Calculus Practice Questions

A vector $w$ bisects the angle between two vectors $u$ and $v$ if it is in the same direction as the sum of their unit vectors. The unit vectors are $\hat{u} = \frac{u}{||u||}$ and $\hat{v} = \frac{v}{||v||}$.
So, $w$ must be proportional to $\hat{u} + \hat{v}$.
$w = k \left( \frac{u}{||u||} + \frac{v}{||v||} \right)$
We are given $w = u + \alpha v$. Equating the two expressions:
$u + \alpha v = \frac{k}{||u||}u + \frac{k}{||v||}v$
By comparing the coefficients of $u$, we get $1 = \frac{k}{||u||}$, so $k = ||u||$.
By comparing the coefficients of $v$, we get $\alpha = \frac{k}{||v||} = \frac{||u||}{||v||}$.
Given that $||u|| = 2||v||$, we can substitute this in:
$\alpha = \frac{2||v||}{||v||} = 2$

To evaluate the limit, we can use a substitution. Let $t = x - 4$. As $x$ approaches 4, $t$ approaches 0. Substituting this into the expression gives us the standard trigonometric limit:
$\lim_{t \to 0} \frac{\sin(t)}{t}$
The value of this well-known limit is 1.

Let $f(x)$ be a polynomial of degree $n$. We can write $f(x)$ as:
$f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$
Then, $f(-x)$ is:
$f(-x) = a_n (-x)^n + a_{n-1} (-x)^{n-1} + \dots + a_1 (-x) + a_0$
The sum $f(x) + f(-x)$ will cancel out terms with odd powers of $x$. For example, if $n$ is odd, the highest degree term $a_n x^n$ will cancel with $a_n(-x)^n = -a_n x^n$. If $n$ is even, $a_n x^n$ will add with $a_n(-x)^n = a_n x^n$ to give $2a_n x^n$.
Given that the degree of $f(x) + f(-x)$ is 10, this implies that the highest odd power term in $f(x)$ must be less than 10, and the highest even power term must be $x^{10}$. Therefore, the degree of $f(x)$, denoted as $n$, must be at least 10, with $a_{10} \neq 0$.

Now, let's consider the derivative $g(x) = f'(x)$:
$g(x) = f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \dots + a_1$
And $g(-x) = f'(-x)$:
$g(-x) = n a_n (-x)^{n-1} + (n-1) a_{n-1} (-x)^{n-2} + \dots + a_1$
We are interested in the degree of $g(x) - g(-x)$. This difference will cancel out terms with even powers of $x$ in $g(x)$ and retain terms with odd powers. Since $f(x)$ has degree 10, $f'(x)$ will have degree $10-1=9$.
So $g(x)$ is a polynomial of degree 9.
$g(x) = 10a_{10}x^9 + 9a_9x^8 + \dots + a_1$
$g(-x) = 10a_{10}(-x)^9 + 9a_9(-x)^8 + \dots + a_1$
$g(x) - g(-x) = (10a_{10}x^9 - 10a_{10}(-x)^9) + (9a_9x^8 - 9a_9(-x)^8) + \dots$
$g(x) - g(-x) = (10a_{10}x^9 - (-10a_{10}x^9)) + (9a_9x^8 - 9a_9x^8) + \dots$
$g(x) - g(-x) = 20a_{10}x^9 + \text{terms of lower odd powers}$
Since $a_{10} \neq 0$, the highest degree term in $g(x) - g(-x)$ is $x^9$.

The final answer is $\boxed{9}$
[CORRECT_OPTION: N/A]

The formula in option D, $\forall x (p(x)\vee q(x))\Rightarrow (\forall x p(x) \vee \forall x q(x))$, is not universally valid. The antecedent $\forall x (p(x)\vee q(x))$ states that for every x, at least one of p(x) or q(x) is true. The consequent $(\forall x p(x) \vee \forall x q(x))$ states that either p(x) is true for all x, or q(x) is true for all x.

Consider a counterexample where the domain is the set of integers. Let p(x) be "x is even" and q(x) be "x is odd". The antecedent is true, as every integer is either even or odd. However, the consequent is false because it's not true that all integers are even, nor is it true that all integers are odd. Since the antecedent can be true while the consequent is false, the implication is not valid.

