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Function - Practice Questions (40)

Question 1: 1. If the function $f ( x ) = \frac { ( 2 x + 1 ) ( x - a ) } { x } ( a \in R )$ is odd, then the re...

1. If the function $f ( x ) = \frac { ( 2 x + 1 ) ( x - a ) } { x } ( a \in R )$ is odd, then the real number $a =$

  • A. A. $\frac { 1 } { 2 }$
  • B. B. 0
  • C. C. - 1
  • D. D. 1

Answer: A

Solution: SOLUTION: The $\because$ function $f ( x ) = \frac { ( 2 x + 1 ) ( x - a ) } { x } ( a \in R )$ is odd. $\therefore f ( - x ) = \frac { ( - 2 x + 1 ) ( - x - a ) } { - x } = - \frac { ( 2 x - 1 ) ( x + a ) } { x } = - f ( x ) = - \frac { ( 2 x + 1 ) ( x - a ) } { x }$, $\therefore f ( - x ) = \frac { ( - 2 x + 1 ) ( - x - a ) } { - x } = - \frac { ( 2 x - 1 ) ( x + a ) } { x } = - f ( x ) = - \frac { ( 2 x + 1 ) ( x - a ) } { x }$, $\therefore f ( - x ) = \frac { ( - 2 x + 1 ) ( - x - a ) } { - x } = - \frac { ( 2 x - 1 ) ( x + a ) } { x } = - f ( x ) = - \frac { ( 2 x + 1 ) ( x - a ) } { x }$ $\therefore ( 2 x - 1 ) ( x + a ) = ( 2 x + 1 ) ( x - a )$, i.e. $2 x ^ { 2 } - a + ( 2 a - 1 ) x = 2 x ^ { 2 } - a + ( 1 - 2 a ) x$. which simplifies to $( 2 a - 1 ) _ { x = 0 }$, which is $a = \frac { 1 } { 2 }$, which is $a = \frac { 1 } { 2 }$, which is $a = \frac { 1 } { 2 }$.

Question 2: 2. The function $f ( x ) = \frac { \sqrt { x + 1 } } { 2 - x }$ has the definition domain $\_\_\_\_$

2. The function $f ( x ) = \frac { \sqrt { x + 1 } } { 2 - x }$ has the definition domain $\_\_\_\_$

  • A. A. $[ - 1,2 ) \cup ( 2 , + \infty )$
  • B. B. $( - 1 , + \infty )$
  • C. C. $[ - 1,2 )$
  • D. D. $[ - 1 , + \infty )$

Answer: A

Solution: For the function $f ( x ) = \frac { \sqrt { x + 1 } } { 2 - x }$, we have $\left\{ \begin{array} { l } x + 1 \geq 0 \\ 2 - x \neq 0 \end{array} \right.$, which solves $x \geq - 1$ and $x \neq 2$. Therefore, the function $f ( x ) = \frac { \sqrt { x + 1 } } { 2 - x }$ is defined by $[ - 1,2 ) \cup ( 2 , + \infty )$.

Question 3: 3. The definition domain of the function $f ( x ) = \log _ { 2 } ( 1 - x ) + \frac { 1 } { x }$ is

3. The definition domain of the function $f ( x ) = \log _ { 2 } ( 1 - x ) + \frac { 1 } { x }$ is

  • A. A. $( - \infty , 1 )$
  • B. B. $( - \infty , 0 ) \cup ( 0,1 )$
  • C. C. $( 0,1 )$
  • D. D. $( - \infty , 1 ]$

Answer: B

Solution:

Question 4: 4. The graph of the function $y = \frac { x - 2 } { x - 1 }$ is

4. The graph of the function $y = \frac { x - 2 } { x - 1 }$ is

  • A. A. ![](/images/questions/function/image-001.jpg)
  • B. B. ![](/images/questions/function/image-002.jpg)
  • C. C. ![](/images/questions/function/image-003.jpg)
  • D. D. ![](/images/questions/function/image-004.jpg)

Answer: B

Solution: Method 1: When $x = 2$, $y = 0$, only B option is satisfied. Method 2 : $y = \frac { x - 2 } { x - 1 } = - \frac { 1 } { x - 1 } + 1$ . The graph of function $y = \frac { x - 2 } { x - 1 }$ is obtained by shifting function $y = - \frac { 1 } { x }$ to the right by 1 unit and then up by 1 unit. Then, the graph of $y = - \frac { 1 } { x }$ is obtained by shifting the function $y = - \frac { 1 } { x }$ to the right by 1 unit and then up by 1 unit.

Question 5: 5. Which of the following functions is an odd function in the domain of definition and an increasing...

