CBSE Class 12 Mathematics Question Paper 2024 (Set 3- 65/4/3) Available - Download Solution PDF with Answer Key

Let ⃗u = 2⃗a and ⃗v = -⃗b. The angle θ between ⃗u and ⃗v is given by:

cos(θ) = (⃗u ⋅ ⃗v) / (|⃗u||⃗v|)

⃗u ⋅ ⃗v = (2⃗a) ⋅ (-⃗b) = -2(⃗a ⋅ ⃗b) = -2√3

|⃗u| = |2⃗a| = 2|⃗a| = 2(1) = 2

|⃗v| = |-⃗b| = |⃗b| = 2

cos(θ) = (-2√3) / (2 * 2) = -√3/2

θ = 5π/6

|⃗a| = √(2² + (-1)² + 1²) = √6

|⃗b| = √(1² + (-3)² + (-5)²) = √35

|⃗c| = √((-3)² + 4² + 4²) = √41

If the triangle is a right-angled triangle, the Pythagorean theorem should hold. We check if |⃗a|² + |⃗b|² = |⃗c|²:

(√6)² + (√35)² = 6 + 35 = 41 = (√41)²

Since |⃗a|² + |⃗b|² = |⃗c|², the vectors represent the sides of a right-angled triangle.

Let ⃗a = a₁ˆi + a₂ˆj + a₃ˆk, where |⃗a| = a = √(a₁² + a₂² + a₃²).

⃗a × ˆi = -a₂ˆk + a₃ˆj => |⃗a ׈i|² = a₂² + a₃²

⃗a × ˆj = a₁ˆk - a₃ˆi => |⃗a × ˆj|² = a₁² + a₃²

⃗a × ˆk = -a₁ˆj + a₂ˆi => |⃗a × ˆk|² = a₁² + a₂²

|⃗a ׈i|² + |⃗a × ˆj|² + |⃗a × ˆk|² = (a₂² + a₃²) + (a₁² + a₃²) + (a₁² + a₂²) = 2(a₁² + a₂² + a₃²) = 2|⃗a|² = 2a²

A² = [[3, 1], [-1, 2]] × [[3, 1], [-1, 2]] = [[8, 5], [-5, 3]]

7I = [[7, 0], [0, 7]]

A² + 7I = [[8+7, 5+0], [-5+0, 3+7]] = [[15, 5], [-5, 10]]

kA = [[3k, k], [-k, 2k]]

Comparing A² + 7I and kA:

3k = 15 => k = 5

k = 5

-k = -5 => k = 5

2k = 10 => k = 5

Therefore, k = 5.

AB = I

Calculating AB:

AB = (1/3) × [[1(-2) + (-1)(9) + 2(6), 1(0) + (-1)(2) + 2(1), 1(1) + (-1)(-3) + 2(λ)], [0(-2) + 2(9) + (-3)(6), 0(0) + 2(2) + (-3)(1), 0(1) + 2(-3) + (-3)(λ)], [3(-2) + (-2)(9) + 4(6), 3(0) + (-2)(2) + 4(1), 3(1) + (-2)(-3) + 4(λ)]]

AB = (1/3) × [[1, 0, 4+2λ], [0, 1, -6-3λ], [0, 0, 9+4λ]]

Since AB = I = [[1, 0, 0], [0, 1, 0], [0, 0, 1]], comparing the elements:

(1/3)(1) = 1 (This is consistent)

(1/3)(4+2λ) = 0 => 4 + 2λ = 0 => λ = -2

(1/3)(9+4λ) = 1 => 9+4λ=3 => 4λ = -6 => λ=-3/2

Comparing the last element of the second row : -6-3λ = 0 => λ = -2

Therefore, λ = -2

Let u = x² and v = x³. We want to find du/dv.

Using the chain rule:

du/dv = (du/dx) / (dv/dx)

du/dx = 2x

dv/dx = 3x²

du/dv = (2x) / (3x²) = 2/(3x)

The function f(x) = |x| + |x-2| is continuous everywhere because the absolute value function is continuous. However, it's not differentiable at the points where the absolute value expressions inside change sign (i.e., at x=0 and x=2).

At x = 0:

Left-hand derivative: lim_h→0⁻ (|0+h| + |0+h-2| - (|0| + |0-2|))/h = lim_h→0⁻ (-h - (h-2) - 2)/h = lim_h→0⁻ (-2h)/h = -2

Right-hand derivative: lim_h→0⁺ (|0+h| + |0+h-2| - (|0| + |0-2|))/h = lim_h→0⁺ (h - (h-2) - 2)/h = lim_h→0⁺ (0)/h = 0

Since the left and right-hand derivatives are not equal, f(x) is not differentiable at x=0.