The given functional equation is $af(x)+bf\left ( \frac{1}{x} \right )=\frac{1}{x}-25$.
To solve for $f(x)$, we can create a system of two equations. Replace $x$ with $1/x$ in the original equation to get the second equation: $af\left ( \frac{1}{x} \right )+bf(x)=x-25$.
Now, solve this system for $f(x)$. Multiplying the first equation by $a$ and the second by $b$ and then subtracting gives:
$(a^{2-}b^2)f(x) = a(\frac{1}{x}-25) - b(x-25)$.
$f(x) = \frac{1}{a^{2-}b^2} [ \frac{a}{x} - bx - 25a + 25b ]$.
Integrating from 1 to 2:
\int_{1}^{2}f(x)dx = \frac{1}{a^{2-}b^2} \[a\ln x - \frac{bx^2}{2} - (25a-25b)x]_1^2 = \frac{1}{a^{2-}b^2}[a(\ln2-25)+\frac{47b}{2}]$.

The function is given by $f(x) = x^{-1/3} = \frac{1}{\sqrt[3]{x}}$.
I) The function is not continuous in the interval [-1, 1]. As x approaches 0, $f(x)$ approaches infinity, creating a discontinuity at x=0. Thus, statement I is false.
II) Since the function's value tends to infinity as x approaches 0, the function is not bounded in the interval [-1, 1]. Thus, statement II is true.
III) The area A is the integral $\int_{-1}^{1} x^{-1/3} dx$. This is an improper integral but it converges to a finite, non-zero value. Thus, statement III is true.
Therefore, only statements II and III are true.

The general term of the series, $\frac{1}{x(x+1)}$, can be decomposed into partial fractions as $\frac{1}{x} - \frac{1}{x+1}$. When the summation is expanded, it becomes a telescoping series:
$\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{99} - \frac{1}{100}\right)$
All the intermediate terms cancel out, leaving only the first term, $1$, and the last term, $-\frac{1}{100}$. The sum is therefore $1 - \frac{1}{100} = \frac{99}{100} = 0.99$.

This limit is of the indeterminate form $\infty ^0$. To solve it, let $y = x^{1/x}$.
Taking the natural logarithm of both sides gives $\ln(y) = \ln(x^{1/x}) = \frac{\ln(x)}{x}$.
Now, we find the limit of $\ln(y)$ as $x \to \infty$:
$\lim_{x\to \infty } \ln(y) = \lim_{x\to \infty } \frac{\ln(x)}{x}$
This is of the form $\frac{\infty }{\infty }$, so we can apply L'Hôpital's Rule:
$\lim_{x\to \infty } \frac{(\ln(x))'}{(x)'} = \lim_{x\to \infty } \frac{1/x}{1} = 0$.
Since $\lim_{x\to \infty } \ln(y) = 0$, the original limit is $\lim_{x\to \infty } y = e^0 = 1$.

This definite integral can be solved using the method of substitution. Let $u = 1/x$. The differential $du$ is then $du = -1/x^2 \, dx$, which means $1/x^2 \, dx = -du$.

We must also change the limits of integration. When $x = 1/\pi$, $u = \pi$. When $x = 2/\pi$, $u = \pi/2$.

The integral transforms to:
$\int_{\pi}^{\pi/2} \cos(u) (-du) = \int_{\pi/2}^{\pi} \cos(u) \, du$
$= [\sin(u)]_{\pi/2}^{\pi} = \sin(\pi) - \sin(\pi/2) = 0 - 1 = -1$

The limit is of the indeterminate form $\infty ^0$. Let $L = \lim_{x\rightarrow \infty }(1+x^{2})^{e^{-x}}$. We can evaluate this by first taking the natural logarithm:
$\ln(L) = \lim_{x\rightarrow \infty } \ln((1+x^{2})^{e^{-x}}) = \lim_{x\rightarrow \infty } \frac{\ln(1+x^2)}{e^x}$
This is now of the form $\frac{\infty }{\infty }$, so we can apply L'Hôpital's Rule.
$\ln(L) = \lim_{x\rightarrow \infty } \frac{2x/(1+x^2)}{e^x} = \lim_{x\rightarrow \infty } \frac{2x}{(1+x^2)e^x} = 0$
Since $\ln(L) = 0$, we have $L = e^0 = 1$.

To evaluate the integral $\int_{0}^{\pi }x^{2}cosxdx$, we can use integration by parts, which states $\int udv = uv - \int vdu$.

