Find the equation of the normal to the curve \r\n5<\/mn>x<\/mi>y<\/mi>2<\/mn><\/msup>–<\/mo>2<\/mn>y<\/mi>2<\/mn><\/msup>=<\/mo>18<\/mn><\/math>\r\n at the point (1, 2) .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ2.\u00a0 \u00a0[N09.P1]- 7 marks<\/h6>\r\nA certain population can be modelled by the differential equation \r\nd<\/mi>y<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac>=<\/mo>k<\/mi> <\/mo>y<\/mi> <\/mo>cos<\/mi>k<\/mi>t<\/mi><\/math>\r\n ,\u00a0 where \r\ny<\/mi><\/math>\r\n is the population at time \r\nt<\/mi><\/math>\r\n hours and \r\nk<\/mi><\/math>\r\n is a positive constant.\r\n(a)\u00a0 Given that \r\ny<\/mi>=<\/mo>y<\/mi>0<\/mn><\/msub><\/math>\r\n when \r\nt<\/mi>=<\/mo>0<\/mn><\/math>\r\n , express \r\ny<\/mi><\/math>\r\n in terms of \r\nk<\/mi><\/math>\r\n , \r\nt<\/mi><\/math>\r\n and \r\ny<\/mi>0<\/mn><\/msub><\/math>\r\n .\r\n(b) Find the ratio of the minimum size of the population to the maximum size of the population.\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ3.\u00a0 \u00a0[M09.P1]-5 marks<\/h6>\r\nThe diagram below shows a curve with equation \r\ny<\/mi>=<\/mo>1<\/mn>+<\/mo>k<\/mi> <\/mo>sin<\/mi>x<\/mi><\/math>\r\n , defined for \r\n0<\/mn>≤<\/mo>x<\/mi>≤<\/mo>3<\/mn>π<\/mi><\/math>\r\n .\r\nThe point lies on the curve and B \r\na<\/mi>,<\/mo>b<\/mi><\/mrow><\/mfenced><\/math>\r\n is the maximum point.\r\n(a) Show that \r\nk<\/mi>=<\/mo>–<\/mo>6<\/mn><\/math>\r\n .\r\n(b) Hence, find the values of \r\na<\/mi><\/math>\r\n and \r\nb<\/mi><\/math>\r\n .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [3 marks]\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ4.\u00a0 \u00a0[N10.P1]- 8 marks<\/h6>\r\nConsider the curve \r\ny<\/mi>=<\/mo>x<\/mi>e<\/mi>x<\/mi><\/msup><\/math>\r\n and the line \r\ny<\/mi>=<\/mo>k<\/mi>x<\/mi>,<\/mo>k<\/mi>∈<\/mo>R<\/mi><\/math>\r\n (a) Let \r\nk<\/mi>=<\/mo>0<\/mn><\/math>\r\n .<\/p>\r\n(l)Show that the curve and the line intersect once.<\/p>\r\n(ll)Find the angle between the tangent to the curve and the line at the point of intersection.<\/p>\r\n(b)Let \r\nk<\/mi>=<\/mo>1<\/mn><\/math>\r\n .\u00a0Show that the line is a tangent to the curve.\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ5.\u00a0 \u00a0[M10.P1]- 8 marks<\/h6>\r\nThe function f is defined by \r\nf<\/mi>x<\/mi><\/mfenced>=<\/mo>e<\/mi>x<\/mi>2<\/mn><\/msup>–<\/mo>2<\/mn>x<\/mi>–<\/mo>1<\/mn>.<\/mo>5<\/mn><\/mrow><\/msup><\/math>\r\n .<\/p>\r\n(a) Find \r\nf<\/mi>‘<\/mo><\/msup>x<\/mi><\/mfenced><\/math>\r\n .