BC MCQ Set 70

Question 1 of 38 | Integration and Accumulation of Change · Level 2

$\displaystyle\int_{0}^{3} (x + 1)^{\dfrac{1}{2}} d x =$

$\dfrac{21}{2}$

$7$

$\dfrac{16}{3}$

$\dfrac{14}{3}$

$-\dfrac{1}{4}$

Question 2 of 38 | Analytical Applications of Differentiation · Level 2

If $f(x) = x + \dfrac{1}{x}$, then the set of values for which $f$ increases is

$(-\infty, -1] \cup [1, \infty)$

$[-1, 1]$

$(-\infty, \infty)$

$(0, \infty)$

$(-\infty, 0) \cup (0, \infty)$

Question 3 of 38 | Differentiation: Definition and Fundamental Properties · Level 3

For what non-negative value of $b$ is the line given by $y = -\dfrac{1}{3} x + b$ normal to the curve $y = x^3$?

$0$

$1$

$\dfrac{4}{3}$

$\dfrac{10}{3}$

$\dfrac{10 \sqrt{3}}{3}$

Question 4 of 38 | Differentiation: Definition and Fundamental Properties · Level 2

If $f(x) = \dfrac{x - 1}{x + 1}$ for all $x \neq -1$, then $f'(1) =$

$-1$

$-\dfrac{1}{2}$

$0$

$\dfrac{1}{2}$

$1$

Question 5 of 38 | Differentiation: Definition and Fundamental Properties · Level 2

If $y = \sin x$ and $y^{(n)}$ means the $n$th derivative of $y$, then the smallest positive integer $n$ for which $y^{(n)} = y$ is

$2$

$4$

$5$

$6$

$8$

Question 6 of 38 | Applications of Integration · Level 3

The length of the curve $y = \ln \sec x$ from $x = 0$ to $x = b$, where $0 < b < \dfrac{\pi}{2}$, may be expressed by which of the following integrals?

$\displaystyle\int_{0}^{b} \sec x d x$

$\displaystyle\int_{0}^{b} \sec^2 x d x$

$\displaystyle\int_{0}^{b} (\sec x \tan x) d x$

$\displaystyle\int_{0}^{b} \sqrt{1 + (\ln \sec x)^2} d x$

$\displaystyle\int_{0}^{b} \sqrt{1 + (\sec^2 x \tan^2 x)} d x$

Question 7 of 38 | Contextual Applications of Differentiation · Level 2

Let $y = x \sqrt{1 + x^2}$. When $x = 0$ and $d x = 2$, the value of $d y$ is

$-2$

$-1$

$0$

$1$

$2$

Question 8 of 38 | Integration and Accumulation of Change · Level 3

If $n$ is a known positive integer, for what value of $k$ is $\displaystyle\int_{1}^{k} x^{n-1} d x = \dfrac{1}{n}$?

$0$

$\left(\dfrac{2}{n}\right)^{\dfrac{1}{n}}$

$((2 n - 1)/n)^{\dfrac{1}{n}}$

$2^{\dfrac{1}{n}}$

$2^n$

Question 9 of 38 | Infinite Sequences and Series · Level 3

A series expansion of $\dfrac{\sin t}{t}$ is

$1 - \dfrac{t^2}{3!} + \dfrac{t^4}{5!} - \dfrac{t^6}{7!} + ...$

$\dfrac{1}{t} - \dfrac{t}{2!} + \dfrac{t^3}{4!} - \dfrac{t^5}{6!} + ...$

$1 + \dfrac{t^2}{3!} + \dfrac{t^4}{5!} + \dfrac{t^6}{7!} + ...$

$\dfrac{1}{t} + \dfrac{t}{2!} + \dfrac{t^3}{4!} + \dfrac{t^5}{6!} + ...$

$t - \dfrac{t^3}{3!} + \dfrac{t^5}{5!} - \dfrac{t^7}{7!} + ...$

Question 10 of 38 | Integration and Accumulation of Change · Level 3

The number of bacteria in a culture is growing at a rate of $3000 e^{2 \dfrac{t}{5}}$ per unit of time $t$. At $t = 0$, the number was \$7,500$. Find the number at $t = 5$.

