 6sin sin cos + cos + 6cos + sin cos + sin = 6 + = (a) a = ; b = ; c = and d = (b) A(8º, ) (c) period is 8º (graph repeats itself after 8º.) (d) Range of y = c cos d is y. (e) Two possible answers: y = cos or y = sin ( + 5º) 7. (a) f()=sin + f()=cos f() g() (b) f() g() < for º < 5º 8. (a) a = ; b =. (b) sin = cos or tan = [Divide both sides by cos ] tan / =.7 + k 8 with k Z. So the solutions are =.7 or =.7 (c) for <.7 or for.7 < 6 (d) y = cos Section Finding optimal solutions 6. (a) 7 = h so h = (b) Total surface area A = ( + h + h) = + h + h = + 6h = + substituting h = 6. 6 after

20 da 6 (c) 8 = or 8 = 6 = 7 so that =. d (d) = so A = = = 8 cm Find d A 8 > for all positive values of so there is minimum value for = for A. d. (a) Given: 8 = h h (b) Surface area of the bo is: A = ( + h + h) = 8 + h(8 + ) = 8 + da 6 (c) 8 d. (a) h (t) = 6 (8 + ) = or d A = 6 8 = so =.. This is a minimum because d 6 = 5 cm The value of the minimum area of cardboard used is: 8 + d h dt (b) h (t) = or t t the value of h () =.75 =.75. So the depth is decreasing. t t = t(.75 t) = so rate of water flowing into the tank, be the same as the rate of water flowing out of the tank at t = (8: h) or at t =.75 = h min so that is at : h. (a) V = (6 ) =. dv (b) (i) = 8 d V d V = or ( ) = for = or for = ; 8 so for = so we d d d have a maimum here. dv [ A sign diagram of indicates as well a maimum: ] d min ma (ii) For =, V() = 6 6 = 6( 6 ) = 8 m dt 5. (a) = T = ½ t dt (b) (i) T = ½ t < for ½ t < multiply by : t > so the temperature drops for < t. (ii) Since t the temperature range is T() T() = = º C. 6. (a) Epress the TSA in terms of and y: TSA = area triangle + area rectangles = ½ + y( + + 5) = + y = 6 as is given. Make y subject: y = 6 divide both sides by : y (b) First epress the volume in terms of : V() = 6 y = 8 6. Then dv d dv d 8 8 dv dv for = this is indeed a maimum value because > for =.5 and < for =.5. d d

21 7. (a) dy d a b c At a turning point: dy = so a +b +c =.* d Use the standard formula: b b a c for this case: a = a; b = b and c =c. a b b ac b b ac b b ac b b a 6a 6a a ac qed. (b) Apply: b b ac since < there is no turning point. a 8. (a) B(t) = + 5t 5t so B(t) = 5 t and B() = 5 =. (b) B(t) = 5 t = for t = 5 h so 5h after adding the bactericide the no of bacteria will decrease. (c) B(t) = or + 5t 5t = or t t = or (t +)(t ) = t =. So hours after adding the bactericide all bacteria will be dead.. (a) Area cardboard used is A = + h * (b) Volume is cm so h = or h = /. Substitute this in * and you get: A = + d (c) Differentiate A and equate to zero: 8 = or 8 = so = 5 or d = 5 7. cm and h = = 5. cm. d. (a) T T.t. 6 ; substitute t = 5: rate of change is. 5.6 =.6.6 =.76. dt (b) T() =.8.6 = 8.6 =. C.. (a) Volume = 5 cm = πr 5 h so h = r (b) TSA = A = πr + πrh = πr 5 + πr = πr + r (c) da d r da d r r 7r for a minimum value so 7 r r 7r = or πr = 7 or r = 7 =.8 cm. Section Solving logarithmic equations. log log. Use the change of base formula for logs: log log log log log8 8 or use the substitution: log = p log log p log8 p multiply both sides by p: log p log log b a log b log a p log8 p or p log8 log