<\/p>\r\n(b) You are given that \r\ny<\/mi>=<\/mo>f<\/mi>x<\/mi><\/mfenced><\/mrow>x<\/mi>–<\/mo>1<\/mn><\/mrow><\/mfrac><\/math>\r\n has a local minimum at \r\nx<\/mi>=<\/mo>a<\/mi><\/math>\r\n , \r\na<\/mi>><\/mo>1<\/mn><\/math>\r\n . Find the value of \r\na<\/mi><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ6.\u00a0 \u00a0[M14.P1]- 9 marks<\/h6>\r\nA curve has equation \r\na<\/mi>r<\/mi>c<\/mi>tan<\/mi>x<\/mi>2<\/mn><\/msup>+<\/mo>a<\/mi>r<\/mi>c<\/mi>tan<\/mi>y<\/mi>2<\/mn><\/msup>=<\/mo>π<\/mi>4<\/mn><\/mfrac>.<\/mo><\/math>\r\n<\/p>\r\n(a) Find \r\nd<\/mi>y<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac><\/math>\r\n in terms of \r\nx<\/mi><\/math>\r\n \u00a0and \r\ny<\/mi><\/math>\r\n .<\/p>\r\n(b) Find the gradient of the curve at the point where \r\nx<\/mi>=<\/mo>1<\/mn>2<\/mn><\/msqrt><\/mfrac><\/math>\r\n and<\/p>\r\n\r\ny<\/mi><<\/mo>0<\/mn><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ7.\u00a0 \u00a0[N14.P1]- 6 marks<\/h6>\r\nA tranquilizer is injected into a muscle from which it enters the bloodstream. The concentration \u00a0<\/span>\r\nC<\/mi><\/math>\r\n in \r\nm<\/mi>g<\/mi>l<\/mi>–<\/mo>1<\/mn><\/mrow><\/msup><\/math>\r\n , of tranquilizer in the bloodstream can be modelled by the function \r\nC<\/mi>t<\/mi><\/mfenced>=<\/mo>
A certain population can be modelled by the differential equation \r\nd<\/mi>y<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac>=<\/mo>k<\/mi> <\/mo>y<\/mi> <\/mo>cos<\/mi>k<\/mi>t<\/mi><\/math>\r\n ,\u00a0 where \r\ny<\/mi><\/math>\r\n is the population at time \r\nt<\/mi><\/math>\r\n hours and \r\nk<\/mi><\/math>\r\n is a positive constant.\r\n(a)\u00a0 Given that \r\ny<\/mi>=<\/mo>y<\/mi>0<\/mn><\/msub><\/math>\r\n when \r\nt<\/mi>=<\/mo>0<\/mn><\/math>\r\n , express \r\ny<\/mi><\/math>\r\n in terms of \r\nk<\/mi><\/math>\r\n , \r\nt<\/mi><\/math>\r\n and \r\ny<\/mi>0<\/mn><\/msub><\/math>\r\n .\r\n(b) Find the ratio of the minimum size of the population to the maximum size of the population.\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ3.\u00a0 \u00a0[M09.P1]-5 marks<\/h6>\r\nThe diagram below shows a curve with equation \r\ny<\/mi>=<\/mo>1<\/mn>+<\/mo>k<\/mi> <\/mo>sin<\/mi>x<\/mi><\/math>\r\n , defined for \r\n0<\/mn>≤<\/mo>x<\/mi>≤<\/mo>3<\/mn>π<\/mi><\/math>\r\n .\r\nThe point lies on the curve and B \r\na<\/mi>,<\/mo>b<\/mi><\/mrow><\/mfenced><\/math>\r\n is the maximum point.\r\n(a) Show that \r\nk<\/mi>=<\/mo>–<\/mo>6<\/mn><\/math>\r\n .\r\n(b) Hence, find the values of \r\na<\/mi><\/math>\r\n and \r\nb<\/mi><\/math>\r\n .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [3 marks]\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ4.\u00a0 \u00a0[N10.P1]- 8 marks<\/h6>\r\nConsider the curve \r\ny<\/mi>=<\/mo>x<\/mi>e<\/mi>x<\/mi><\/msup><\/math>\r\n and the line \r\ny<\/mi>=<\/mo>k<\/mi>x<\/mi>,<\/mo>k<\/mi>∈<\/mo>R<\/mi><\/math>\r\n (a) Let \r\nk<\/mi>=<\/mo>0<\/mn><\/math>\r\n .