JEE MAIN - Mathematics (2019 - 12th January Morning Slot - No. 21)
Considering only the principal values of inverse functions, the set
A = { x $$ \ge $$ 0: tan$$-$$1(2x) + tan$$-$$1(3x) = $${\pi \over 4}$$}
A = { x $$ \ge $$ 0: tan$$-$$1(2x) + tan$$-$$1(3x) = $${\pi \over 4}$$}
contains two elements
contains more than two elements
is an empty set
is a singleton
Explanation
tan$$-$$1(2x) + tan$$-$$1(3x) = $$\pi $$/4
$$ \Rightarrow \,\,{{5x} \over {1 - 6{x^2}}}$$ = 1
$$ \Rightarrow $$ 6x2 + 5x $$-$$ 1 = 0
x = $$-$$1 or x = $${1 \over 6}$$
x = $${1 \over 6}$$
$$ \because $$ x > 0
$$ \Rightarrow \,\,{{5x} \over {1 - 6{x^2}}}$$ = 1
$$ \Rightarrow $$ 6x2 + 5x $$-$$ 1 = 0
x = $$-$$1 or x = $${1 \over 6}$$
x = $${1 \over 6}$$
$$ \because $$ x > 0
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