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  • Let <img alt="x=2$ be a root of the equation $x^2+p x+q=0" src="https://entrancecorner.oncodecogs.com/gif.latex?x%3D2%24%20be%20a%20root%20of%20the%20equation%20%24x%5E2&plus;p%20x&plus;q%3

Let x=2$ be a root of the equation $x^2+p x+q=0\mathrm{ and}
f(x)=\left\{\begin{array}{cl} \frac{1-\cos \left(x^2-4 p x+q^2+8 q+16\right)}{(x-2 p)^4} & , x \neq 2 p \\ 0, & , x=2 p \end{array}\right.

\lim _{x \rightarrow 2 p^{+}}[f(x)]
where\left [ \cdot \right ] denotes greatest integer function, is

Option: 1

0


Option: 2

-1


Option: 3

2


Option: 4

1


Answers (1)

best_answer

\begin{aligned} & f(x)=\left\{\begin{array}{l} \frac{1-\cos \left(x^2-4 p x+q^2+8 q+16\right)}{(x-2 p)^4}, x \neq 2 p \\ 0 \quad, \quad x=2 p \end{array}\right. \\\end{aligned}
\because x=2 \text { is a root of equation } x^2+p x+q=0
\mathrm{\therefore 4+2 p+q=0}
\mathrm{\Rightarrow 2 p=-q-4 }
\mathrm{ \Rightarrow 4 p^2=(q+4)^2=q^2+8 q+16 }         ...............(1)
\text { Now } \lim _{x \rightarrow 2 p^{+}} f(x)=\lim _{x \rightarrow 2 p^{+}} \frac{1-\cos \left(x^2-4 p x+4 p^2\right)}{(x-2 p)^4}    (From (1))    
=\lim _{x \rightarrow 2 p^{+}}\left[\frac{1-\cos (x-2 p)^2}{\left\{(x-2 p)^2\right\}^2}\right]
=\frac{1}{2}\left\{\because \lim _{x \rightarrow 0} \frac{1-\cos \theta}{\theta^2}=\frac{1}{2}\right\}
\therefore \lim _{x \rightarrow 2 p^{+}}[f(x)]=\left[\frac{1}{2}\right]=0
 

Posted by

Deependra Verma

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