Let S be the set of all real roots of the equation, 3x(3x−1)+2=|3x−1|+|3x−2|. Then S :
A
is a singleton.
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B
is an empty set.
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C
contains at least four elements.
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D
contains exactly two elements.
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Solution
The correct option is A is a singleton. 3x(3x−1)+2=|3x−1|+|3x−2| Let 3x=t Then t(t−1)+2=|t−1|+|t−2| ⇒t2−t+2=|t−1|+|t−2| We plot t2−t+2 and |t−1|+|t−2| As t=3x is always positive, therefore only positive values of t will be the solution.
The graphs intersect for only positive value of t. So, there is single value of x given by x=log3t Therefore, we have only one solution.