1. - Jazzy Maths - A2 Doubles F block Friday 9th July 20
A1 Doubles Tracking Test 4 Part A(36 marks: 44 minutes)center141766001.46722775022. Figure 2 shows part of the curve with equation y=fx. The curve passes through the points P(-1.5, 0) and Q(0, 5) as shown. On separate diagrams, sketch the curve with equation y=fx (2)y=f(x) (2)Indicate clearly on each sketch the coordinates of the points at which the curve crosses or meets the axes.3. Find the integral cosxsin2x dx (3)4. Prove, from first principles, that the derivative of cos3x is -3sin3x.You may assume that as h→0,sin3hh→3 and cos3h-1h→0 (5)3581400005. (8)655321299720006. (5)END OF TESTMark Scheme TT4 Part A1.2. cosx (sinx)2 dxy=(sinx)3dydx=3cosx (sinx)2M1A1=13sin3x+cA13. Let fx=cos3xf'x=limh→0fx+h-fxh=limh→0cos?(3x+3h)-cos?(3x)h=limh→0cos3xcos3h-sin3xsin3h-cos?3xh=limh→0(cos3h-1hcos3x-sin3hhsin3x)As h→0, sin3hh→3 and cos3h-1h→0,So the expression in side the limit tends to0xcos3x-3xsin3x=-3sin3xHence the derivative of cos3xis-3sin?(3x)M1A1M1M1A14.1+sin2x dx=x-12cos2x0π3 =π3+14-(-12 ) =π3+34When x=π3, y=2+32Area of triangle =12 x π3 x 2+32 = 2π+3π12Shaded region = π3+34 - 2π+3π12 = 4π+9-2π+3π12 =1122π+9-π3 *M1 A1M1A1B1M1M1 A15.y=kx2-akx2+adydx=kx2+a2kx-(kx2-a)(2kx)kx2+a2dydx=4akxkx2+a2Turning point where dydx=0, therefore 4akx=0So when x=0, y=-aa=-1Single turning point occurs at (0,-1)M1A1A1M1A16. ................
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