22 Use the formula to solve this quadratic equation: p = ( )log log [This is quite long better to go over to log ( log8) log log log log so p = log = log 8 this means = 8 or p = log = log ½ this means = ½ log log : log. multiply with log log (log ) log so that (log )(log ). Solutions are: log = 8 and log = ½.]. (a) Write log as a based log: log log log so the equation becomes: log log log Multiply both sides with log : log log = [log = ] For easy writing: log = p then the equation changes to: p p = or (p + )(p ) = so that p = or p = ½ log = or = the other solution: log = ½ means =. (b) y = ab take log: log y = log a + log b log b = so b = and log a = [the y intercept] so a =.. y A B. (a) > so > and at the same time: + > or > ½. So P only eists for >. (b) log ( ) log () = or log ( )( ) log so that ( )( + ) = work out brackets: = or 5 = or = or ( 5) +( 5) = ( 5)( + ) = so that = ½ [for =, P does not eist see (a)] 5. log( + ) > + log( + ) log( + ) log( + ) > log or log log so or + > ( + ) > 6 so > 8 [for these values of both logs eist and + > ] 6. (a) log p log 5. Use the th law of log to rewrite the second logarithm: p log log p 5. Let log p = q then the equation changes into: q 5. Multiply with q: log p q q 5q + = ; factorise: (q )(q ) = so q = with p = or q = with p =.6 (b) Take log from both sides: log y = log a b or log y = log a + b log so b = ½. Point (, 5) is a point of the line, so 5 = log a + ½ log a = then a =

23 7. Given: log 5log 8. 8 (a) > and (b) log 5log 8. Use the th law: 8 log 8 8 log 5 or log 5 so let log 8 = p. 8 8 log log 8 p 5 = /p this changes into p 5p = factorize: (p + )(p ) = so p = then log 8 = so = 6 p = or log8 = then = ½ 8 8. (a) Two methods are used: I let the base of the log remain ½; II go over to as a base for the log. Working Comment I log log ( ) Apply law to write the left side as one log and write the right side as well in logarithmic form. log ( ) log Since the base no is less than the sign changes. ( ) or or ( )( ) Factorize and conclude. solution for Working Comment II log log ( ) Use law to go over to base : log log log log log log log ( ) log ( ) or ( ) log ( ) log solution for log ( ) log log so or ( )( ) But log = so the eq. can be rewritten: Multiply both sides with ; the sign changes! But log log log From here on it works out the same as I. b b ac 8 ( ) 8 (b) (i) Use the formula: y which means y =. or y =. a (ii) Hence solve the equation + = rewrite using = p we get p p = So =. take log: = log. log =. or =. no solution. [A power of cannot be negative]. (a) + 8 = becomes ( ) + 8 = ; apply substitution k = and you get: k k + 8 = or (k 6)(k ) = so that k = this means = or k = 6 or = 6 or = log 6 log =.6 (b) log ( ) log ()

24 log ( ) use the change of base formula for the second term and the eq. changes into: log ( ) log this can be rewritten as: log ( ) log ( ) apply law : log log so or + = ( ) so that =.8 Section Relation between variables (a) Take the natural logarithm from both sides: ln y = ln ae b or ln y = ln a + b... The graph of ln y against is a straight line graph; with as independent variable and dependant variable ln y. The gradient is b and the ln y intercept: ln a. (b) Line passes through (,.6) and (,.6) replace (,.6) in :.6 = ln a so a = e.6 =.8. replace (,.6) in :.6 =.6 + b this means b = ½. (c) Substitute = in with a and b replaced by the values found in (b): ln y = ln y =.6 + ln y =.6 so y = e.6 =.5. (a) y = e 5 take natural log form both sides: ln y = ln + 5 ln y so m = 5 and c = ln. (b) (i) and (ii) see diagram. (c) (i) = y = e 5 =. (, ln ) (ii) y = = e 5 or e 5 = take natural log from both sides: 5 = ln so = (ln ) 5 =.68.. (a) + > > and > > so y eist for >. (b) = log ( ) ( ) or log 6 log ( )( ) so ( + )( ) = 6 6 = 5 = so = 5 or = 5 this means = 5 because >. (c) Answer: > 5 both logs in log ( ) log ( ) are increasing functions.. (a) R = ka take log from both sides: log R = log ka or log R = log a + log k..*) so log k =.6 or k =. using the y-intercept of *). Substitute (,.) in *) you get:. = log a +.6 log a =. or log a =.5 so a =.5 =.8 (b) Again substitute in *): log R = (.5) +.6 =. so R =.6. Section Functions, inverse and composite functions; completing the squares; quadratic theory. (a) + 8 = ( + 5) = [ + + 5] = [( + + ) 5] = [( + ) ] After working out the square brackets you get: + 8 = ( + ) 8 so a = ; b = and c = 8. (b) Least value of y is 8 for =. (c) + 8 = k + 8 k = Discriminant > or b ac > substitute: 6 ( k) > k > 8k > k > 8. (d) g is a one to one function; f is not. Only one to one functions have an inverse. (e) g() = + 8 or y = + 8 or y = ( + ) 8 swop and y: = (y + ) 8 Make y subject of the equation: (y + ) = + 8 (y + ) = ½ + y So g () = + (f) Domain is ½ + > ½ > 8 and the range is y.

