VCAA Mathematical Methods Calculus

This sequence is incorrect because the values are moving away from $0$ ($-0.05, -0.1, -0.2, \dots$), which does not help estimate the limit as $h \to 0$.

This sequence includes $0$, where the expression $\frac{a^h - 1}{h}$ is undefined (division by zero), and subsequent terms move away from $0$.

This sequence consists of increasing integers moving away from $0$, which would be used to investigate the limit as $h \to \infty$, not as $h \to 0$.

The circle's domain is $-1 \le x \le 1$, so $x = -\sqrt{3}$ is outside the circle and cannot be a point of tangency.

At $x = -1$, the tangent line to the circle is vertical, meaning its slope is undefined rather than $-1$.

This value would correspond to a different tangent slope, likely resulting from using an incorrect angle such as $120^\circ$ or $150^\circ$.

At $x = 0$, the tangent lines to the circle are horizontal (slope of 0), which corresponds to an angle of $0^\circ$ or $180^\circ$, not $135^\circ$.

This option only includes the points where $g(0)$ reaches the global maximum of $f(x)$, which is $\frac{1}{4}$. It misses the intervals where $g(0)$ is a local maximum that remains the absolute maximum on the restricted interval $[0, 1]$.

This incorrectly assumes $a$ can be any value between the critical points. For $a \in \left(-\frac{1}{2}, \frac{1}{2}\right)$, the shifted interval $[-a, 1-a]$ will contain values of $x$ where $f(x) > f(-a)$, meaning $g(0)$ is not the maximum.

This option incorrectly includes $a \ge \frac{\sqrt{2}}{2}$. For these values, $-a \le -\frac{\sqrt{2}}{2}$, and the function $f(x)$ is strictly increasing immediately to the right of $-a$, so $g(0)$ cannot be the maximum on $[0, 1]$.

This incorrectly includes $a \in \left[-\frac{\sqrt{2}}{2}, -\frac{1}{2}\right]$. For these negative values of $a$, the function $f(x)$ increases as $x$ moves to the right of $-a$, meaning $g(x)$ will exceed $g(0)$ on the interval $[0, 1]$.

This option fails to apply the reverse chain rule (u-substitution). The integral of $\sin(kx)$ is $-\frac{1}{k}\cos(kx)$, so you must divide by the coefficient $2$.

This option uses the wrong sign for the antiderivative and misses the chain rule factor. The integral of $\sin(x)$ is $-\cos(x)$, not $\cos(x)$, and the result must be divided by $2$.

This option has the wrong sign for the cosine term. Since the derivative of $\cos(x)$ is $-\sin(x)$, the antiderivative of $\sin(x)$ must be negative.

Incorrect. Evaluating the antiderivative $g(x) = \frac{x^4}{4} - \frac{x^2}{2} + 5$ at $x=2$ yields 7, not 2.

Incorrect. This might result from ignoring the constant of integration $C=5$ and calculating $g(2) = 4 - 2 = 2$, then making an arithmetic error.

Incorrect. This is the value of the initial condition $g(0)$, not the requested value $g(2)$.

Incorrect. This value does not make the total area under the probability density function equal to 1, likely resulting from an error in evaluating the trigonometric integrals.

Incorrect. This might result from incorrectly rationalizing the denominator or making an arithmetic error when solving $k(2 - \sqrt{2}) = 1$.

Incorrect. This is the value of the integral when $k=1$. Since the total area must be 1, $k$ must be the reciprocal of this value.

Incorrect. This might result from confusing the x-coordinate of the y-intercept ($x=0$) with the gradient itself.

Incorrect. This value does not match the derivative evaluated at the y-intercept.

Incorrect. This is the y-coordinate of the y-intercept ($y=e^0=1$), not the gradient. It could also result from forgetting the chain rule and incorrectly assuming the derivative is $e^{3x}$.

Incorrect. This value does not correspond to the derivative evaluated at $x=0$.

This is the value of the function $N(3) = 15 \ln(22) \approx 46$. This represents the number of koalas (or the population increase) at year 3, rather than the rate at which the population is changing.

This value appears to be the result of calculating $N(3) - 20 \approx 26$. This subtracts the initial population from the model's value at $t=3$, which does not represent the instantaneous rate of change.

This is an incorrect value. It does not correspond to the derivative at $t=3$ or the function value, likely resulting from a calculation error.

This is the minimum value of the function on the interval, which occurs at the critical point $x = 1$ since $h(1) = -e$.

This is the $x$-value of the critical point (found by setting $h'(x) = (x-1)e^x = 0$), not the maximum value of the function itself.

This is the $x$-value at which the maximum occurs, but the question asks for the maximum value of the function, which is $h(2) = 0$.

This value does not correspond to the function evaluated at any critical point or endpoint within the given interval $[0, 2]$.

Incorrect. This option incorrectly subtracts the new exponent (4) from the coefficient 10.4 instead of dividing by it.

Incorrect. This option incorrectly adds the new exponent (4) to the coefficient 10.4 instead of dividing by it.

Incorrect. This option incorrectly multiplies the coefficient 10.4 by the new exponent (4) instead of dividing by it, confusing the integration rule with the differentiation rule.