A Level Maths Practice Questions with Solutions

(a) & (b) Converting to Standard Form:

To identify geometric properties of circles, reorganize the equation into the standard form: \[ (x-h)^2 + (y-k)^2 = r^2 \] where \( (h,k) \) matches the center, and \( r \) represents the radius. Complete the square independently for both variables:

\[ (x^2 - 4x) + (y^2 + 6y) = 12 \] \[ \left[(x-2)^2 - 4\right] + \left[(y+3)^2 - 9\right] = 12 \] \[ (x-2)^2 + (y+3)^2 = 12 + 4 + 9 \] \[ (x-2)^2 + (y+3)^2 = 25 \quad \text{--- (Equation I)} \]

Extract parameters by direct comparison with our standard template:

• \( r^2 = 25 \implies r = 5 \) units.
• Area: \( \pi r^2 = 25\pi \) square units.
• Center (b): The coordinates of center \( C \) are \( (2, -3) \).

(c) Determining Coordinates of Endpoint B:

Because line segment \( AB \) constitutes a geometric diameter, the circle's center \( C(2,-3) \) serves explicitly as the midpoint of endpoints \( A(6,-6) \) and \( B(x_0, y_0) \). Use the midpoint theorem:

\[ \dfrac{x_0 + 6}{2} = 2 \implies x_0 + 6 = 4 \implies x_0 = -2 \] \[ \dfrac{y_0 - 6}{2} = -3 \implies y_0 - 6 = -6 \implies y_0 = 0 \]

The coordinates for diametric opposite point \( B \) are \( (-2, 0) \).

We work from the more complex Left-Hand Side (LHS) and resolve to the Right-Hand Side (RHS) through identity substitution.

Step 1: Expand compound angles. Substitute the double-angle identity \( \sin(2x) = 2\sin x\cos x \) directly into the numerator expression:

\[ \text{LHS} = \dfrac{ 2 \cos x + 2 \sin x \cos x}{2\sin x + 2 \sin^2 x} \]

Step 2: Factor algebraic groups. Pull out structural common factors in both halves of the fraction bar (\( 2\cos x \) up top, \( 2\sin x \) below):

\[ \text{LHS} = \dfrac{ 2 \cos x (1 + \sin x)}{2 \sin x (1 + \sin x)} \]

Step 3: Simplify. Given \( 1 + \sin x \neq 0 \), eliminate matching scalar fractions and components across boundaries:

\[ \text{LHS} = \dfrac{\cos x}{\sin x} = \cot x = \text{RHS} \quad \blacksquare \]

(a) Factoring the Cubic Equation:

Since \( f(1) = 0 \), the factor theorem dictates that \( (x - 1) \) acts as a structural factor of the expression. Use polynomial division to reduce degrees:

\[ \dfrac{x^3 - 2x^2 - 5x + 6}{x - 1} = x^2 - x - 6 \]

Factor the resultant quadratic expression by finding values summing to \( -1 \) that yield a product of \( -6 \) (\(-3\) and \(2\)):

\[ x^2 - x - 6 = (x - 3)(x + 2) \]

Thus, written completely as linear products: \( f(x) = (x - 1)(x - 3)(x + 2) \).

(b) Finding the Roots:

Set linear product partitions cleanly to zero: \( (x - 1)(x - 3)(x + 2) = 0 \).
Solutions: \( x = 1, \; x = 3, \; x = -2 \).

(c) Transforming the Exponential Equation:

Analyze reciprocal properties to clarify exponential states: rewrite \( \frac{5}{e^{2x}} \) as \( 5e^{-2x} \). The entire sequence reads: \[ e^{-6x} - 2e^{-4x} - 5e^{-2x} + 6 = 0 \]

Define a dummy variable mapping parameters: let \( z = e^{-2x} \). Squaring gives \( z^2 = e^{-4x} \) and cubing yields \( z^3 = e^{-6x} \). Substitute these in:

\[ z^3 - 2z^2 - 5z + 6 = 0 \]

This matches our exact cubic template solved during part (b). Thus, possible values for \( z \) are \( 1, 3, \) or \( -2 \). Re-evaluate back to structural variables:

1. \( e^{-2x} = 1 \implies -2x = \ln(1) \implies x = 0 \)
2. \( e^{-2x} = 3 \implies -2x = \ln(3) \implies x = -\dfrac{\ln 3}{2} \approx -0.5493 \)
3. \( e^{-2x} = -2 \implies \) No real solutions, since real natural exponents remain strictly positive (\( e^{-2x} > 0 \)).

