MCQ
Solve $\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1} \frac{x-y}{x+y}$ is equal to
- ✓$\frac{\pi}{4}$
- B$\frac{-3 \pi}{4}$
- C$\frac{\pi}{2}$
- D$\frac{\pi}{3}$
$=\tan ^{-1}\left[\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\left(\frac{x}{y}\right)\left(\frac{x-y}{x+y}\right)}\right]$
$=\tan ^{-1}\left[\frac{\frac{x(x+y)-y(x-y)}{y(x+y)}}{\frac{y(x+y)+x(x-y)}{y(x+y)}}\right]$
$=\tan ^{-1}\left(\frac{x^{2}+x y-x y+y^{2}}{x y+y^{2}+x^{2}-x y}\right)$
$=\tan ^{-1}\left(\frac{x^{2}+y^{2}}{x^{2}+y^{2}}\right)=\tan ^{-1} 1=\frac{\pi}{4}$
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$f(x) = \sqrt {\left| {{{\sin }^{ - 1}}\left| {\sin x} \right|} \right| - {{\cos }^{ - 1}}\left| {\cos x} \right|} $ is
$D^*f(x) =\mathop {Limit}\limits_{h \to 0} \frac{{{f^2}(x + h) - {f^2}(x)}}{h}$ where $f^2(x)$ means $[f(x)]^2.$ If $f(x) = x lnx$ then
${\left. {D^*f(x)} \right|_{x = e}}$ has the value