Soal latihan

Nullity 1 by rank-nullity. Its kernel is a one-dimensional subspace.

8πi: one enclosed simple pole at −1 has residue 4.

y=5e^(−2t); derivative is −10e^(−2t)=−2y and initial value is 5.

The order is 24/gcd(9,24)=8. The subgroup is {0,3,6,9,12,15,18,21}; it has index 24/8=3.

The kernel is {0,3,6,9,12}, the image is {0,5,10}, and the quotient has three elements. The codomain has fifteen elements; the map is not onto.

Units are 1,2,4,5,7,8. Nonzero zero divisors are 3 and 6. Since 3·3=0 modulo 9 and 3 has no inverse, not every nonzero element is a unit.

2Z is closed under integer scaling and addition, so it is a Z-submodule. It is not a Q-vector subspace because rational scalar 1/2 sends 2 to 1, which is not in 2Z.

Over Q and R it has no root and is an irreducible quadratic. Over F2 it is (x+1)² and is reducible. The real field differs from the complex field, over which it splits.

No. The intermediate field K(alpha) would have degree 3, and the tower law would require 3 to divide 10.

gcd(8,12)=4, so reduction gives 2x≡1 modulo 3. Thus x≡2 modulo 3 and the four original residues are 2,5,8,11.

Write x=3+5k; 5k≡1 modulo 7 gives k≡3. Therefore x≡18 modulo 35; 18 is the least nonnegative solution.

On [0,2], the supremum error is 2/n, so convergence is uniform. At each fixed real x the limit is still zero, but on R the error is unbounded for every n, so convergence is not uniform there.

Each absolute term is at most 1/n² and that numerical series converges, so the function series converges uniformly and absolutely. The bound 1/n has a divergent sum, making this M-test inconclusive; that failure alone is not a proof of divergence.

The interior is empty because every interval contains irrational points; the closure is R because rationals are dense. Thus the boundary is all of R.

The whole space is both open and closed relative to itself. It is not compact; the sequence 1/n for n≥2 has no subsequence converging to a point in X. Relative closedness alone does not provide Euclidean closedness in R.

Radius 2. At −2 the alternating harmonic series converges, while at 2 the harmonic series diverges; the interval is [−2,2).

The polynomial is x−x³/6. The fourth derivative is sin x and has absolute value at most 1, so the Lagrange bound is |x|⁴/24≤0.000066667. A sharper bound is possible, but is not needed to justify this one.

Both fail. The integral diverges positively on each side of zero; symmetric cutoffs give 2/epsilon−2, which tends to infinity rather than cancelling.

Outer radius 2 and inner radius x give V=π integral 0..2 (4−x²) dx=π(8−8/3)=16π/3.

For f, f_xx=f_yy=2 and f_xy=4, so D=4−16=−12; it is a saddle. For g the Hessian test is inconclusive, but g≥0 with equality only at the origin proves a strict global minimum.

(x+y)²≤2(x²+y²)=16, so the range is [−4,4]. Maximum 4 occurs at (2,2), minimum −4 at (−2,−2). The compact circle ensures extrema are attained.

div F=1+2+3=6. The closed cube has volume 1, so total outward flux is 6. Opposite face orientations must be respected in a direct calculation.

For the first field, counterclockwise circulation is 2π; clockwise is −2π. The second field is undefined at the origin and lacks the smoothness hypothesis on the full disk, even though its curl vanishes away from zero.

The inverse is (y+4)/3. It is not an involution: f(f(x))=9x−16 is not identically x.

g(g(x))=2/(2/x)=x with every input nonzero. Odd iterates equal g, so g⁹(4)=1/2.

Use f(x)=x², A={−3}, B={3}; both images are {9}. Injectivity ensures equal outputs require equal inputs, so disjoint input sets then have disjoint images.

The negation is: there exists an integer x such that every integer y satisfies y≤x. The original is true because y=x+1 is always an integer; its negation is false, and no counterexample to the original exists.

P(A|B)=0.6; P(B|A)=0.4. The joint probability is the product 0.6·0.4, so the events are independent.