\$1,200 e^2$

\$3,000 e^2$

\$7,500 e^2$

\$7,500 e^5$

$\dfrac{15,000}{7} e^7$

Question 11 of 38 | Limits and Continuity · Level 3

Let $g$ be a continuous function on $[0,1]$. Let $g(0)=1$ and $g(1)=0$. Which of the following is NOT necessarily true?

There exists $h$ in $[0,1]$ such that $g(h) \geq g(x)$ for all $x$ in $[0,1]$.

For all $a, b$ in $[0,1]$, if $a = b$, then $g(a) = g(b)$.

There exists $h$ in $[0,1]$ such that $g(h) = \dfrac{1}{2}$.

There exists $h$ in $[0,1]$ such that $g(h) = \dfrac{3}{2}$.

For all $h$ in $(0, 1)$, $\operatorname*{lim}\limits_{x \rightarrow h} g(x) = g(h)$.

Question 12 of 38 | Infinite Sequences and Series · Level 3

Which of the following series converge?
I. $\sum 1/n^2$
II. $\sum \dfrac{1}{n}$
III. $\sum (-1)^n/\sqrt{n}$

I only

III only

I and II only

I and III only

I, II, and III

Question 13 of 38 | Integration and Accumulation of Change · Level 2

$\int x \sqrt{4 - x^2} d x =$

$\dfrac{(4 - x^2)^{\dfrac{3}{2}}}{3} + C$

$-(4 - x^2)^{\dfrac{3}{2}} + C$

$\dfrac{x^2 (4 - x^2)^{\dfrac{3}{2}}}{3} + C$

$-\dfrac{x^2 (4 - x^2)^{\dfrac{3}{2}}}{3} + C$

$-\dfrac{(4 - x^2)^{\dfrac{3}{2}}}{3} + C$

Question 14 of 38 | Integration and Accumulation of Change · Level 3

$\displaystyle\int_{0}^{1} (x + 1) e^{x^2 + 2 x} d x =$

$\dfrac{e^3}{2}$

$\dfrac{e^3 - 1}{2}$

$\dfrac{e^4 - e}{2}$

$e^3 - 1$

$e^4 - e$

Question 15 of 38 | Applications of Integration · Level 3

A particle moves on the curve $y = \ln x$ so that the x-component has velocity $x'(t) = t + 1$ for $t \geq 0$. At time $t = 0$, the particle is at $(1, 0)$. At time $t = 1$, the particle is at

$(2, \ln 2)$

$(e^2, 2)$

$\left(\dfrac{5}{2}, \ln \dfrac{5}{2}\right)$

$(3, \ln 3)$

$\left(\dfrac{3}{2}, \ln \dfrac{3}{2}\right)$

Question 16 of 38 | Differentiation: Definition and Fundamental Properties · Level 3

$\operatorname*{lim}\limits_{h \rightarrow 0} \dfrac{1}{h} \ln\left(\dfrac{2 + h}{2}\right)$ is

$e^2$

$1$

$\dfrac{1}{2}$

$0$

nonexistent

Question 17 of 38 | Limits and Continuity · Level 3

Let $f(x) = 3 x + 1$ for all real $x$ and let $\epsilon > 0$. For which of the following choices of $\delta$ is $|f(x) - 7| < \epsilon$ whenever $|x - 2| < \delta$?

$\dfrac{\epsilon}{4}$

$\dfrac{\epsilon}{2}$

$\dfrac{\epsilon}{\epsilon + 1}$

$\dfrac{\epsilon + 1}{\epsilon}$

$3 \epsilon$

Question 18 of 38 | Integration and Accumulation of Change · Level 3

$\displaystyle\int_{0}^{\dfrac{\pi}{4}} \tan^2 x d x =$

$\dfrac{\pi}{4} - 1$

$1 - \dfrac{\pi}{4}$

$\dfrac{1}{3}$

$\sqrt{2} - 1$

$\dfrac{\pi}{4} + 1$

Question 19 of 38 | Analytical Applications of Differentiation · Level 3

Which of the following is true about the graph of $y = \ln|x^2 - 1|$ in the interval $(-1, 1)$?

