Statement -1: Any function f(x) is an even fu | vedclass

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(D) Statement $-1$ is the standard definition of an even function,which is true.For Statement $-2$,the domain of $f(x) = \frac{1}{\sqrt{1 - x^2}} + \l

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Both statements are true,and Statement $-1$ is the correct explanation for Statement $-2$.

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(D) Statement $-1$ is the standard definition of an even function,which is true.For Statement $-2$,the domain of $f(x) = \frac{1}{\sqrt{1 - x^2}} + \left[ \frac{x^2 + x + 1}{4} \right]$ is determined by $1 - x^2 > 0$,which implies $x \in (-1, 1)$.Within the interval $x \in (-1, 1)$,the expression $g(x) = \frac{x^2 + x + 1}{4}$ ranges from $\frac{(-1)^2 + (-1) + 1}{4} = 0.25$ to $\frac{1^2 + 1 + 1}{4} = 0.75$.Since $0 \le g(x) Thus,for $x \in (-1, 1)$,$f(x) = \frac{1}{\sqrt{1 - x^2}} + 0 = \frac{1}{\sqrt{1 - x^2}}$.Since $f(-x) = \frac{1}{\sqrt{1 - (-x)^2}} = \frac{1}{\sqrt{1 - x^2}} = f(x)$,the function is even.Both statements are true,and Statement $-1$ provides the definition used to verify Statement $-2$.

Statement $-1$: Any function $f(x)$ is an even function if $f(-x) = f(x)$ for all $x$ in its domain.
Statement $-2$: The function $f(x) = \frac{1}{\sqrt{1 - x^2}} + \left[ \frac{x^2 + x + 1}{4} \right]$,where $[.]$ denotes the greatest integer function,is an even function.

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Statement $-1$: Any function $f(x)$ is an even function if $f(-x) = f(x)$ for all $x$ in its domain.Statement $-2$: The function $f(x) = \frac{1}{\sqrt{1 - x^2}} + \left[ \frac{x^2 + x + 1}{4} \right]$,where $[.]$ denotes the greatest integer function,is an even function.