At x = 2:

Left-hand derivative: lim_h→0⁻ (|2+h| + |2+h-2| - (|2| + |2-2|))/h = lim_h→0⁻ (2+h - h - 2)/h = 0

Right-hand derivative: lim_h→0⁺ (|2+h| + |2+h-2| - (|2| + |2-2|))/h = lim_h→0⁺ (2+h + h - 2)/h = lim_h→0⁺(2h)/h = 2

Since the left and right-hand derivatives are not equal, f(x) is not differentiable at x=2.

∫₀^π/3 tan²(θ) dθ = ∫₀^π/3 (sec²(θ) - 1) dθ

= [tan(θ) - θ]₀^π/3

= (tan(π/3) - π/3) - (tan(0) - 0)

= √3 - π/3

The original solution incorrectly multiplied the final answer by 3. The correct answer is √3 - π/3.

The integrating factor (IF) for a first-order linear differential equation of the form dy/dx + P(x)y = Q(x) is given by:

IF = e^∫P(x)dx

In this case, P(x) = 2/x.

IF = e^∫(2/x)dx = e^2ln|x| = e^ln|x²| = x²

The direction ratios of the first line are <-2, 3, 1> (Note the negative sign due to 1-x).

The direction ratios of the second line are <2p,-1,7>.

For the lines to be perpendicular, the dot product of their direction ratios must be zero:

(-2)(2p) + (3)(-1) + (1)(7) = 0

-4p - 3 + 7 = 0

-4p + 4 = 0

-4p = -4

p = -1/2

To find the maximum value of Z = 4x + y, we evaluate Z at each corner point of the feasible region:

• At A(0, 50): Z = 4(0) + 50 = 50
• At B(20, 30): Z = 4(20) + 30 = 80 + 30 = 110
• At C(30, 0): Z = 4(30) + 0 = 120

The maximum value of Z is 120 at point C(30, 0).

The sum of all probabilities in a probability distribution must equal 1:

0.1 + k + 2k + k + 0.1 = 1

4k + 0.2 = 1

4k = 0.8

k = 0.2

The probability that X takes the value 2 is P(X=2) = 2k = 2(0.2) = 0.4 = 2/5.

For a function to be strictly increasing, its derivative must be greater than 0.

f'(x) = k - cos(x)

For f(x) to be strictly increasing, f'(x) > 0 for all x.

k - cos(x) > 0

k > cos(x)

Since the maximum value of cos(x) is 1, k must be greater than 1 for the inequality to hold for all x.

The given line ⃗r = ˆi + ˆk + μ(2ˆi − ˆj) has direction vector <2, -1, 0>.

The required line passes through the point (1, -1, 0) and is parallel to the given line, so it has the same direction vector <2, -1, 0>.

The equation of the line is:

(x - 1)/2 = (y - (-1))/(-1) = (z - 0)/0

(x - 1)/2 = (y + 1)/(-1) = z/0

A scalar matrix is a diagonal matrix where all diagonal elements are equal. Therefore, in the given matrix A:

• a = d = 5 (diagonal elements are equal)
• b = c = 0 (off-diagonal elements are zero)

a + 2b + 3c + 4d = 5 + 2(0) + 3(0) + 4(5) = 5 + 0 + 0 + 20 = 25

If A⁻¹ = (1/7)[[2, 1], [-3, 2]], then to find A, we need to find the inverse of A⁻¹. For a 2x2 matrix M = [[a, b], [c, d]], the inverse is given by (1/det(M))[[d, -b], [-c, a]], where det(M) = ad - bc.

In our case, A⁻¹ = (1/7)[[2, 1], [-3, 2]]. So, det(A⁻¹) = (2*2) - (1*-3) = 4+3 = 7

First, find the inverse of [[2, 1], [-3, 2]] which will be (1/7)[[2,-1],[3,2]]

Then multiply with 7 to get A. We have, A = 7 * (1/7)[[2,-1],[3,2]] = [[2,-1],[3,2]]

The given series is a geometric series with the first term as 'I' and common ratio '-A'. The sum of an infinite geometric series is a/(1-r), where 'a' is the first term and 'r' is the common ratio, when |r|<1

In matrix terms. This becomes

S = I - A + A² - A³ + .... = (I+A)^-1, provided the series converges. (I+A) = [[3, 1],[-4, -1]]. det(I+A) = (3)(-1) - (1)(-4) = 1

Therefore, (I+A)^-1 = [[ -1, -1],[4,3]].

