JEE MAIN - Mathematics (2016 - 9th April Morning Slot - No. 8)
For x $$ \in $$ R, x $$ \ne $$ 0, Let f0(x) = $${1 \over {1 - x}}$$ and
fn+1 (x) = f0(fn(x)), n = 0, 1, 2, . . . .
Then the value of f100(3) + f1$$\left( {{2 \over 3}} \right)$$ + f2$$\left( {{3 \over 2}} \right)$$ is equal to :
fn+1 (x) = f0(fn(x)), n = 0, 1, 2, . . . .
Then the value of f100(3) + f1$$\left( {{2 \over 3}} \right)$$ + f2$$\left( {{3 \over 2}} \right)$$ is equal to :
$${8 \over 3}$$
$${5 \over 3}$$
$${4 \over 3}$$
$${1 \over 3}$$
Explanation
As fn+1(x) = f0(fn(x))
$$ \therefore $$ f1(x) = f0+1(x) = f0(f0(x)) = $${1 \over {1 - {1 \over {1 - x}}}}$$ = $${{x - 1} \over x}$$
f2(x) = f1+1(x) = f0(f1(x)) = $${1 \over {1 - {{x - 1} \over x}}}$$ = x
f3(x) = f2+1(x) = f0(f2(x)) = $${1 \over {1 - x}}$$
f4(x) = f3+1(x) = f0(f3(x)) = $${1 \over {1 - {1 \over {1 - x}}}}$$ = $${{x - 1} \over x}$$
$$ \therefore $$ f0(x) = f3(x) = f6(x) = . . . . . = $${1 \over {1 - x}}$$
f1(x) = f4(x) = f7(x) = . . . . . .= $${{x - 1} \over x}$$
f2(x) = f5(x) = f8(x) = . . . . . . = x
So, f100(3) = $${{3 - 1} \over 3}$$ = $${2 \over 3}$$
f1$$\left( {{2 \over 3}} \right)$$ = $${{{2 \over 3} - 1} \over {{2 \over 3}}}$$ = $$-$$ $${1 \over 2}$$
f2$$\left( {{3 \over 2}} \right)$$ = $${{3 \over 2}}$$
$$ \therefore $$ f100(3) + f1$$\left( {{2 \over 3}} \right)$$ + f2$$\left( {{3 \over 2}} \right)$$
= $${2 \over 3}$$ $$-$$ $${1 \over 2}$$ + $${3 \over 2}$$
= $${5 \over 3}$$
$$ \therefore $$ f1(x) = f0+1(x) = f0(f0(x)) = $${1 \over {1 - {1 \over {1 - x}}}}$$ = $${{x - 1} \over x}$$
f2(x) = f1+1(x) = f0(f1(x)) = $${1 \over {1 - {{x - 1} \over x}}}$$ = x
f3(x) = f2+1(x) = f0(f2(x)) = $${1 \over {1 - x}}$$
f4(x) = f3+1(x) = f0(f3(x)) = $${1 \over {1 - {1 \over {1 - x}}}}$$ = $${{x - 1} \over x}$$
$$ \therefore $$ f0(x) = f3(x) = f6(x) = . . . . . = $${1 \over {1 - x}}$$
f1(x) = f4(x) = f7(x) = . . . . . .= $${{x - 1} \over x}$$
f2(x) = f5(x) = f8(x) = . . . . . . = x
So, f100(3) = $${{3 - 1} \over 3}$$ = $${2 \over 3}$$
f1$$\left( {{2 \over 3}} \right)$$ = $${{{2 \over 3} - 1} \over {{2 \over 3}}}$$ = $$-$$ $${1 \over 2}$$
f2$$\left( {{3 \over 2}} \right)$$ = $${{3 \over 2}}$$
$$ \therefore $$ f100(3) + f1$$\left( {{2 \over 3}} \right)$$ + f2$$\left( {{3 \over 2}} \right)$$
= $${2 \over 3}$$ $$-$$ $${1 \over 2}$$ + $${3 \over 2}$$
= $${5 \over 3}$$
Comments (0)
