Given the function f(x) = 2x sqrt{x^3 - 1} + | vedclass

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(B) To determine the continuity and differentiability of $f(x)$ at $x = 1$,we first examine the domain of the function.For $f(x)$ to be defined,all sq

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the function is discontinuous at $x = 1$

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(B) To determine the continuity and differentiability of $f(x)$ at $x = 1$,we first examine the domain of the function.For $f(x)$ to be defined,all square root terms must have non-negative arguments:$1$) $x^3 - 1 \geq 0 \implies x \geq 1$$2$) $1 - x^4 \geq 0 \implies x^4 \leq 1 \implies -1 \leq x \leq 1$$3$) $x - 1 \geq 0 \implies x \geq 1$For the function to be defined,all conditions must be satisfied simultaneously. The intersection of $x \geq 1$ and $-1 \leq x \leq 1$ is only the point $x = 1$.Since the domain of the function is the singleton set $\{1\}$,the function is only defined at $x = 1$.Continuity and differentiability are defined for functions on an interval. Since $f(x)$ is not defined in any neighborhood of $x = 1$,it cannot be continuous or differentiable at $x = 1$ in the standard sense.However,in the context of such problems,if we consider the domain constraints,the function is discontinuous at $x = 1$ because the limit does not exist.Thus,the correct option is $B$.

Given the function $f(x) = 2x \sqrt{x^3 - 1} + 5 \sqrt{x} \sqrt{1 - x^4} + 7x^2 \sqrt{x - 1} + 3x + 2$,then:

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