Calculus Practice Questions

Let's define the problem formally. The weight of a sequence $b = (b_0, b_1, ..., b_{k-1})$ is $W(b) = \sum_{i=0}^{k-1} \frac{b_i}{2^i}$.
$X$ is the maximum weight of a subsequence of $S = (a_0, a_1, ..., a_{n-1})$.
$Y$ is the maximum weight of a subsequence of $S' = (a_1, a_2, ..., a_{n-1})$.

To find the maximum weight subsequence for $S$, we can consider two possibilities regarding the first element, $a_0$.

Case 1: The optimal subsequence does not include $a_0$.
If the subsequence with the maximum weight does not contain $a_0$, it must be formed entirely from the elements of $S' = (a_1, a_2, ..., a_{n-1})$. By definition, the maximum weight that can be achieved from a subsequence of $S'$ is $Y$.

Case 2: The optimal subsequence includes $a_0$.
If we include $a_0$, it will be the first term of the subsequence. Its contribution to the total weight will be $a_0/2^0 = a_0$. The rest of the subsequence will be chosen from $S' = (a_1, a_2, ..., a_{n-1})$. Let this be a subsequence $S'_{sub}$. The total weight will be $a_0$ plus the weight contribution from $S'_{sub}$. When elements of $S'_{sub}$ follow $a_0$, their positions in the weight calculation are shifted by one. For example, an element $a_k$ which was at index $k-1$ in $S'$ is now at index $k$ in the full subsequence. Its weight contribution changes from $a_k/2^{k-1}$ to $a_k/2^k$. This means the total weight contribution from $S'_{sub}$ is halved.
To maximize the total weight, we should choose $S'_{sub}$ to be the subsequence from $S'$ with the maximum possible weight, which is $Y$. The weight of this part will be $Y/2$. So, the maximum total weight in this case is $a_0 + Y/2$.

Conclusion:
The overall maximum weight $X$ is the greater of the values from the two cases.
Therefore, $X = \max(Y, a_0 + Y/2)$.

This limit can be evaluated using the standard definition of the exponential function, $e$. One of the fundamental forms of this limit is:

$\lim_{x\to \infty } (1 + \frac{a}{x})^{bx} = e^{ab}$

The given expression is $\lim_{n\rightarrow \infty }(1-\frac{1}{n})^{2n}$.

We can rewrite this expression to match the standard form:

$\lim_{n\rightarrow \infty }(1+\frac{-1}{n})^{2n}$

By comparing this to the general formula $(1 + \frac{a}{x})^{bx}$, we can see that $a = -1$ and $b = 2$.

Therefore, the value of the limit is $e^{a \times b} = e^{-1 \times 2} = e^{-2}$.

The integrand can be simplified using the tangent subtraction formula:
$\frac{1-\tan x}{1+\tan x} = \frac{\tan(\pi/4)-\tan x}{1+\tan(\pi/4)\tan x} = \tan(\pi/4 - x)$
Let $u = \pi/4 - x$. Then $du = -dx$.
When $x=0$, $u=\pi/4$. When $x=\pi/4$, $u=0$.
$\int_{0}^{\pi/4} \tan(\pi/4 - x) dx = \int_{\pi/4}^{0} \tan(u) (-du) = \int_{0}^{\pi/4} \tan(u) du$
Now, we integrate $\tan(u)$:
$\Big[\ln|\sec u|\Big]_0^{\pi/4} = \ln|\sec(\pi/4)| - \ln|\sec(0)|$
$= \ln(\sqrt{2}) - \ln(1) = \ln(2^{1/2}) - 0$
$= \frac{1}{2}\ln(2)$

The final answer is $\boxed{\text{D}}$.

The set $X = \{x \in R^3 \mid x_1 + x_2 + x_3 = 0\}$ is a subspace of $R^3$ because it's defined by a homogeneous linear equation. Its dimension is $3 - 1 = 2$.
The two given vectors are $v_1 = [1,-1,0]^T$ and $v_2 = [1,0,-1]^T$.

1. Membership in X: For $v_1$, $1 + (-1) + 0 = 0$. For $v_2$, $1 + 0 + (-1) = 0$. Both vectors belong to X.
2. Linear Independence: $v_1$ and $v_2$ are linearly independent because $v_1$ is not a scalar multiple of $v_2$ (e.g., $1 \ne k \cdot 0$ for the second component if $v_1=k v_2$).
3. Spanning X: Since X is a 2-dimensional subspace and we have found two linearly independent vectors within X, this set of two vectors forms a basis for X.

Therefore, $\\{[1,-1,0]^{T},[1,0,-1]^{T}\\}$ is a basis for the subspace X.

Statement P: A function is continuous if its graph can be drawn without lifting the pen. The graph of $f(x)=|x|$ has no breaks or jumps, including at $x=0$.
We evaluate the limit at $x=0$: $\lim_{x \to 0^-} |x| = \lim_{x \to 0^-} (-x) = 0$ and $\lim_{x \to 0^+} |x| = \lim_{x \to 0^+} x = 0$. Since $f(0)=0$, $\lim_{x \to 0} f(x) = f(0)$. Thus, $f(x)$ is continuous for all real $x$. P is TRUE.

Statement Q: A function is differentiable at a point if its graph has a unique tangent line at that point. At $x=0$, the graph of $f(x)=|x|$ has a sharp corner (a "cusp").
The left-hand derivative at $x=0$ is $\lim_{h \to 0^-} \frac{|0+h|-|0|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1$.
The right-hand derivative at $x=0$ is $\lim_{h \to 0^+} \frac{|0+h|-|0|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1$.
Since the left-hand derivative ($\ne -1$) and right-hand derivative ($\ne 1$) are not equal, $f(x)$ is not differentiable at $x=0$. Thus, $f(x)$ is not differentiable for all real values of $x$. Q is FALSE.

Therefore, P is true and Q is false.