Final Real Solution Set: \( \left\{ 0, \; -\dfrac{\ln 3}{2} \right\} \).

(a) Solving the Quadratic Equation:

Solve via the quadratic formula where \( a = 1, \; b = \frac{\sqrt{2}-1}{2}, \; c = -\frac{\sqrt{2}}{4} \):

\[ \Delta = b^2 - 4ac = \left(\frac{\sqrt{2}-1}{2}\right)^2 - 4(1)\left(-\frac{\sqrt{2}}{4}\right) \] \[ \Delta = \frac{2 - 2\sqrt{2} + 1}{4} + \frac{4\sqrt{2}}{4} = \frac{3 + 2\sqrt{2}}{4} = \left(\frac{\sqrt{2}+1}{2}\right)^2 \]

Extract roots tracking radical changes cleanly:

\[ x = \dfrac{-\frac{\sqrt{2}-1}{2} \pm \frac{\sqrt{2}+1}{2}}{2} \] \[ x_1 = \dfrac{1}{2}, \quad x_2 = -\dfrac{\sqrt{2}}{2} \]

(b) Proving the Double-Angle Statement:

Recall standard cosine double-angle transformations: \( \cos(2x) = 1 - 2\sin^2 x \). Substitute directly into our standard formulation:

\[ \text{LHS} = \cos(2x) + 1 = (1 - 2\sin^2 x) + 1 = 2 - 2\sin^2 x = \text{RHS} \quad \blacksquare \]

(c) Analyzing the Transformed Periodic Equation:

Substitute our part (b) verified identity into the compound problem structure:

\[ 4 - 2(2 - 2\sin^2 x) + 2(\sqrt{2}-1)\sin x - \sqrt{2} = 0 \] \[ 4 - 4 + 4\sin^2 x + 2(\sqrt{2}-1)\sin x - \sqrt{2} = 0 \] \[ 4\sin^2 x + 2(\sqrt{2}-1)\sin x - \sqrt{2} = 0 \]

Divide down completely by structural factor \( 4 \) to establish standard structures:

\[ \sin^2 x + \dfrac{\sqrt{2}-1}{2}\sin x - \dfrac{\sqrt{2}}{4} = 0 \]

Let \( z = \sin x \). The equation perfectly tracks the quadratic structural baseline solved during part (a). Hence:

\[ \sin x = \dfrac{1}{2} \quad \text{or} \quad \sin x = -\dfrac{\sqrt{2}}{2} \]

Evaluate these within domain requirements across periodic loops \( [0, 2\pi) \):

• \( \sin x = \frac{1}{2} \implies x = \dfrac{\pi}{6}, \; \dfrac{5\pi}{6} \)
• \( \sin x = -\frac{\sqrt{2}}{2} \implies x = \dfrac{5\pi}{4}, \; \dfrac{7\pi}{4} \)

Combined Solution Set: \( \left\{ \dfrac{\pi}{6}, \; \dfrac{5\pi}{6}, \; \dfrac{5\pi}{4}, \; \dfrac{7\pi}{4} \right\} \).

(a) Graphing Key Periodic Loops:

The standard period for transformations sizing systems scales explicitly: \( T = \frac{2\pi}{2} = \pi \). Calculate values across core structural intervals:

\( x \)	\( 0 \)	\( \frac{\pi}{4} \)	\( \frac{\pi}{2} \)	\( \frac{3\pi}{4} \)	\( \pi \)
\( \sin(2x) \)	\( 0 \)	\( 1 \)	\( 0 \)	\( -1 \)	\( 0 \)

(b) Solving the Exact Statement:

Evaluate trigonometric properties directly over scaling angles: \[ \sin(2x) = \dfrac{1}{2} \implies 2x = \dfrac{\pi}{6} + 2k\pi \quad \text{or} \quad 2x = \dfrac{5\pi}{6} + 2k\pi \]

Isolate structural metrics inside the domain parameters \( [0, \pi) \): \( x = \dfrac{\pi}{12} \) and \( x = \dfrac{5\pi}{12} \).