1. First application of integration by parts:
   Let $u = x^2$ and $dv = \cos x dx$.
   Then $du = 2x dx$ and $v = \sin x$.
   $\int_{0}^{\pi }x^{2}cosxdx = x^2 \sin x \Big|_0^\pi - \int_{0}^{\pi} 2x \sin x dx$
   Evaluating $x^2 \sin x \Big|_0^\pi$: $(\pi^2 \sin \pi) - (0^2 \sin 0) = 0 - 0 = 0$.
   So, the integral becomes $- \int_{0}^{\pi} 2x \sin x dx$

2. Second application of integration by parts for $- \int_{0}^{\pi} 2x \sin x dx$:
   Let $u = 2x$ and $dv = \sin x dx$.
   Then $du = 2 dx$ and $v = -\cos x$.
   - \Big\$[ 2x (-\cos x) \Big|_0^\pi - \int_{0}^{\pi} (-\cos x) (2 dx) \Big]\$
   = - \Big\$[ -2x \cos x \Big|_0^\pi + 2 \int_{0}^{\pi} \cos x dx \Big]\$
   Evaluating $-2x \cos x \Big|_0^\pi$: $(-2\pi \cos \pi) - (-2 \cdot 0 \cos 0) = (-2\pi \cdot -1) - 0 = 2\pi$.
   Evaluating $2 \int_{0}^{\pi} \cos x dx$: $2 \sin x \Big|_0^\pi = 2(\sin \pi - \sin 0) = 2(0 - 0) = 0$.
   Substituting these values back: $- [2\pi + 0] = -2\pi$

Thus, the value of the integral is $-2\pi$.

To find the value of $t$, we first need to determine the first and second derivatives of $f(x) = x \sin x$.

1. First Derivative: Using the product rule $(uv)' = u'v + uv'$, where $u=x$ and $v=\sin x$:
   $f'(x) = (1)(\sin x) + (x)(\cos x) = \sin x + x \cos x$

2. Second Derivative: Differentiate $f'(x)$ using the sum and product rules:
   $f''(x) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(x \cos x)$
   $f''(x) = \cos x + [(1)(\cos x) + (x)(-\sin x)] = \cos x + \cos x - x \sin x = 2 \cos x - x \sin x$

3. Substitute into the given equation:
   Substitute $f''(x)$, $f(x)$ into $f''(x)+f(x)+t\cos x=0$:
   $(2 \cos x - x \sin x) + (x \sin x) + t \cos x = 0$

4. Simplify and Solve for t:
   $2 \cos x + t \cos x = 0$
   $\cos x (2 + t) = 0$
   For this equation to hold for all $x$ where $\cos x \neq 0$, we must have:
   $2 + t = 0 \Rightarrow t = -2$

The value of $t$ is -2.

To evaluate the integral $\int_{0}^{2\pi}|x \sin x| dx$, we must consider the sign of the integrand. For $x \in [0, 2\pi]$, the term $x$ is non-negative, so the sign of $x \sin x$ is determined by the sign of $\sin x$. We split the integral at $x=\pi$, where $\sin x$ changes from positive to negative.

$\int_{0}^{2\pi}|x \sin x| dx = \int_{0}^{\pi} x \sin x \, dx + \int_{\pi}^{2\pi} (-x \sin x) \, dx$

Using integration by parts, we find the antiderivative: $\int x \sin x \, dx = -x \cos x + \sin x$. Now we evaluate the two definite integrals.

$\int_{0}^{\pi} x \sin x \, dx = [-x \cos x + \sin x]_{0}^{\pi} = (-\pi \cos \pi + \sin \pi) - (0) = \pi$

$-\int_{\pi}^{2\pi} x \sin x \, dx = -[-x \cos x + \sin x]_{\pi}^{2\pi} = -[(-2\pi) - (-\pi)] = 3\pi$

Summing the results, the total value of the integral is $\pi + 3\pi = 4\pi$. Given that the integral equals $k\pi$, we have $4\pi = k\pi$, which implies $k=4$.

The final answer is $\boxed{4}$.

The function $f(\theta)$ is given as a determinant. Since the second and third rows are constants, $f(\theta)$ can be expressed as:
$f(\theta) = A \sin\theta + B \cos\theta + C \tan\theta$
where A, B, and C are constants determined by the cofactors of the first row. This function $f(\theta)$ is continuous and differentiable on the interval $[\frac{\pi}{6}, \frac{\pi}{3}]$.

We observe that $f(\frac{\pi}{6}) = 0$ because the first row becomes identical to the second row, making the determinant zero.
Similarly, $f(\frac{\pi}{3}) = 0$ because the first row becomes identical to the third row, making the determinant zero.

