Editorial - Matlan - Aplikasi Turunan
Problem 1
x3-3*x2=0
x={0,+3}
(ƒ^′)(x)=3*x2-6*x
(ƒ^′)(3)=3⋅32-6⋅3=9
(B)
Problem 2
(ƒ^′)(x)=1/2*(x+a2)(-1/2)
(ƒ^′)(3*a2)=1/2*(4*a2)(-1/2)=1/(2⋅2*a)=1/(4*a)
Find the target gradient.
2*y-2*a*x+5=0
2*y=2*a*x-5
y=a*x-5/2
The gradient is a
1/(4*a)=a
1/4=a2
a=1/2
Let's find the coordinate of the tangent.
ƒ(3*a2)=√(,4*a2)=2*a=2⋅1/2=1
Substitute.
y=m*x+b
y=1/2*x+b
1=1/2⋅3⋅1/(22)+b
b=5/8
(C)
Problem 3
ƒ(-1)=-2-4=-6
(ƒ^′)(-1)=2
g(-1)=-1/2⋅(-1)+3/2=2
(g^′)(-1)=-1/2
h(x)=ƒ(x)⋅g(x)
(h^′)(x)=(ƒ^′)(x)⋅g(x)+ƒ(x)⋅(g^′)(x)
(h^′)(-1)=2⋅2+(-6)⋅(-1/2)=7
(D)
Problem 4
ƒ(2)=6-2=4
(ƒ^′)(2)=3
g(2)=-4+1=-3
(g^′)(2)=-2
(h^′)(2)=3⋅(-3)+4⋅(-2)=-17
(E)
Problem 5
y=m*x+c
(ƒ^′)(x)=(1⋅(x+2)-x⋅1)/((x+2)2)=2/(2)
(ƒ^′)(a)=2/((a+2)2)=m
Identity of the gradient
m=((y_1)-(y_0))/((x_1)-(x_0))
2/((a+2)2)=(b-0)/(a+2)
2/(a+2)=b
ƒ(a)=a/(a+2)=b
Substitute.
2/(a+2)=a/(a+2)
∴a=2
2/(2+2)=b
∴b=1/2
∴a2-4*b=22-4⋅1/2=2
(C)
Problem 6
(ƒ^′)(x)=6*x2-4
(ƒ^′)(-1)=6-4=2=m
ƒ(-1)=-2+4+3=5
y=m*x+c
5=-1⋅2+c
c=7
∴y=2*x+7
(B)
Problem 7
y=x3+6*x2+9*x+7
(y^′)=3*x2+12*x+9
(y^′*′)=6*x+12
6*x+12=0
x=-2
Substitute back to y
y=(-2)3+6⋅(-2)2+9⋅(-2)+7=5
(-2,5)
(C)
Problem 8
ƒ(x)=(x+1)/(x2+x+1)
(ƒ^′)(x)=(x⋅(x2+x+1)-(x+1)*(2*x+1))/((x2+x+1)2)=0
(x2+x+1)-(x+1)*(2*x+1)=-x2-2*x
x={0,-2}
Substitute.
x=0
y=1/1=1
(0,1)
(B)
Problem 9
Typo, ƒ(x)=-1/3*x3+p*x2+2*p*x+5
Selalu turun, i.e monoton turun
(ƒ^′)(x)≤0 for all x
(ƒ^′)(x)=-x2+2*p*x+2*p≤0
Definite negative for a*x2+b*x+c:
D=(2*p)2-4⋅(-1)⋅(2*p)=4*p2+8*p≤0
4*p*(p+2)≤0
Bounds: p=-2,0
Test p=-1 4⋅-1*(-1+2)=-4≤0 (true)
∴-2≤p≤0
(E)
Problem 10
(ƒ^′)(x)=3*x2+6*k*x-9*k2<0
(x+3*k)*(x-k)<0
x=-3*k,k
Karena sistem pertidaksamaan, bisa asumsi hasil akhirnya pasti -2<x<6
-3*k=6
k=-2
dan sesuai dengan soal, maka k=-2
(B)
Problem 11
(ƒ^′)(x)=(-5*x+10)/√(3,x)=0
Local maxima at x=2
Undefined derivative at x=0 (divided by 0)
So, the bounds are x=0 and x=2
Let's test x=1
(ƒ^′)(1)=-5+10=5
That means the function increases between x=0 and x=2.
