For the Function defined by , among the two statements: (I) The set contains exactly two elements, and (II) The set is an empty set, [2026]

Question

For the Function $f : [1, \infty) \to [1, \infty)$ defined by $f (x) = {(x - 1)}^{4} + 1$ , among the two statements:

(I) The set $S = {x \in [1, \infty) : f (x) = f^{- 1} (x)}$ contains exactly two elements, and

(II) The set $S = {x \in [1, \infty) : f (x) = f^{- 1} (x + 1)}$ is an empty set, [2026]

Smart Achievers · Accepted Answer

(1)

$f (x) = f^{- 1} (x)$

$\Rightarrow f (x) = x \Rightarrow {(x - 1)}^{4} + 1 = x$

$\Rightarrow {(x - 1)}^{4} - (x - 1) = 0$

$\Rightarrow {(x - 1) [(x - 1)}^{3} - 1] = 0$

$\Rightarrow x = 1, x = 2$

So, set S has exactly 2 elements

$∴$ (I) is true.

Now, $f (x) = {(x - 1)}^{4} + 1$

$\Rightarrow y - 1 = {(x - 1)}^{4} \Rightarrow x = {(y - 1)}^{1 / 4} + 1$

$\Rightarrow f^{- 1} (y) = {(y - 1)}^{1 / 4} + 1$

i.e., $f^{- 1} (x) = {(x - 1)}^{1 / 4} + 1$

$∴ f^{- 1} (x + 1) = x^{1 / 4} + 1$

Now, $f (x) = f^{- 1} (x + 1)$

$\Rightarrow {(x - 1)}^{4} + 1 = x^{1 / 4} + 1$

$\Rightarrow {(x - 1)}^{4} = x^{1 / 4}$

$∴$ (II) is not true.

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