First calculating A²: A² = [[2,1],[-4,-2]] * [[2,1],[-4,-2]] = [[0,0],[0,0]] = 0 (zero matrix)

So, the given series simplifies considerably. S = I - A + 0 - 0 + .... = I - A = [[1,0],[0,1]] - [[2,1],[-4,-2]] = [[-1,-1],[4,3]]

Since A² is zero, higher power terms in the series becomes zero. Therefore, we get I-A.

Therefore, I - A + A² - A³ + ... = [[1,0],[0,1]] - [[2,1],[-4,-2]] = [[-1, -1],[4, 3]].

Rewrite the equation in the standard form of a linear differential equation:

dy/dx + P(x)y = Q(x)

x(dy/dx) + 2y²(dy/dx) = y

x(dy/dx) - y = -2y²(dy/dx)

dx/dy - (x/y) = (-2y²(dy/dx))/y(dy/dx) = -2y

dx/dy -(1/y)x = -2y

Here the integrating factor is e^{∫-1/y dy} = e^-lny = 1/y.

The given differential equation is not in standard linear form with x as independent variable.

It can be rearranged as:

x + 2y²(dy/dx) = y

x + 2y²(dy/dx) = y

dx/dy = (x + 2y²)/y = x/y + 2y

dx/dy - x/y = 2y

This is a linear differential equation in x (treating x as the dependent variable and y as independent).

The integrating factor is e^{∫-1/y dy} = e^{-ln y} = 1/y.

Assertion:

For a relation to be reflexive, (x, x) must be in R for all x. Here, x + x = 2x. If x > 1, then 2x is not prime (it's composite). Therefore, R is not reflexive. The assertion is true.

Reason:

The number 2n is not composite for all n. For example, when n = 1, 2n = 2, which is prime. The reason is false.

Assertion:

If the objective function is parallel to an edge of the feasible region, then the maximum value can occur at infinitely many points along that edge. The assertion is true.

Reason:

The optimal solution of an LPP with a bounded feasible region usually occurs at the corner points. However, it's possible for the solution to occur along an entire edge if the objective function line is parallel to that edge. The reason is true, but it doesn't explain why the maximum occurs at infinite points in the specific case described in the assertion.

y = cos³(sec²(2t))

Using the chain rule:

dy/dt = 3cos²(sec²(2t)) * [-sin(sec²(2t))] * [2sec(2t) * sec(2t)tan(2t) * 2]

dy/dt = -12cos²(sec²(2t)) * sin(sec²(2t)) * sec²(2t)tan(2t)

x^y = e^x²−y

Take the natural logarithm of both sides:

y ln(x) = x² - y

Differentiate both sides with respect to x using the product rule and chain rule:

(dy/dx)ln(x) + y(1/x) = 2x - dy/dx

(dy/dx)(ln(x) + 1) = 2x - y/x

dy/dx = (2x - y/x) / (ln(x) + 1)

From the original equation, y ln(x) = x² - y => y(1+ln x) = x² => y= x²/(1+ln x)

Substitute this value of y into the expression for dy/dx:

dy/dx = [2x - (x²/(1+ln x))/x]/(ln x + 1)

dy/dx = [2x - x/(1+ln x)]/(ln x + 1)

dy/dx = [2x(1+ln x) - x]/ (1+ln x)²

dy/dx = [x + 2xln x]/ (1+ln x)²

dy/dx = x(1+2ln x)/(1+ln x)²

Consider ylnx = x²-y

ylnx + y = x²

y(1+lnx) = x²

y = x²/(1+lnx)

dy/dx = [(1+lnx)2x - x² * (1/x)]/(1+lnx)²

= [2x + 2xlnx -x]/(1+lnx)²

= x(1+2lnx)/(1+lnx)²

= [lnx(1+lnx) + lnx]/(1+lnx)²

= [lnx + ln²x+lnx]/(1+lnx)²

=lnx(1+lnx)/(1+lnx)²= lnx/(1+lnx)

Let 'a' be the side length of the cube.

