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QJEEM19Macd6e: JEE Main 2019 Mathematics - Class 11 JEEM19Macd6e Solution: For , Let f 1 (x) = , f 2 (x) = 1 – x and f 3 (x) = be three…

For x∈R−{0,1}x \in R - \left\{ {0,1} \right\} , Let f1(x) = 1x1\over x , f2 (x) = 1 – x

and f3 (x) = 11−x1 \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 :

Multiple Choice Options

  • A f1 (x)
  • B 1x1 \over x f3 (x)
  • C f2 (x)
  • D f3 (x)

Step-by-Step Verified Solution

Given,

f1(x) = 1x{1 \over x}

f2(x) = 1 −- x

f3(x) = 11−x{1 \over {1 - x}}

(f2 ⋅\cdot J ⋅\cdot f1) (x) = f3(x)

⇒\Rightarrow    f2 {J(f1(x))} = f3(x)

⇒\Rightarrow    f2{J ( 1x{1 \over x} )} = 11−x{1 \over {1 - x}}

⇒\Rightarrow    1 −- J( 1x{1 \over x} ) = 11−x{1 \over {1 - x}}

⇒\Rightarrow   J( 1x{{1 \over x}} ) = 1 −- 11−x{{1 \over {1 - x}}}

⇒\Rightarrow    J ( 1x{{1 \over { x}}} ) = −x1−x{{ - x} \over {1 - x}} = xx−1{x \over {x - 1}}

Put x inplace of 1x{1 \over x}

∴\therefore    J(x) = 1x1x−1{{{1 \over x}} \over {{1 \over x} - 1}}

= 11−x=f3(x){1 \over {1 - x}} = {f_3}\left( x \right)

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