JEE MAIN - Mathematics (2020 - 8th January Evening Slot - No. 10)
Let S be the set of all real roots of the equation,
3x(3x – 1) + 2 = |3x – 1| + |3x – 2|. Then S :
3x(3x – 1) + 2 = |3x – 1| + |3x – 2|. Then S :
contains exactly two elements.
is an empty set.
is a singleton.
contains at least four elements.
Explanation
Let 3x = t ; t $$>$$ 0
t(t – 1) + 2 = |t – 1| + |t – 2|
t2 – t + 2 = |t – 1| + |t – 2|
Case-I : t $$<$$ 1
t2 – t + 2 = 1 – t + 2 – t
$$ \Rightarrow $$ t2 + 2 = 3 – t
$$ \Rightarrow $$ t2 + t – 1 = 0
$$ \Rightarrow $$ t = $${{ - 1 \pm \sqrt 5 } \over 2}$$
$$ \Rightarrow $$ t = $${{\sqrt 5 - 1} \over 2}$$ [ As t $$>$$ 0]
Case-II : 1 $$ \le $$ t $$<$$ 2
$$ \Rightarrow $$ t2 – t + 2 = t – 1 + 2 – t
$$ \Rightarrow $$ t2 – t + 1 = 0
D $$<$$ 0 so no real solution.
Case-III : t $$ \ge $$ 2
$$ \Rightarrow $$ t2 – t + 2 = t – 1 + t – 2
$$ \Rightarrow $$ t2 – 3t - 5 = 0
$$ \Rightarrow $$ D $$<$$ 0 so no real solution.
t(t – 1) + 2 = |t – 1| + |t – 2|
t2 – t + 2 = |t – 1| + |t – 2|
Case-I : t $$<$$ 1
t2 – t + 2 = 1 – t + 2 – t
$$ \Rightarrow $$ t2 + 2 = 3 – t
$$ \Rightarrow $$ t2 + t – 1 = 0
$$ \Rightarrow $$ t = $${{ - 1 \pm \sqrt 5 } \over 2}$$
$$ \Rightarrow $$ t = $${{\sqrt 5 - 1} \over 2}$$ [ As t $$>$$ 0]
Case-II : 1 $$ \le $$ t $$<$$ 2
$$ \Rightarrow $$ t2 – t + 2 = t – 1 + 2 – t
$$ \Rightarrow $$ t2 – t + 1 = 0
D $$<$$ 0 so no real solution.
Case-III : t $$ \ge $$ 2
$$ \Rightarrow $$ t2 – t + 2 = t – 1 + t – 2
$$ \Rightarrow $$ t2 – 3t - 5 = 0
$$ \Rightarrow $$ D $$<$$ 0 so no real solution.
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