Find the domain and range of the function that is the solution to the differential equation dy/dx = e^(x+y) and whose graph passes through the origin. (b) For what values of the constant k would the function f(x) = kx/(1+x^2) + arctan(x) have an inverse? (c) (i) Explain why the equation tan^{-1}(3x) + tan^{-1}(10x) = θ, where -π < θ < π, has exactly one solution. (ii) Solve tan^{-1}(3x) + tan^{-1}(10x) = 3π/4. (d) A particle is projected from the origin, with initial speed V at an angle of θ to the horizontal. The position vector of the particle, r(t), where t is the time after projection and g is the acceleration due to gravity, is given by r(t) = (Vcosθ)t i + (Vsinθ - (g/2)t^2) j (DO NOT prove this). Let D(t) be the distance of the particle from the origin at time t, so D(t) = |r(t)|. Show that for 0 0.

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Find the domain and range of the function that is the solution to the differential equation dy/dx = e^(x+y) and whose graph passes through the origin - HSC - SSCE Mathematics Extension 1 - Question 14 - 2024 - Paper 1

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Find the domain and range of the function that is the solution to the differential equation dy/dx = e^(x+y) and whose graph passes through the origin. (b) For wha... show full transcript

Worked Solution & Example Answer:Find the domain and range of the function that is the solution to the differential equation dy/dx = e^(x+y) and whose graph passes through the origin - HSC - SSCE Mathematics Extension 1 - Question 14 - 2024 - Paper 1

Step 1

Find the domain and range of the function that is the solution to the differential equation

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Answer

To solve the differential equation $rac{dy}{dx} = e^{x+y}$ we can separate the variables:

$e^{-y}dy = e^{x}dx$ .

Integrating both sides: $\int e^{-y} dy = \int e^{x} dx$ which gives us $-e^{-y} = e^{x} + C$ or $e^{-y} = -e^{x} - C$ .

We rewrite this as: $y = -ln(-e^{x} - C).$

To ensure the function passes through the origin (0,0): We set: $0 = -ln(-C)$ which implies that $C = -1$ .

Thus the specific solution can be written as: $y = -ln(-e^{x} + 1).$

Now, to determine the domain: For the logarithm to be defined,

ightarrow e^{x} < 1 ightarrow x < 0.$$ So the domain is: $$(-\infty, 0).$$ For the range, as x approaches 0, y approaches infinity, while for x approaching -infinity, y approaches negative infinity. Hence, the range is: $$(-\infty, \infty).$$

Step 2

For what values of the constant k would the function f(x) = kx/(1+x^2) + arctan(x) have an inverse?

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Answer

The function f(x) is defined as: $f(x) = \frac{kx}{1+x^2} + arctan(x).$ To have an inverse, f(x) must be either strictly increasing or strictly decreasing.

Finding f'(x) for the monotonicity:

$f'(x) = \frac{k(1+x^2) - kx(2x)}{(1+x^2)^2} + \frac{1}{1+x^2} = \frac{k(1-x^2) + 1}{(1+x^2)^2}$ .

Setting f'(x) = 0: This expression has to maintain the same sign for all x. Case 1: If (k > 0), then (1 - x^2 \geq 0) implies ( -1 < x < 1). To maintain monotonicity for all x, we need (k \leq 1). Case 2: If (k < 0), then (1 - x^2 \leq 0) implies no valid x. Thus, k must satisfy (-1 \leq k \leq 1) for f(x) to have an inverse.

Step 3

Explain why the equation tan^{-1}(3x) + tan^{-1}(10x) = θ, where -π < θ < π, has exactly one solution.

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Answer

The function tan^{-1}(x) is strictly increasing and continuous. Therefore, the sum of two such functions, tan^{-1}(3x) + tan^{-1}(10x), will also be strictly increasing. Since the range of a strictly increasing function is the entire real line, there will be exactly one intersection point with any horizontal line, confirming that the equation tan^{-1}(3x) + tan^{-1}(10x) = θ has exactly one solution, for -π < θ < π.

Step 4

Solve tan^{-1}(3x) + tan^{-1}(10x) = 3π/4.

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Answer

To solve this equation, we can use the identity for the sum of arctangents:

$tan(a + b) = \frac{tan(a) + tan(b)}{1 - tan(a)tan(b)}$ .

Setting: $a = tan^{-1}(3x), b = tan^{-1}(10x)$ Thus: $tan(3π/4) = -1$ . Substituting:

ightarrow -1 + 30x^2 = 13x$$. Rearranging gives: $$30x^2 - 13x - 1 = 0$$. Using the quadratic formula: discriminant: D = b^2 - 4ac = (-13)^2 - 4*30*(-1) = 169 + 120 = 289\ roots: x = \frac{13 \pm 17}{60} \rightarrow x = \frac{30}{60} = 0.5 \text{ or } x = \frac{-4}{60} = -\frac{1}{15}.$$ The only solution is x = \frac{1}{2}.

Step 5

Show that for 0 < sin(θ) < (√(8/9)), the distance, D(t), is increasing for all t > 0.

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Answer

We have the distance: $D(t) = |r(t)| = \sqrt{(Vcosθt)^2 + [Vsinθt - \frac{gt^2}{2}]^2}$ . To show that D(t) is increasing, we need to show that $\frac{dD}{dt} > 0.$ Using the derivatives, $\frac{dD}{dt} = \frac{1}{D} \frac{d|r(t)|}{dt},$ we find $\frac{d|r(t)|}{dt}$ where the signs of the terms depend on θ: $\frac{d|r(t)|}{dt} > 0 \implies 0 < Vsin(θ) - \frac{gt}{2} < Vsin(θ).$ Thus confirming that D(t) is increasing when (0 < sin(θ) < (\sqrt{8/9}).$$

Find the domain and range of the function that is the solution to the differential equation dy/dx = e^(x+y) and whose graph passes through the origin - HSC - SSCE Mathematics Extension 1 - Question 14 - 2024 - Paper 1

Worked Solution & Example Answer:Find the domain and range of the function that is the solution to the differential equation dy/dx = e^(x+y) and whose graph passes through the origin - HSC - SSCE Mathematics Extension 1 - Question 14 - 2024 - Paper 1

Find the domain and range of the function that is the solution to the differential equation

For what values of the constant k would the function f(x) = kx/(1+x^2) + arctan(x) have an inverse?

Explain why the equation tan^{-1}(3x) + tan^{-1}(10x) = θ, where -π < θ < π, has exactly one solution.

Solve tan^{-1}(3x) + tan^{-1}(10x) = 3π/4.

Show that for 0 < sin(θ) < (√(8/9)), the distance, D(t), is increasing for all t > 0.

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