It is increasing.

It attains a relative minimum at $(0, 0)$.

It has a range of all real numbers.

It is concave down.

It has an asymptote of $x = 0$.

Question 20 of 38 | Analytical Applications of Differentiation · Level 3

If $f(x) = \dfrac{1}{3} x^3 - 4 x^2 + 12 x - 5$ on $[0, 9]$, the absolute maximum value occurs at $x =$

$0$

$2$

$4$

$6$

$9$

Question 21 of 38 | Integration and Accumulation of Change · Level 3

If the substitution $\sqrt{x} = \sin y$ is made in $\displaystyle\int_{0}^{\dfrac{1}{2}} \dfrac{\sqrt{x}}{\sqrt{1 - x}} d x$, the resulting integral is

$\displaystyle\int_{0}^{\dfrac{1}{2}} \sin^2 y d y$

$2 \displaystyle\int_{0}^{\dfrac{1}{2}} \dfrac{\sin^2 y}{\cos y} d y$

$2 \displaystyle\int_{0}^{\dfrac{\pi}{4}} \sin^2 y d y$

$\displaystyle\int_{0}^{\dfrac{\pi}{4}} \sin^2 y d y$

$2 \displaystyle\int_{0}^{\dfrac{\pi}{6}} \sin^2 y d y$

Question 22 of 38 | Differential Equations · Level 4

If $y'' = 2 y'$ and if $y = y' = e$ when $x = 0$, then when $x = 1$, $y =$

$\dfrac{e}{2}(e^2 + 1)$

$e$

$\dfrac{e^3}{2}$

$\dfrac{e}{2}$

$\dfrac{e^3 - e}{2}$

Question 23 of 38 | Integration and Accumulation of Change · Level 2

$\displaystyle\int_{1}^{2} \dfrac{x - 4}{x^2} d x =$

$-\dfrac{1}{2}$

$\ln 2 - 2$

$\ln 2$

$2$

$\ln 2 + 2$

Question 24 of 38 | Differentiation: Composite, Implicit, and Inverse Functions · Level 2

If $f(x) = \ln(\ln x)$, then $f'(x) =$

$\dfrac{1}{x}$

$\dfrac{1}{\ln x}$

$\dfrac{\ln x}{x}$

$x$

$\dfrac{1}{x \ln x}$

Question 25 of 38 | Differentiation: Composite, Implicit, and Inverse Functions · Level 4

If $y = x^{\ln x}$, then $y' =$

$\dfrac{x^{\ln x} \ln x}{x^2}$

$x^{\dfrac{1}{x}} \ln x$

$\dfrac{2 x^{\ln x} \ln x}{x}$

$\dfrac{x^{\ln x} \ln x}{x}$

None of the above

Question 26 of 38 | Differentiation: Composite, Implicit, and Inverse Functions · Level 3

Suppose $f$ is an odd function and $f'(x_0)$ exists. Which of the following must equal $f'(-x_0)$?

$f'(x_0)$

$-f'(x_0)$

$\dfrac{1}{f'(x_0)}$

$\dfrac{-1}{f'(x_0)}$

None of the above

Question 27 of 38 | MCQ · Level 2

The average (mean) value of $\sqrt{x}$ over the interval $0 \leq x \leq 2$ is

$\dfrac{1}{3} \sqrt{2}$

$\dfrac{1}{2} \sqrt{2}$

$\dfrac{2}{3} \sqrt{2}$

$1$

$\dfrac{4}{3} \sqrt{2}$

Question 28 of 38 | Applications of Integration · Level 3

The region in the first quadrant bounded by $y = \sec x$, $x = \dfrac{\pi}{4}$, and the axes is rotated about the x-axis. Volume?