<\/p>\r\n(l)Show that the curve and the line intersect once.<\/p>\r\n(ll)Find the angle between the tangent to the curve and the line at the point of intersection.<\/p>\r\n(b)Let \r\nk<\/mi>=<\/mo>1<\/mn><\/math>\r\n .\u00a0Show that the line is a tangent to the curve.\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ5.\u00a0 \u00a0[M10.P1]- 8 marks<\/h6>\r\nThe function f is defined by \r\nf<\/mi>x<\/mi><\/mfenced>=<\/mo>e<\/mi>x<\/mi>2<\/mn><\/msup>–<\/mo>2<\/mn>x<\/mi>–<\/mo>1<\/mn>.<\/mo>5<\/mn><\/mrow><\/msup><\/math>\r\n .<\/p>\r\n(a) Find \r\nf<\/mi>‘<\/mo><\/msup>x<\/mi><\/mfenced><\/math>\r\n .<\/p>\r\n(b) You are given that \r\ny<\/mi>=<\/mo>f<\/mi>x<\/mi><\/mfenced><\/mrow>x<\/mi>–<\/mo>1<\/mn><\/mrow><\/mfrac><\/math>\r\n has a local minimum at \r\nx<\/mi>=<\/mo>a<\/mi><\/math>\r\n , \r\na<\/mi>><\/mo>1<\/mn><\/math>\r\n . Find the value of \r\na<\/mi><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ6.\u00a0 \u00a0[M14.P1]- 9 marks<\/h6>\r\nA curve has equation \r\na<\/mi>r<\/mi>c<\/mi>tan<\/mi>x<\/mi>2<\/mn><\/msup>+<\/mo>a<\/mi>r<\/mi>c<\/mi>tan<\/mi>y<\/mi>2<\/mn><\/msup>=<\/mo>π<\/mi>4<\/mn><\/mfrac>.<\/mo><\/math>\r\n<\/p>\r\n(a) Find \r\nd<\/mi>y<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac><\/math>\r\n in terms of \r\nx<\/mi><\/math>\r\n \u00a0and \r\ny<\/mi><\/math>\r\n .<\/p>\r\n(b) Find the gradient of the curve at the point where \r\nx<\/mi>=<\/mo>1<\/mn>2<\/mn><\/msqrt><\/mfrac><\/math>\r\n and<\/p>\r\n\r\ny<\/mi><<\/mo>0<\/mn><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ7.\u00a0 \u00a0[N14.P1]- 6 marks<\/h6>\r\nA tranquilizer is injected into a muscle from which it enters the bloodstream. The concentration \u00a0<\/span>\r\nC<\/mi><\/math>\r\n in \r\nm<\/mi>g<\/mi>l<\/mi>–<\/mo>1<\/mn><\/mrow><\/msup><\/math>\r\n , of tranquilizer in the bloodstream can be modelled by the function \r\nC<\/mi>t<\/mi><\/mfenced>=<\/mo>
The diagram below shows a curve with equation \r\ny<\/mi>=<\/mo>1<\/mn>+<\/mo>k<\/mi> <\/mo>sin<\/mi>x<\/mi><\/math>\r\n , defined for \r\n0<\/mn>≤<\/mo>x<\/mi>≤<\/mo>3<\/mn>π<\/mi><\/math>\r\n .\r\nThe point lies on the curve and B \r\na<\/mi>,<\/mo>b<\/mi><\/mrow><\/mfenced><\/math>\r\n is the maximum point.\r\n(a) Show that \r\nk<\/mi>=<\/mo>–<\/mo>6<\/mn><\/math>\r\n .\r\n(b) Hence, find the values of \r\na<\/mi><\/math>\r\n and \r\nb<\/mi><\/math>\r\n .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [3 marks]\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ4.