5. Which of the following functions is an odd function in the domain of definition and an increasing function on ${ } ^ { ( 0 , + \infty ) }$ ( )

  • A. A. $y = - 3 x$
  • B. B. $y = 3 ^ { x }$
  • C. C. $y = x + \frac { 1 } { x }$
  • D. D. $y = x - \frac { 1 } { x }$

Answer: D

Solution: ${ } ^ { y = - 3 x }$ is odd but decreasing on ${ } ^ { ( 0 , + \infty ) }$, A is wrong; $y = 3 ^ { x }$ is a non-singular and non-even function, B is wrong; $\mathrm { y } = \mathrm { x } + \frac { 1 } { \mathrm { x } }$ is odd in the definition domain, but decreasing in $( 0,1 )$ and increasing in $( 1 , + \infty )$, C is wrong; $y = x - \frac { 1 } { x }$ is an odd function on the domain of definition and an increasing function on $( 0 , + \infty )$, D is correct;

Question 6: 6. The function $f ( x ) = \frac { \sqrt { 1 - x } } { x }$ has the domain of definition ().

6. The function $f ( x ) = \frac { \sqrt { 1 - x } } { x }$ has the domain of definition ().

  • A. A. $( 0,1 ]$
  • B. B. $( - \infty , 0 )$
  • C. C. $( - \infty , 1 ]$
  • D. D. $( - \infty , 0 ) \cup ( 0,1 ]$

Answer: D

Solution: From $\left\{ \begin{array} { l } 1 - x \geq 0 \\ x \neq 0 \end{array} \right.$, solve : $x \leq 1$ and $x \neq 0$. $\therefore$ The function $f ( x ) = \frac { \sqrt { 1 - x } } { x }$ has the domain of definition: $( - \infty , 0 ) \cup ( 0,1 ]$.

Question 7: 7. The function $f ( x ) = \frac { 1 } { \sqrt [ 3 ] { x - 1 } } + \sqrt { x }$ has the domain of de...

7. The function $f ( x ) = \frac { 1 } { \sqrt [ 3 ] { x - 1 } } + \sqrt { x }$ has the domain of definition ( )

  • A. A. $[ 0 , + \infty )$
  • B. B. $( 1 , + \infty )$
  • C. C. $[ 0,1 )$
  • D. D. $[ 0,1 ) \cup ( 1 , + \infty )$

Answer: D

Solution: $\because \left\{ \begin{array} { l } x - 1 \neq 0 \\ x . .0 \end{array} \Rightarrow x \geq 0 \right.$ and $x \neq 1$ , . $\therefore$ function has a domain of definition of $[ 0,1 ) \cup ( 1 , + \infty )$ .

Question 8: 8. Part of the graph of the function $y = \left( x ^ { 2 } - x ^ { - 2 } \right) \sin x$ may be ( )

8. Part of the graph of the function $y = \left( x ^ { 2 } - x ^ { - 2 } \right) \sin x$ may be ( )

  • A. A. ![](/images/questions/function/image-005.jpg)
  • B. B. ![](/images/questions/function/image-006.jpg)
  • C. C. ![](/images/questions/function/image-007.jpg)
  • D. D. ![](/images/questions/function/image-008.jpg)

Answer: B

Solution: Remember $f ( x ) = \left( x ^ { 2 } - x ^ { - 2 } \right) \sin x$, then $f ( - x ) = - \left( x ^ { 2 } - x ^ { - 2 } \right) \sin x$, so $f ( x ) = - f ( - x ) , f ( x )$ is an odd function. Therefore, $f ( x ) = - f ( - x ) , f ( x )$ is an odd function, so the image is symmetric about the origin. At this point, we can exclude A and C, and take $x = \frac { \pi } { 2 } , f \left( \frac { \pi } { 2 } \right) = \left( \frac { \pi } { 2 } \right) ^ { 2 } - \left( \frac { \pi } { 2 } \right) ^ { - 2 } > 0$ to exclude D.

Question 9: 9. The following functions are even functions

9. The following functions are even functions

  • A. A. $y = \sin x$
  • B. B. $y = \ln \left( \sqrt { x ^ { 2 } + 1 } - x \right)$
  • C. C. $y = e ^ { x }$
  • D. D. $y = \ln \sqrt { x ^ { 2 } + 1 }$

Answer: D

Solution: TEST ANALYSIS: $y = \sin x$ is an odd function, $f ( x ) = \ln \left( \sqrt { x ^ { 2 } + 1 } - x \right)$, then $$ \begin{aligned} & f ( - x ) + f ( x ) \quad = \ln \left( \sqrt { x ^ { 2 } + 1 } - x \right) \\ & + \ln \left( \sqrt { x ^ { 2 } + 1 } + x \right) \quad = \ln \left( \sqrt { x ^ { 2 } + 1 } - x \right) \left( \sqrt { x ^ { 2 } + 1 } + x \right) \mid \ln 1 = 0 \quad , \quad \therefore f ( - x ) = - f ( x ) \end{aligned} $$ Therefore, the function $f ( x ) = \ln \left( \sqrt { x ^ { 2 } + 1 } - x \right)$ is odd and $y = e ^ { x } \quad$ is non odd and non even. $$ f ( x ) = \ln \sqrt { x ^ { 2 } + 1 } \quad , \quad f ( - x ) = \sqrt { ( - x ) ^ { 2 } + 1 } = f ( x ) \quad \text { 是偶函数, } $$ Therefore, the answer is D. POINT: Judgment of parity of a function.

