Question

If $\underset{x\to \infty }{lim}(\sqrt{x^2-x+1}-ax-b)=0,$ then for $k\ge 2,$ $\underset{n\to \infty }{lim}{\sec }^{2n}(k!\pi b)$ is equal to

• A. $-a$
• B. $a$
• C. $-b$
• D. $b$

Answer 1

The correct option is B $a$

$\underset{x\to \infty }{lim}(\sqrt{x^2-x+1}-ax-b)=0$
Here, $a>0$ because if $a\le 0$ then $lim\to \infty$

On rationalizing,
$\underset{x\to \infty }{lim}\frac{(\sqrt{x^2-x+1}-ax-b)\times (\sqrt{x^2-x+1}+ax+b)}{\sqrt{x^2-x+1}+ax+b}=0\Rightarrow \underset{x\to \infty }{lim}\frac{x^2-x+1-(ax+b{)}^2}{\sqrt{x^2-x+1}+ax+b}=0\Rightarrow \underset{x\to \infty }{lim}\frac{(1-a^2)x^2-(1+2ab)x+1-b^2}{\sqrt{x^2-x+1}+ax+b}=0$
This is possible only when $1-a^2=0$ and $1+2ab=0$
$\Rightarrow a=\pm 1$
$\Rightarrow a=1(\because a>0)$ and $b=\frac{-1}{2}$

Now, $k!\pi b=-\frac{\pi k!}{2}=$ an integral multiple of $\pi (\because k\ge 2)$
$\therefore {\sec }^2(k!\pi b)=(\pm 1{)}^2=1$
$\Rightarrow \underset{n\to \infty }{lim}{\sec }^{2n}(k!\pi b)=\underset{n\to \infty }{lim}{1}^n=1=a$

Alternate Solution:
$\underset{x\to \infty }{lim}(\sqrt{x^2-x+1}-ax-b)=0...\left(1\right)$
$\Rightarrow \underset{x\to \infty }{lim}(x\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}-ax-b)=0$
$\Rightarrow \underset{x\to \infty }{lim}(x-ax-b)=0$
$\Rightarrow \underset{x\to \infty }{lim}(1-a)x-b=0$
This is possible only when $a=1$
Substitute $a=1$ in equation $\left(1\right),$ we get
$\Rightarrow \underset{x\to \infty }{lim}(\sqrt{x^2-x+1}-x-b)=0$
$\Rightarrow \underset{x\to \infty }{lim}\sqrt{x^2-x+1}=\underset{x\to \infty }{lim}(x+b)$
Squaring both the sides,
$\underset{x\to \infty }{lim}\left(x^2-x+1\right)=\underset{x\to \infty }{lim}(x^2+2bx+b^2)$
$\Rightarrow \underset{x\to \infty }{lim}(-x+1)=\underset{x\to \infty }{lim}(2bx+b^2)$
$\Rightarrow \underset{x\to \infty }{lim}(-x)=\underset{x\to \infty }{lim}2bx$
$\Rightarrow b=-\frac{1}{2}$

​​​​​​​Now, $k!\pi b=-\frac{\pi k!}{2}=$ an integral multiple of $\pi (\because k\ge 2)$
$\therefore {\sec }^2(k!\pi b)=(\pm 1{)}^2=1$
$\Rightarrow \underset{n\to \infty }{lim}{\sec }^{2n}(k!\pi b)=\underset{n\to \infty }{lim}{1}^n=1=a$