25 ln. (a) fg(5) = f(g(5)) = f(ln ) =.5 ln (b) ff() = f(f()) = f = ( ) ( ) 5 ( ) ( ) (c) f() = and g is only defined for > so gf() cannot be evaluated. (d) f y ( y) () y = swop and y and make y subject: = ( y) y rewrite: y y + y = y( + ) = f () = g () y = ln( ) write as a power: e y = swop the and the y: e = y so g () = e +. (a) (b) (i) (ii) h() = log h() = log f() (c) f() = log or y = log or y = log so y = or = y so f () = [Etra: Check: f (f ()) = f( ) = log ( )= indeed]. (a) range of f is y e (b) y = e + so ln y = + or = ln y swop and y : y = ln so f - ln. (c) fg() = f(g()) = f(7) = e 8 5. (a) fg(a) = or f(g(a)) = f(a ) = a = a = a = 5 so a = ± 5 (b) first f : y = or.5 y.5 = so f :.5.5 So gf (b) = 8 becomes g(.5 b.5) = 8 or (.5 b.5) = 8 this means.5 b.5 = ± Multiply with and you get: b = ± 6 so b = 7 or b = 5 (c) ff(c) = 7 becomes f( ) = 7 or ( ) = = 7 so that 6 = or =. (d) No g has no inverse because g is not a one to one function. Eample g() = 5 has two solutions: ±5. 6. (a) a > so a >. [Logarithm can only be taken from a positive number.] (b) (i) Range of f is R (ii) f() = log ( ) or y = log ( ); write as a power: y = swop and y: = y so y = + f () so f () = + (c) (d) gf(7) = g( log ) = g() = (e) fg() = f() = log () cannot be done. y f() 5

26 ln. 7. (a) fg() = f(ln ) =. ln. (b) f() = =.5 yet g only defined for >. y (c) (i) y swop and y: or (y + ) = y y or y + y = or y( ) = So that we get for f : (ii) g: ln( ) for > or y = ln( + ) swop and y and go over to eponential notation: = ln(y + ) or e = y + so that y = e and g : e (d) See diagram y y = e y = ln( + ) 8. (a) (i) Since e > for all values of ; the range of f is y > (ii) y = e + so = e y + = e y take ln form both sides: ln( ) = y so f ()= ln( ) (iii) Domain of f is >. y (iv) See diagram to the right: (b) (i) gf() = g(e + ) = (e + ) = e + 7 = 7.8 (ii) f g( ) = f ( 5) = ln( 8) this does not eist since the answer for any power of e is always positive. y = f() y = f - () Section Sequences & patterns and series. (5 r ) = = = 5. r Or using the formula for an arithmetical sum ( Take care there are terms here): S n = ½ n [a + (n )d] = ½ [ 5 + ()] = [ 6] = = 5. (a) Decrease of % means the yield decreases with a factor.8. So the production is a geometrical sequence: 7; 7.8; 7 (.8) ; 7 (.8). This is a geometrical sequnce with a ratio <. a 7 7 Sum is: S 5 kg of copper. r.8 5 (b) a = 7; r =.8 so t n = ar n = 7.8 n < first solve 7.8 n = or.8 n = / 7 take log from both sides: (n ) log.8 = log / 7 n log.8 = log.8 + log / 7 log.8 log 7 n.7 years so the mine will be abandonded after years. [T = = 7 kg] log.8 But T = 7.8 = kg] So in the th year the production will fall below kg. 6