(c) Resolving Inequalities Dynamically:

Isolate algebraic entities: \( \frac{1}{2} + \sin(2x) \ge 1 \implies \sin(2x) \ge \dfrac{1}{2} \).

By mapping boundaries from our exact values evaluated in part (b), the sine function remains strictly above the threshold level between its entry and exit intersections.

Interval Output: \( \left[ \dfrac{\pi}{12}, \; \dfrac{5\pi}{12} \right] \).

A composite function evaluates to zero if and only if its inner argument matches an established root of the base function. Thus, the inner polynomial must equal either \( 1 \) or \( -3 \).

Case 1: Evaluate argument matching the root at 1:

\[ -2x^2 + x + 3 = 1 \implies 2x^2 - x - 2 = 0 \] \[ x = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-2)}}{2(2)} = \dfrac{1 \pm \sqrt{17}}{4} \]

Case 2: Evaluate argument matching the root at -3:

\[ -2x^2 + x + 3 = -3 \implies 2x^2 - x - 6 = 0 \] \[ (2x + 3)(x - 2) = 0 \implies x = -\dfrac{3}{2}, \quad x = 2 \]

Combined Solution Set for structural composite zeros: \( \left\{ -\dfrac{3}{2}, \; 2, \; \dfrac{1 - \sqrt{17}}{4}, \; \dfrac{1 + \sqrt{17}}{4} \right\} \).

(a) Tracking Simultaneous Coefficient Equations:

Apply the factor theorem: set intercept coordinates directly to zero values.

\[ f(1) = 0 \implies 1 + a + b + 4 = 0 \implies a + b = -5 \] \[ f(-1) = 0 \implies 1 - a + b + 4 = 0 \implies -a + b = -5 \]

Adding the simultaneous linear pairs isolates the constants cleanly: \( 2b = -10 \implies b = -5 \). Substituted back, it resolves to \( a = 0 \). Therefore, parameters are \( a = 0, \; b = -5 \).

(b) Complete Factorization Breakdown:

Substitute calculated metrics to reveal the true polynomial expression: \[ f(x) = x^4 - 5x^2 + 4 \]

This expression functions as a quadratic in terms of \( x^2 \). Factoring directly yields:

\[ f(x) = (x^2 - 1)(x^2 - 4) \]

Apply difference-of-squares expansions across both quadratic groups to find the linear products: \( f(x) = (x - 1)(x + 1)(x - 2)(x + 2) \).

Curve Analysis: Because the function features purely even exponents (\( f(-x) = f(x) \)), the graph exhibits perfect reflectional symmetry across the vertical y-axis. The curve crosses the horizontal boundary at \( \pm 1 \) and \( \pm 2 \), intercepts the vertical axis at \( (0, 4) \), and rises infinitely on both outer limbs due to its positive leading quartic term.

Step 1: Isolate the radical line passing through AB. Expand both expressions completely to eliminate quadratic groupings:

\[ \text{Circle 1: } x^2 + y^2 - 4y + 4 = 16 \implies x^2 + y^2 - 4y = 12 \] \[ \text{Circle 2: } x^2 + 2x + 1 + y^2 = 9 \implies x^2 + y^2 + 2x = 8 \]

Subtract the second simplified equation from the first to eliminate quadratic terms: \[ (x^2 + y^2 - 4y) - (x^2 + y^2 + 2x) = 12 - 8 \]

\[ -2x - 4y = 4 \implies y = -\dfrac{1}{2}x - 1 \quad \text{--- (Line Equation I)} \]

Step 2: Find the line tracing center points C1 to C2. Extract coordinate values from the standard circle equations: center \( C_1 = (0, 2) \) and center \( C_2 = (-1, 0) \). Determine the slope gradient:

\[ m = \dfrac{2 - 0}{0 - (-1)} = 2 \]

Using point-slope form with \( C_2(-1,0) \), the linear tracking equation is: \[ y - 0 = 2(x + 1) \implies y = 2x + 2 \quad \text{--- (Line Equation II)} \]

Step 3: Solve the simultaneous system for point M. Equate Equation I and Equation II:

\[ 2x + 2 = -\dfrac{1}{2}x - 1 \implies \dfrac{5}{2}x = -3 \implies x = -\dfrac{6}{5} \] \[ y = 2\left(-\dfrac{6}{5}\right) + 2 = -\dfrac{12}{5} + \dfrac{10}{5} = -\dfrac{2}{5} \]

Intersection coordinates for point \( M \) are \( \left( -\dfrac{6}{5}, \; -\dfrac{2}{5} \right) \).

