VCAA Mathematical Methods Algebra, number and structure

Incorrect. Although the algebraic expression for the inverse is correct, the domain of the inverse function must exclude $x = -2$ to avoid division by zero, not $x = 3$.

Incorrect. Simplifying this expression yields $\frac{3x+11}{x+2}$, which does not match the correct inverse function $f^{-1}(x) = \frac{3x-1}{x+2}$.

Incorrect. This expression has the wrong sign in the numerator. The correct inverse function is $f^{-1}(x) = \frac{3x-1}{x+2}$.

Incorrect. While this point lies on the graph of $f(x)$ (since $x \approx \ln(5)$ gives $f(x) = 15$), $g(1.609) \approx 0.6$. The intersection point must satisfy both equations.

Incorrect. This point lies on the graph of $g(x)$ (since $x \approx \ln(3)$ gives $g(x) = 1$), but $f(1.099) \approx 11$. The functions are not equal at this x-value.

Incorrect. This point lies on the graph of $g(x)$ (since $x \approx \ln(1.5)$ gives $g(x) = 2$), but $f(0.4065) \approx 8$. The y-values differ significantly.

When $k = \frac{3}{2}$, the ratio of the constants equals the ratio of the coefficients, meaning the lines are coincident and the system has infinitely many solutions, not zero.

While both values make the lines parallel, $k = \frac{3}{2}$ makes the lines coincident (infinitely many solutions). Only $k = -2$ makes them distinct and parallel (no solutions).

For all values of $k$ other than $-2$ and $\frac{3}{2}$, the lines have different slopes and intersect at exactly one point, yielding a unique solution rather than no solution.

This incorrectly assumes the minimum of $f(x)$ occurs at the boundary $x=1$ (giving $f(1)=-3$), missing the true minimum at the vertex $x=2$.

This incorrectly uses $f(1)=-3$ as the minimum value of the inner function, failing to account for the vertex of the parabola at $x=2$.

The maximum value $e^4$ is achieved at $x=2$, which is included in the domain $(1, \infty)$, so the interval must be closed at $e^4$.

This formula is incorrect. It does not match the roots of the derivative of the volume function, $V'(x) = 12x^2 - 4(a+b)x + ab = 0$.

While this is a root of the derivative $V'(x) = 0$, it corresponds to the local minimum of the volume function and falls outside the valid domain $0 < x < \min(a/2, b/2)$.

This formula is incorrect. It does not match the roots of the derivative of the volume function, $V'(x) = 12x^2 - 4(a+b)x + ab = 0$.

This option has an incorrect discriminant. The term under the square root should be $a^2 - ab + b^2$, which comes from simplifying the discriminant of the quadratic equation $V'(x) = 0$.

While $k = -5$ does yield infinite solutions, substituting $k = 4$ creates the equations $4x + 5y = 9$ and $4x + 5y = 0$, which are parallel lines with no solutions.

Substituting $k = 4$ results in the equations $4x + 5y = 9$ and $4x + 5y = 0$. These represent parallel lines that never intersect, meaning the system has zero solutions.

Values of $k$ other than $-5$ and $4$ make the determinant of the coefficient matrix non-zero ($k(k+1) - 20 \neq 0$), which means the system will have exactly one unique solution.

This set includes values that yield a single unique solution (any $k$ other than $-5$ or $4$) and no solutions ($k = 4$), rather than infinite solutions.

This results from a sign error when solving the equation $-11 - 8a = 5$, incorrectly yielding $8a = 16$ instead of $-8a = 16$.

This value stems from an arithmetic error when evaluating the polynomial at $x = -2$.

This results from incorrectly applying the Remainder Theorem by evaluating $p(2) = 5$ instead of $p(-2) = 5$.

This value comes from an arithmetic error, such as incorrectly evaluating the linear and constant terms of $p(-2)$.

This is the y-coordinate of the vertex (the minimum value of the function), not the x-coordinate required for the domain.

This is one of the roots of the quadratic equation ($x=-5$), not the x-coordinate of the vertex that determines where the function changes direction.

The interval $(-\infty, 0]$ includes the vertex at $x = -1.5$, meaning the function decreases and then increases within this domain, so it is not one-to-one.

This is the other root of the quadratic equation ($x=2$). The interval $(-\infty, 2]$ includes the vertex, so the function is not one-to-one.

This interval is closed at $-\frac{1}{\sqrt{2}}$, but $f\left(-\frac{1}{\sqrt{2}}\right)$ is undefined because it requires taking the natural logarithm of zero.

While this interval satisfies the domain of $(f \circ g)(x)$, it includes values between $-\frac{\pi}{4}$ and $-\frac{1}{\sqrt{2}}$ where $(g \circ f)(x)$ is undefined.

This interval includes values where $f(x)$ is undefined, and the closed endpoints would result in taking the logarithm of zero or a negative number.

This interval contains values where $f(x)$ is undefined, as the domain of $(g \circ f)(x)$ strictly requires $x > -\frac{1}{\sqrt{2}}$.