$\dfrac{\pi^2}{4}$

$\pi - 1$

$\pi$

$2 \pi$

$\dfrac{8 \pi}{3}$

Question 29 of 38 | Integration and Accumulation of Change · Level 3

$\displaystyle\int_{0}^{1} \dfrac{x + 1}{x^2 + 2 x - 3} d x$ is

$-\ln \sqrt{3}$

$-\dfrac{\ln \sqrt{3}}{2}$

$\dfrac{1 - \ln \sqrt{3}}{2}$

$\ln \sqrt{3}$

divergent

Question 30 of 38 | Limits and Continuity · Level 3

$\operatorname*{lim}\limits_{x \rightarrow 0} \dfrac{1 - \cos^2(2 x)}{x^2} =$

$-2$

$0$

$1$

$2$

$4$

Question 31 of 38 | Integration and Accumulation of Change · Level 2

If $\displaystyle\int_{1}^{2} f(x - c) d x = 5$, then $\displaystyle\int_{1-c}^{2-c} f(x) d x =$

$5 + c$

$5$

$5 - c$

$c - 5$

$-5$

Question 32 of 38 | Differentiation: Composite, Implicit, and Inverse Functions · Level 2

Let $f$ and $g$ be differentiable: $f(1)=2, f'(1)=3, f'(2)=-4, g(1)=2, g'(1)=-3, g'(2)=5$. If $h(x) = f(g(x))$, then $h'(1) =$

$-9$

$-4$

$0$

$12$

$15$

Question 33 of 38 | Parametric Equations, Polar Coordinates, and Vector-Valued Functions · Level 3

The area of the region enclosed by the polar curve $r = 1 - \cos \theta$ is

$\dfrac{3 \pi}{4}$

$\pi$

$\dfrac{3 \pi}{2}$

$2 \pi$

$3 \pi$

Question 34 of 38 | Integration and Accumulation of Change · Level 3

Given $\begin{cases} f(x) = x + 1 & \quad \text{if } x < 0 \\ f(x) = \cos \pi x & \quad \text{if } x \geq 0 \end{cases}$, $\displaystyle\int_{-1}^1 f(x) d x =$

$\dfrac{1}{2} + \dfrac{1}{\pi}$

$-\dfrac{1}{2}$

$\dfrac{1}{2} - \dfrac{1}{\pi}$

$\dfrac{1}{2}$

$-\dfrac{1}{2} + \pi$

Question 35 of 38 | Integration and Accumulation of Change · Level 3

Calculate the approximate area under $y = x^2$ from $x = 1$ to $x = 2$ by trapezoidal rule with divisions at $x = \dfrac{4}{3}$, $\dfrac{5}{3}$.

$\dfrac{50}{27}$

$\dfrac{251}{108}$

$\dfrac{7}{3}$

$\dfrac{127}{54}$

$\dfrac{77}{27}$

Question 36 of 38 | Integration and Accumulation of Change · Level 3

$\int \arcsin x d x =$

$\sin x - \int \dfrac{x d x}{\sqrt{1 - x^2}}$

$\dfrac{(\arcsin x)^2}{2} + C$

$\arcsin x + \int \dfrac{d x}{\sqrt{1 - x^2}}$

$x \arccos x - \int \dfrac{x d x}{\sqrt{1 - x^2}}$

$x \arcsin x - \int \dfrac{x d x}{\sqrt{1 - x^2}}$

Question 37 of 38 | Differential Equations · Level 4

If $f$ is the solution of $x f'(x) - f(x) = x$ such that $f(-1) = 1$, then $f(e^{-1}) =$

$-2 e^{-1}$

$0$

$e^{-1}$

$-e^{-1}$

$2 e^{-2}$

Question 38 of 38 | Integration and Accumulation of Change · Level 4

Suppose $g'(x) < 0$ for all $x \geq 0$ and $F(x) = \displaystyle\int_{0}^{x} t g'(t) d t$ for $x \geq 0$. Which is FALSE?

$F$ takes on negative values.

$F$ is continuous for $x > 0$.

$F(x) = x g(x) - \displaystyle\int_{0}^{x} g(t) d t$

$F'(x)$ exists for $x > 0$.

$F$ is an increasing function.

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