(a) Let H be a subspace of R⁴ spanned by the vectors v₁ = (1, –2, 5, –3), v₂ = (2, 3, 1, –4), v₃ = (3, 8, –3, –5). Then find a…

(a) Let c₁v₁ + c₂v₂ + c₃v₃ = 0. Using the coordinates of v₁, v₂, v₃:

c₁ + 2c₂ + 3c₃ = 0 –2c₁ + 3c₂ + 8c₃ = 0 5c₁ + c₂ – 3c₃ = 0 –3c₁ – 4c₂ – 5c₃ = 0

From the first equation, c₁ = –2c₂ – 3c₃. Substitute in the second:

–2(–2c₂ – 3c₃) + 3c₂ + 8c₃ = 0 ⇒ 4c₂ + 6c₃ + 3c₂ + 8c₃ = 0 ⇒ 7c₂ + 14c₃ = 0 ⇒ c₂ = –2c₃.

Then c₁ = –2(–2c₃) – 3c₃ = c₃. The third and fourth equations are then automatically satisfied. Hence

c₃v₁ – 2c₃v₂ + c₃v₃ = 0 ⇒ v₁ – 2v₂ + v₃ = 0 ⇒ v₃ = 2v₂ – v₁.

So v₃ is dependent on v₁ and v₂, while v₁ and v₂ are not scalar multiples, hence independent. Therefore

Basis of H = {v₁, v₂} = {(1, –2, 5, –3), (2, 3, 1, –4)}, dim H = 2.

To extend this basis to R⁴, take e₃ = (0, 0, 1, 0), e₄ = (0, 0, 0, 1). Consider

αv₁ + βv₂ + γe₃ + δe₄ = 0.

Comparing the first two coordinates:

α + 2β = 0 –2α + 3β = 0.

The determinant is 1·3 – 2(–2) = 7 ≠ 0, so α = β = 0. Then the third and fourth coordinates give γ = 0 and δ = 0. Thus the four vectors are independent, and since R⁴ has dimension 4,

Basis of R⁴ = {v₁, v₂, e₃, e₄}.

(b) Since B = {v₁, v₂, v₃} is a basis of R³, every vector is of the form a v₁ + b v₂ + c v₃. By linearity,

T(a v₁ + b v₂ + c v₃) = aTv₁ + bTv₂ + cTv₃ = a(1, 1, 0) + b(1, 0, –1) + c(2, 1, –1) = (a + b + 2c, a + c, –b – c).

The range of T is spanned by Tv₁, Tv₂, Tv₃. But

Tv₃ = (2, 1, –1) = (1, 1, 0) + (1, 0, –1) = Tv₁ + Tv₂.

Hence Tv₃ is dependent on Tv₁ and Tv₂, which are independent. Therefore

Basis of range T = {(1, 1, 0), (1, 0, –1)}, rank T = 2.

For the null space, set T(a v₁ + b v₂ + c v₃) = 0:

a + b + 2c = 0 a + c = 0 –b – c = 0.

From a + c = 0, a = –c. From –b – c = 0, b = –c. The first equation is then automatically satisfied. Thus

kernel T = {–c v₁ – c v₂ + c v₃ : c ∈ R} = span{v₃ – v₁ – v₂}.

Therefore

Basis of null space T = {v₃ – v₁ – v₂}, nullity T = 1.

Rank-nullity theorem checks: 2 + 1 = 3 = dim R³.

(c) For x ≠ 0, the function is a quotient of continuous functions,

f(x) = 1/(1 – e^(–1/x)).

The denominator vanishes only if e^(–1/x) = 1, i.e. –1/x = 0, which is impossible for finite x ≠ 0. Hence f is continuous for every x ≠ 0.

Now examine x = 0.

Right-hand limit:

x → 0⁺ ⇒ –1/x → –∞ ⇒ e^(–1/x) → 0.

Therefore

lim_x→0⁺ f(x) = 1/(1 – 0) = 1.

Left-hand limit:

x → 0⁻ ⇒ –1/x → +∞ ⇒ e^(–1/x) → +∞.

Therefore, since 1 – e^(–1/x) → –∞,

lim_x→0⁻ f(x) = 0 (approached from below).

The left and right limits are different, so lim_x→0 f(x) does not exist. Since f(0) = 0, the function is discontinuous at x = 0. It is continuous on (–∞, 0) and (0, ∞), and is left-continuous at 0 but not right-continuous.

Hence f is continuous for all x ≠ 0 and has a jump discontinuity at x = 0.

(d) Let f(x) = ln x. By Taylor’s theorem about x = 1,

f(x) = f(1) + f′(1)(x – 1) + f″(1)(x – 1)²/2! + f‴(1)(x – 1)³/3! + …

Now f(1) = 0. Also,

f′(x) = 1/x ⇒ f′(1) = 1, f″(x) = –1/x² ⇒ f″(1) = –1, f‴(x) = 2/x³ ⇒ f‴(1) = 2.

In general,

fⁿ(x) = (–1)^(n–1)(n – 1)!/xⁿ, so fⁿ(1) = (–1)^(n–1)(n – 1)!.

Hence

ln x = (x – 1) – (x – 1)²/2 + (x – 1)³/3 – (x – 1)⁴/4 + … = Σ (n=1 to ∞) (–1)^(n+1)(x – 1)ⁿ/n.

This expansion is valid for |x – 1| < 1, and at x = 2 it converges conditionally.

Put x = 1.1, so x – 1 = 0.1. Then

ln(1.1) = 0.1 – 0.1²/2 + 0.1³/3 – 0.1⁴/4 + 0.1⁵/5 – 0.1⁶/6 + 0.1⁷/7 – …

= 0.1 – 0.005 + 0.0003333333 – 0.000025 + 0.000002 – 0.0000001667 + 0.0000000143 – …

Summing,

ln(1.1) ≈ 0.0953101798.

Therefore, correct up to four decimal places,

ln(1·1) = 0.0953.

(e) The given circle is the intersection of the sphere

x² + y² + z² = 9

and the plane

x – y + z = 3.

The plane has normal vector n = (1, –1, 1). The distance of the plane from the origin is

|3|/√(1² + (–1)² + 1²) = 3/√3 = √3.

The sphere has radius 3, so the radius of the circle of intersection is

r = √(3² – (√3)²) = √(9 – 3) = √6.

The centre of this circle is the foot of the perpendicular from the origin to the plane:

c = (3/3)(1, –1, 1) = (1, –1, 1).

For a right circular cylinder to contain this circle, the plane of the circle must be perpendicular to the cylinder axis. Hence the axis is parallel to n = (1, –1, 1), and it passes through c = (1, –1, 1). The radius of the cylinder is √6.

Let X = (x, y, z). The perpendicular distance from X to the axis line through c in direction n is given by

d² = |(X – c) × n|² / |n|².

Now

X – c = (x – 1, y + 1, z – 1),

and

(X – c) × n = (y + z, z – x, –x – y).

Also |n|² = 3. Since d = √6, we get

(y + z)² + (z – x)² + (x + y)² = 6 × 3 = 18.

Simplifying,

2(x² + y² + z² + xy + yz – xz) = 18,

so

x² + y² + z² + xy + yz – xz = 9.