Question 10: 10. The interval ${ } ^ { [ 2 a - 1,11 ] }$ is known, and the range of values of the real number ${ ...

10. The interval ${ } ^ { [ 2 a - 1,11 ] }$ is known, and the range of values of the real number ${ } ^ { a }$ is ( ).

  • A. A. $( - \infty , 6 )$
  • B. B. $( 6 , + \infty )$
  • C. C. $( 1,6 )$
  • D. D. $( - \infty , 6 ]$

Answer: A

Solution: According to the definition of interval, we can find $2 a - 1 < 11$ and get $a < 6$.

Question 11: 11. The partial graph of the function $f ( x ) = x ^ { 3 } \cdot 3 ^ { x }$ is approximately ( )

11. The partial graph of the function $f ( x ) = x ^ { 3 } \cdot 3 ^ { x }$ is approximately ( )

  • A. A. ![](/images/questions/function/image-009.jpg)
  • B. B. ![](/images/questions/function/image-010.jpg)
  • C. C. ![](/images/questions/function/image-011.jpg)
  • D. D. ![](/images/questions/function/image-012.jpg)

Answer: A

Solution: When ${ } ^ { x > 0 }$, $f ( x ) > 0$, so exclude $B , D$; when ${ } ^ { x < 0 }$, $f ( x ) < 0$, so exclude $C$. FORMULA_5]].

Question 12: 12. If the function $f ( x ) = x + \frac { 1 } { x }$ is known, the value of $f ( 2 ) + f ( - 2 )$ i...

12. If the function $f ( x ) = x + \frac { 1 } { x }$ is known, the value of $f ( 2 ) + f ( - 2 )$ is ( ).

  • A. A. - 1
  • B. B. 0
  • C. C. 1
  • D. D. 2

Answer: B

Solution: Because $f ( x ) = x + \frac { 1 } { x }$, therefore $f ( 2 ) + f ( - 2 ) = 2 + \frac { 1 } { 2 } - 2 - \frac { 1 } { 2 } = 0$.

Question 13: 13. The image of the function $f ( x ) = \frac { \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } } {...

13. The image of the function $f ( x ) = \frac { \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } } { 16 ( | x | - 1 ) }$ is approximately

  • A. A. ![](/images/questions/function/image-013.jpg)
  • B. B. ![](/images/questions/function/image-014.jpg)
  • C. C. ![](/images/questions/function/image-015.jpg)
  • D. D. ![](/images/questions/function/image-016.jpg)

Answer: B

Solution: Since $f ( - x ) = \frac { \mathrm { e } ^ { - x } - \mathrm { e } ^ { x } } { 16 ( | x | - 1 ) } = \frac { \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } } { 16 ( | x | - 1 ) } = - f ( x )$, $f ( x )$ is an odd function, and the graph of the function is symmetric about the origin, A can be ruled out; when ${ } ^ { x > 1 }$, C and D can be ruled out from $f ( x ) > 0$.

Question 14: 14. Function $f ( x ) = a x ^ { 2 } + ( b - 3 ) x + 3 , x \in \left[ a ^ { 2 } - 2 , a \right] _ { \...

14. Function $f ( x ) = a x ^ { 2 } + ( b - 3 ) x + 3 , x \in \left[ a ^ { 2 } - 2 , a \right] _ { \text {是偶函数 } } , a + b = ( )$

  • A. A. 4
  • B. B. 1
  • C. C. 4 or 1
  • D. D. Other values

Answer: A

Solution: Since $f ( x ) = a x ^ { 2 } + ( b - 3 ) x + 3 , x \in \left[ a ^ { 2 } - 2 , a \right]$ is an even function. so $\left\{ \begin{array} { l } a ^ { 2 } - 2 = - a \\ a > a ^ { 2 } - 2 \\ a x ^ { 2 } + ( b - 3 ) x + 3 = a ( - x ) ^ { 2 } - ( b - 3 ) x + 3 \end{array} \right.$ solves for $\left\{ \begin{array} { l } a = 1 \\ b = 3 \end{array} \right.$. so $a + b = 4$ , and

Question 15: 15. Given the function $f ( x ) = \ln \left( 1 + x ^ { 2 } \right) - \frac { 1 } { 1 + | x | }$, the...

