Question

If $L=\underset{x\to \frac{\pi }{2}}{\mathrm{Lim}}\underset{\frac{\pi }{2}}{\overset{x}{\int }}\frac{{\cos }^{2}t\sqrt{a^2-{\cos }^{2}t}}{bx+a\cot x-\frac{b\pi }{2}}dt$ exists finitely, then

• A. $L=0$ if $a\ne b$
• B. $L=1$ if $a=b,a<0$
• C. $L=-1$ if $a\ne b$
• D. $L=-1$ if $a=b,a>0$

Answer 1

Answer: (A), (B), (D)

$L=\underset{x\to \frac{\pi }{2}}{\mathrm{Lim}}\underset{\frac{\pi }{2}}{\overset{x}{\int }}\frac{{\cos }^{2}t\sqrt{a^2-{\cos }^{2}t}}{bx+a\cot x-\frac{b\pi }{2}}dt(\frac{0}{0}$ form $)$
$=\underset{x\to \frac{\pi }{2}}{\mathrm{Lim}}\frac{{\cos }^{2}x\sqrt{a^2-{\cos }^{2}x}}{b-a{\mathrm{cosec}}^{2}x}=0\text{ if }a\ne b$
when $a=b$
$L=\underset{x\to \frac{\pi }{2}}{\mathrm{Lim}}\frac{{\cos }^{2}x\sqrt{a^2-{\cos }^{2}x}}{a\left(1-{\mathrm{cosec}}^{2}x\right)}=\underset{x\to \frac{\pi }{2}}{\mathrm{Lim}}\frac{{\cos }^{2}x\sqrt{a^2-{\cos }^{2}x}}{-a{\cot }^{2}x}$
$=\underset{x\to \frac{\pi }{2}}{\mathrm{Lim}}\frac{{\sin }^{2}x\sqrt{a^2-{\cos }^{2}x}}{-a}=\frac{|a|}{-a}=\{\begin{array}{ll}-1;& a>0\\ 1;& a<0\end{array}$