(a) Radicand Restructuring:

Halve the linear parameter value, square it, and insert balance elements to complete the square: \[ x^2 - 2x - 3 = (x - 1)^2 - 1 - 3 = (x - 1)^2 - 4 \]

(b) Tracking Inverse Algebra:

Equate the target variable to the restructured functional statement, then swap variables to solve for the inverse relation:

\[ x = \sqrt{(y - 1)^2 - 4} \]

Square both sides and isolate the squared term containing \( y \):

\[ x^2 = (y - 1)^2 - 4 \implies (y - 1)^2 = x^2 + 4 \]

Take the square root of both sides, noting the dual sign possibilities: \[ y - 1 = \pm\sqrt{x^2 + 4} \implies y = 1 \pm\sqrt{x^2 + 4} \]

Since our primary definition enforces domain rules where variables remain restricted above critical limits (\( x \ge 3 \)), the inverse function must map onto outputs greater than or equal to 3. Therefore, select the positive root option:

\[ g^{-1}(x) = 1 + \sqrt{x^2 + 4} \]

(c) Deducing the Range:

By the fundamental definitions of inverse relationships, the range of an inverse function \( g^{-1} \) is identically equal to the domain constraints of its parent function \( g \).

Range Output: \( [3, \; \infty) \).

Step 1: Take derivatives across boundaries implicitly. Differentiate both sides with respect to \( x \), applying the chain rule to the terms containing \( y \):

\[ \dfrac{d}{dx}(-2x^2 + 3y^2 - 2x) = \dfrac{d}{dx}(1) \]

\[ -4x + 6y\left(\dfrac{dy}{dx}\right) - 2 = 0 \]

Step 2: Substitute the target condition. To find horizontal tangents, set \( \frac{dy}{dx} = 0 \):

\[ -4x - 2 = 0 \implies x = -\dfrac{1}{2} \]

Step 3: Solve for the corresponding y-coordinates. Substitute \( x = -\frac{1}{2} \) back into the original curve equation to find the valid points on the graph:

\[ -2\left(-\dfrac{1}{2}\right)^2 + 3y^2 - 2\left(-\dfrac{1}{2}\right) = 1 \] \[ -2\left(\dfrac{1}{4}\right) + 3y^2 + 1 = 1 \implies -\dfrac{1}{2} + 3y^2 = 0 \] \[ 3y^2 = \dfrac{1}{2} \implies y^2 = \dfrac{1}{6} \implies y = \pm\sqrt{\dfrac{1}{6}} \]

The target coordinates matching horizontal boundaries are \( \left(-\dfrac{1}{2}, \; \sqrt{\dfrac{1}{6}}\right) \) and \( \left(-\dfrac{1}{2}, \; -\sqrt{\dfrac{1}{6}}\right) \).

(a) Difference Identity Expansion:

Apply the sine difference identity \( \sin(A - B) = \sin A\cos B - \cos A\sin B \):

\[ \sin\left(x - \dfrac{\pi}{3}\right) = \sin x \cos\dfrac{\pi}{3} - \cos x \sin\dfrac{\pi}{3} \]

Substitute known exact values (\( \cos\frac{\pi}{3} = \frac{1}{2}, \; \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} \)):

\[ \sin\left(x - \dfrac{\pi}{3}\right) = \dfrac{1}{2}\sin x - \dfrac{\sqrt{3}}{2}\cos x \]

(b) Reorganizing Linear Equations:

Equate definitions across core terms:

\[ 2\sin x + 3\cos x = \dfrac{1}{2}\sin x - \dfrac{\sqrt{3}}{2}\cos x \]

Group matching variable sets on opposite sides of the equation:

\[ 2\sin x - \dfrac{1}{2}\sin x = -3\cos x - \dfrac{\sqrt{3}}{2}\cos x \] \[ \dfrac{3}{2}\sin x = \left( -3 - \dfrac{\sqrt{3}}{2} \right)\cos x \]

(c) Evaluating Exact Intercept Metrics:

Divide by \( \cos x \) to form a tangent expression, then solve for \( x \):

\[ \tan x = \dfrac{-3 - \frac{\sqrt{3}}{2}}{\frac{3}{2}} = -2 - \dfrac{\sqrt{3}}{3} \approx -2.5774 \]