\u00a0 \u00a0[N10.P1]- 8 marks<\/h6>\r\nConsider the curve \r\ny<\/mi>=<\/mo>x<\/mi>e<\/mi>x<\/mi><\/msup><\/math>\r\n and the line \r\ny<\/mi>=<\/mo>k<\/mi>x<\/mi>,<\/mo>k<\/mi>∈<\/mo>R<\/mi><\/math>\r\n (a) Let \r\nk<\/mi>=<\/mo>0<\/mn><\/math>\r\n .<\/p>\r\n(l)Show that the curve and the line intersect once.<\/p>\r\n(ll)Find the angle between the tangent to the curve and the line at the point of intersection.<\/p>\r\n(b)Let \r\nk<\/mi>=<\/mo>1<\/mn><\/math>\r\n .\u00a0Show that the line is a tangent to the curve.\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ5.\u00a0 \u00a0[M10.P1]- 8 marks<\/h6>\r\nThe function f is defined by \r\nf<\/mi>x<\/mi><\/mfenced>=<\/mo>e<\/mi>x<\/mi>2<\/mn><\/msup>–<\/mo>2<\/mn>x<\/mi>–<\/mo>1<\/mn>.<\/mo>5<\/mn><\/mrow><\/msup><\/math>\r\n .<\/p>\r\n(a) Find \r\nf<\/mi>‘<\/mo><\/msup>x<\/mi><\/mfenced><\/math>\r\n .<\/p>\r\n(b) You are given that \r\ny<\/mi>=<\/mo>f<\/mi>x<\/mi><\/mfenced><\/mrow>x<\/mi>–<\/mo>1<\/mn><\/mrow><\/mfrac><\/math>\r\n has a local minimum at \r\nx<\/mi>=<\/mo>a<\/mi><\/math>\r\n , \r\na<\/mi>><\/mo>1<\/mn><\/math>\r\n . Find the value of \r\na<\/mi><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ6.\u00a0 \u00a0[M14.P1]- 9 marks<\/h6>\r\nA curve has equation \r\na<\/mi>r<\/mi>c<\/mi>tan<\/mi>x<\/mi>2<\/mn><\/msup>+<\/mo>a<\/mi>r<\/mi>c<\/mi>tan<\/mi>y<\/mi>2<\/mn><\/msup>=<\/mo>π<\/mi>4<\/mn><\/mfrac>.<\/mo><\/math>\r\n<\/p>\r\n(a) Find \r\nd<\/mi>y<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac><\/math>\r\n in terms of \r\nx<\/mi><\/math>\r\n \u00a0and \r\ny<\/mi><\/math>\r\n .<\/p>\r\n(b) Find the gradient of the curve at the point where \r\nx<\/mi>=<\/mo>1<\/mn>2<\/mn><\/msqrt><\/mfrac><\/math>\r\n and<\/p>\r\n\r\ny<\/mi><<\/mo>0<\/mn><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ7.\u00a0 \u00a0[N14.P1]- 6 marks<\/h6>\r\nA tranquilizer is injected into a muscle from which it enters the bloodstream. The concentration \u00a0<\/span>\r\nC<\/mi><\/math>\r\n in \r\nm<\/mi>g<\/mi>l<\/mi>–<\/mo>1<\/mn><\/mrow><\/msup><\/math>\r\n , of tranquilizer in the bloodstream can be modelled by the function \r\nC<\/mi>t<\/mi><\/mfenced>=<\/mo>
Consider the curve \r\ny<\/mi>=<\/mo>x<\/mi>e<\/mi>x<\/mi><\/msup><\/math>\r\n and the line \r\ny<\/mi>=<\/mo>k<\/mi>x<\/mi>,<\/mo>k<\/mi>∈<\/mo>R<\/mi><\/math>\r\n (a) Let \r\nk<\/mi>=<\/mo>0<\/mn><\/math>\r\n .<\/p>\r\n(l)Show that the curve and the line intersect once.<\/p>\r\n(ll)Find the angle between the tangent to the curve and the line at the point of intersection.<\/p>\r\n(b)Let \r\nk<\/mi>=<\/mo>1<\/mn><\/math>\r\n .\u00a0Show that the line is a tangent to the curve.\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ5.