0<x<2
(B)
Problem 12
(ƒ^′)(x)=(3*x-4)/(2√(,x-2))
3*x-4=0
∴x>4/3
x>2 also has to be true due to the original function.
So, x>2
(E)
Problem 13
(ƒ^′)(x)=12*x3-12*x2
(ƒ^′)(-2)=12⋅(-2)3-12⋅(-2)2=-144
(ƒ^′)(2)=12⋅23-12⋅22=48
The function (OVERALL) decreases then increases in the interval, therefore the minimum is at (ƒ^′)(0)
12*x3-12*x2=0
x={0,+1}
All minimum points are stationary points, but not vice versa, so let's check each and find the minimum.
ƒ(0)=2
ƒ(1)=3-4+2=1
The minimum is at y=1
"Because the function decreases then increases within the interval, the maximum must be at the edge." This is false.
We have to prove that the function doesn't increase within the bounds.
(ƒ^′)(x)=12*x3-12*x2=12*x2*(x-1)
x2≥0 is always true, while (x-1) needs more rigorous proof.
(x-1) is always negative for all x<1
So for the interval of -2≤x≤1, (ƒ^′)(0)≤0
That means, we don't have to worry about the function increasing in that half. It WILL increase after x=1 though.
So, the maximum has to be at the ends of the interval.
ƒ(-2)=3⋅(-2)4-4⋅(-2)3+2=82
ƒ(2)=3⋅24-4⋅23+2=18
∴M+m=82+1=83
(D)
Problem 14
(g^′)(x)=12*x2-3
12*x2=3
x={-1/2,+1/2}
Minimum di salah satu dari keduanya.
g(1/2)=4⋅1/(23)-3⋅1/2+3=2
(g^)(-1/2)=4⋅1/((-2)3)-3⋅1/(-2)+3=4
Titik minimumnya di (1/2,2)
Substitute.
(ƒ^′)(x)=2*a*x+b
2*a⋅1/2+b=0=a+b
=0 karena turunannya harus 0
(C)
Problem 15
(ƒ^′)(x)=3*x2-27
3*x2=27
x={-3,+3}
Minimum at x=3
ƒ(3)=33-27⋅3=-54
(?)
Problem 16
P(x)=(50-x/2)⋅x-(x2)/4-35*x-25=-25+15*x-3/4*x2
(P^′)(x)=-3/2*x+15=0
x=10
(B)
Problem 17
V(x)=(30-2*x)*(30-2*x)*x
(V^′)(x)=12*x2-240*x+900
x2-20*x+75=0
x={5,+15}
V(15)=(30-30)*(30-30)⋅15=0
V(5)=(30-10)*(30-10)⋅5=2000
(A)
Problem 18
p=2*x+2*y+1/2*2*π*x
y=p/2-x+(-π*x)/2
A(x)=2*x⋅y+1/2*π*x2
A(x)=2*x⋅((p-2*x-π*x)/2)+1/2*π*x2
A(x)=p*x-(2+π/2)*x2
(A^′)(x)=p-(4+π)*x=0
x=p/(4+π)
(C)
Problem 19
(ƒ^′)(x)=(-sin(x)⋅sin(x)-(2+cos(x))⋅cos(x))/(sin^2)(x)=-/sin2(x)
(ƒ^′)(π/2)=-(1+2⋅0)/(12)=-1
ƒ(π/2)=(2+0)/1=2
y=-x+b
2=-π/2+b
b=2+π/2
(E)
Problem 20
g(x)=√(,(sin^2)(2*x-1/12*π)-1/4)
g(π/6)=1/2
The tangent is at (π/6,1/2)
(g^′)(x)=sin(4*x-1/6*π)/√(,(sin^2)(2*x-1/12*π)-1/4)
(g^′)(π/6)=1/1/2=2
y=2*x+b
1/2=2⋅π/6+b
b=1/2+(-π)/3
y=2*x+1/2-1/3*π
(D)
Problem 21
Identity
-2*sin(A)*sin(B)=cos(A+B)-cos(A-B)
ƒ(x)=cos(x)+1/2
(ƒ^′)(x)=-sin(x)