15. Given the function $f ( x ) = \ln \left( 1 + x ^ { 2 } \right) - \frac { 1 } { 1 + | x | }$, the range of $x$ that makes $f ( x ) > f ( 2 x - 1 )$ hold is

  • A. A. $\left( \frac { 1 } { 3 } , 1 \right)$
  • B. B. $\left( - \infty , - \frac { 1 } { 3 } \right) \cup ( 1 , + \infty )$
  • C. C. $\left( - \frac { 1 } { 3 } , \frac { 1 } { 3 } \right)$
  • D. D. $\left( - \infty , - \frac { 1 } { 3 } \right) \cup \left( \frac { 1 } { 3 } , + \infty \right)$

Answer: A

Solution: From the question, the function $f ( x )$ is defined by $R$ and $f ( - x ) = \ln \left[ 1 + ( - x ) ^ { 2 } \right] - \frac { 1 } { 1 + | - x | } = f ( x )$ , so the function is even. When $x \geq 0$, the function $y = \ln \left( 1 + x ^ { 2 } \right) , y = - \frac { 1 } { 1 + x }$ is monotonically increasing, so the function $f ( x )$ is monotonically increasing. And $f ( x ) > f ( 2 x - 1 )$, so $| x | > | 2 x - 1 |$, solve $x \in \left( \frac { 1 } { 3 } , 1 \right)$.

Question 16: 16. As shown in the figure, the area of the rectangle ${ } _ { A O B C }$ is ${ } _ { 4 }$, and the ...

16. As shown in the figure, the area of the rectangle ${ } _ { A O B C }$ is ${ } _ { 4 }$, and the image of the inverse proportional function $y = \frac { k } { x } ( k \neq 0 )$ has a branch passing through the intersection of the diagonals of the rectangle $P$, then the analytic formula of the inverse proportional function is ( ) ![](/images/questions/function/image-001.jpg)

  • A. A. $y = - \frac { 1 } { x }$
  • B. B. $y = \frac { 1 } { x }$
  • C. C. $y = - \frac { 2 } { x }$
  • D. D. $y = \frac { 2 } { x }$

Answer: A

Solution: Let the length of the rectangle be ${ } _ { a }$ and the width of the rectangle be $\frac { 4 } { a }$. Combining the graphs, we know that the coordinates of the point $P$ are $\stackrel { \text { ซ } } { \in } \frac { a } { \mathrm { e } } \frac { a } { 2 } , \frac { 2 \ddot { \boldsymbol { a } } } { a \dot { \boldsymbol { \varphi } } }$ . Since the point ${ } _ { P }$ is on the inverse function $y = \frac { k } { x } ( k \neq 0 )$, the point $y = \frac { k } { x } ( k \neq 0 )$ is on the inverse function $\frac { 2 } { a } = \frac { k } { - \frac { a } { 2 } }$. So $\frac { 2 } { a } = \frac { k } { - \frac { a } { 2 } }$, which solves $k = - 1 , y = - \frac { 1 } { x }$, is the inverse function of $y = \frac { k } { x } ( k \neq 0 )$.

Question 17: 17. If the set $A = \{ - 2 , - 1,0,1,2 \} , B = \left\{ x \mid y = \sqrt { x ^ { 2 } - 4 } \right\}$...

17. If the set $A = \{ - 2 , - 1,0,1,2 \} , B = \left\{ x \mid y = \sqrt { x ^ { 2 } - 4 } \right\}$ is known, then $A \cap B =$ ()

  • A. A. $\{ - 2 \}$
  • B. B. $\{ - 2,2 \}$
  • C. C. $\{ 2 \}$
  • D. D. $\varnothing$

Answer: B

Solution: For the set $B , x ^ { 2 } - 4 \geq 0 \Rightarrow x \geq 2$ or $^ { x \leq - 2 }$, i.e., $B = ( - \infty , - 2 ] \cup [ 2 , + \infty )$, then $A \cap B = \{ - 2,2 \}$.

Question 18: 18. The following four functions do not have parity ()

18. The following four functions do not have parity ()

  • A. A. $f ( x ) = x ^ { 2 } - 1$
  • B. B. $f ( x ) = x ^ { 2 } + x$
  • C. C. $f ( x ) = x + \sqrt [ 3 ] { x }$
  • D. D. $f ( x ) = 0$

