Consider the function f(x) = x^3 - 8x^2 + 20x | vedclass

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

Question

(C) Given $f(x) = x^3 - 8x^2 + 20x - 13$. By testing small values or using synthetic division,we find that $x=1$ is a root of $f(x) = 0$,so $(x-1)$ is

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Given $f(x) = x^3 - 8x^2 + 20x - 13$. By testing small values or using synthetic division,we find that $x=1$ is a root of $f(x) = 0$,so $(x-1)$ is a factor. Performing division,we get $f(x) = (x - 1)(x^2 - 7x + 13)$. For $f(x)$ to be a prime number,one of the factors must be $1$ (since the other factor must be a prime number). Case $1$: $x - 1 = 1 \implies x = 2$. If $x = 2$,$f(2) = (2 - 1)(2^2 - 7(2) + 13) = 1 	imes (4 - 14 + 13) = 1 	imes 3 = 3$,which is prime. Case $2$: $x^2 - 7x + 13 = 1 \implies x^2 - 7x + 12 = 0 \implies (x - 3)(x - 4) = 0 \implies x = 3$ or $x = 4$. If $x = 3$,$f(3) = (3 - 1)(1) = 2 	imes 1 = 2$,which is prime. If $x = 4$,$f(4) = (4 - 1)(1) = 3 	imes 1 = 3$,which is prime. Thus,the values of $x$ are $2, 3, 4$. The number of such positive integers is $3$.

Consider the function $f(x) = x^3 - 8x^2 + 20x - 13$. The number of positive integers $x$ for which $f(x)$ is a prime number is:

Similar Questions

If $f(x) = \frac{x - |x|}{|x|}$,then $f(-1) = $

If $f(x)=2\{x\}+5x$,where $\{x\}$ is the fractional part function,then $f(-1.4)$ is

For some $a, b, c \in N$,let $f(x)=ax-3$ and $g(x)=x^b+c$,$x \in R$. If $(fog)^{-1}(x)=\left(\frac{x-7}{2}\right)^{1/3}$,then $(fog)(ac) + (gof)(b)$ is equal to $..........$

Let $[x]$ denote the integral part of $x \in R$. Let $g(x) = x - [x]$. Let $f(x)$ be any continuous function with $f(0) = f(1)$. Then the function $h(x) = f(g(x))$:

Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = 2x + |x|$,then $f(2x) + f(-x) - f(x) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module