Find the primary reference solution via the arctangent function: \[ x = \arctan(-2.5774) \approx -1.20 \text{ radians} \]

Add periodic iterations of \( \pi \) to find the valid values within our domain requirement \( [0, 2\pi) \):

• \( x_1 = -1.20 + \pi \approx \mathbf{1.94} \text{ rad} \)
• \( x_2 = -1.20 + 2\pi \approx \mathbf{5.08} \text{ rad} \)

(a) Double-Angle Identity Manipulation:

Combine the double-angle formula with the fundamental Pythagorean identity to isolate the squared sine term:

\[ \cos(2x) = \cos^2 x - \sin^2 x = (1 - \sin^2 x) - \sin^2 x = 1 - 2\sin^2 x \]

Rearranging the equation yields the standard power-reducing identity: \( \sin^2 x = \dfrac{1 - \cos(2x)}{2} \).

(b) Evaluating the Definite Integral:

Substitute the identity from part (a) into the integral expression and split it into two distinct components:

\[ I = \int_0^{\pi} x \left( \dfrac{1 - \cos(2x)}{2} \right) dx = \dfrac{1}{2}\int_0^{\pi} x dx - \dfrac{1}{2}\int_0^{\pi} x \cos(2x) dx \]

The first linear integral evaluates directly via the power rule:

\[ \dfrac{1}{2}\left[ \dfrac{x^2}{2} \right]_0^{\pi} = \dfrac{\pi^2}{4} \]

Solve the remaining product integral using Integration by Parts for the section \( \int_0^{\pi} x \cos(2x) dx \):

Let \( u = x \implies u' = 1 \). Let \( v' = \cos(2x) \implies v = \dfrac{1}{2}\sin(2x) \).

\[ \int_0^{\pi} x \cos(2x) dx = \left[ \dfrac{1}{2}x\sin(2x) \right]_0^{\pi} - \int_0^{\pi} \dfrac{1}{2}\sin(2x) dx \] \[ = \left[ \dfrac{1}{2}x\sin(2x) \right]_0^{\pi} - \left[ -\dfrac{1}{4}\cos(2x) \right]_0^{\pi} \]

Evaluating this boundary set from \( 0 \) to \( \pi \) results in a net value of zero, as shown below:

\[ \left(0 - 0\right) + \left( \dfrac{1}{4}\cos(2\pi) - \dfrac{1}{4}\cos(0) \right) = \dfrac{1}{4}(1) - \dfrac{1}{4}(1) = 0 \]

Combining the results from both components yields the final evaluated value: \( I = \dfrac{\pi^2}{4} \).

The region is bounded above by the parabola \( y_{top} = x^2 + k \) and below by the line \( y_{bottom} = x - 1 \), spanning the interval from \( x = 1 \) to \( x = 2 \). Set up the area integral:

\[ A = \int_1^2 \left( (x^2 + k) - (x - 1) \right) dx \] \[ A = \int_1^2 (x^2 - x + k + 1) dx \]

Integrate each term using the power rule:

\[ A = \left[ \dfrac{1}{3}x^3 - \dfrac{1}{2}x^2 + (k + 1)x \right]_1^2 \]

Evaluate the expression at the upper bound (\( x = 2 \)) and subtract the evaluation at the lower bound (\( x = 1 \)):

\[ \text{At } x=2: \quad \dfrac{8}{3} - 2 + 2k + 2 = 2k + \dfrac{8}{3} \] \[ \text{At } x=1: \quad \dfrac{1}{3} - \dfrac{1}{2} + k + 1 = k + \dfrac{5}{6} \] \[ A = \left(2k + \dfrac{8}{3}\right) - \left(k + \dfrac{5}{6}\right) = k + \dfrac{11}{6} \]

Set the evaluated area expression equal to the target value of 5 and solve for \( k \):

\[ k + \dfrac{11}{6} = 5 \implies k = 5 - \dfrac{11}{6} = \dfrac{19}{6} \]

The first-principles definition of a derivative states: \[ f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h} \]

Substitute \( f(x) = \sin x \) and evaluate at the target point \( x = \dfrac{\pi}{4} \):

\[ \sin'\left(\dfrac{\pi}{4}\right) = \lim_{h \to 0} \dfrac{\sin\left(\frac{\pi}{4}+h\right) - \sin\left(\frac{\pi}{4}\right)}{h} \]