Answer: B

Solution: It is easy to know that the domains of the functions in option ABCD are R, i.e., their domains are symmetric about the origin. For A, since $f ( x ) = x ^ { 2 } - 1$, so $f ( - x ) = ( - x ) ^ { 2 } - 1 = x ^ { 2 } - 1 = f ( x )$, so $f ( x ) _ { \text {是偶函数,故 } }$ A does not meet the requirements, and is wrong; For B, let $x = 1$, then $y = 1 ^ { 2 } + 1 = 2$, i.e., $( 1,2 )$ is the point on $f ( x ) = x ^ { 2 } + x$, it is easy to know that the point of symmetry of ${ } ^ { ( 1,2 ) }$ about the origin is [[]], and the point of symmetry is ${ } ^ { ( 1,2 ) }$. INLINE_FORMULA_8]], and it is clear that the point on ${ } ^ { ( - 1 , - 2 ) _ { \text {不是 } } } { } ^ { f ( x ) }$ is ${ } ^ { ( - 1 , - 2 ) _ { \text {不是 } } } { } ^ { f ( x ) }$, while ${ } ^ { ( 1,2 ) }$ is $y _ { \text {轴对称的点为 } } { } ^ { ( - 1,2 ) }$ about $y _ { \text {轴对称的点为 } } { } ^ { ( - 1,2 ) }$, and it is clear that ${ } ^ { ( - 1,2 ) }$ is not ${ } ^ { f ( x ) }$. FORMULA_13]], i.e., there is at least one point ${ } ^ { ( 1,2 ) }$ on the image of ${ } ^ { f ( x ) }$ that is not symmetric about the origin and not symmetric about the axis of ${ } ^ { y }$, so $f ( x ) _ { \text {is neither odd nor even function, so B meets the requirement and is correct;} }$ For C, since $f ( x ) = x + \sqrt [ 3 ] { x }$, so $f ( - x ) = - x + \sqrt [ 3 ] { ( - x ) } = - ( x + \sqrt [ 3 ] { x } ) = - f ( x )$, so $f ( x )$ is an odd function, so C does not meet the requirements, error ; For D, since $f ( x ) = 0$, so $f ( - x ) = 0 = - f ( x ) = f ( x )$, so $f ( x ) _ { \text {既是奇函数,也是偶 } }$ function, so D does not meet the requirements, error.

Question 19: 19. The image of the function $y = 2 x ^ { 2 } - 2 ^ { | x | }$ in ${ } ^ { [ - 2,2 ] }$ is approxim...

19. The image of the function $y = 2 x ^ { 2 } - 2 ^ { | x | }$ in ${ } ^ { [ - 2,2 ] }$ is approximately

  • A. A. ![](/images/questions/function/image-017.jpg)
  • B. B. ![](/images/questions/function/image-018.jpg)
  • C. C. ![](/images/questions/function/image-019.jpg)
  • D. D. ![](/images/questions/function/image-020.jpg)

Answer: C

Solution: It is easy to know that the function $y = 2 x ^ { 2 } - 2 ^ { | x | } \left( { } ^ { x \in [ - 2,2 ] } \right)$ is an even function and the graph is symmetric about the ${ } ^ { y }$ axis, which can be ruled out BD , $x = 2$ when $y = 2 \times 2 ^ { 2 } - 2 ^ { 2 } = 4 > 0$ , which can be ruled out A.

Question 20: 20. The partial image of the function $f ( x ) = \frac { 3 \mathrm { e } ^ { x } \cos x } { \mathrm ...

20. The partial image of the function $f ( x ) = \frac { 3 \mathrm { e } ^ { x } \cos x } { \mathrm { e } ^ { 2 x } - 1 }$ is roughly ()

  • A. A. ![](/images/questions/function/image-021.jpg)
  • B. B. ![](/images/questions/function/image-022.jpg)
  • C. C. ![](/images/questions/function/image-023.jpg)
  • D. D. ![](/images/questions/function/image-024.jpg)

Answer: C

Solution: Since the function's domain of definition is $\{ x \mid x \neq 0 \}$, ABD can be eliminated.

Question 21: 21. The function $f ( x ) = \frac { 1 } { 2 } \log _ { 2 } x - 3 \left( \frac { 1 } { 2 } \right) ^ ...

21. The function $f ( x ) = \frac { 1 } { 2 } \log _ { 2 } x - 3 \left( \frac { 1 } { 2 } \right) ^ { x }$ has zeros in the interval ( ).

  • A. A. $( 0,1 )$
  • B. B. $( 1,2 )$
  • C. C. $( 2,3 )$
  • D. D. $( 3,4 )$

Answer: C

Solution: From the analytic equation, we know that the function is increasing on $f ( x ) _ { \text {在 } } ( 0 , + \infty )$ and $f ( 2 ) = \frac { 1 } { 2 } - \frac { 3 } { 4 } = - \frac { 1 } { 4 } < 0 , f ( 3 ) = \frac { 1 } { 2 } \log _ { 2 } 3 - \frac { 3 } { 8 } > \frac { 1 } { 2 } - \frac { 3 } { 8 } = \frac { 1 } { 8 } > 0$ . So the interval of the zeros of $f ( x )$ is $( 2,3 )$.

Question 22: 22. The function $f ( x ) = \frac { \left( x ^ { 2 } - 1 \right) ^ { 0 } } { \sqrt { 3 x - x ^ { 2 }...

22. The function $f ( x ) = \frac { \left( x ^ { 2 } - 1 \right) ^ { 0 } } { \sqrt { 3 x - x ^ { 2 } } }$ has a domain of definition ( ).

  • A. A. $[ 0,3 ]$
  • B. B. $( 0,3 )$
  • C. C. $[ 0,1 ) \cup ( 1,3 ]$
  • D. D. $( 0,1 ) \cup ( 1,3 )$

Answer: D

Solution: From the question $\left\{ \begin{array} { l } x ^ { 2 } - 1 \neq 0 \\ 3 x - x ^ { 2 } > 0 \end{array} \right.$ , solve $0 < x < 3$ and $x \neq 1$ .

