Math  /  Algebra

Question\begin{tabular}{|c|c|c|c|} \hline Q1 & \begin{tabular}{l} The curve of y2y+x=1\frac{y^{2}}{y+x}=1 is symmetric about: \\ A) Origin \\ B) xx-axis \end{tabular} & C) y-axis & D) Not symmetric \\ \hline Q2 & \begin{tabular}{l} limx+1+2x2x+2=\lim _{x \rightarrow+\infty} \frac{\sqrt{1+2 x^{2}}}{x+2}= \\ A) 2-\sqrt{2} \\ B) 2 \end{tabular} & C) 2\sqrt{2} & D) -2 \\ \hline Q3 & \begin{tabular}{l} The function f(x)=x2sin(x)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+xx1f(x)=4+\frac{x}{x-1}, then f1(x)=f^{-1}(x)= \\ A) 2x3x\frac{2-x}{3-x} \\ B) 3x4x\frac{3-x}{4-x} \end{tabular} & C) 4x5x\frac{4-x}{5-x} & D) 5x6x\frac{5-x}{6-x} \\ \hline Q5 & \begin{tabular}{l} For all real numbers xx, a function f(x)f(x) satis \\ A) 5 \\ B) -5 \end{tabular} & \begin{tabular}{l} f(x)+1x|f(x)+1| \leq \mid x \\ C) -1 \end{tabular} & \begin{tabular}{l} 5, then limx5f(x)\lim _{x \rightarrow-5} f(x) \\ D) 1 \end{tabular} \\ \hline \end{tabular}

Studdy Solution
Evaluate the limit for limx5f(x) \lim_{x \rightarrow -5} f(x) given f(x)+1x |f(x)+1| \leq |x| .
5.1: Analyze the inequality: - f(x)+1x |f(x)+1| \leq |x| implies xf(x)+1x -|x| \leq f(x) + 1 \leq |x| . - Rearrange: x1f(x)x1 -|x| - 1 \leq f(x) \leq |x| - 1 .
5.2: Evaluate the limit as x5 x \rightarrow -5 : - As x5 x \rightarrow -5 , x=5 |x| = 5 . - The inequality becomes 51f(x)51 -5 - 1 \leq f(x) \leq 5 - 1 , or 6f(x)4 -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 1 .
Answer for Q5: D) 1

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