JEE Main 2016 · 9 Apr · Shift 1 · Mathematics Q61

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MathematicsSingle correct+4 / -1Question 61 of 90

For $x\in\mathbb R$ , $x\ne0$ , $x\ne1$ , let $f_0(x)=\dfrac1{1-x}$ and $f_{n+1}(x)=f_0(f_n(x))$ , n = 0, 1, 2, . . . Then the value of $f_{100}(3)+f_1\left(\dfrac23\right)+f_2\left(\dfrac32\right)$ is equal to :

\dfrac83

\dfrac53

\dfrac43

\dfrac13

Answer and worked solution

Official answer: B

$f_1(x)=\frac{x-1}x$ and $f_2(x)=x$ , so the functions repeat with period 3 and $f_{100}=f_1$ . Then
$f_1(3)+f_1\left(\tfrac23\right)+f_2\left(\tfrac32\right)=\tfrac23-\tfrac12+\tfrac32=\tfrac53.$

Topics:Composition and inverse functions

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JEE Main 2016 · 9 Apr · Shift 1, Mathematics question 61