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Worksheet 11A
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Solutions
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(a)
ddx(x20)=20x19
(b)
ddx(1x4)=ddx(x−4)=−4x−5=−(41)(1x5)=−4x5
(c)
ddx√x=ddx(x12)=12x−12=(12)(1√x)=12√x
(d)
ddxx−35=−35x−85=−35(15√x8)=−355√x8
(e)
ddx(88)=0
(f)
ddx(−2.9)=0
(g)
ddx(45x3+2√x−7)=ddx(45x3+2x−12−7)=45(3)x2+2(−12)x−32−0=125x2−x−32=125x2−1√x3
(h)
ddx(63√x−πx2+x)=ddx(6x13−πx−2+x)=6(13)x−23−π(−2)x−3+1=2x−23+2πx−3+1=(21)(13√x2)+(2π1)(1x3)+1=23√x2+2πx3+1
(i)
ddx(x2−6x)=ddx(x2x−6x)=ddx(x−6x−1)=1−6(−1)x−2=1+6x−2=1+(61)(1x2)=1+6x2
(j)
ddx[(x2+1)(x−1)]=ddx(x3−x2+x−1)=3x2−2x+1
(a)
dydx=13(2)x−1=23x−1When x=6,dydx=23(6)−1=3
(b)
y=4xy=4x−1dydx=4(−1)x−2=−4x−2=(−41)(1x2)=−4x2000000[Need to find value of x to substitute in]Substitute y=−2 into eqn of curve,−2=4x−2x=4x=4−2x=−2Substitute x=−2 into dydx,dydx=−4(−2)2=−1
(c)
y=(5x−1)(2x+1)y=10x2+5x−2x−1y=10x2+3x−1dydx=10(2)x+3=20x+3When x=0,000000[y-intercept]dydx=20(0)+3=3
(d)
y=x+1x2y=xx2+1x2y=1x+1x2y=x−1+x−2dydx=(−1)x−2+(−2)x−3=−1x2−2x3000000[Need to find value of x to substitute in]Substitute y=0 into eqn of curve,000000[x-intercept]0=x+1x20=x+1−1=xSubstitute x=−1 into dydx,dydx=−1(−1)2−2(−1)3=1
y=(10−x2)2y=102−2(10)(x2)+(x2)2000000[(a−b)2=a2−2ab+b2]y=100−20x2+x4dydx=−20(2)x+4x3=−40x+4x3When dydx=0,0=−40x+4x30=4x(−10+x2)
4x=0 or −10+x2=0x=0x2=10x=±√10Substitute into eqn of curve,Substitute into eqn of curve,y=[10−(0)2]2y=[10−(±√10)2]2y=100y=0∴.(0,100)∴.(√10,0),(−√10,0)
y=ax+bx2Using (−1,−9),−9=a(−1)+b(−1)2−9=−a+b1−9=−a+ba=b+900--- (1)y=ax+bx2y=ax+bx−2dydx=a+b(−2)x−3=a−2bx−3=a−(2b1)(1x3)=a−2bx3When x=1 and dydx=12,12=a−2b(1)312=a−2b112=a−2b00--- (2)Substitute (1) into (2),12=b+9−2b−b+2b=9−12b=−3Substitute into (1),a=−3+9a=6∴a=6,b=−3
y=ax+bx2y=ax−1+bx−2dydx=a(−1)x−2+b(−2)x−3=−ax−2−2bx−3=−(a1)(1x2)−(2b1)(1x3)=−ax2−2bx3When x=14 and dydx=−3,−3=−a(14)2−2b(14)3−3=−a116−2b164−3=−16a−128b3=16a+128bLet a=1916,3=16(1916)+128b3=19+128b−16=128b−16128=b−18=b∴a=1916,b=−18
(i)
y=3−14x2dydx=−14(2)x=−12xWhen x=−4,dydx=−12(−4)=2
(ii)
Gradient at new point=−12000000[m1×m2=−1]=−12Substitute dydx=−12 into dydx=−12x,−12=−12x1=xSubstitute x=1 into eqn of curve,y=3−14(1)2y=114∴.(1,234)
(i)
y=ax3−x2+bUsing (1,14),14=a(1)3−(1)2+b14=a(1)−1+b14=a−1+b54=a+b00--- (1)dydx=a(3)x2−2x=3ax2−2xWhen x=1 and dydx=−54,−54=3a(1)2−2(1)−54=3a(1)−234=3a14=aSubstitute a=14 into (1),54=14+b1=b∴a=14,b=1
(ii)
dydx=3ax2−2x=3(14)x2−2x=34x2−2xWhen dydx=−54,−54=34x2−2x0=34x2−2x+540=3x2−8x+50=(x−1)(3x−5)
x−1=0 or 3x−5=0x=1 (Provided point)3x=5x=53
y=ax3−x2+by=14x3−x2+1When x=53,y=14(53)3−(53)2+1y=−67108∴.(123,−67108)
f′(x)=2(3)x2−15(2)x−84=6x2−30x−84f′(x)<06x2−30x−84<0x2−5x−14<0(x+2)(x−7)<0

−2<x<7
(i)
y=x2(5x−1)y=5x3−x2dydx=5(3)x2−2x=15x2−2xy−x=0.3y=x+0.3000000[y=mx+c]Gradient of line=1Substitute dydx=1 into dydx=15x2−2x,1=15x2−2x0=15x2−2x−10=(5x+1)(3x−1)
5x+1=0 or 3x−1=05x=−13x=1x=−15x=13Substitute into eqn of curve,Substitute into eqn of curve,y=(−15)2[5(−15)−1]y=(13)2[5(13)−1]y=−225y=227∴.(−15,−225)∴.(13,227)
(ii)
dydx=15x2−2xWhen dydx=−2,−2=15x2−2x0=15x2−2x+2b2−4ac=(−2)2−4(15)(2)=−116<0⟹0=15x2−2x+2 has no solutions∴No points on curve such that gradient of tangent=−2
(iii)
dydx=15x2−2x=15(x2−215x)=15[x2−215x+(115)2−(115)2]000000[Complete the square]=15[(x−115)2−1225]=15(x−115)2−115Minimum value of dydx=−115
Note: I don't think this is relevant for O levels
(i)
Let f(x)=x3f′(x)=limh→0f(x+h)−f(x)h=limh→0(x+h)3−x3h=limh→0(x2+2hx+h2)(x+h)−x3h=limh→0x3+hx2+2hx2+2h2x+h2x+h3−x3h=limh→03hx2+3h2x+h3h=limh→03x2+3hx+h2=3x2+3(0)x+02=3x2
(ii)
Let f(x)=(x+1)(x−1)=x2−1f′(x)=limh→0f(x+h)−f(x)h=limh→0(x+h)2−1−(x2−1)h=limh→0x2+2hx+h2−1−x2+1h=limh→02hx+h2h=limh→02x+h=2x+0=2x