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Mathematics Advanced Solutions GitHub Pages
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1 A 3 D 5 B 7 C 9 D,2 C 4 A 6 B 8 B 10 A,Working Justification. Question 1,Substitute the values in to obtain the equation. 16000 1000 1,Looking at the choices available the answer is A. Question 2, A is not necessarily true because we could construct an isosceles triangle that has angles 45 67 5. and 67 5 which is not a right angled triangle, B is not necessarily true because from the requirement of SAS if the corresponding angle is not the.
included angle of the two equal corresponding sides then the two triangles are not always congruent. D is not necessarily true because if the hexagon is irregular the interior angle is not necessarily 120. C is always correct as any pairs of diagonals which bisect each other in a quadrilateral will define. two sets of congruent triangles where we can deduce that equal sides must be parallel due to alternate. Question 3, Note that f 0 x g 0 x is equivalent to f x g x c i e f x g x c In other words if the. difference between f x and g x is a constant then they both must have the same derivative. For A f x g x sin x cos x 2 2 sin x cos x sin2 x cos2 x 1. For B f x g x tan2 x sec2 x 1,For C f x g x loge x loge 5x loge 5. For D f x g x sin2 x cos2 x 2 cos2 x 1 which cannot be simplified further into a con. stant Hence the answer is D,Question 4, First find the gradient of the secant between the points x0 f x0 and x0 h f x0 h then let h 0. to find the derivative by first principles Hence the expression is. f x0 h f x0,f 0 x0 lim,h 0 x0 h x0,f x0 h f x0,So the correct answer is A. Question 5, The quadratic polynomial has the general equation f x ax2 bx c Stationary points occur when.
f 0 x 0 which means,Hence f 0 0 so the answer is B. Question 6,Note that the gradients of the two lines are and. Rearranging A suggests that so the lines must be parallel which is consistent with the. conclusion, Rearranging C suggests that 1 so the lines must intersect at right angles which is consis. tent with the conclusion, If the coefficients are exactly equal on both lines then the lines have the same equation and must co. incide so D is consistent with the conclusion, If the coefficients are not equal this does not necessarily mean the two lines cannot coincide The lines can.
coincide if the coefficients are in the same ratio but not necessarily equal For example x y 1 0 and. 2x 2y 2 0 have coefficients which are not equal yet the lines coincide Hence B is not necessarily. Question 7,x 2 2 x2 dx which then simplifies to the. From sketching the region defined the area of R is. integral 4x 4 dx From sketching each of the choices it can be seen that A and B have no. resemblance to the region R From the remaining Z 1choices C and D we see that C is the correct. answer because the area of the region is given by 4x 4 dx. Question 8, A is not necessarily correct because the integral f 0 t dt is actually the amount of water that has. leaked from the bucket not the water remaining in the bucket. C is not necessarily correct because whilst the rate is f 0 t at time t the volume that has leaked is. f t c for some constant c which could possibly be non zero. D is not necessarily correct because when t T the value of f 0 T is not necessarily zero. B is correct because the volume that has leaked out of the bucket completely is given by f 0 t dt. which is f T f 0,Question 9, From a sketch it is can be seen that P S is in fact a vertical line and therefore has the same x coordinate. as S which is x a substitute this back into the equation y 2 4ax to get y 2a or y 2a which. suggests that D is correct,Question 10, If A and B are mutually exclusive then it is impossible for outcomes A and B to occur together Hence. the answer is A Note that C is actually the probability that either outcome A OR outcome B occurs. when they are mutually exclusive,Section II,Question 11.
a Using the change of base law to common base 4,log2 x log3 x log4 x. log4 x log4 x,log4 2 log4 3,log4 x 1 0,log4 x 0 or log4 x log4 3. b Using the fact that cos 90 x sin x,sin 0 sin 1 sin 2 sin 90 sin 0 sin 1 sin 2 sin 90. cos 0 cos 1 cos 2 cos 90 sin 90 sin 89 sin 88 sin 0. sin 0 sin 1 sin 2 sin 90,sin 0 sin 1 sin 2 sin 90,sin3 x cos3 x sin x cos x. sin x cos x sin2 x sin x cos x cos2 x sin x cos x 0. sin x cos x sin2 x sin x cos x cos2 x 1 0,sin x cos x sin x cos x 0.
Consider the cases,3 sin x cos x, Combining cases 1 2 and 3 to get full set of solutions x 0 2. d i Given that,Hence P P0 ekt satisfies kP,ii When t t0 P aP0 so. aP0 P0 ekt0,Substitute this back into P, e The shortest distance to the lines ax by 0 and bx ay 0 for the point P x y must be equal so. ax by bx ay,a2 b2 a2 b 2,ax by bx ay,ax by bx ay,a b x a b y. OR ax by bx ay,a b x a b y, Hence the locus is y x which is shown in the sketch below.
f If x2 xy y 2 0 then x y x2 xy y 2 0 or equivalently x3 y 3 1 Also note that. x2015 y 2015,x y x y x y 2015,x2015 y 2015,2015 substitute results 1 and 2. Question 12,a loga b by change of base,loga x loga b. b By distance formula, Note that OB BA as 4ABO is an isosceles triangle Using Pythagoras s Theorem. OB 2 OA2 40,OB 2 OA2 20,Using distance formula for OB and BA. OB 2 x2 y 2,BA2 x 6 2 y 2 2,But since OB 2 OA2 20,x2 y 2 12x 4y 40 20.
This implies that,y 10 3x sub this into x2 y 2 20,x2 10 3x 2 20. 10x2 60x 80 0,When x 2 y 4,When x 4 y 2, However from the diagram the point B lies in the first quadrant hence B has coordinates 2 4. ii Let VC be the volume of the cylinder with radius. x 1 and height 2 and VX be the volume of the solid. formed by rotating the region bounded by y tan the line x 2 and the x axis about the x axis. VX tan2 dx,Volume of solid VC VX, d i Substituting the equation of the tangent y mx b into the parabola x2 4ay. x2 4a mx b,x2 4amx 4ab 0, Since the line y mx b is a tangent to the parabola then 0. 16a2 m2 4 1 4ab, ii For the y intercept to lie on the directrix then b a.
Equations of possible tangents are y x a,Question 13. p q 2 p q 2,p2 2pq q 2 p2 2 p q q 2,This means that pq 0. ii If we let p x a and q b x then,x a x b a b,From part i this implies that pq 0 hence. b i The angle that the line joining 0 and T p q makes with the positive x axis is 3. So the gradient of that line is tan 3 3,The equation of this line is. y q 3 x p sub 0, Similarly the line joining 0 and T p q makes with the positive x axis is 3.
ii Noting that, Since the quadratic polynomial is monic then P x x2 2px p2. Since ABZ is equilateral then AZ BZ 1,Also AD BC opposite sides of parallelogram ABCD. Since X and Y are the centres of equilateral triangles P AD and QBC respectively then. The distance from the centres X and Y to the vertices of P AD and QBC respectively are equal. XA bisects P AD and Y B bisects CBQ, From the first point it can be deduced that AX BY 2. From the second point it can be deduced that,Let ABC x Since ABCD is a parallelogram.


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