Population standard deviation is 4√9=12. Solve 12/√n=1 to obtain n=144; a fourfold precision gain needs sixteen times as many observations.

Length factor 0.8 gives area factor 0.64 and volume factor 0.512. The decreases are 36% and 48.8%, respectively.

The first is an ellipse with a=7,c=5,b²=49−25=24: x²/49+y²/24=1. The second is a hyperbola x²/9−y²/16=1 restricted to x≤−3, because the signed difference is negative.

Amplitude 3; period π/2; right shift π/4. One maximum is x=π/4, where the argument is zero.

Only the parabola segment y=x², −1≤x≤1, is traced. It moves from (−1,1) through (0,0) to (1,1). The slope is 2t, hence −1 at t=−1/2.

Coefficient is (2+1)/(1+1)=3/2, so projection is (3/2,3/2). Remainder (1/2,−1/2) has dot product zero with (1,1).

The scalar triple product is 6 and volume is 6. Exchanging the last two edges changes the signed result to −6, while geometric volume remains 6.

Integrating factor e^(2x) gives y=3+C e^(−2x); C=−2, so y=3−2e^(−2x). Its derivative is 4e^(−2x), and y′+2y=6.

The repeated characteristic root is 1, giving (C₁+C₂x)e^x. A particular solution is x²e^x/2, because (D−1)²(e^x z)=e^x z″ and z″=1. Add both contributions.

Subtract the first-box lower bound 2. There are 8 remaining stars and 2 bars, so C(10,2)=45 allocations.

Before an iteration B=2m+1. Updating S gives m²+2m+1=(m+1)², then B becomes 2(m+1)+1. After four iterations S=16,B=9; B is the next addend, not the last one.

J=[[1,2],[3,1]] has determinant −5; inverse is [[−1/5,2/5],[3/5,−1/5]]. Therefore v_x=3/5 and u_y=2/5.

It is bijective on R² with inverse (x,y)↦(real cube root of x,y). Its determinant 3u² vanishes at u=0, where the inverse fails to be differentiable. Nonzero determinant is a sufficient differentiable-local-inverse condition, not a necessary condition for mere invertibility.

Forward y₁=7,y₂=20−14=6. Backward x₂=3,x₁=(7−3)/3=4/3. Verify Ux=(7,6) and Ly=(7,20).

Rank B=4. The intersection of im B (dimension 4) with ker A (dimension 3) has dimension at least 4+3−5=2 and at most 3. Therefore nullity AB ranges from 1+2=3 to 1+3=4; the ambient dimension forces overlap.

Region 0≤x≤y≤1 gives ∫₀¹y e^(2y)dy. Integration by parts gives [y e^(2y)/2−e^(2y)/4]₀¹=(e²+1)/4. The factor y comes from the inner x-interval length.

Reflection x↦π−x reverses the integrand sign. The endpoint limits are 8 and −8, so singularities are removable and the integral is 0.

Current section area is π(4−1)=3π, so signed dV/dt=3π/2, positive for filling.

The first is f′(0)/g′(0)=4/5. The second is −f′(0)/(2g′(0))=−2/5 after factoring, with all nearby denominators checked.

Closure is {12,24,36,…}. The point 36 belongs; 6 does not, because U_6 excludes 12.

No. Its image is n, not a Cauchy sequence in ordinary R. The space is complete, but completeness does not make every sequence Cauchy.

For a triple pole, residue is g″(1)/2!=4e²/2=2e². The contour integral is 2πi times the residue, or 4πi e².

The partitions are 5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1 and 1+1+1+1+1. There are seven classes. Orders alone do not distinguish every type.

Apply idempotence to a+a: (a+a)²=a+a. Expanding gives 4a²=2a; since a²=a, subtract 2a to obtain 2a=0. Only distributivity and additive-ring operations were used.

Their fourth power is −1, so their eighth power is 1 and their order cannot divide 4. The divisors of 8 are 1,2,4,8, so their order is 8. Vieta gives sum 0 and product 1.