\u00a0 \u00a0[M10.P1]- 8 marks<\/h6>\r\nThe function f is defined by \r\nf<\/mi>x<\/mi><\/mfenced>=<\/mo>e<\/mi>x<\/mi>2<\/mn><\/msup>–<\/mo>2<\/mn>x<\/mi>–<\/mo>1<\/mn>.<\/mo>5<\/mn><\/mrow><\/msup><\/math>\r\n .<\/p>\r\n(a) Find \r\nf<\/mi>‘<\/mo><\/msup>x<\/mi><\/mfenced><\/math>\r\n .<\/p>\r\n(b) You are given that \r\ny<\/mi>=<\/mo>f<\/mi>x<\/mi><\/mfenced><\/mrow>x<\/mi>–<\/mo>1<\/mn><\/mrow><\/mfrac><\/math>\r\n has a local minimum at \r\nx<\/mi>=<\/mo>a<\/mi><\/math>\r\n , \r\na<\/mi>><\/mo>1<\/mn><\/math>\r\n . Find the value of \r\na<\/mi><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ6.\u00a0 \u00a0[M14.P1]- 9 marks<\/h6>\r\nA curve has equation \r\na<\/mi>r<\/mi>c<\/mi>tan<\/mi>x<\/mi>2<\/mn><\/msup>+<\/mo>a<\/mi>r<\/mi>c<\/mi>tan<\/mi>y<\/mi>2<\/mn><\/msup>=<\/mo>π<\/mi>4<\/mn><\/mfrac>.<\/mo><\/math>\r\n<\/p>\r\n(a) Find \r\nd<\/mi>y<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac><\/math>\r\n in terms of \r\nx<\/mi><\/math>\r\n \u00a0and \r\ny<\/mi><\/math>\r\n .<\/p>\r\n(b) Find the gradient of the curve at the point where \r\nx<\/mi>=<\/mo>1<\/mn>2<\/mn><\/msqrt><\/mfrac><\/math>\r\n and<\/p>\r\n\r\ny<\/mi><<\/mo>0<\/mn><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ7.\u00a0 \u00a0[N14.P1]- 6 marks<\/h6>\r\nA tranquilizer is injected into a muscle from which it enters the bloodstream. The concentration \u00a0<\/span>\r\nC<\/mi><\/math>\r\n in \r\nm<\/mi>g<\/mi>l<\/mi>–<\/mo>1<\/mn><\/mrow><\/msup><\/math>\r\n , of tranquilizer in the bloodstream can be modelled by the function \r\nC<\/mi>t<\/mi><\/mfenced>=<\/mo>
(l)Show that the curve and the line intersect once.<\/p>\r\n
(ll)Find the angle between the tangent to the curve and the line at the point of intersection.<\/p>\r\n
(b)Let \r\nk<\/mi>=<\/mo>1<\/mn><\/math>\r\n .\u00a0Show that the line is a tangent to the curve.\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ5.\u00a0 \u00a0[M10.P1]- 8 marks<\/h6>\r\nThe function f is defined by \r\nf<\/mi>x<\/mi><\/mfenced>=<\/mo>e<\/mi>x<\/mi>2<\/mn><\/msup>–<\/mo>2<\/mn>x<\/mi>–<\/mo>1<\/mn>.<\/mo>5<\/mn><\/mrow><\/msup><\/math>\r\n .<\/p>\r\n(a) Find \r\nf<\/mi>‘<\/mo><\/msup>x<\/mi><\/mfenced><\/math>\r\n .<\/p>\r\n(b) You are given that \r\ny<\/mi>=<\/mo>f<\/mi>x<\/mi><\/mfenced><\/mrow>x<\/mi>–<\/mo>1<\/mn><\/mrow><\/mfrac><\/math>\r\n has a local minimum at \r\nx<\/mi>=<\/mo>a<\/mi><\/math>\r\n , \r\na<\/mi>><\/mo>1<\/mn><\/math>\r\n . Find the value of \r\na<\/mi><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ6.\u00a0 \u00a0[M14.P1]- 9 marks<\/h6>\r\nA curve has equation \r\na<\/mi>r<\/mi>c<\/mi>tan<\/mi>x<\/mi>2<\/mn><\/msup>+<\/mo>a<\/mi>r<\/mi>c<\/mi>tan<\/mi>y<\/mi>2<\/mn><\/msup>=<\/mo>π<\/mi>4<\/mn><\/mfrac>.