JEE MAIN - Mathematics (2021 - 16th March Morning Shift - No. 4)
The number of elements in the set {x $$\in$$ R : (|x| $$-$$ 3) |x + 4| = 6} is equal to :
4
2
3
1
Explanation
Case 1 :
x $$ \le $$ $$-$$4
($$-$$x $$-$$ 3)($$-$$x $$-$$ 4) = 6
$$ \Rightarrow $$ (x + 3)(x + 4) = 6
$$ \Rightarrow $$ x2 + 7x + 6 = 0
$$ \Rightarrow $$ x = $$-$$1 or $$-$$6
but x $$ \le $$ $$-$$4
x = $$-$$6
Case 2 :
x $$\in$$ ($$-$$4, 0)
($$-$$x $$-$$ 3)(x + 4) = 6
$$ \Rightarrow $$ $$-$$x2 $$-$$ 7x $$-$$ 12 $$-$$ 6 = 0
$$ \Rightarrow $$ x2 + 7x + 18 = 0
D < 0 No solution
Case 3 :
x $$ \ge $$ 0
(x $$-$$ 3)(x + 4) = 6
$$ \Rightarrow $$ x2 + x $$-$$ 12 $$-$$ 6 = 0
$$ \Rightarrow $$ x2 + x $$-$$ 18 = 0
x = $${{ - 1 \pm \sqrt {1 + 72} } \over 2}$$
$$ \therefore $$ x = $${{\sqrt {73} - 1} \over 2}$$ only
x $$ \le $$ $$-$$4
($$-$$x $$-$$ 3)($$-$$x $$-$$ 4) = 6
$$ \Rightarrow $$ (x + 3)(x + 4) = 6
$$ \Rightarrow $$ x2 + 7x + 6 = 0
$$ \Rightarrow $$ x = $$-$$1 or $$-$$6
but x $$ \le $$ $$-$$4
x = $$-$$6
Case 2 :
x $$\in$$ ($$-$$4, 0)
($$-$$x $$-$$ 3)(x + 4) = 6
$$ \Rightarrow $$ $$-$$x2 $$-$$ 7x $$-$$ 12 $$-$$ 6 = 0
$$ \Rightarrow $$ x2 + 7x + 18 = 0
D < 0 No solution
Case 3 :
x $$ \ge $$ 0
(x $$-$$ 3)(x + 4) = 6
$$ \Rightarrow $$ x2 + x $$-$$ 12 $$-$$ 6 = 0
$$ \Rightarrow $$ x2 + x $$-$$ 18 = 0
x = $${{ - 1 \pm \sqrt {1 + 72} } \over 2}$$
$$ \therefore $$ x = $${{\sqrt {73} - 1} \over 2}$$ only
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