Since $f(\theta)$ is continuous on $[\frac{\pi}{6}, \frac{\pi}{3}]$ and differentiable on $(\frac{\pi}{6}, \frac{\pi}{3})$, and $f(\frac{\pi}{6}) = f(\frac{\pi}{3}) = 0$, by Rolle's Theorem, there must exist at least one $\theta \in (\frac{\pi}{6}, \frac{\pi}{3})$ such that $f'(\theta) = 0$. Thus, statement (I) is TRUE.

As $f(\theta)$ is not identically zero (e.g., $f(\frac{\pi}{4}) \neq 0$), its derivative $f'(\theta)$ cannot be identically zero on the interval. If $f'(\theta)$ were always zero, then $f(\theta)$ would be a constant, which is not true. Therefore, there must exist at least one $\theta \in (\frac{\pi}{6}, \frac{\pi}{3})$ such that $f'(\theta) \neq 0$. Thus, statement (II) is also TRUE.

To check for continuity of a function $f(x)$ at $x=c$, we need to verify three conditions:

1. $f(c)$ is defined.
2. $\lim_{x \to c^-} f(x)$ exists.
3. $\lim_{x \to c^+} f(x)$ exists.
4. $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.

For option A, $f(3) = 2$.
The left-hand limit is $\lim_{x \to 3^-} \frac{x+3}{3} = \frac{3+3}{3} = 2$.
The right-hand limit is $\lim_{x \to 3^+} (x-1) = 3-1 = 2$.
Since $f(3) = 2$ and both limits equal 2, the function is continuous at $x=3$.

Problem Analysis

We are given the function $f(x) = \sin(x)$ in the interval $x \in [\frac{\pi}{4}, \frac{7\pi}{4}]$. We need to find the number and location of local minima.

Step-by-Step Derivation

1. Find Critical Points:
   To find the extrema, we take the first derivative and set it to zero:
   $f'(x) = \cos(x)$
   Setting $f'(x) = 0$ within the interval $[\frac{\pi}{4}, \frac{7\pi}{4}]$:

   • $x = \frac{\pi}{2}$
   • $x = \frac{3\pi}{2}$

2. Second Derivative Test:
   Calculate the second derivative to determine the nature of these points:
   $f''(x) = -\sin(x)$

   • At $x = \frac{\pi}{2}$: $f''(\frac{\pi}{2}) = -\sin(\frac{\pi}{2}) = -1$. Since $f''(x) < 0$, this is a local maximum.
   • At $x = \frac{3\pi}{2}$: $f''(\frac{3\pi}{2}) = -\sin(\frac{3\pi}{2}) = -(-1) = 1$. Since $f''(x) > 0$, this is a local minimum.

3. Check Boundary Points:
   In a closed interval, we must also check the endpoints for local extrema:

   • At $x = \frac{\pi}{4}$: The function is increasing ($f'(x) > 0$), so the left boundary acts as a local minimum.
   • At $x = \frac{7\pi}{4}$: The function is increasing toward the boundary ($f'(x) > 0$), so the right boundary acts as a local maximum.

Conclusion

There are two local minima:

1. One at the left boundary: $x = \frac{\pi}{4}$
2. One at the critical point: $x = \frac{3\pi}{2}$

Final Result

The number and locations of local minima are two, at $\pi/4$ and $3\pi/2$.

Correct Answer: D

We can simplify the integrand using Euler's formula, $e^{ix} = \cos x + i \sin x$.
Therefore, $\frac{\cos x + i \sin x}{\cos x - i \sin x} = \frac{e^{ix}}{e^{-ix}} = e^{ix - (-ix)} = e^{2ix}$.

Now, we need to evaluate the definite integral:
$\int_{0}^{\pi/2} e^{2ix} dx$
The integral of $e^{ax}$ is $\frac{1}{a}e^{ax}$. In this case, $a=2i$.
$\left[\frac{e^{2ix}}{2i}\right]_{0}^{\pi/2}$
Substitute the limits of integration:
$\frac{e^{2i(\pi/2)}}{2i} - \frac{e^{2i(0)}}{2i} = \frac{e^{i\pi}}{2i} - \frac{e^0}{2i}$
Using Euler's formula again, $e^{i\pi} = \cos \pi + i \sin \pi = -1 + 0i = -1$, and $e^0 = 1$.
$\frac{-1}{2i} - \frac{1}{2i} = \frac{-2}{2i} = \frac{-1}{i}$
To simplify $\frac{-1}{i}$, multiply the numerator and denominator by $i$:
$\frac{-1}{i} \times \frac{i}{i} = \frac{-i}{i^2} = \frac{-i}{-1} = i$