27 a( r ) 7(.8 ) (c) S 5(.8 ) kg r.8. (a) (i) t = S = = = a (ii) S = 6 = t + t yet t = so t = 5 so the difference d = (b) T = ar and T 8 = ar 7 then T 8 T = r 5 yet T 8 T = 8 = so r = ½. k n n n (c) 5 ( ) = 86 or 5 ( ) = 86 divide both sides by 5: ( ) = 8857 use the sum formula for gp: n k k k n a( r ) ( ) 8857 = 77 = ( k ) divide by : 58 = k or 5 = k r log 5 This equation can be solved with logaritm: k = =. log. (a) ( r ) = = () =. r (b) (i) t = 6. (c) (ii) So the sequence is like:, 6, 6,.. n a S r use k n n a( r ) ( 6 ) 5 Sn so S 6 = 56 letters. r 5. (a) (i) Given t 8 = t a + 7d = (a + d ) rewrite: = a d. and t = so t = a + d =. gives: d = so d =. Using you get as well a = (ii) Using you get as well a = then t 8 + t +. t + t = S S 7. Use S n = ½ n [a + (n )d] so S S 7 = {½ [ + ]} {½ 7[ + 6 ]} = 56 = 6 (b) (i) (t ) = = 6 = 5. t (ii) 7p + 7p + 7p a 7 p (iii) S r p 7 p (iv) 5 or 7p = 5 5p so 8p = 5 or p = 6 so that p = /. p 6. (a) t 5 = S 5 S = = (b) k k = + + = a r (a) (i) T = (m ) ; T = (m ) etc.so this is a geometric sequence with first term (m ) and ratio (m ). (ii) A series converges if the ratio r is: < r < or < m < add to all three sides: < m < 7

28 a The formula for the sum is: S in which a is T and r is the ratio. Substitute: r n k (b) 5 86 this forms a geometric sequence with a = 5 and r =.. k The sum formula of a geometric sequence is: S n n ( m ) ( m ) S ( m ) m a( r ) 5( ) substitute: 86 simplify: r 86 5 = n or 5 = n take log n = log 5 log n = (c) Arithmetic progression then 5 = y or = ½(5 + y). 8 y Geometric progression then y cross multiplication: 8 = y. Substitute into : 8 ½(5 + y) = y rewrite this quadratic equation to standard form: y 8y 5 = [Usually it takes time to find the proper number combination that factorizes the eq, so let us use the formula] y so y = 5 or y =.5. y = 5 gives = 5 and y =.5 gives = (a) S n =½ n[a + (n )d] = a and for the last term: t n = a + (n )d = a Write ½ n[a + (n )d] = a as ½ n[a + a + (n )d] = a or substituting the underlined part: ½n [a + a] = a or ½n a = a divide by a: ½n = so n =. (b) (i) T n = (m ) n+ so T = (m ) ; T = (m ) and T = (m ) and so on. So this is a geometric sequence with first term: a = (m ) and r = (m ). (ii) For < r < or < m < add : < m < the sum to infinity is S = a ( r ) = = a [ (m )] = a [m ] = ( m ) m. [Paper 5 Q7] a. (a) So a =. and r =.8 then S r. (b) a = and d = 6 so t n a( n) d = + 6(n ) > 5 or 6(n ) > 5 or n > 8.8 so n = 85 n k (c) 678. So n = 678 this is a GP with a = 6 and r =. k n n a( r ) 6( ) The sum formula is S n 678 or n = 86 so that n = 87 r Use logs to solve this eponential equation: n = log 87 log = 5.. (a) (i) T = ar 5 = 6 = 88 cm n 5 a( r ) [ ] 6 5 (ii) S = 6[ ] =.8 m 5 r 6 6 (b) There are terms in,, 7,, 68. [One can see this by subtracting 8 from each term.] So = ; a = and b = 8. 6 a m (c) t = m and t = 6 so r = and S 5 use cross multiplication: 5 m 6 r m 5m 5 = m or m 5m + 5 = or (m )(m 5) = so m = or m = 5. n 5 = m; multiply with m: m 8

