Question

If $\underset{x\to \infty }{lim}4x(\frac{\pi }{4}-{\tan }^{-1}\frac{x+1}{x+2})=y^2+4y+5$,

then $y$ can be equal to?

• A. $1$
• B. $-1$
• C. $-4$
• D. $-3$

Answer 1

Answer: (B), (D)

The correct options are
B $-1$
D $-3$

${lim}_{x\to \infty }4x(\frac{\pi }{4}-{\tan }^{-1}\frac{x+1}{x+2})=y^2+4y+5$

$={lim}_{x\to \infty }4x({\tan }^{-1}1-{\tan }^{-1}\frac{x+1}{x+2})$

$={lim}_{x\to \infty }4x\left(\frac{1-\frac{x+1}{x+2}}{1+1\left(\frac{x+1}{x+2}\right)}\right)={lim}_{x\to \infty }4x\left(\frac{1}{2x+3}\right)$

$={lim}_{x\to \infty }\frac{4}{2+\frac{3}{x}}=\frac{4}{2+0}=2$

$y^2+4y+5=2\Rightarrow y^2+4y+3=0$

$(y+3)(y+1)=0$

$\Rightarrow y=-1,-3$