<\/mo><\/math>\r\n<\/p>\r\n(a) Find \r\nd<\/mi>y<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac><\/math>\r\n in terms of \r\nx<\/mi><\/math>\r\n \u00a0and \r\ny<\/mi><\/math>\r\n .<\/p>\r\n(b) Find the gradient of the curve at the point where \r\nx<\/mi>=<\/mo>1<\/mn>2<\/mn><\/msqrt><\/mfrac><\/math>\r\n and<\/p>\r\n\r\ny<\/mi><<\/mo>0<\/mn><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ7.\u00a0 \u00a0[N14.P1]- 6 marks<\/h6>\r\nA tranquilizer is injected into a muscle from which it enters the bloodstream. The concentration \u00a0<\/span>\r\nC<\/mi><\/math>\r\n in \r\nm<\/mi>g<\/mi>l<\/mi>–<\/mo>1<\/mn><\/mrow><\/msup><\/math>\r\n , of tranquilizer in the bloodstream can be modelled by the function \r\nC<\/mi>t<\/mi><\/mfenced>=<\/mo>
The function f is defined by \r\nf<\/mi>x<\/mi><\/mfenced>=<\/mo>e<\/mi>x<\/mi>2<\/mn><\/msup>–<\/mo>2<\/mn>x<\/mi>–<\/mo>1<\/mn>.<\/mo>5<\/mn><\/mrow><\/msup><\/math>\r\n .<\/p>\r\n(a) Find \r\nf<\/mi>‘<\/mo><\/msup>x<\/mi><\/mfenced><\/math>\r\n .<\/p>\r\n(b) You are given that \r\ny<\/mi>=<\/mo>f<\/mi>x<\/mi><\/mfenced><\/mrow>x<\/mi>–<\/mo>1<\/mn><\/mrow><\/mfrac><\/math>\r\n has a local minimum at \r\nx<\/mi>=<\/mo>a<\/mi><\/math>\r\n , \r\na<\/mi>><\/mo>1<\/mn><\/math>\r\n . Find the value of \r\na<\/mi><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ6.\u00a0 \u00a0[M14.P1]- 9 marks<\/h6>\r\nA curve has equation \r\na<\/mi>r<\/mi>c<\/mi>tan<\/mi>x<\/mi>2<\/mn><\/msup>+<\/mo>a<\/mi>r<\/mi>c<\/mi>tan<\/mi>y<\/mi>2<\/mn><\/msup>=<\/mo>π<\/mi>4<\/mn><\/mfrac>.<\/mo><\/math>\r\n<\/p>\r\n(a) Find \r\nd<\/mi>y<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac><\/math>\r\n in terms of \r\nx<\/mi><\/math>\r\n \u00a0and \r\ny<\/mi><\/math>\r\n .<\/p>\r\n(b) Find the gradient of the curve at the point where \r\nx<\/mi>=<\/mo>1<\/mn>2<\/mn><\/msqrt><\/mfrac><\/math>\r\n and<\/p>\r\n\r\ny<\/mi><<\/mo>0<\/mn><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ7.\u00a0 \u00a0[N14.P1]- 6 marks<\/h6>\r\nA tranquilizer is injected into a muscle from which it enters the bloodstream. The concentration \u00a0<\/span>\r\nC<\/mi><\/math>\r\n in \r\nm<\/mi>g<\/mi>l<\/mi>–<\/mo>1<\/mn><\/mrow><\/msup><\/math>\r\n , of tranquilizer in the bloodstream can be modelled by the function \r\nC<\/mi>t<\/mi><\/mfenced>=<\/mo>
(a) Find \r\nf<\/mi>‘<\/mo><\/msup>x<\/mi><\/mfenced><\/math>\r\n .<\/p>\r\n(b) You are given that \r\ny<\/mi>=<\/mo>f<\/mi>x<\/mi><\/mfenced><\/mrow>x<\/mi>–<\/mo>1<\/mn><\/mrow><\/mfrac><\/math>\r\n has a local minimum at \r\nx<\/mi>=<\/mo>a<\/mi><\/math>\r\n , \r\na<\/mi>><\/mo>1<\/mn><\/math>\r\n . Find the value of \r\na<\/mi><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ6.