JEE MAIN - Mathematics (2019 - 9th January Morning Slot - No. 10)
For $$x \in R - \left\{ {0,1} \right\}$$, Let f1(x) = $$1\over x$$, f2 (x) = 1 – x
and f3 (x) = $$1 \over {1 - x}$$ be three given
functions. If a function, J(x) satisfies
(f2 o J o f1) (x) = f3 (x) then J(x) is equal to :
and f3 (x) = $$1 \over {1 - x}$$ be three given
functions. If a function, J(x) satisfies
(f2 o J o f1) (x) = f3 (x) then J(x) is equal to :
f1 (x)
$$1 \over x$$ f3 (x)
f2 (x)
f3 (x)
Explanation
Given,
f1(x) = $${1 \over x}$$
f2(x) = 1 $$-$$ x
f3(x) = $${1 \over {1 - x}}$$
(f2 $$ \cdot $$ J $$ \cdot $$ f1) (x) = f3(x)
$$ \Rightarrow $$ f2 {J(f1(x))} = f3(x)
$$ \Rightarrow $$ f2{J ($${1 \over x}$$)} = $${1 \over {1 - x}}$$
$$ \Rightarrow $$ 1 $$-$$ J($${1 \over x}$$) = $${1 \over {1 - x}}$$
$$ \Rightarrow $$ J($${{1 \over x}}$$) = 1 $$-$$ $${{1 \over {1 - x}}}$$
$$ \Rightarrow $$ J ($${{1 \over { x}}}$$) = $${{ - x} \over {1 - x}}$$ = $${x \over {x - 1}}$$
Put x inplace of $${1 \over x}$$
$$ \therefore $$ J(x) = $${{{1 \over x}} \over {{1 \over x} - 1}}$$
= $${1 \over {1 - x}} = {f_3}\left( x \right)$$
f1(x) = $${1 \over x}$$
f2(x) = 1 $$-$$ x
f3(x) = $${1 \over {1 - x}}$$
(f2 $$ \cdot $$ J $$ \cdot $$ f1) (x) = f3(x)
$$ \Rightarrow $$ f2 {J(f1(x))} = f3(x)
$$ \Rightarrow $$ f2{J ($${1 \over x}$$)} = $${1 \over {1 - x}}$$
$$ \Rightarrow $$ 1 $$-$$ J($${1 \over x}$$) = $${1 \over {1 - x}}$$
$$ \Rightarrow $$ J($${{1 \over x}}$$) = 1 $$-$$ $${{1 \over {1 - x}}}$$
$$ \Rightarrow $$ J ($${{1 \over { x}}}$$) = $${{ - x} \over {1 - x}}$$ = $${x \over {x - 1}}$$
Put x inplace of $${1 \over x}$$
$$ \therefore $$ J(x) = $${{{1 \over x}} \over {{1 \over x} - 1}}$$
= $${1 \over {1 - x}} = {f_3}\left( x \right)$$
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