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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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^

Vedclass · Accepted Answer

Both $I$ and $III$

Vedclass · Answer

(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.

Similar Questions

Let $f(x) = (1 + b^2)x^2 + 2bx + 1$ and $m(b)$ be the minimum value of $f(x)$ for a given $b$. As $b$ varies,the range of $m(b)$ is

The number of roots of the equation $|x|^2 - 7|x| + 12 = 0$ is

The roots of the equation $|x - 2|^2 + |x - 2| - 6 = 0$ are:

If $l, m, n$ are real and $l \neq m$,then the roots of the equation $(l - m)x^2 - 5(l + m)x - 2(l - m) = 0$ are

$\alpha, \beta, \gamma, 2, \varepsilon$ are the roots of the equation $x^5+4 x^4-13 x^3-52 x^2+36 x+144=0$. If $\alpha < \beta < \gamma < 2 < \varepsilon$,then find the value of $\alpha+2 \beta+3 \gamma+5 \varepsilon$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module