MCQ
$\int_0^\infty {\frac{{{x^2}\,dx}}{{({x^2} + {a^2})({x^2} + {b^2})}}} = $
- A$\frac{\pi }{{2(a - b)}}$
- B$\frac{\pi }{{2(b - a)}}$
- C$\frac{\pi }{{(a + b)}}$
- ✓$\frac{\pi }{{2(a + b)}}$
$ = \int_0^\infty {\frac{1}{{{x^2} + {b^2}}}dx - {a^2}\int_0^\infty {\frac{1}{{({x^2} + {a^2})({x^2} + {b^2})}}{\rm{ }}} dx} $
$ = \left[ {\frac{1}{b}{{\tan }^{ - 1}}\frac{x}{b}} \right]_0^\infty - \frac{{{a^2}}}{{({a^2} - {b^2})}}\int_0^\infty {\left( {\frac{1}{{{x^2} + {b^2}}} - \frac{1}{{{x^2} + {a^2}}}} \right){\rm{ }}} dx$
$ = \frac{1}{b}.\frac{\pi }{2} - \frac{{{a^2}}}{{({a^2} - {b^2})}}\left[ {\frac{1}{b}{{\tan }^{ - 1}}\frac{x}{b} - \frac{1}{a}{{\tan }^{ - 1}}\frac{x}{a}} \right]_0^\infty $
$= \frac{\pi }{{2(a + b)}}$.
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$[A]$ $x=-1$ $[B]$ $x=0$ $[C]$ $x=2$ $[D] x=1$
Statement $-1 :$$S=\{x:f(x)=f^{-1}(x)\}=$$\left\{ {1,2} \right\}$
Statement $-2 :$ $f $ is a bijection and ${f^{ - 1}}\left( x \right) = 1 + \sqrt {x - 1} \;,x \ge 1$