33. If two positive real numbers $x , y$ satisfy $x ( 1 + \ln x ) = y \mathrm { e } ^ { y - 1 }$, give the following inequalities: (1) $y < x < 1$; (2) $1 < x < y$ ;
(3) $1 < y < x$ ; (4) $y < 1 < x$ . The number of these that may hold is ( )
- A. A. 0
- B. B. 1
- C. C. 2
- D. D. 3
Answer: C
Solution: $x ( 1 + \ln x ) = y \mathrm { e } ^ { y - 1 } \Rightarrow y \mathrm { e } ^ { y - 1 } = ( 1 + \ln x ) \mathrm { e } ^ { 1 + \ln x - 1 }$, the
constructor $f ( y ) = y \mathrm { e } ^ { y } ( y > 0 ) \Rightarrow f ^ { \prime } ( y ) = ( y + 1 ) \mathrm { e } ^ { y } > 0$, so the function $f ( y )$ is increasing on the set of positive real numbers.
Since $x$ is positive real, it follows from $x ( 1 + \ln x ) = y \mathrm { e } ^ { y - 1 } \Rightarrow 1 + \ln x > 0$ that
Therefore by ${ } ^ { y \mathrm { e } ^ { y - 1 } } = ( 1 + \ln x ) \mathrm { e } ^ { 1 + \ln x - 1 } \Rightarrow f ( y ) = f ( 1 + \ln x ) \Rightarrow y = 1 + \ln x$ , the
Let $g ( y ) = 1 + \ln y - y \Rightarrow g ^ { \prime } ( y ) = \frac { 1 - y } { y }$, when $y > 1$, $g ^ { \prime } ( y ) < 0 , g ( y )$ monotonically decreases, when $0 < y < 1 _ { \text {时 } , ~ } g ^ { \prime } ( y ) > 0 , g ( y )$ monotonically increases, so $g ( y ) _ { \text {max } } = g ( 1 ) = 0$, so we have $g ( y ) _ { \text {max } } = g ( 1 ) = 0$. INLINE_FORMULA_11]], and $y = 1 + \ln x$, so $y \leq x$, taking the equal sign when and only when $y = x = 1$, and when $x \neq 1$, , and from the above, $y < x < 1$, or $1 < y < x$.
[点睛]Key point : The deformation of the equation to construct a function and the use of the nature of the derivative is the key to solving the problem. 34 The D
[Knowledge Points] Determine the parameters from the solution of the quadratic inequality, solve the quadratic inequality without parameters
[Analysis]the inequality is $( x - a ) ( x - 1 ) < 0$, divided into $a > 1$ and $a < 1$ two cases of discussion, to find the solution set of the inequality, combined with the meaning of the problem and the representation of the set, can be solved.
By the meaning of the question, the inequality $x ^ { 2 } - ( a + 1 ) x + a < 0$ can be reduced to $( x - a ) ( x - 1 ) < 0$, and when $a > 1$, the solution set of the inequality $x ^ { 2 } - ( a + 1 ) _ { x + a < 0 }$ will be $( 1 , a )$. To have exactly 3 integers in the solution set of the inequality $x ^ { 2 } - ( a + 1 ) x + a < 0$, $4 < a \leq 5$; when $a < 1$ is $a < 1$ the solution set of the inequality $x ^ { 2 } - ( a + 1 ) x + a < 0$ is $( a , 1 )$. 31]]. If there are exactly 3 integers in the solution set of the inequality $x ^ { 2 } - ( a + 1 ) x + a < 0$, then $- 3 \leq a < - 2$ , and the range of the real number ${ } ^ { a }$ is $[ - 3 , - 2 ) \cup ( 4,5 ]$. Therefore, the choice: D.
[Eyesight] This question mainly examines the solution of quadratic inequality and its application, in which the answer to memorize the solution of quadratic inequality, combined with the relationship between the elements and the set of solutions is the key to the answer, focusing on the examination of arithmetic and the ability to solve.