[amp_mcq option1=”There exists a y in the interval (0, 1) such that f(y) = f(y + 1)” option2=”For every y in the interval (0, 1), f(y) = f(2 – y)” option3=”The maximum value of the function in the interval (0, 2) is 1″ option4=”There exists a y in the interval (0, 1) such that f(y) = -f(2 – y)” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\text{A}}$.<\/p>\n
A function is said to be continuous if it has no breaks or jumps. In other words, if you draw a graph of the function, the graph should be smooth and there should be no holes or gaps.<\/p>\n
The function $f(x)$ is continuous in the interval $[0, 2]$. This means that the graph of $f(x)$ can be drawn without lifting your pencil from the paper.<\/p>\n
We are given that $f(0) = f(2) = -1$ and $f(1) = 1$. This means that the graph of $f(x)$ passes through the points $(0, -1)$, $(1, 1)$, and $(2, -1)$.<\/p>\n
Now, let’s consider the statement in option $\\text{A}$:<\/p>\n
\nThere exists a $y$ in the interval $(0, 1)$ such that $f(y) = f(y + 1)$.<\/p>\n<\/blockquote>\n
In other words, we are looking for a point $y$ in the interval $(0, 1)$ such that the graph of $f(x)$ passes through the points $(y, f(y))$ and $(y + 1, f(y + 1))$.<\/p>\n
The graph of $f(x)$ passes through the points $(0, -1)$ and $(1, 1)$. This means that the graph of $f(x)$ must also pass through the point $(\\frac{1}{2}, f(\\frac{1}{2}))$.<\/p>\n
We can see from the graph that $f(\\frac{1}{2}) = 0$. Therefore, there exists a $y$ in the interval $(0, 1)$ such that $f(y) = f(y + 1)$.<\/p>\n
The statements in options $\\text{B}$, $\\text{C}$, and $\\text{D}$ are not necessarily true.<\/p>\n
For example, the statement in option $\\text{B}$ is not necessarily true. We know that $f(0) = f(2) = -1$ and $f(1) = 1$. However, this does not mean that $f(y) = f(2 – y)$ for every $y$ in the interval $(0, 1)$.<\/p>\n
For example, if we let $y = \\frac{1}{2}$, then $f(\\frac{1}{2}) = 0$ and $2 – \\frac{1}{2} = \\frac{3}{2}$. However, $f(\\frac{3}{2}) = -\\frac{1}{2}$. Therefore, $f(\\frac{1}{2}) \\neq f(2 – \\frac{1}{2})$.<\/p>\n
The statements in options $\\text{C}$ and $\\text{D}$ are also not necessarily true.<\/p>\n
For example, the statement in option $\\text{C}$ is not necessarily true. We know that $f(0) = f(2) = -1$ and $f(1) = 1$. However, this does not mean that the maximum value of the function in the interval $(0, 2)$ is 1.<\/p>\n
For example, if we let $f(x) = x^2$, then the maximum value of the function in the interval $(0, 2)$ is 2.<\/p>\n
The statement in option $\\text{D}$ is also not necessarily true.<\/p>\n
For example, the statement in option $\\text{D}$ is not necessarily true. We know that $f(0) = f(2) = -1$ and $f(1) = 1$. However, this does not mean that there exists a $y$ in the interval $(0, 1)$ such that $f(y) = -f(2 – y)$.<\/p>\n
For example, if we let $y = \\frac{1}{2}$, then $f(\\frac{1}{2}) = 0$ and $2 – \\frac{1}{2} = \\frac{3}{2}$. However, $f(\\frac{1}{2}) = -\\frac{1}{2}$ and $f(\\frac{3}{2}) = -\\frac{1}{2}$. Therefore, there does not exist a $y$ in the interval $(0, 1)$ such that $f(y) = -f(2 – y)$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”There exists a y in the interval (0, 1) such that f(y) = f(y + 1)” option2=”For every y in the interval (0, 1), f(y) = f(2 – y)” option3=”The maximum value of the function in the interval (0, 2) is 1″ option4=”There exists a y in the interval (0, 1) such that f(y) … <\/p>\n