Question 23: 23. If the function $f ( x ) = \left\{ \begin{array} { l l } 2 ^ { x } , & x \leq 3 \\ x - 3 , & x >...

23. If the function $f ( x ) = \left\{ \begin{array} { l l } 2 ^ { x } , & x \leq 3 \\ x - 3 , & x > 3 \end{array} \right.$ is known, then the value of $f ( f ( 1 ) - f ( 5 ) )$ is

  • A. A. 1
  • B. B. 2
  • C. C. 3
  • D. D. - 3

Answer: A

Solution: From the analytic formula of the function : $f ( 1 ) = 2 ^ { 1 } = 2 , f ( 5 ) = 5 - 3 = 2$ $\therefore f ( f ( 1 ) - f ( 5 ) ) = f ( 0 ) = 2 ^ { 0 } = 1$ The correct option for this question: A This question is about solving the value of a segmented function, which is a basic question.

Question 24: 24. The following groups of functions represent the same function ( )

24. The following groups of functions represent the same function ( )

  • A. A. $f ( x ) = x , g ( x ) = ( \sqrt { x } ) ^ { 2 }$
  • B. B. $f ( x ) = \frac { x ^ { 2 } - 1 } { x - 1 } , g ( x ) = x + 1$
  • C. C. $f ( t ) = | t | , g ( x ) = \sqrt { x ^ { 2 } }$
  • D. D. $f ( x ) = \frac { | x | } { x } , g ( x ) = \left\{ \begin{array} { l } 1 , x \geq 0 \\ - 1 , x < 0 \end{array} \right.$

Answer: C

Solution: A: $f ( x )$ is defined by $\mathrm { R } , g ( x )$ is defined by $[ 0 , + \infty )$, then A is wrong; B: The domain of definition of $f ( x )$ is R for $\{ x \mid x \neq 1 \} , g ( x )$, then B is wrong; C: The domains of $f ( t ) _ { \text {和 } } g ( x )$ are all R and $g ( x ) = \sqrt { x ^ { 2 } } = | x |$, then C is correct; D: The domain of definition of $f ( x )$ is R for $\{ x \mid x \neq 0 \} , g ( x )$, then D is incorrect.

Question 25: 25. If the function $f ( x ) = 2 \left( x ^ { 3 } + x + 1 \right) + \sin x$ is known, then the solut...

25. If the function $f ( x ) = 2 \left( x ^ { 3 } + x + 1 \right) + \sin x$ is known, then the solution set of $f ( - x ) + f ( 3 x - 2 ) < 4$ is ( ).

  • A. A. $( - \infty , 1 )$
  • B. B. $( 1 , + \infty )$
  • C. C. $( - \infty , 2 )$
  • D. D. $( 2 , + \infty )$

Answer: A

Solution: Let $g ( x ) = f ( x ) - 2 = 2 x ^ { 3 } + 2 x + \sin x$ , which gives the function ${ } ^ { g ( x ) }$ as an odd function. $g ^ { \prime } ( x ) = 6 x ^ { 2 } + 2 + \cos x > 0$, so that the function ${ } ^ { g ( x ) }$ is monotonically increasing on ${ } ^ { R }$. $f ( - x ) + f ( 3 x - 2 ) < 4 \Rightarrow f ( - x ) - 2 < - f ( 3 x - 2 ) + 2 \Rightarrow g ( - x ) < - g ( 3 x - 2 ) \Rightarrow g ( - x ) < g ( - 3 x + 2 )$ So $- x < - 3 x + 2 \Rightarrow x < 1$.

Question 26: 26. $f ( x ) = a x ^ { 5 } + b x ^ { 3 } + x ^ { 2 } + x + 1 ( a , b$ is known to be the constant $)...

26. $f ( x ) = a x ^ { 5 } + b x ^ { 3 } + x ^ { 2 } + x + 1 ( a , b$ is known to be the constant $)$, if $f ( 2 ) = 11$, then $f ( - 2 ) =$ ( )

  • A. A. - 11
  • B. B. - 1
  • C. C. 0
  • D. D. 1

Answer: B

Solution: Since $f ( x ) = a x ^ { 5 } + b x ^ { 3 } + x ^ { 2 } + x + 1 ( a , b$ is the constant $)$, from $f ( 2 ) = 11$: $32 a + 8 b + 4 + 2 + 1 = 11$ which is $32 a + 8 b = 4$; Hence $f ( - 2 ) = - 32 a - 8 b + 4 - 2 + 1 = - 4 + 3 = - 1$.

Question 27: 27. The function $f ( x ) = \left\{ \begin{array} { l } \frac { a } { x ^ { 2 } + 1 } , x \leq 0 ; \...