\u00a0 \u00a0[M14.P1]- 9 marks<\/h6>\r\nA curve has equation \r\na<\/mi>r<\/mi>c<\/mi>tan<\/mi>x<\/mi>2<\/mn><\/msup>+<\/mo>a<\/mi>r<\/mi>c<\/mi>tan<\/mi>y<\/mi>2<\/mn><\/msup>=<\/mo>π<\/mi>4<\/mn><\/mfrac>.<\/mo><\/math>\r\n<\/p>\r\n(a) Find \r\nd<\/mi>y<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac><\/math>\r\n in terms of \r\nx<\/mi><\/math>\r\n \u00a0and \r\ny<\/mi><\/math>\r\n .<\/p>\r\n(b) Find the gradient of the curve at the point where \r\nx<\/mi>=<\/mo>1<\/mn>2<\/mn><\/msqrt><\/mfrac><\/math>\r\n and<\/p>\r\n\r\ny<\/mi><<\/mo>0<\/mn><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ7.\u00a0 \u00a0[N14.P1]- 6 marks<\/h6>\r\nA tranquilizer is injected into a muscle from which it enters the bloodstream. The concentration \u00a0<\/span>\r\nC<\/mi><\/math>\r\n in \r\nm<\/mi>g<\/mi>l<\/mi>–<\/mo>1<\/mn><\/mrow><\/msup><\/math>\r\n , of tranquilizer in the bloodstream can be modelled by the function \r\nC<\/mi>t<\/mi><\/mfenced>=<\/mo>
(b) You are given that \r\ny<\/mi>=<\/mo>f<\/mi>x<\/mi><\/mfenced><\/mrow>x<\/mi>–<\/mo>1<\/mn><\/mrow><\/mfrac><\/math>\r\n has a local minimum at \r\nx<\/mi>=<\/mo>a<\/mi><\/math>\r\n , \r\na<\/mi>><\/mo>1<\/mn><\/math>\r\n . Find the value of \r\na<\/mi><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ6.\u00a0 \u00a0[M14.P1]- 9 marks<\/h6>\r\nA curve has equation \r\na<\/mi>r<\/mi>c<\/mi>tan<\/mi>x<\/mi>2<\/mn><\/msup>+<\/mo>a<\/mi>r<\/mi>c<\/mi>tan<\/mi>y<\/mi>2<\/mn><\/msup>=<\/mo>π<\/mi>4<\/mn><\/mfrac>.<\/mo><\/math>\r\n<\/p>\r\n(a) Find \r\nd<\/mi>y<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac><\/math>\r\n in terms of \r\nx<\/mi><\/math>\r\n \u00a0and \r\ny<\/mi><\/math>\r\n .<\/p>\r\n(b) Find the gradient of the curve at the point where \r\nx<\/mi>=<\/mo>1<\/mn>2<\/mn><\/msqrt><\/mfrac><\/math>\r\n and<\/p>\r\n\r\ny<\/mi><<\/mo>0<\/mn><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ7.\u00a0 \u00a0[N14.P1]- 6 marks<\/h6>\r\nA tranquilizer is injected into a muscle from which it enters the bloodstream. The concentration \u00a0<\/span>\r\nC<\/mi><\/math>\r\n in \r\nm<\/mi>g<\/mi>l<\/mi>–<\/mo>1<\/mn><\/mrow><\/msup><\/math>\r\n , of tranquilizer in the bloodstream can be modelled by the function \r\nC<\/mi>t<\/mi><\/mfenced>=<\/mo>
A curve has equation \r\na<\/mi>r<\/mi>c<\/mi>tan<\/mi>x<\/mi>2<\/mn><\/msup>+<\/mo>a<\/mi>r<\/mi>c<\/mi>tan<\/mi>y<\/mi>2<\/mn><\/msup>=<\/mo>π<\/mi>4<\/mn><\/mfrac>.<\/mo><\/math>\r\n<\/p>\r\n(a) Find \r\nd<\/mi>y<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac><\/math>\r\n in terms of \r\nx<\/mi><\/math>\r\n \u00a0and \r\ny<\/mi><\/math>\r\n .