• Volume (V) = a³
• Surface Area (S) = 6a²

• dV/dt = 3a²(da/dt) = 6 cm³/s
• dS/dt = 12a (da/dt)

A function is strictly decreasing in an interval if its derivative is negative in that interval.

f(x) = sin x + cos x

f'(x) = cos x - sin x

For f(x) to be strictly decreasing, f'(x) < 0:

cos x - sin x < 0

cos x < sin x

This inequality is true in the interval (π/4, 5π/4).

tan⁻¹[cos x / (1 − sin x)]

= tan⁻¹ [(cos²(x/2)-sin²(x/2))/(cos²(x/2)+sin²(x/2)-2sin(x/2)cos(x/2))]

= tan⁻¹ [(cos²(x/2)-sin²(x/2))/(cos(x/2)-sin(x/2))²] = tan⁻¹ [(cos(x/2)-sin(x/2))(cos(x/2)+sin(x/2))/(cos(x/2)-sin(x/2))²]

= tan⁻¹[(cos(x/2)+sin(x/2))/(cos(x/2)-sin(x/2))] = tan⁻¹[(1+tan(x/2))/(1-tan(x/2))] = tan⁻¹[tan(π/4+x/2)] = π/4 + x/2

Alternatively:

cos x / (1 - sin x) = (sin(π/2 - x))/(sin²(x/2) + cos²(x/2) - 2sin(x/2)cos(x/2))

=(sin(π/2 - x))/(cos x/2 - sin x/2)² = (cos(x/2) + sin(x/2)) * (cos(x/2) - sin(x/2))/(cos x/2 - sin x/2)²

= (cos(x/2) + sin(x/2))/(cos x/2 - sin x/2)) = (1+tan(x/2))/(1-tan(x/2))

Thus, tan⁻¹[cos x / (1 − sin x)] = tan⁻¹(cot(π/4 + x/2)) = tan⁻¹(tan(π/2 - π/4 -x/2)) = π/4-x/2

tan⁻¹(1) = π/4 (Principal value)

cos⁻¹(-1/2) = 2π/3 (Principal value)

sin⁻¹(-1/√2) = -π/4 (Principal value)

Therefore, tan⁻¹(1) + cos⁻¹(-1/2) + sin⁻¹(-1/√2) = π/4 + 2π/3 - π/4 = 2π/3

I = ∫ (2x) / [(x² + 1)(x² − 4)] dx

Let u = x², then du=2xdx

I = ∫du/[(u+1)(u-4)]

We use partial fractions:

1/[(u+1)(u-4)] = A/(u+1) + B/(u-4)

1 = A(u-4) + B(u+1)

If u=-1, 1 = -5A => A = -1/5

If u=4, 1 = 5B => B = 1/5

I = ∫[(-1/5)/(u+1) + (1/5)/(u-4)]du

I = (-1/5)ln|u+1| + (1/5)ln|u-4| + C

I = (1/5)ln|(u-4)/(u+1)| + C

Substituting back u = x²:

I = (1/5)ln|(x²-4)/(x²+1)| + C

Let y = y₁ + y₂, where y₁ = (cos x)^x and y₂ = cos⁻¹(√x).

Finding dy₁/dx:

Take the natural logarithm of both sides of y₁ = (cos x)^x:

ln(y₁) = x ln(cos x)

Differentiate both sides with respect to x using the product rule and chain rule:

(1/y₁)(dy₁/dx) = ln(cos x) + x(-sin x / cos x) = ln(cos x) - x tan x

dy₁/dx = y₁(ln(cos x) - x tan x) = (cos x)^x(ln(cos x) - x tan x)

Finding dy₂/dx:

y₂ = cos⁻¹(√x)

Using the chain rule:

dy₂/dx = (-1/√(1 - (√x)²)) * (1/(2√x)) = -1 / (2√(x(1-x)))

Finding dy/dx:

dy/dx = dy₁/dx + dy₂/dx = (cos x)^x(ln(cos x) - x tan x) - 1 / (2√(x(1-x)))

dy/dx = y cot(2x)

Separate variables:

dy/y = cot(2x) dx

Integrate both sides:

∫(1/y) dy = ∫cot(2x) dx

ln|y| = (1/2)ln|sin(2x)| + C

ln|y| = ln(√(sin(2x))) + C

y = e^C√(sin(2x)) = k√(sin(2x))

Using the initial condition y(π/4) = 2:

2 = k√(sin(2π/4)) = k√(sin(π/2)) = k

k=2

Therefore, the particular solution is y = 2√(sin(2x)).