-sin(x)=-√(,3)/2
∴x=1/3*π,2/3*π
Karena di kuadran 2, x=(2*π)/3
(Yang di kuadran 2 sudutnya)
ƒ((2*π)/3)=cos((2*π)/3)+1/2=0
The tangent is at ((2*π)/3,0)
y=-√(,3)/2*x+b
0=-√(,3)/2⋅(2*π)/3+b
b=(π√(,3))/3
y=-√(,3)/2+(π√(,3))/3=-√(,3)/2*(x-2/3*π)
(A)
Problem 22
√(,1/2+sin(x))=1
sin(x)=1/2
x=π/6
(ƒ^′)(x)=cos(x)/(2√(,1/2+sin(x)))
(ƒ^′)(π/6)=√(,3)/2/(2√(,1/2+1/2))=√(,3)/4
y-1=√(,3)/4*(x-1/6*π)
(A)
Problem 23
(ƒ^′)(x)=(sec^2)(x)
(ƒ^′)(π/4)=(sec^2)(π/4)=2
m=-1/2
1=-1/2⋅π/4+b
b=1+π/8
y=-1/2*x+π/8+1
(E)
Problem 24
(ƒ^′)(x)=(-sin(x)⋅(1+sin(x))-cos(x)*(cos(x)))/((1+sin(x))2)=-((sin^2)(x)+(cos^2)(x)+sin(x))/((1+sin(x))2)
(ƒ^′)(x)=-(1+sin(x))/((1+sin(x))2)=-1/(1+sin(x))
(ƒ^′)(π/2)=-1/(1+1)=-1/2
m=-1/(-1/2)=2
ƒ(π/2)=0
0=2⋅π/2+b
b=-π
y=2*x-π=2*(x-π/2)
(E)
Problem 25
(ƒ^′)(x)=-(sin(2*x)-1/2)/√(,(sin^2)(x)-x/2+2*π)<0
sin(2*x)-1/2>0
(flip karena dikali -1)
sin(2*x)>1/2
1/12*π<x<5/12*π
(D)
Problem 26
(ƒ^′)(x)=3*(sin^2)(x)*cos(x)
3*sin2(x)*cos(x)>0
Hanya cos(x) bisa negatif. Yang lain selalu positif.
cos(x)>0
0<x<90
270<x<360
Easy.
(C)
Problem 27
ƒ(x)=sin(x)*(1+cos(x)+(cos^2)(x)+…)
sin(x) increases as x→π
Problem 28
(ƒ^′)(x)=3*(sin^2)(x)*cos(x)+cos(x)
(ƒ^′)(x)=cos(x)*(3*(sin^2)(x)+1)
3*(sin^2)(x)+1 will always be positive.
cos(x) can be negative or positive. That means ƒ(x) has extreme points of maximum and minimum. ƒ(x) also changes direction at x=π/2, which means the change happens within the bounds.
(E)
Problem 29
ƒ(x)=(2-(2*(sin^2)(x))/(cos^2)(x))/1/(cos^2)(x)=2*(cos^2)(x)-2*(sin^2)(x)
ƒ(x)=2*cos(2*x)
(ƒ^′)(x)=-4*sin(2*x)
sin(2*x)=0
x=0,π/2
ƒ(0)=(2-0)/1=2
ƒ(π/2)=2⋅cos(π)=-2
(A)
Problem 30
ƒ(x)=cot(x+π/3)-cot(x-π/6)
(ƒ^′)(x)=-(csc^2)(x+π/3)+(csc^2)(x-π/6)=0
(csc^2)(x-π/6)=(csc^2)(x+π/3)
|csc(x-π/6)|=|csc(x+π/3)|
x-π/6=±(x+π/3)+k*π
Only negative yields a result.
x=-π/12+(k*π)/2
For all integer k, the only one that satisfies is k=0 which leads to x=-π/12
(D)
Problem 31
Using 2 identities
(sin^2)(x)-(cos^2)(x)=-cos(2*x)
2*sin(x)*cos(x)=sin(2*x)
ƒ(x)=sin(2*x)-cos(2*x)+4√(,2)
(ƒ^′)(x)=2*cos(2*x)+2*sin(2*x)=0
2*cos(2*x)=-2*sin(2*x)
cos(2*x)=-sin(2*x)
x=0,(3*π)/8,(7*π)/8,(11*π)/8
ƒ(0)=0-1+4√(,2)=4√(,2)-1
ƒ((3*π)/8)=sin((3*π)/4)-cos((3*π)/4)+4√(,2)=√(,2)/2+√(,2)/2+4√(,2)=5√(,2)
(C)
Problem 32
k=√(,82+152)=17
Harmonic addition....!!!!
