MCQ
The value of $\int {\frac{{2\,\,dx}}{{\sqrt {1 - 4{x^2}} }}} $ is
- A${\tan ^{ - 1}}(2x) + c$
- B${\cot ^{ - 1}}(2x) + c$
- C${\cos ^{ - 1}}(2x) + c$
- ✓${\sin ^{ - 1}}(2x) + c$
Put $2x = \sin \theta $ ==> $2dx = \cos \theta \,\,d\theta $
$ \Rightarrow I = \int {\frac{{\cos \theta \,\,d\theta }}{{\sqrt {1 - {{\sin }^2}\theta } }} = \int {\frac{{\cos \theta }}{{\cos \theta }}d\theta = \int {d\theta + c = \theta + c} } } $.
Therefore, $I = {\sin ^{ - 1}}(2x) + c.$
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$(a-c) x^2+(b-a) x+(c-b)=0$ where $a, b, c$ are distinct real numbers such that the matrix
$\left[\begin{array}{ccc}\alpha^2 & \alpha & 1 \\1 & 1 & 1 \\a & b & c\end{array}\right]$
is singular. Then the value of
$\frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)}$