Solutions
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Question 1
(a)
−4<x<7
(b)
x<−0.4 or x>3
(c)
2x2+7x≥42x2+7x−4≥0(x+4)(2x−1)≥0
x≤−4 or x≥0.5
(d)
3x2≤x2−x+32x2+x−3≤0(2x+3)(x−1)≤0
−1.5≤x≤1
Question 2 - Expression is always positive/negative
(i)
2px2+7x+18pa>000000[Minimum curve ∪]2p>0p>02p>000000[Condition #1]b2−4ac=(7)2−4(2p)(18p)=49−8p(18p)=49−144p2b2−4ac<000000[No real roots]49−144p2<0144p2−49>0(12p)2−(7)2>0(12p+7)(12p−7)>0
p<−712 or p>71200000[Condition #2]∴p>71200000[Satisfy conditions #1 & #2]
(ii)
2px2+7x+18pa<000000[Maximum curve ∩]2p<0p<02p<000000[Condition #1]b2−4ac<000000[No real roots][Same as part (i)]p<−712 or p>71200000[Condition #2]∴p<−712
Question 3 - Expression is always negative
−x2+qx+(3q+8)a=−1<000000[Maximum curve ∩]b2−4ac=(q)2−4(−1)(3q+8)=q2+4(3q+8)=q2+12q+32b2−4ac<000000[No real roots]q2+12q+32<0(q+8)(q+4)<0
−8<q<−4
∴Smallest integer value of q=−7
Question 4
(i)
y=x+k000 --- (1)y=7x−kx2000 --- (2)Substitute (1) into (2),x+k=7x−kx2kx2−6x+k=0b2−4ac=(−6)2−4(k)(k)=36−4k2b2−4ac≥000000[Real roots since line intersects the curve - can be once or twice]36−4k2≥04k2−36≤0k2−9≤0(k+3)(k−3)≤0
−3≤k≤3
(ii)
b2−4ac=000000[1 real root since line is tangent to curve]36−4k2=0−4k2=−36k2=−36−4k2=9k=±√9k=±3
Question 5
x(5x−3)<2x(x−4)+25x2−3x<2x2−8x+23x2+5x−2<0(x+2)(3x−1)<0
−2<x<13
Question 6
Consider (3x−5)2−(32)2=0(3x−5)2=(32)23x−5=±√(32)23x−5=±1.53x=±1.5+53x=6.5 or 3.5x=136 or 76∴(3x−5)2−(32)2≥0(x−76)(x−136)≥0
x≤76 or x≥136
Question 7 - Solve simultaneous inequality
Question 8 - Expression is always positive
x2+2k+10=x−3kxx2+3kx−x+2k+10=0x2+(3k−1)x+(2k+10)=0a=1>000000[Minimum curve ∪]b2−4ac=(3k−1)2−4(1)(2k+10)=(3k)2−2(3k)(1)+(1)2−4(2k+10)=9k2−6k+1−8k−40=9k2−14k−39b2−4ac<000000[No real roots]9k2−14k−39<0(9k+13)(k−3)<0
−139<k<3
Question 9
(i)
(p+1)x2+4px=8x−2p(p+1)x2+4px−8x+2p=0(p+1)x2+(4p−8)x+2p=0b2−4ac=(4p−8)2−4(p+1)(2p)=(4p)2−2(4p)(8)+(8)2−(4p+4)(2p)=16p2−64p+64−(8p2+8p)=16p2−64p+64−8p2−8p=8p2−72p+64b2−4ac<000000[No real roots]8p2−72p+64<0p2−9p+8<0(p−1)(p−8)<0
1<p<8
(ii)
b2−4ac=0(p−1)(p−8)=0p=1 or 8
Question 10 - Line touches a curve
y=12x+p000 --- (1)x2+y2=8p000 --- (2)Substitute (1) into (2),x2+(12x+p)2=8px2+(12x)2+2(12x)(p)+(p)2=8px2+14x2+px+p2=8p54x2+px+p2−8p=05x2+4px+4p2−32p=0b2−4ac=(4p)2−4(5)(4p2−32p)=16p2−20(4p2−32p)=16p2−80p2+640p=−64p2+640pb2−4ac=000000[Real and equal roots since line touches the curve]−64p2+640p=0−64p(p−10)=0p=00 or 0p−10=000000000p=10
Question 11
(i)
y=mx+c000 --- (1)x2+y2=4000 --- (2)Substitute (1) into (2),x2+(mx+c)2=4x2+(mx)2+2(mx)(c)+(c)2=4x2+m2x2+2mcx+c2=4(1+m2)x2+2mcx+(c2−4)=0b2−4ac=(2mc)2−4(1+m2)(c2−4)=4m2c2−4(c2−4+m2c2−4m2)=4m2c2−4c2+16−4m2c2+16m2=16m2−4c2+16b2−4ac=000000[Real and equal roots since line is tangent to curve]16m2−4c2+16=016m2=4c2−164m2=c2−400000 (Proven)
(ii)
4m2=c2−4Let c=2,4m2=(2)2−44m2=0m2=0m=±√0m=0y=mx+cy=(0)x+(2)y=2∴Equation of line: .y=2
Question 12 - Real-life problem
Profit=Revenue−Cost=R(x)−C(x)=5−3x−(17−8x−2x2)=5−3x−17+8x+2x2=2x2+5x−12Profit≤000000[No profit]2x2+5x−12≤0(x+4)(2x−3)≤0
−4≤x≤1.5
∴No. of years=1.5No. of months=1.5 years×12=18
Question 13
(i)
y=x2−2(a+b)x+2b000 --- (1)y=2c000 --- (2)Substitute (2) into (1),2c=x2−2(a+b)x+2b0=x2+(−2a−2b)x+2b−2cb2−4ac=(−2a−2b)2−4(1)(2b−2c)=(−2a)2−2(−2a)(2b)+(2b)2−4(2b−2c)=4a2+8ab+4b2−8b+8cb2−4ac<000000[No real roots since line does not intersect curve]4a2+8ab+4b2−8b+8c<0a2+2ab+b2−2b+2c<0(a+b)2−2b+2c<0(a+b)2<2b−2c
(ii)
(a+b)2<2b−2cLet a=1 and b=2,(1+2)2<2(2)−2c(3)2<4−2c9<4−2c2c<4−92c<−5c<−52c<−2.5c=−3Line: .y=−6Curve: .y=x2−6x+4