Let p(x) = x^2 + ax + b have two distinct rea | vedclass

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(B) Given $p(x) = x^2 + ax + b$ has two distinct real roots,say $r_1$ and $r_2$. Thus,$p(x) = (x - r_1)(x - r_2)$.Then $g(x) = p(x^3) = (x^3 - r_1)(x^

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Both $I$ and $III$

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(B) Given $p(x) = x^2 + ax + b$ has two distinct real roots,say $r_1$ and $r_2$. Thus,$p(x) = (x - r_1)(x - r_2)$.Then $g(x) = p(x^3) = (x^3 - r_1)(x^3 - r_2)$.Since $x^3 - r = 0$ has exactly one real root for any real $r$,the roots of $g(x)$ are $x = \sqrt[3]{r_1}$ and $x = \sqrt[3]{r_2}$.Since $r_1 
eq r_2$,we have $\sqrt[3]{r_1} 
eq \sqrt[3]{r_2}$. Thus,$g(x)$ has exactly two distinct real roots. Statement $I$ is true and $II$ is false.For statement $III$,$g(x) = (x^3)^2 + a(x^3) + b = x^6 + ax^3 + b$. As $x 	o \infty$,$g(x) 	o \infty$. Since $g(x)$ is a continuous polynomial of even degree $(6)$,it must have a global minimum value. Thus,there exists a real number $\alpha$ such that $g(x) \geq \alpha$ for all real $x$. Statement $III$ is true.Therefore,both $I$ and $III$ are true.

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