Expand the compound angle in the numerator using the sine addition identity:

\[ \sin\left(\frac{\pi}{4}\right)\cosh + \cos\left(\frac{\pi}{4}\right)\sinh - \sin\left(\frac{\pi}{4}\right) \]

Reorganize and split the expression into two standard trigonometric limits:

\[ = \sin\left(\dfrac{\pi}{4}\right) \lim_{h \to 0} \left(\dfrac{\cos h - 1}{h}\right) + \cos\left(\dfrac{\pi}{4}\right) \lim_{h \to 0} \left(\dfrac{\sin h}{h}\right) \]

Apply the fundamental geometric limit theorems: \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \) and \( \lim_{h \to 0} \frac{\cos h - 1}{h} = 0 \).

\[ \sin'\left(\dfrac{\pi}{4}\right) = \sin\left(\dfrac{\pi}{4}\right) \cdot (0) + \cos\left(\dfrac{\pi}{4}\right) \cdot (1) = \cos\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2} \]

(a) Derivative and Integral Framework:

Differentiate \( f(x) = [u(x)]^{-1} \) using the power and chain rules:

\[ f'(x) = -1[u(x)]^{-2} \cdot u'(x) = -\dfrac{u'(x)}{u^2(x)} \]

Integrating both sides yields the corresponding integration template:

\[ \int \dfrac{u'(x)}{u^2(x)} dx = -\dfrac{1}{u(x)} + C \]

(b) Applying the Template to the Core Integral:

Reorganize the given integral to match the structural template found in part (a):

\[ \int_2^4 \dfrac{1}{x \ln^2 x} dx = \int_2^4 \dfrac{\left(\frac{1}{x}\right)}{(\ln x)^2} dx \]

Let \( u(x) = \ln x \), which implies its derivative is exactly \( u'(x) = \frac{1}{x} \). This fits the pattern perfectly. Integrate using the template:

\[ \int_2^4 \dfrac{u'(x)}{u^2(x)} dx = \left[ -\dfrac{1}{\ln x} \right]_2^4 = \left( -\dfrac{1}{\ln 4} \right) - \left( -\dfrac{1}{\ln 2} \right) \]

Simplify the result using the logarithm power rule, noting that \( \ln 4 = \ln(2^2) = 2\ln 2 \):

\[ = -\dfrac{1}{2\ln 2} + \dfrac{1}{\ln 2} = \dfrac{-1 + 2}{2\ln 2} = \dfrac{1}{2\ln 2} \]

Step 1: Write the circle's equation and find its derivative. Write the standard equation using the given radius and center, then differentiate implicitly:

\[ (x-3)^2 + (y+1)^2 = 4 \quad \text{--- (Equation I)} \]

\[ 2(x-3) + 2(y+1)\dfrac{dy}{dx} = 0 \implies (x-3) + (y+1)\dfrac{dy}{dx} = 0 \]

Step 2: Substitute the target slope condition. Set the derivative equal to the desired slope, \( \frac{dy}{dx} = 2 \):

\[ (x-3) + 2(y+1) = 0 \implies (x-3) = -2(y+1) \quad \text{--- (Equation II)} \]

Step 3: Solve the simultaneous system. Substitute the expression for \( (x-3) \) from Equation II back into Equation I:

\[ \left(-2(y+1)\right)^2 + (y+1)^2 = 4 \] \[ 4(y+1)^2 + (y+1)^2 = 4 \implies 5(y+1)^2 = 4 \] \[ (y+1)^2 = \dfrac{4}{5} \implies y+1 = \pm\dfrac{2}{\sqrt{5}} \implies y = -1 \pm\dfrac{2}{\sqrt{5}} \]

Find the corresponding x-coordinates by substituting these y-values back into Equation II:

\[ x - 3 = -2\left(\pm\dfrac{2}{\sqrt{5}}\right) \implies x = 3 \mp \dfrac{4}{\sqrt{5}} \]

The two points on the circle with a tangent slope of 2 are:

\( \left(3 - \dfrac{4}{\sqrt{5}}, \; -1 + \dfrac{2}{\sqrt{5}}\right) \) and \( \left(3 + \dfrac{4}{\sqrt{5}}, \; -1 - \dfrac{2}{\sqrt{5}}\right) \).