27. The function $f ( x ) = \left\{ \begin{array} { l } \frac { a } { x ^ { 2 } + 1 } , x \leq 0 ; \\ ( 2 - a ) x + 3 a - 1 , x > 0 . \end{array} \right.$ is known to be increasing on $( - \infty , + \infty )$, then the range of the real number $a$ is

  • A. A. $( 2 , + \infty )$
  • B. B. $( 0,2 ]$
  • C. C. $\left[ \frac { 1 } { 2 } , 2 \right)$
  • D. D. $\left( 0 , \frac { 1 } { 2 } \right]$

Answer: C

Solution: The $\because$ function $f ( x ) = \left\{ \begin{array} { l } \frac { a } { x ^ { 2 } + 1 } , x \leq 0 ; \\ ( 2 - a ) x + 3 a - 1 , x > 0 . \end{array} \right.$ is increasing on $( - \infty , + \infty )$, the $( - \infty , + \infty )$ $\therefore \left\{ \begin{array} { l } a > 0 \\ 2 - a > 0 \\ 3 a - 1 \geq a \end{array} \right.$ to find $\frac { 1 } { 2 } \leq a < 2$.

Question 28: 28. It is known that the function $f ( x ) = \left\{ \begin{array} { l } 3 ^ { x } + 1 , x < 1 \\ 2 ...

28. It is known that the function $f ( x ) = \left\{ \begin{array} { l } 3 ^ { x } + 1 , x < 1 \\ 2 x ^ { 2 } - x , x > 1 \end{array} \right.$, then $f ( f ( 0 ) ) = ( \quad )$

  • A. A. 6
  • B. B. 4
  • C. C. 2
  • D. D. 1

Answer: A

Solution: From the question, $f ( 0 ) = 3 ^ { 0 } + 1 = 2$ , then $f ( f ( 0 ) ) = f ( 2 ) = 2 \times 2 ^ { 2 } - 2 = 6$ , the

Question 29: 29. If the function $f ( x ) = \left\{ \begin{array} { c } 2 ^ { x } + a + 2 ( x \leq 1 ) \\ - \log ...

29. If the function $f ( x ) = \left\{ \begin{array} { c } 2 ^ { x } + a + 2 ( x \leq 1 ) \\ - \log _ { 2 } ( x + 1 ) , ( x > 1 ) \end{array} \right.$ has a maximum value, then the range of values of the real number $a$ is ( ).

  • A. A. $[ 0 , \infty ]$
  • B. B. $[ - 5,1 ]$
  • C. C. $( \infty , - 5 )$
  • D. D. $[ - 5 , + \infty )$

Answer: D

Solution: SOLUTION: Since $f ( x ) = 2 ^ { x } + a + 2$ is monotonically increasing on $( - \infty , 1 ]$, $f ( x ) \leq f ( 1 ) = a + 4$ , the Since $f ( x ) = - \log _ { 2 } ( x + 1 )$ is monotonically decreasing on ${ } ^ { ( 1 , + \infty ) }$, $f ( x ) < f ( 1 ) = - 1$ , . Since the function $f ( x ) = \left\{ \begin{array} { c } 2 ^ { x } + a + 2 ( x \leq 1 ) \\ - \log _ { 2 } ( x + 1 ) , ( x > 1 ) \end{array} \right.$ has a maximum value. So $a + 4 \geq - 1$, which solves for $a \geq - 5$, has a maximum value. So the real number $a$ is in the range $[ - 5 , + \infty )$ , and the real number $[ - 5 , + \infty )$ is in the range $[ - 5 , + \infty )$ .

Question 30: 30. Given the function $f ( x ) = x \cdot \sin x$ , if $x _ { 1 } , x _ { 2 } \in \left[ - \frac { \...

30. Given the function $f ( x ) = x \cdot \sin x$ , if $x _ { 1 } , x _ { 2 } \in \left[ - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right]$ and $f \left( x _ { 1 } \right) > f \left( x _ { 2 } \right)$ , then the following inequality holds constant

  • A. A. $x _ { 1 } > x _ { 2 }$
  • B. B. $x _ { 1 } < x _ { 2 }$
  • C. C. $x _ { 1 } + x _ { 2 } > 0$
  • D. D. $x _ { 1 } ^ { 2 } > x _ { 2 } ^ { 2 }$

Answer: D

Solution: Test analysis :From what is known: $f ( x )$ is an even function, $y = x , y = \sin x$ is increasing on the interval $\left[ 0 , \frac { \pi } { 2 } \right]$ and $y = x , y = \sin x$ is positive on the interval $\left[ 0 , \frac { \pi } { 2 } \right]$, so $f ( x )$ is increasing on the interval $\left[ 0 , \frac { \pi } { 2 } \right]$, since $f ( x )$ is increasing on the interval $\left[ 0 , \frac { \pi } { 2 } \right]$, since $\left[ 0 , \frac { \pi } { 2 } \right]$ is not a positive number. positive, so $f ( x )$ is increasing on the interval $\left[ 0 , \frac { \pi } { 2 } \right]$ because $f \left( x _ { 1 } \right) > f \left( x _ { 2 } \right)$, so $\left| x _ { 1 } \right| > \left| x _ { 2 } \right|$, so $x _ { 1 } ^ { 2 } > x _ { 2 } ^ { 2 }$, so [[]] should be chosen. INLINE_FORMULA_10]]. 考点:1,函数的单调性;2,函数的奇偶性.

