The function f(x) = intlimits_0^x sqrt{1 - t^ | vedclass

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(D) $1$. Domain: The integrand $\sqrt{1 - t^4}$ is defined when $1 - t^4 \ge 0$,which implies $t^4 \le 1$,so $t \in [-1, 1]$. Thus,$f(x)$ is defined o

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All of the Above

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(D) $1$. Domain: The integrand $\sqrt{1 - t^4}$ is defined when $1 - t^4 \ge 0$,which implies $t^4 \le 1$,so $t \in [-1, 1]$. Thus,$f(x)$ is defined on $[-1, 1]$.$2$. Monotonicity: By the Fundamental Theorem of Calculus,$f'(x) = \sqrt{1 - x^4}$. For $x \in (-1, 1)$,$f'(x) > 0$,which means $f(x)$ is an increasing function.$3$. Parity: $f(-x) = \int\limits_0^{-x} \sqrt{1 - t^4} \, dt$. Let $t = -u$,then $dt = -du$. When $t=0, u=0$ and when $t=-x, u=x$. Thus,$f(-x) = \int\limits_0^x \sqrt{1 - (-u)^4} (-du) = -\int\limits_0^x \sqrt{1 - u^4} \, du = -f(x)$. Therefore,$f(x)$ is an odd function.Since all statements are correct,the correct option is $D$.

The function $f(x) = \int\limits_0^x \sqrt{1 - t^4} \, dt$ is such that

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