MCQ
Find equation of line joining $(1,2)$ and $(3,6)$ using determinates
- ✓$y=2 x$
- B$y=3 x$
- C$y=4 x+7$
- D$y=2 x+9$
$\therefore \frac{1}{2}\left|\begin{array}{lll}1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1\end{array}\right|=0$
$\Rightarrow \frac{1}{2}[1(6-y)-2(3-x)+1(3 y-6 x)]=0$
$\Rightarrow 6-y-6+2 x+3 y-6 x=0$
$\Rightarrow 2 y-4 x=0$
$\Rightarrow y=2 x$
Hence, the equation of the line joining the given points is $y=2 x$
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Statement $-2$ : $f(x) = \frac{1}{{\sqrt {1 - {x^2}} }} + \left[ {\frac{{{x^2} + x + 1}}{4}} \right]$ , where $[.]$ is greatest integer function. Function $f(x)$ is even function
$(I)$ $g$ is an increasing function in $(0,1)$
$(II)$ $g$ is one-one in $(0,1)$ Then,
$\left| {1 - {{\log }_{\frac{1}{6}}}x} \right| + \left| {{{\log }_2}x} \right| + 2 = \left| {3 - {{\log }_{\frac{1}{6}}}x + {{\log }_{\frac{1}{2}}}x} \right|$ is $\left[ {\frac{a}{b},a} \right],a,b, \in N,$ then the value of $(a + b)$ is