(xe^x + y)dx = x dy

Rearrange to get in the standard linear form:

dy/dx - y/x = e^x

Integrating Factor (IF) = e^∫(-1/x)dx = e^-lnx = 1/x

Multiply the equation by the IF:

(1/x)(dy/dx) - (1/x²)y = e^x/x

d(y/x)/dx = e^x/x

Integrate both sides:

∫d(y/x) = ∫(e^x/x)dx

y/x = ∫(e^x/x)dx + C

There is no elementary function to represent the given integral ∫(e^x/x)dx. So, it has to be represented as an infinite series. If you mean x(e^x+y)dx = xdy, then proceed as follows.

Given: x(e^x+y)dx = xdy

x(e^x+y) = x(dy/dx) ; dy/dx - y = e^x

IF = e^∫-1dx = e^-x

y * e^-x = ∫e^x * e^-xdx = x+c

When x=1, y=1: 1 * e^-1 = 1+c ; c=e^-1 - 1

y * e^-x = x + e^-1 - 1

This integral can be solved using integration by parts:

Let u = sec θ and dv = sec²θ dθ

du = sec θ tan θ dθ and v = tan θ

∫ sec³(θ) dθ = sec θ tan θ - ∫sec θ tan²θ dθ

= sec θ tan θ - ∫sec θ (sec²θ - 1) dθ

= sec θ tan θ - ∫sec³θ dθ + ∫sec θ dθ

2∫ sec³(θ) dθ = sec θ tan θ + ln|sec θ + tan θ|

∫ sec³(θ) dθ = (1/2)sec θ tan θ + (1/2)ln|sec θ + tan θ| + C

Let A be the event that the lost card is a king, and B be the event that a king is drawn from the remaining cards.

We want to find P(A|B) = probability that lost card is a king, given that a king was drawn.

Using Bayes' theorem:

P(A|B) = [P(B|A) * P(A)] / P(B)

P(A) = 4/52 = 1/13 (Probability of losing a king initially)

P(B|A) = 3/51 (Probability of drawing a king given a king was lost)

P(B) = P(B|A)P(A) + P(B|A')P(A') = (3/51)(4/52) + (4/51)(48/52) = 1/13

P(A|B) = [(3/51) * (1/13)] / (1/13) = 3/51 = 1/17

Let P(Odd) = p. Then P(Even) = 2p.

Since the sum of probabilities is 1: 3p + 3(2p) = 9p = 1 => p = 1/9

P(1) = P(3) = P(5) = 1/9

P(2) = P(4) = P(6) = 2/9

Let X be the number of sixes in two throws.

P(X=0) = (7/9)² = 49/81

P(X=1) = 2 * (2/9) * (7/9) = 28/81

P(X=2) = (2/9)² = 4/81

Probability Distribution:

X | 0 | 1 | 2

P(X) | 49/81 | 28/81 | 4/81

Mean (µ) = Σ xP(x) = 0*(49/81) + 1*(28/81) + 2*(4/81) = (28+8)/81 = 36/81 = 4/9

Evaluate Z at each corner point:

• Z(A) = 3(0) - 4(8) = -32
• Z(B) = 3(4) - 4(10) = -28
• Z(C) = 3(6) - 4(8) = -14
• Z(D) = 3(6) - 4(5) = -2
• Z(E) = 3(4) - 4(0) = 12

The maximum value of Z is 12 at point E(4,0).

For the maximum value of Z to occur at both B(4, 10) and C(6, 8), the objective function line px + qy = k (where k is a constant) must coincide with the line segment BC. This is because all points along that edge would then give the same maximum value.

The slope of BC is (10-8)/(4-6) = 2/(-2) = -1

Also the objective function line must be parallel to the line segment BC. The equation of line BC is: y-10 = -1(x-4) or x+y = 14

The slope of objective function px + qy = k is -p/q

The slopes must be equal : -p/q = -1 => p=q

∫₀^π/4 x / (1 + cos 2x + sin 2x) dx

1+cos2x + sin2x = 2cos²x + sin2x = 1+cos²x - sin²x + 2sinxcosx = 1+cos2x + sin2x

= 1+ cos2x + sin2x = 1+√2 sin(2x + π/4)

Substitute u = 2x + π/4, so du = 2dx or dx = du/2

Limits: x=0 => u = π/4 and x = π/4 => u = 3π/4

(1/2)∫_π/4^3π/4 x/(1+√2sinu) du = (1/4)∫_π/4^3π/4 (u-π/4)/(1+√2sinu) du

Let I = ∫ e^x [ 1/(1 + x²)^3/2 + x/√(1 + x²) ] dx

Let the integrand e^x [ 1/(1 + x²)^3/2 + x/√(1 + x²) ] be f(x)

