(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,
Solution
Statement (I) Analysis
The function is given by $f(x) = (x-1)^4 + 1$ for $x \in [1,\infty)$.
- Find the inverse function: Let $y = (x-1)^4 + 1$. Solving for $x$, we get $y-1 = (x-1)^4$. Since $x \ge 1$, $x-1 \ge 0$. Thus, $x-1 = (y-1)^{1/4}$, which yields $x = (y-1)^{1/4} + 1$. So, $f^{-1}(x) = (x-1)^{1/4} + 1$.
- Check monotonicity: The derivative is $f'(x) = 4(x-1)^3$. For $x \in [1,\infty)$, $f'(x) \ge 0$, meaning $f(x)$ is strictly increasing.
- Solve $f(x) = f^{-1}(x)$: For a strictly increasing function, solutions to $f(x) = f^{-1}(x)$ coincide with solutions to $f(x) = x$. $ (x-1)^4 + 1 = x $ $ (x-1)^4 = x-1 $ Let $v = x-1$. Then $v^4 = v$, or $v^4 - v = 0$. Factoring gives $v(v^3 - 1) = 0$. The solutions are $v=0$ or $v=1$.
- Find $x$ values:
- If $v=0$, $x-1=0 \implies x=1$.
- If $v=1$, $x-1=1 \implies x=2$.
- Conclusion for (I): The set $\{x \in [1,\infty) : f(x) = f^{-1}(x)\}$ contains exactly two elements, $\{1, 2\}$. Statement (I) is TRUE.
Statement (II) Analysis
We analyze the set $S = \{x \in [1,\infty) : f(x) = f^{-1}(x+1)\}$.
- Set up the equation: $ f(x) = f^{-1}(x+1) $ $ (x-1)^4 + 1 = ((x+1)-1)^{1/4} + 1 $ $ (x-1)^4 = x^{1/4} $
- Substitution: Let $u = x-1$. Since $x \in [1,\infty)$, $u \in [0,\infty)$. Substituting $x = u+1$ into the equation gives: $ u^4 = (u+1)^{1/4} $
- Solve for $u$: Raising both sides to the power of 4 yields: $ (u^4)^4 = ((u+1)^{1/4})^4 $ $ u^{16} = u+1 $ Rearranging, we get $u^{16} - u - 1 = 0$. Let $k(u) = u^{16} - u - 1$.
- Evaluate $k(u)$ at key points:
- $k(0) = 0^{16} - 0 - 1 = -1$.
- $k(1) = 1^{16} - 1 - 1 = -1$.
- $k(2) = 2^{16} - 2 - 1 = 65536 - 3 = 65533$.
- Existence of a solution: Since $k(u)$ is continuous, and $k(1) < 0$ while $k(2) > 0$, the Intermediate Value Theorem guarantees at least one root $u_0$ in the interval $(1, 2)$.
- Conclusion for (II): This root $u_0$ corresponds to a solution $x_0 = u_0 + 1$ in the interval $(2, 3)$. Thus, the set $S$ is not empty. Statement (II) is FALSE.
Final Conclusion
Statement (I) is TRUE and Statement (II) is FALSE. Therefore, only statement (I) is TRUE.
Similar Questions
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
- The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______
- Let $z = (1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni)$, where $i = \sqrt{-1}$. If $|z|^2 = 44200$, then $n$ is equal to _______
- Let $S = \{(m, n) : m, n \in \{1, 2, 3, \dots, 50\}\}$. If the number of elements $(m, n)$ in S such that $6^m + 9^n$ is a multiple of 5 is p and the number of elements $(m, n)$ in S such that $m + n$ is a square of a prime number is q, then $p + q$ is equal to ____________.
- Let $a_1 = 1$ and for $n \ge 1, a_{n+1} = \frac{1}{2} a_n + \frac{n^2 - 2n - 1}{n^2(n+1)^2}$. Then $\left| \sum_{n=1}^\infty \left( a_n - \frac{2}{n^2} \right) \right|$ is equal to ____________.
If $ \left(\frac{1}{^{15}C_0} + \frac{1}{^{15}C_1}\right)$ $\left(\frac{1}{^{15}C_1} + \frac{1}{^{15}C_2}\right)$ $\dots$ $ \left(\frac{1}{^{15}C_{12}} + \frac{1}{^{15}C_{13}}\right)$ = $\frac{\alpha^{13}}{^{14}C_0 ^{14}C_1 \dots ^{14}C_{12}} $, then $30\alpha$ is equal to __________.
- Three persons enter in a lift at the ground floor. The lift will go upto 10$^{\text{th}}$ floor. The number of ways, in which the three persons can exit the lift at three different floors, if the lift does not stop at first, second and third floors, is equal to _________.
- The number of 4-letter words, with or without meaning, which can be formed using the letters PQR PQRSTUVP, is ___________.
- Let $|A| = 6$, where A is a $3 \times 3$ matrix. If $|\text{adj}(3\text{adj}(A^2 \cdot \text{adj}(2A)))| = 2^m \cdot 3^n$, $m, n \in \mathbf{N}$, then $m + n$ is equal to ___________.
Important Questions from Algebra
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
- The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______
- Let $z = (1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni)$, where $i = \sqrt{-1}$. If $|z|^2 = 44200$, then $n$ is equal to _______
- Let $S = \{(m, n) : m, n \in \{1, 2, 3, \dots, 50\}\}$. If the number of elements $(m, n)$ in S such that $6^m + 9^n$ is a multiple of 5 is p and the number of elements $(m, n)$ in S such that $m + n$ is a square of a prime number is q, then $p + q$ is equal to ____________.