<\/p>\r\n(b) Find the gradient of the curve at the point where \r\nx<\/mi>=<\/mo>1<\/mn>2<\/mn><\/msqrt><\/mfrac><\/math>\r\n and<\/p>\r\n\r\ny<\/mi><<\/mo>0<\/mn><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ7.\u00a0 \u00a0[N14.P1]- 6 marks<\/h6>\r\nA tranquilizer is injected into a muscle from which it enters the bloodstream. The concentration \u00a0<\/span>\r\nC<\/mi><\/math>\r\n in \r\nm<\/mi>g<\/mi>l<\/mi>–<\/mo>1<\/mn><\/mrow><\/msup><\/math>\r\n , of tranquilizer in the bloodstream can be modelled by the function \r\nC<\/mi>t<\/mi><\/mfenced>=<\/mo>
(a) Find \r\nd<\/mi>y<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac><\/math>\r\n in terms of \r\nx<\/mi><\/math>\r\n \u00a0and \r\ny<\/mi><\/math>\r\n .<\/p>\r\n(b) Find the gradient of the curve at the point where \r\nx<\/mi>=<\/mo>1<\/mn>2<\/mn><\/msqrt><\/mfrac><\/math>\r\n and<\/p>\r\n\r\ny<\/mi><<\/mo>0<\/mn><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ7.\u00a0 \u00a0[N14.P1]- 6 marks<\/h6>\r\nA tranquilizer is injected into a muscle from which it enters the bloodstream. The concentration \u00a0<\/span>\r\nC<\/mi><\/math>\r\n in \r\nm<\/mi>g<\/mi>l<\/mi>–<\/mo>1<\/mn><\/mrow><\/msup><\/math>\r\n , of tranquilizer in the bloodstream can be modelled by the function \r\nC<\/mi>t<\/mi><\/mfenced>=<\/mo>
(b) Find the gradient of the curve at the point where \r\nx<\/mi>=<\/mo>1<\/mn>2<\/mn><\/msqrt><\/mfrac><\/math>\r\n and<\/p>\r\n\r\ny<\/mi><<\/mo>0<\/mn><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ7.\u00a0 \u00a0[N14.P1]- 6 marks<\/h6>\r\nA tranquilizer is injected into a muscle from which it enters the bloodstream. The concentration \u00a0<\/span>\r\nC<\/mi><\/math>\r\n in \r\nm<\/mi>g<\/mi>l<\/mi>–<\/mo>1<\/mn><\/mrow><\/msup><\/math>\r\n , of tranquilizer in the bloodstream can be modelled by the function \r\nC<\/mi>t<\/mi><\/mfenced>=<\/mo>
\r\ny<\/mi><<\/mo>0<\/mn><\/math>\r\n .\r\nSolution<\/span><\/a><\/div>\r\n\r\n<\/div><\/p>\r\nQ7.\u00a0 \u00a0[N14.P1]- 6 marks<\/h6>\r\nA tranquilizer is injected into a muscle from which it enters the bloodstream. The concentration \u00a0<\/span>\r\nC<\/mi><\/math>\r\n in \r\nm<\/mi>g<\/mi>l<\/mi>–<\/mo>1<\/mn><\/mrow><\/msup><\/math>\r\n , of tranquilizer in the bloodstream can be modelled by the function \r\nC<\/mi>t<\/mi><\/mfenced>=<\/mo>
A tranquilizer is injected into a muscle from which it enters the bloodstream. The concentration \u00a0<\/span>\r\nC<\/mi><\/math>\r\n in \r\nm<\/mi>g<\/mi>l<\/mi>–<\/mo>1<\/mn><\/mrow><\/msup><\/math>\r\n , of tranquilizer in the bloodstream can be modelled by the function \r\nC<\/mi>t<\/mi><\/mfenced>=<\/mo>
![\"\"](\"http:\/\/ibalmaths.com\/wp-content\/uploads\/2020\/02\/6A..jpg\")
![\"\"](\"http:\/\/ibalmaths.com\/wp-content\/uploads\/2020\/02\/6B-1.jpg\")