Consider g(x) = e^x /√(1 + x²)

g'(x) = e^x /√(1 + x²) + e^x (-1/2)(1+x²)^-3/2(2x)

g'(x) = e^x /√(1 + x²) - xe^x/(1+x²)^3/2

g'(x) = e^x(1+x²-x)/√(1 + x²)*(1 + x²) = f(x)

Therefore I = ∫f(x)dx = g(x)+c = e^x /√(1 + x²)+c

One-one (Injective):

Assume f(x₁) = f(x₂)

(x₁ - 3)/(x₁ - 5) = (x₂ - 3)/(x₂ - 5)

(x₁ - 3)(x₂ - 5) = (x₂ - 3)(x₁ - 5)

x₁x₂ - 5x₁ - 3x₂ + 15 = x₁x₂ - 5x₂ - 3x₁ + 15

-5x₁ - 3x₂ = -5x₂ - 3x₁

2x₁ = 2x₂

x₁ = x₂

Therefore, f is one-one.

Onto (Surjective):

Let y ∈ B (i.e., y ≠ 1). We need to find x ∈ A such that f(x) = y.

y = (x - 3)/(x - 5)

y(x - 5) = x - 3

yx - 5y = x - 3

x(y - 1) = 5y - 3

x = (5y - 3)/(y - 1)

Since y ≠ 1, the denominator is not zero, so x is defined. Also if we assume y belongs to B and substitute x=5 we have 5-3/5-5 = y => 2/0 = y which means y cannot belong to B. Thus x≠5. So x ∈ A. Therefore, f is onto.

Reflexive:

For all a ∈ R, (a, a) ∈ S if a - a + √2 is irrational.

a - a + √2 = √2, which is irrational. So, S is reflexive.

Symmetric:

If (a, b) ∈ S, then a - b + √2 is irrational. For symmetry, we need to check if (b, a) ∈ S, i.e., if b - a + √2 is irrational.

Let a = 1 and b = 1+√2

a - b + √2 = 1 - (1+√2) + √2 = 0 which is rational. So (a,b) does not belong to S. So not symmetric.

Now consider a=1, b=2 => a-b+√2 = -1+√2 which is irrational

b-a+√2 = 1+√2 which is irrational

Transitive:

If (a, b) ∈ S and (b, c) ∈ S, then a - b + √2 and b - c + √2 are irrational. We need to check if (a, c) ∈ S, i.e., if a - c + √2 is irrational.

Let a - b + √2 = I₁ (irrational) and b - c + √2 = I₂ (irrational)

There's no general rule that the sum or difference of two irrational numbers is always irrational.

Consider a-c+√2 = a - b + √2 + b - c + √2 - √2 = I₁ + I₂ - √2

For example, if I₁ = 1+√2 and I₂ = 1-√2, then I₁ + I₂ - √2 = 2-√2. which is irrational In general since √2 is irrational if a-b+√2 is irrational => a-b is irrational and b-c is irrational. So if a-b+√2 is irrational => a-b is irrational and b-c+√2 is irrational => b-c is irrational. But this does not imply a-c is irrational. If we take a-b = 1-√2, b-c = √3, a-c = 1-√2+√3 which is an irrational number.

But if a-b = √2 and b-c = 1-√2 then a-c = 1 which is rational

Consider the case where a - b + √2 = √3 and b - c + √2 = -√3 + r, where r is rational.
Then, a - c + √2 = √3 + ( -√3 + r - √2 + √2) + √2 - √2 = √3 - √3 + r = r. Which is rational. Hence not transitive.

1. Find a point on the given line:

Let x/2 = (2y-6)/4 = (1-z)/-1 = k

If we set k=0, we get x = 0, 2y-6 = 0 => y=3 and 1-z = 0 => z=1

So, a point on the given line is P(0, 3, 1).

2. Find the direction vector of the lines:

The direction ratios of the given line are 2, 2, -1. So, the direction vector is ⃗d = <2, 2, -1>.

