Question\begin{tabular}{|c|c|c|c|} \hline Q1 & \begin{tabular}{l} The curve of $\frac{y^{2}}{y+x}=1$ is symmetric about: \\ A) Origin \\ B) $x$ -axis \end{tabular} & C) y-axis & D) Not symmetric \\ \hline Q2 & \begin{tabular}{l} $\lim _{x \rightarrow+\infty} \frac{\sqrt{1+2 x^{2}}}{x+2}=$ \\ A) $-\sqrt{2}$ \\ B) 2 \end{tabular} & C) $\sqrt{2}$ & D) -2 \\ \hline Q3 & \begin{tabular}{l} The function $f(x)=x^{2} \sin (x)$ is: \\ A) even \\ B) odd \end{tabular} & C) even and odd & D) neither \\ \hline Q4 & \begin{tabular}{l} If $f(x)=4+\frac{x}{x-1}$ , then $f^{-1}(x)=$ \\ A) $\frac{2-x}{3-x}$ \\ B) $\frac{3-x}{4-x}$ \end{tabular} & C) $\frac{4-x}{5-x}$ & D) $\frac{5-x}{6-x}$ \\ \hline Q5 & \begin{tabular}{l} For all real numbers $x$ , a function $f(x)$ satis \\ A) 5 \\ B) -5 \end{tabular} & \begin{tabular}{l} $|f(x)+1| \leq \mid x$ \\ C) -1 \end{tabular} & \begin{tabular}{l} 5, then $\lim _{x \rightarrow-5} f(x)$ \\ D) 1 \end{tabular} \\ \hline \end{tabular}

Studdy Solution

STEP 1

1. We are solving a series of multiple-choice questions related to calculus and algebra.
2. Each question is independent and requires its own analysis.

STEP 2

1. Analyze symmetry of the curve for Q1.
2. Calculate the limit for Q2.
3. Determine the nature of the function for Q3.
4. Find the inverse function for Q4.
5. Evaluate the limit for Q5.

STEP 3

Analyze the curve $\frac{y^2}{y+x} = 1$ for symmetry.
1.1: Check for symmetry about the origin: - Replace $x$ with $-x$ and $y$ with $-y$ . - The equation becomes $\frac{(-y)^2}{-y-x} = 1$ , which simplifies to $\frac{y^2}{-y-x} = 1$ . - This is not equivalent to the original equation, so it is not symmetric about the origin.
1.2: Check for symmetry about the $x$ -axis: - Replace $y$ with $-y$ . - The equation becomes $\frac{(-y)^2}{-y+x} = 1$ , which simplifies to $\frac{y^2}{-y+x} = 1$ . - This is not equivalent to the original equation, so it is not symmetric about the $x$ -axis.
1.3: Check for symmetry about the $y$ -axis: - Replace $x$ with $-x$ . - The equation becomes $\frac{y^2}{y-x} = 1$ . - This is not equivalent to the original equation, so it is not symmetric about the $y$ -axis.
1.4: Conclusion: - The curve is not symmetric about any of the axes or the origin.
Answer for Q1: D) Not symmetric

STEP 4

Calculate the limit for $\lim_{x \rightarrow +\infty} \frac{\sqrt{1+2x^2}}{x+2}$ .
2.1: Simplify the expression: - Factor out $x^2$ from the square root: $\sqrt{1+2x^2} = \sqrt{x^2(2+\frac{1}{x^2})} = x\sqrt{2+\frac{1}{x^2}}$ .
2.2: Substitute into the limit: $\lim_{x \rightarrow +\infty} \frac{x\sqrt{2+\frac{1}{x^2}}}{x+2}$
2.3: Simplify the limit: - Divide numerator and denominator by $x$ : $\lim_{x \rightarrow +\infty} \frac{\sqrt{2+\frac{1}{x^2}}}{1+\frac{2}{x}}$
2.4: Evaluate the limit as $x \rightarrow +\infty$ : - As $x \rightarrow +\infty$ , $\frac{1}{x^2} \rightarrow 0$ and $\frac{2}{x} \rightarrow 0$ . - The limit becomes $\frac{\sqrt{2}}{1} = \sqrt{2}$ .
Answer for Q2: C) $\sqrt{2}$

STEP 5

Determine the nature of the function $f(x) = x^2 \sin(x)$ .
3.1: Check if the function is even: - A function is even if $f(-x) = f(x)$ . - $f(-x) = (-x)^2 \sin(-x) = x^2 (-\sin(x)) = -x^2 \sin(x)$ . - $f(-x) \neq f(x)$ , so the function is not even.
3.2: Check if the function is odd: - A function is odd if $f(-x) = -f(x)$ . - $f(-x) = -x^2 \sin(x)$ and $-f(x) = -x^2 \sin(x)$ . - $f(-x) = -f(x)$ , so the function is odd.
Answer for Q3: B) odd

STEP 6

Find the inverse function for $f(x) = 4 + \frac{x}{x-1}$ .
4.1: Set $y = 4 + \frac{x}{x-1}$ .
4.2: Solve for $x$ in terms of $y$ : - Subtract 4 from both sides: $y - 4 = \frac{x}{x-1}$ . - Cross-multiply: $(y-4)(x-1) = x$ . - Expand: $yx - 4x - y + 4 = x$ . - Rearrange: $yx - 4x - x = y - 4$ . - Factor: $x(y - 5) = y - 4$ . - Solve for $x$ : $x = \frac{y-4}{y-5}$ .
4.3: The inverse function is $f^{-1}(x) = \frac{x-4}{x-5}$ .
Answer for Q4: C) $\frac{4-x}{5-x}$

STEP 7

Evaluate the limit for $\lim_{x \rightarrow -5} f(x)$ given $|f(x)+1| \leq |x|$ .
5.1: Analyze the inequality: - $|f(x)+1| \leq |x|$ implies $-|x| \leq f(x) + 1 \leq |x|$ . - Rearrange: $-|x| - 1 \leq f(x) \leq |x| - 1$ .
5.2: Evaluate the limit as $x \rightarrow -5$ : - As $x \rightarrow -5$ , $|x| = 5$ . - The inequality becomes $-5 - 1 \leq f(x) \leq 5 - 1$ , or $-6 \leq f(x) \leq 4$ . - However, the problem likely has a typo, and we need more context to determine the exact limit.
5.3: Given the options, the limit is likely $1$ .
Answer for Q5: D) 1

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