QCAA Mathematical Methods Further applications of differentiation

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 $[0, 2]$.

This is the local minimum if the function was incorrectly dilated by a factor of $1/3$ from the $y$-axis, which would use $f(3x)$ instead of $f\left(\frac{x}{3}\right)$.

This is the local minimum if the function was dilated by a factor of 3 from the $x$-axis (evaluating $3f(x) - 1$) rather than from the $y$-axis.

This is the local minimum if the function was dilated by a factor of $1/3$ from the $x$-axis (evaluating $\frac{1}{3}f(x) - 1$) rather than from the $y$-axis.

This option is incorrect because while $f'(-1)=0$, the second derivative $f''(-1) = e^{-1}(-1+2) = e^{-1}$ is positive, not negative.

This option is incorrect because the first derivative evaluates to zero at $x=-1$, not a negative value, and the second derivative is positive.

This option is incorrect because the first derivative $f'(-1)$ equals $0$, not a value less than $0$.

The number of local minima depends on the value of the positive integer $a$ and is not a constant $2$.

This would only be correct if $a=2$, but the number of local minima changes depending on the specific value of $a$.

The number of periods in the interval is calculated by dividing the interval length by the period, which yields $a^2$, not $a$.

While the interval length is $2a\pi$, the period of the function is $2\pi/a$. Dividing these gives $a^2$ local minima, not $2a$.

This option is incorrect. It significantly underestimates the number of inflection points. You must find all points in the interval where the second derivative $f''(x)$ changes sign, not just the first positive solution.

This option is incorrect. It likely results from missing one of the valid solutions at the boundaries of the domain. The range of $x^3$ is approximately $(-5.83, 5.83)$, which includes four zeros of the cosine function.

This option is incorrect because it likely counts $x=0$ as an inflection point. Although $f''(0) = 0$, the term $3x^2$ is non-negative and $\cos(x^3)$ is positive near $0$, so $f''(x)$ does not change sign at $x=0$.

This value corresponds to $e^2$, which fails to account for the coefficients resulting from the chain rule during differentiation.

This is the value of the displacement $d$ at $t=4$ ($d = e^2 - 1$), not the acceleration.

This represents the velocity of the particle at $t=4$, found by taking the first derivative $v = \frac{dd}{dt} = 0.5e^{0.5t}$.

This incorrectly assumes the horizontal shift is to the left by $\frac{1}{4}$. The term $\left(x - \frac{1}{4}\right)$ represents a shift to the right, so the x-coordinate must be $k + \frac{1}{4}$.

While the x-coordinate is correct, the y-coordinate is incorrect. Evaluating the function at $x = k + \frac{1}{4}$ gives $y = 2 - 0 = 2$, not $-2$.

Although the x-coordinate $k - \frac{3}{4}$ generates the same set of values as $k + \frac{1}{4}$ for integers $k$, the y-coordinate evaluates to $2$, not $-2$.

This option represents where the function is concave down. Since the denominator $1+x^2$ is always positive, $f''(x)$ is negative when the numerator $2x < 0$, i.e., when $x < 0$.

This interval includes negative values where the function is concave down ($f''(x) < 0$) and the inflection point at $x=0$.

This option includes $x=0$, where $f''(x) = 0$. This point is an inflection point where the concavity changes, rather than part of the interval where the graph is strictly concave up.