Which of the following is an improper rationa | vedclass

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

Question

(D) rational fraction $\frac{p(x)}{q(x)}$ is called an improper rational fraction if the degree of the numerator $p(x)$ is greater than or equal to th

Vedclass · Accepted Answer

$\frac{x^2+1}{x^2-1}$

Vedclass · Answer

(D) rational fraction $\frac{p(x)}{q(x)}$ is called an improper rational fraction if the degree of the numerator $p(x)$ is greater than or equal to the degree of the denominator $q(x)$.$(a)$ For $\frac{x^2+1}{(x^2+2)(x^2+x+1)}$,degree of $p(x) = 2$ and degree of $q(x) = 4$. Since $2 $(b)$ For $\frac{x^2+1}{(x+3)(x^2-x+1)}$,degree of $p(x) = 2$ and degree of $q(x) = 3$. Since $2 $(c)$ For $\frac{x}{x^2+3x+1}$,degree of $p(x) = 1$ and degree of $q(x) = 2$. Since $1 $(d)$ For $\frac{x^2+1}{x^2-1}$,degree of $p(x) = 2$ and degree of $q(x) = 2$. Since the degree of the numerator is equal to the degree of the denominator,it is an improper rational fraction.

Which of the following is an improper rational fraction?

Similar Questions

If $\frac{x^4}{(x-1)(x-2)}=f(x)+\frac{A}{x-1}+\frac{B}{x-2}$,then $f(-2)+A+B=$

If $\frac{x^4}{(x-1)^2(x+1)}=Ax+B+\frac{P}{(x-1)}+\frac{Q}{(x-1)^2}+\frac{R}{x+1}$,then $2AP-BQ+R=$

If we resolve the rational fraction $\frac{1}{(1-2x)^2(1-3x)}$ into partial fractions of the form $\frac{A}{1-3x} + \frac{B}{1-2x} + \frac{C}{(1-2x)^2}$,then what is $\min \{A, B, C\} = $

If $\frac{3x + a}{x^2 - 3x + 2} = \frac{A}{x - 2} - \frac{10}{x - 1}$,then:

The partial fractions of $\frac{3x - 1}{(1 - x + x^2)(2 + x)}$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module