A) 2
B) 3
C) 4
D) 8
Correct Answer: A
Solution :
[a] $-|3x-1|={{\alpha }_{1}}$ $\Rightarrow \,\,\,\,3x-1=\pm {{\alpha }_{1}}$ $\Rightarrow \,\,\,\,x=\frac{1\pm {{\alpha }_{1}}}{3}$ Similarly, $x=\frac{1\pm {{\alpha }_{2}}}{3}$ and $x=\frac{1\pm {{\alpha }_{3}}}{3}$. $x={{\beta }_{1}},{{\beta }_{2}},.....{{\beta }_{6}}$. $\sum\limits_{i=1}^{n}{{{\beta }_{i}}=\left( \frac{1-{{\alpha }_{1}}}{3} \right)+\left( \frac{1+{{\alpha }_{1}}}{3} \right)+\left( \frac{1-{{\alpha }_{2}}}{3} \right)}$ $+\left( \frac{1+{{\alpha }_{2}}}{3} \right)+\left( \frac{1-{{\alpha }_{3}}}{3} \right)+\left( \frac{1+{{\alpha }_{3}}}{3} \right)=2$You need to login to perform this action.
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