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(A) Given $f(x) = x^5 - 5x + a$.To find the number of real roots,we analyze the function $g(x) = x^5 - 5x$,where $f(x) = g(x) + a = 0$,which implies $

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$(B, D)$

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(A) Given $f(x) = x^5 - 5x + a$.To find the number of real roots,we analyze the function $g(x) = x^5 - 5x$,where $f(x) = g(x) + a = 0$,which implies $g(x) = -a$.First,find the critical points of $g(x)$ by setting $g'(x) = 0$:$g'(x) = 5x^4 - 5 = 5(x^4 - 1) = 5(x^2 - 1)(x^2 + 1) = 5(x - 1)(x + 1)(x^2 + 1) = 0$.The critical points are $x = 1$ and $x = -1$.The local maximum value is $g(-1) = (-1)^5 - 5(-1) = -1 + 5 = 4$.The local minimum value is $g(1) = (1)^5 - 5(1) = 1 - 5 = -4$.For $f(x) = 0$,we need $g(x) = -a$.$1$. If $-a > 4$ (i.e.,$a $2$. If $-a 4$),the line $y = -a$ is below the local minimum,so there is only $1$ real root.$3$. If $-4 Thus,$(B)$ is correct ($a > 4$ implies $1$ root) and $(D)$ is correct ($-4 < a < 4$ implies $3$ roots).

Let $a \in R$ and let $f: R \rightarrow R$ be given by $f(x)=x^5-5x+a$. Then
$(A)$ $f(x)$ has three real roots if $a > 4$
$(B)$ $f(x)$ has only one real root if $a > 4$
$(C)$ $f(x)$ has three real roots if $a < -4$
$(D)$ $f(x)$ has three real roots if $-4 < a < 4$

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Let $a \in R$ and let $f: R \rightarrow R$ be given by $f(x)=x^5-5x+a$. Then$(A)$ $f(x)$ has three real roots if $a > 4$$(B)$ $f(x)$ has only one real root if $a > 4$$(C)$ $f(x)$ has three real roots if $a < -4$$(D)$ $f(x)$ has three real roots if $-4 < a < 4$