Since both lines are parallel, the second line also has the same direction vector: ⃗d = <2, 2, -1>

3. Vector connecting the given point to a point on the line:

The parallel line passes through Q(4, 0, -5). The vector connecting P to Q is:

⃗PQ = <4-0, 0-3, -5-1> = <4, -3, -6>

4. Distance between the lines:

The shortest distance between two parallel lines is given by the length of the projection of the vector ⃗PQ onto a vector perpendicular to the lines. This can be written as |⃗PQ x ⃗d|/|⃗d|

⃗PQ x ⃗d = <(-3)(-1) - (-6)(2), (-6)(2) - (4)(-1), (4)(2) - (-3)(2)> = <3+12,-12+4,8+6> = <15,-8,14>

|⃗PQ x ⃗d| = √(15² + (-8)² + 14²) = √(225+64+196) = √485

|⃗d| = √(2²+2²+(-1)²) = √9 = 3

Distance = √485/3

1. Direction vectors:

The direction vector of the first line is ⃗d₁ = <-3, 2, 2k>.

The direction vector of the second line is ⃗d₂ = <3k, 1, -5>

2. Condition for perpendicularity:

For the lines to be perpendicular, the dot product of their direction vectors must be zero:

⃗d₁ ⋅ ⃗d₂ = (-3)(3k) + (2)(1) + (2k)(-5) = 0

-9k + 2 - 10k = 0

-19k = -2

k = 2/19

3. Direction vector of the perpendicular line:

A vector perpendicular to both ⃗d₁ and ⃗d₂ is given by their cross product ⃗d₁ × ⃗d₂.

⃗d₁= <-3, 2, 4/19>

⃗d₂ = <6/19,1,-5>

⃗d₁ × ⃗d₂ = <(2)(-5) - (4/19)(1), (4/19)(6/19) - (-3)(-5), (-3)(1) - (2)(6/19)> = <-10 - 4/19, 24/361-15,-3-12/19> = <-194/19, -5391/361,-69/19>

Since this vector is parallel to the required line it can also be written as = <-388, -361*3+24 , -138> = <-388,-1059,-138> = <-194/19,-5391/361,-69/19>

4. Vector equation of the line:

The line passes through the point (3, -4, 7). The vector equation of the line is:

⃗r = <3, -4, 7> + λ⃗n

where ⃗n = <-194/19,-5391/361,-69/19> or <-388,-1059,-138>

1. Finding A⁻¹:

|A| = 1(3 - 0) - 2(2+1) + 1(0-3) = 3 - 6 -3 = -6

Cofactor matrix of A:

[[3,-3,-3],[-2,0,2],[7,3,-1]]

Adjoint of A (transpose of cofactor matrix):

[[3,-2,7],[-3,0,3],[-3,2,-1]]

A⁻¹ = (Adjoint of A)/|A| = (-1/6)[[3,-2,7],[-3,0,3],[-3,2,-1]] = [[-(1/2),1/3,-(7/6)],[1/2,0,-(1/2)],[1/2,-(1/3),1/6]]

2. Solving the system of equations:

The system of equations can be written as AX = B, where

A = [[1, 2, 1], [2, 3, -1], [1, 0, 1]]

X = [[x], [y], [z]]

B = [[5], [1], [8]]

X = A⁻¹B = (-1/6)[[3,-2,7],[-3,0,3],[-3,2,-1]] * [[5],[1],[8]] = (-1/6)[[15-2+56],[-15+0+24],[-15+2-8]] = (-1/6)[[69],[9],[-21]] = [[-23/2],[-3/2],[7/2]]

Thus, x = 2 , y = -1 and z = 3

Given the equations are:

x + 2y + z = 5

2x + 3y - z = 1

x - y + z = 8

From A^-1, we have

x = (-1/2)*5 + (1/3)*1 + (-7/6)*8 = -5/2 + 1/3 - 28/3 = -15+2-56/6 = -69/6 = -23/2

y = (1/2)*5 - (1/2)*8 = 5/2 -4 = -3/2

z = (1/2)*5 + (-1/3)*1 + (1/6)*8 = 5/2 -1/3 + 4/3 = 1/2 + 1 = 3/2

y = x|x| can be defined piecewise as:

y = x² for x ≥ 0

y = -x² for x < 0

The graph is symmetric about the origin. The area bounded by the curve, the x-axis, and x=-2 and x=2 is given by:

Area = ∫_-2⁰ -x² dx + ∫₀² x² dx

= [-x³/3]_-2⁰ + [x³/3]₀²

= (0 - (-(-2)³/3)) + (2³/3 - 0) = 8/3 + 8/3 = 16/3 square units.

