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 $f(x) = \frac{1}{\sqrt{x + 2\sqrt{2x - 4}}} + \frac{1}{\sqrt{x - 2\sqrt{2x - 4}}}$ for $x > 2$,then $f(11) = $

Let $A = \{1, 3, 4, 6, 9\}$ and $B = \{2, 4, 5, 8, 10\}$. Let $R$ be a relation defined on $A \times B$ such that $R = \{((a_1, b_1), (a_2, b_2)) : a_1 \leq b_2 \text{ and } b_1 \leq a_2\}$. Then the number of elements in the set $R$ is

Let $f(x)$ and $g(x)$ be two continuous functions defined from $R \rightarrow R$,such that $f(x_1) > f(x_2)$ and $g(x_1) < g(x_2)$ for all $x_1 > x_2$. Then the solution set of $f(g(\alpha^2 - 2\alpha)) > f(g(3\alpha - 4))$ is

If $e^{f(x)}=\frac{10+x}{10-x}, x \in(-10,10)$ and $f(x)=k f\left(\frac{200 x}{100+x^2}\right)$, then $k$ is equal to

Let $A = \{1, 2, 3, 4, 5\}$ and $B = \{1, 2, 3, 4, 5, 6\}$. Then the number of functions $f: A \rightarrow B$ satisfying $f(1) + f(2) = f(4) - 1$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module