You can also solve the previous problem via this method.
a*cos(x)+b*sin(x)=k*cos(x-α)
a*cos(x)+b*sin(x)=k*sin(x+β)
k=√(,a2+b2)
ƒ(x)=17*cos(2*x-α)+17
Maximum of cos is always 1
17⋅1+17=34
(D)
Problem 33
d(x)/d(t)=-12*sin(4*t)=-12*sin(4⋅π/2)=-12
d(y)/d(t)=-6*sin(3*t)=-6*sin(3⋅π/2)=6
v=√(,144+36)=√(,180)=6√(,5)
(D)
Problem 34
h=m*sin(2*α)
For the square in the middle, the area is m⋅m*sin(2*α)=m2*sin(2*α)
For the combined are of the triangles, the area is a square: m*cos(2*α)⋅m*sin(2*α)=m2*sin(2*α)*cos(2*α)=(m2)/2*sin(4*α)
A(α)=m2*sin(2*α)+(m2)/2*sin(4*α)
Derive with respect to α.
(A^′)(α)=2*m2*cos(2*α)+2*m2*cos(4*α)=2*m2*(cos(2*α)+cos(4*α))
cos(2*α)+cos(4*α)=0
cos(4*α)=-cos(2*α)=cos(π-2*α)
4*α=±(π-2*α)+2*π*k
Try positive.
4*α=π-2*α
6*α=π
α=1/6*π=30
Try negative.
4*α=-π+2*α
2*α=-π
α=-90
Not acute. So, we take α=π/6
A=m2*sin(π/3)+(m2)/2*sin(2/3*π)=m2⋅√(,3)/2+(m2)/2⋅√(,3)/2
A=m2⋅3/4⋅√(,3)
(?)
Problem 35
a adalah titik tengah.
a=(12+18)/2=15
b adalah amplitudo.
b=(18-12)/2=3
a-2*b=15-6=9
(B)
Problem 36
Let DA be x=2
A(α)=2*sin(α)⋅2*cos(α)+2⋅2*sin(α)=2*sin(2*α)+4*sin(α)
(A^′)(α)=4*cos(2*α)+4*cos(α)
cos(2*α)+cos(α)=0
2*(cos^2)(α)-1+cos(α)=0
(2*cos(α)-1)*(cos(α)+1)=0
∴cos(α)=1/2
∴a=60
(C)
Problem 37
(ƒ^′)(x)=sin(2*x)+sin(2*x)+5=2*sin(2*x)+5
(ƒ^′*′)(x)=4*cos(2*x)
4*cos(2*x)<0
cos(2*x)<0
π/2<2*x<3/2*π
1/4*π<x<3/4*π
(D)
Problem 38
(ƒ^′)(x)=2*cos(x-π/3)
(ƒ^′*′)(x)=-2*sin(x-π/3)>0
sin(x-π/3)<0
π<x-π/3<2*π
(4*π)/3<x<2*π (boundnya di2*π)
-π<x-π/3<0
-(2*π)/3<x<π/3
0≤x<π/3
So:
0≤x<π/3
(4*π)/3<x≤2*π
(B)
Problem 39
(ƒ^′)(x)=2*(sec^2)(x)
(ƒ^′*′)(x)=(8*sin(2*x))/(cos^3)(2*x)
8*sin(2*x)=0
sin(2*x)=0
2*x=0,π
x=0,π/2
ƒ(π/2)=tan(π)=0
(π/2,0)
(C)
Problem 40
The frequency doesn't change, which makes this easy. Originally, the inflection points are at wherever y=0 for sin(π*x), which is 0, 1, 2, etc.
Then, the function gets transformed. The amplitude doesn't matter here, as the original y coordinate is 0, but the function does get translated along the y axis, which means we have to add the y coordinate by 2.
(0,2)
(1,2)
(2,2)
(E)