The equation of the ellipse can be rewritten as:

x²/25 + y²/9 = 1

Solving for y:

y² = 9(1 - x²/25) = (9/25)(25 - x²)

y = ±(3/5)√(25 - x²)

The area bounded by the ellipse, x=-2, x=2 and the x-axis is:

Area = 2∫₀² (3/5)√(25 - x²) dx = (6/5)∫₀² √(25 - x²) dx

= (6/5)[(x/2)√(25-x²) + (25/2)sin⁻¹(x/5)]₀²

=(6/5)[√21 + (25/2)sin⁻¹(2/5)]

= (6√21)/5 + 15sin⁻¹(2/5)

Part (i): Probability that at least one of them is selected

P(at least one) = 1 - P(none selected).

P(none) = (1 - 1/5) * (1 - 1/3) * (1 - 1/4) = (4/5) * (2/3) * (3/4) = 2/5.

Thus, P(at least one) = 1 - 2/5 = 3/5.

Final Answer: 3/5.

Part (ii): P(G | H)

P(G | H) = P(G ∩ H) / P(H).

P(H) = 4/5, P(G) = 1/3.

P(G ∩ H) = (1/3) * (4/5) = 4/15.

P(G | H) = (4/15) / (4/5) = 1/3.

Final Answer: 1/3.

Part (iii): Probability that exactly one of them is selected

P(exactly one) = (1/10) + (1/5) + (2/15) = 13/30.

Final Answer: 13/30.

Part (iii) OR: Probability that exactly two of them are selected

P(exactly two) = (1/20) + (1/30) + (1/15) = 3/20.

Final Answer: 3/20.

(i) Determine the number of units x that should be sold to maximize the revenue R(x) = x·p(x). Also, verify the result.

The demand function is given as:

.

The revenue function R(x) is the product of price and quantity sold:.

Simplifying the revenue function:.

To maximize revenue, we take the derivative of R(x) with respect to x, set it equal to zero, and solve for x:.

Set the derivative equal to zero:.

To verify if this is a maximum, we check the second derivative of R(x):.

Since the second derivative is negative, the function R(x) is concave down, confirming that x = 450 maximizes the revenue.

Final Answer: The number of units x that should be sold to maximize revenue is 450.

From the demand function:.

Substitute x = 450 into the demand function to find the price at the maximum revenue point:.

The original price was 350, so the rebate is:.

Final Answer: The rebate in the price of the calculator that the store should give to maximize revenue is 125.

Part (i): Distance between star V and star A

The distance formula is given by:

|VA| = |A - V|

Compute A - V:

A - V = (7 - (-3), 5 - 7, 8 - 11) = (10, -2, -3).

The magnitude is:

|A - V| = sqrt(10² + (-2)² + (-3)²) = sqrt(100 + 4 + 9) = sqrt(113).

Final Answer: Distance between V and A is sqrt(113) units.

Part (ii): Unit vector in the direction of DA

The vector DA is given by: DA = A - D.

Compute A - D:

A - D = (7 - 2, 5 - 3, 8 - 4) = (5, 2, 4).

The magnitude of DA is:

|DA| = sqrt(5² + 2² + 4²) = sqrt(25 + 4 + 16) = sqrt(45) = 3 sqrt(5).

The unit vector is:

uDA = (1 / 3 sqrt(5)) (5, 2, 4).

Simplify:

uDA = (5 / 3 sqrt(5), 2 / 3 sqrt(5), 4 / 3 sqrt(5)).

Final Answer: Unit vector in the direction of DA is (5 / 3 sqrt(5), 2 / 3 sqrt(5), 4 / 3 sqrt(5)).

Part (iii): Measure of angle VDA

The cosine of the angle is given by:

cos(θ) = (DV · DA) / (|DV| |DA|).

First, compute DV:

DV = V - D = (-3 - 2, 7 - 3, 11 - 4) = (-5, 4, 7).

The dot product DV · DA is:

DV · DA = (-5)(5) + (4)(2) + (7)(4) = -25 + 8 + 28 = 11.

The magnitude of DV is:

|DV| = sqrt((-5)² + (4)² + (7)²) = sqrt(25 + 16 + 49) = sqrt(90) = 3 sqrt(10).

The magnitude of DA is |DA| = 3 sqrt(5) (calculated earlier).

Substituting:

cos(θ) = 11 / (3 sqrt(10) * 3 sqrt(5)) = 11 / (9 sqrt(50)) = 11 / (45 sqrt(2)).

Final Answer: θ = cos⁻¹(11 / 45 sqrt(2)).

Part (iii) OR: Projection of DV on DA

The projection formula is:

Projection = (DV · DA) / |DA|.

Substituting the values:

Projection = 11 / (3 sqrt(5)).

Final Answer: Projection of DV on DA is 11 / (3 sqrt(5)).