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 $f(x) = (x - a)(x - b) - (\frac{a + b}{2})$. If $f(x) = 0$ has both non-negative roots,then the minimum value of $f(x)$ is:

Let $S$ be the set of all real roots of the equation $3^{x}(3^{x}-1)+2=|3^{x}-1|+|3^{x}-2|$. Then $S$

If the difference of the roots of the equation $x^2-7x+10=0$ is same as the difference of the roots of the equation $x^2-17x+k=0$,then a divisor of $k$ is

The roots of the quadratic equation $(a + b - 2c)x^2 - (2a - b - c)x + (a - 2b + c) = 0$ are

Let $\theta$ be an acute angle such that the equation $x^3+4 x^2 \cos \theta+x \cot \theta=0$ has multiple roots. Then the value of $\theta$ (in radians) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module