Application of Derivatives: questions with answers

Line 1 is correct. Line 2 is the first wrong line: the top of the rectangle is the diameter of the semicircle and lies inside the window, so it is not part of the perimeter. The constraint is $2x + 2y + \pi x = 10$. Lines 3 and 4 describe the right method but rest on the wrong constraint.

Lines 1 and 2 are both true statements: the rule is right and the area really is $16\pi$ when $r = 4$. Line 3 is the first wrong line -- it differentiates a number. The rate must come from differentiating the rule $\pi r^2$ while $r$ is a variable, giving $2\pi r$, and only then substituting: the rate is $8\pi$, not $0$.

Lines 1 to 3 are all true: they compute $\frac{dy}{dt}$, the rate at which $y$ changes. Line 4 is the first wrong line -- it reports that as the rate of the slope. The slope is $m = 5 - 6x^2$, and $\frac{dm}{dt} = (-12x)\cdot2 = -48$ at $x = 2$.

Write $\frac{dV}{dt} = -kS$ with $k > 0$. But $V = \frac{4}{3}\pi r^3$ gives $\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} = S\frac{dr}{dt}$, so $S\frac{dr}{dt} = -kS$ and $\frac{dr}{dt} = -k$: the radius decreases at a constant rate. Its size is $k$, whatever $k$ is -- not necessarily 1; and the two rates do not cancel, they are the same rate written two ways.

Lines 1 to 3 are correct. Line 4 is the first wrong line: $f$ is increasing on each of the two intervals separately, but not on their union, because it decreases on $(0, 2)$ in between -- a point of the left piece can be higher than a point of the right piece. The answer must be stated as two intervals.

$|x| > 1$ means $x > 1$ or $x < -1$: the two outer rays $(-\infty, -1)$ and $(1, \infty)$, everything outside the middle interval $(-1, 1)$. Reading it as the middle interval reverses the set; keeping only one ray loses half of it.

A trigonometric equation has a solution in every period, and the interval $[0, 2\pi]$ is long enough to contain more than the principal one; every solution inside $[0, 2\pi]$ is a cut point. Solutions beyond the end of $[0, 2\pi]$ do not belong: the given interval bounds the analysis.

The largest value of $f'(x) = a - \cos x$ occurs at $\cos x = -1$. It must be negative, giving $a < -1$. Checking the smallest derivative value instead gives $a < 1$. Replacing the trigonometric term by zero gives $a < 0$.

A is true: for $x_1 < x_2$ one has $f(x_1) < f(x_2)$, and $f'(x) = 5x^4 \ge 0$ with a single zero confirms it. R is false as stated: at $x = 0$, $f'(0) = 0$, so $f'$ is not positive for all $x$. A true assertion can have a false reason, because increasing does not require $f' > 0$ everywhere; and the zero of $f'$ does not make A false.

Yes. 'Local' means largest or smallest among nearby values only. Near $x = -2$ every value of $f$ is at most $-4$, and near $x = 2$ every value is at least $4$; the two neighbourhoods lie on opposite sides of the gap at $x = 0$, so the values need not compare the way the words suggest. A local maximum can be smaller than a local minimum.

A is true: the squared factor keeps $f'$ negative on both sides of $x = -2$, so there is no sign change and no extremum. R is true as a computation. But R does not explain A -- $f'(-2) = 0$ is exactly what makes $x = -2$ a candidate; the reason there is no extremum is the absence of a sign change. So both are true and R is not the correct explanation of A.

Lines 1 and 2 are correct. Line 3 is the first wrong line: $x$ is simply the variable of the given function and nothing restricts it to positive values, so $x = -\frac{5}{2}$ is a legitimate critical point. With $p''(-\frac{5}{2}) = -16 < 0$ it is a maximum, value $p(-\frac{5}{2}) = 75$; Line 4 is false.

A local minimum is an interior point lower than its neighbours on both sides -- a valley. The graph has two valleys, so two local minima, even though only one of them is the absolute minimum. End points are not local extrema.

$|x| \ge 0$ with equality at $x = 0$, so $f$ has a minimum value $-3$ at the corner $x = 0$. On the other side $|x|$ grows without bound, so $f$ can be made as large as we please and has no maximum value. The inner expression $x$ has no bound at all, but the modulus supplies one end; and the corner's $x$-coordinate is not the value there.

Lines 1 and 2 are correct. Line 3 is the first wrong line: $y$ is not a constant -- it is $15 - x$ -- and the constraint must be used to write $P = f(x) = x^2 + (15 - x)^2$ before differentiating. Then $f'(x) = 4x - 30 = 0$ at $x = \frac{15}{2}$ and Line 4 is false.