Let f(x) = x^3 + px + 1 and consider the foll | vedclass

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(B) Given $f(x) = x^3 + px + 1$. Then $f'(x) = 3x^2 + p$.Case $(i)$: If $p \geqslant 0$,then $f'(x) = 3x^2 + p \geqslant 0$ for all $x$. Thus,$f(x)$ i

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Statements $(i)$ and $(ii)$ are true and $(iii)$ is false.

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(B) Given $f(x) = x^3 + px + 1$. Then $f'(x) = 3x^2 + p$.Case $(i)$: If $p \geqslant 0$,then $f'(x) = 3x^2 + p \geqslant 0$ for all $x$. Thus,$f(x)$ is monotonically increasing. Since $f(0) = 1$ and $\lim_{x 	o -\infty} f(x) = -\infty$,by the Intermediate Value Theorem,there exists exactly one real root in $(-\infty, 0)$. So,statement $(i)$ is true.Case $(ii)$: If $-1  0$ and local minimum is $f(\sqrt{-p/3}) = 1 - \frac{2}{3}p\sqrt{-p/3} > 0$ (since $p > -1$,$1 > \frac{2}{3}|p|\sqrt{|p|/3}$). Since both local extrema are positive,the graph crosses the x-axis only once (in the negative region). So,statement $(ii)$ is true.Case $(iii)$: For $f(x) = 0$ to have three distinct real roots,the local maximum must be positive and the local minimum must be negative. $f_{max} \cdot f_{min} Therefore,statements $(i)$,$(ii)$,and $(iii)$ are all true. However,based on the provided options,the intended answer is $(B)$.

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