29 Section 5 Inequalities; algebraic fractions and completion of the square.. > or > first solve = ( )( + ) = so = or = The solutions for > are now: < or >, because we are dealing here with a parabola.. (a) + 5 < first solve + 5 = or = or ( + ) 5( + ) = so that ( + )( 5) = so + 5 < for < <.5 because we are dealing here with a parabola. (b) (i) y = + = ( + + ) + = ( + + ) + + = ( + ) + so A = ; B = and C =. (ii) Range y.. (a) (i) y = + = [ + + ] + = [ + + ] + + = ( + ) + 5. So a = ; b = and c = 5 (ii) Turning point is (, 5) (iii) The coefficient of is negative so this turning point is a maimum. (b) (i) f(5) = and f() = so the range is y 5 [The coordinate of the turning point lays within the given interval!] (ii) f - does not eist because f is not a one to one function on the given domain.. (a) A B A( ) B( ) ( A B) A B ( )( ) ( )( ) (b) From (a) A B =.. and A B = then gives: A B = A B = 5A = 5 So A = and by substituting A = in you get: B = 5. (i) y y = or y b b ac y = use the formula: y 5. a The + will give y =. and the will give y =.6 (ii) + =., let p = then becomes: p p = Using the positive solution from (i) we get: = + 5 take logarithm: log( 5).8 log 6. (a) + rewrite: + factorize: ( )( + ) so for or ½ (b) (i) + = ( + + ) + = {( + ) } + = ( + ) +. So a = ; B = and C =. (ii) Turning point of y= + is (; ) y ½ 7. Algebraic solution by equating the two equations: c + = + Rewrite to standard form: c + = or ( + c) + 6 =. No solutions or D < or b ac < or ( + c) 6 < or + c + c 6 < Rewrite in standard form: c + c 6 < factorize: (c + )(c 6) <. No intersection with the curve for: < c < 6. d Using calculus: Find the derivative: dy =

30 In general the point with -coordinate a, has a tangent with gradient a. This point has y-coordinate y = a a +. So the point of touch is (a, a a + ) and the eq. of the tangent y = (a ) + c (a, a a + ) is a point on this line: a a + = (a )a + c Make c subject: c = a + and c = so a + = or a = ±. So the gradients of the line of touch to the parabola are a = m = 6 and a = m = So the line will not intersect with the curve for < c < 6. [This last conclusion can only be drawn realizing the turning point is (, ) and the line is passing through (, ).If you cannot see this, sketch the parabola] 8. (a) (i) y = + 6 = [ 6] + 6 = [( ) ] + 6 = ( ). So a = ; b = and c =. (ii) So the turning point is (, ) d y d (iii) a > so it is a minimum. Or check = which is positive. So it is a minimum. (b) (i) f(5) = = 56. So the range is y 56 (ii) f does not eist because f() = f() =. So f maps on and.. (a) < 5 or + 5 < or ( )( 5) <. Now think of the graph of y = ( )( 5); it has zeros in = ½ and = 5 so < 5 for ½ < < 5. (b) (i) The epression + is identical to a( + p) + q = a + ap + ap + q. This means a = [Both epressions are identical so the coefficient in front of are the same.] As well: ap = or p = 6 or p =. Finally = ap + q or = ( ) ( ) + q or q = 5. (ii) The epression can be written as ( ) + 5 so the turning point is (; 5).