Let x=2 be a root of the equation x^2+px+q=0 | vedclass

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

Question

(C) Given $x=2$ is a root of $x^2+px+q=0$,we have $4+2p+q=0$,so $q = -2p-4$. Substituting $q$ into the expression inside the cosine: $x^2-4px+q^2+8q+1

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Given $x=2$ is a root of $x^2+px+q=0$,we have $4+2p+q=0$,so $q = -2p-4$. Substituting $q$ into the expression inside the cosine: $x^2-4px+q^2+8q+16 = x^2-4px+(-2p-4)^2+8(-2p-4)+16 = x^2-4px+4p^2+16p+16-16p-32+16 = x^2-4px+4p^2 = (x-2p)^2$. Thus,$f(x) = \frac{1-\cos((x-2p)^2)}{(x-2p)^4}$ for $x 
eq 2p$. Using the limit $\lim_{	heta 	o 0} \frac{1-\cos 	heta}{	heta^2} = \frac{1}{2}$,we have $\lim_{x 	o 2p} f(x) = \lim_{x 	o 2p} \frac{1-\cos((x-2p)^2)}{((x-2p)^2)^2} = \frac{1}{2}$. Since $\lim_{x 	o 2p^+} f(x) = \frac{1}{2}$,the greatest integer function $[f(x)]$ for $x$ near $2p$ (where $0 < f(x) < 1$) is $[f(x)] = 0$.

Let $x=2$ be a root of the equation $x^2+px+q=0$ and $f(x)=\begin{cases} \frac{1-\cos(x^2-4px+q^2+8q+16)}{(x-2p)^4}, & x \neq 2p \\ 0, & x=2p \end{cases}$. Then $\lim _{x \rightarrow 2p^{+}}[f(x)]$,where $[.]$ denotes the greatest integer function,is $........$

Similar Questions

If $f(x) = \begin{cases} \frac{5}{2} - x, & x < 2 \\ 1, & x = 2 \\ x - \frac{3}{2}, & x > 2 \end{cases}$,then:

If $f(x) = \begin{cases} e^{1/x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$,then:

If $f(x)= \begin{cases}-2 \sin x & , \quad x \leqslant-\frac{\pi}{2} \\ a \sin x+b & , \quad \frac{-\pi}{2} < x < \frac{\pi}{2} \\ \cos x & , \quad x \geqslant \frac{\pi}{2}\end{cases}$ is continuous at $x=-\frac{\pi}{2}$ and $x=\frac{\pi}{2}$,then the value of $2a+b$ is

Discuss the continuity of the function $f$,where $f$ is defined by $f(x) = \begin{cases} -2, & \text{if } x \le -1 \\ 2x, & \text{if } -1 < x \le 1 \\ 2, & \text{if } x > 1 \end{cases}$. Is it continuous at $x=3$?

If $f(x) = \begin{cases} \frac{|x - a|}{x - a}, & x \neq a \\ 1, & x = a \end{cases}$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module