Question 31: 31. The graph of function $y = f ( x )$ is shown in the figure. Observing the graph, we can see that...

31. The graph of function $y = f ( x )$ is shown in the figure. Observing the graph, we can see that the definition domain and value domain of function $y = f ( x )$ are respectively ![](/images/questions/function/image-002.jpg)

  • A. A. $[ - 5,0 ] \cup [ 2,6 ] , [ 0,5 ]$
  • B. B. $[ - 5,6 ] , [ 0 , + \infty )$
  • C. C. $[ - 5,0 ] \cup [ 2,6 ) , [ 0 , + \infty )$
  • D. D. $[ - 5 , + \infty ) , [ 2,5 ]$

Answer: C

Solution: The definition domain of a function is the range of values of the independent variable $x$, from the figure we can see that the independent variable $[ - 50 ] \cup [ 26 )$ of this function, and the value domain of the function is the range of values of the function value, from the figure we can see that the value domain of this function is $[ 0 , + \infty )$. Therefore, choose: $C$.

Question 32: 32. If $f \left( x - \frac { 1 } { x } \right) = x ^ { 2 } + \frac { 1 } { x ^ { 2 } }$ is known, th...

32. If $f \left( x - \frac { 1 } { x } \right) = x ^ { 2 } + \frac { 1 } { x ^ { 2 } }$ is known, then the analytic formula of $f ( x + 1 )$ is

  • A. A. $f ( x + 1 ) = ( x + 1 ) ^ { 2 } + \frac { 1 } { ( x + 1 ) ^ { 2 } }$
  • B. B. $f ( x + 1 ) = \left( x - \frac { 1 } { x } \right) ^ { 2 } + \frac { 1 } { \left( x - \frac { 1 } { x } \right) ^ { 2 } }$
  • C. C. $f ( x + 1 ) = ( x + 1 ) ^ { 2 } + 2$
  • D. D. $f ( x + 1 ) = ( x + 1 ) ^ { 2 } + 1$

Answer: C

Solution: $\mathrm { Q } \left( x - \frac { 1 } { x } \right) ^ { 2 } = x ^ { 2 } + \frac { 1 } { x ^ { 2 } } - 2 , \therefore x ^ { 2 } + \frac { 1 } { x ^ { 2 } } = \left( x - \frac { 1 } { x } \right) ^ { 2 } + 2$ . $\mathrm { Q } f \left( x - \frac { 1 } { x } \right) = x ^ { 2 } + \frac { 1 } { x ^ { 2 } } = \left( x - \frac { 1 } { x } \right) ^ { 2 } + 2 , \therefore f ( x ) = x ^ { 2 } + 2$, therefore, $f ( x + 1 ) = ( x + 1 ) ^ { 2 } + 2$, so choose C. [Eyes on] This question examines the analytic formula of the function, belongs to the medium problem, seek function analytic formula common problem type by the following kinds: (1) according to the practical application of the analytic formula of the function; (2) commutative method for the analytic formula of the function, the use of commutative method must pay attention to the range of parameters after the commutation; (3) to be determined coefficient method for the analytic formula, this method is suitable for both known function name of the function analytic formula; (4) elimination method for the analytic formula of the function, this method is suitable for seeking the inverse or opposite of the function of the analytic formula of the independent variables.

Question 33: 33. If $2023 ^ { x } - 2023 ^ { y } < 2024 ^ { - x } - 2024 ^ { - y } ( x , y \in \mathrm { R } )$, ...

33. If $2023 ^ { x } - 2023 ^ { y } < 2024 ^ { - x } - 2024 ^ { - y } ( x , y \in \mathrm { R } )$, then ( )

  • A. A. $\ln ( y - x + 1 ) > 0$
  • B. B. $\ln ( y - x + 1 ) < 0$
  • C. C. $\ln | x - y | > 0$
  • D. D. $\ln | x - y | < 0$

Answer: A

Solution: From $2023 ^ { x } - 2023 ^ { y } < 2024 ^ { - x } - 2024 ^ { - y }$, we get $2023 ^ { x } - 2024 ^ { - x } < 2023 ^ { y } - 2024 ^ { - y }$. such that $f ( x ) = 2023 ^ { x } - 2024 ^ { - x }$ , it is clear that the function $f ( x )$ is monotonically increasing on R and $f ( x ) < f ( y )$ , so that ${ } ^ { x < y }$, i.e., $y - x > 0$ , then , so $\ln ( y - x + 1 ) > \ln 1 = 0$, A is correct, B is wrong; From $y - x > 0$, it is clear that when $y - x = 1$, $\ln | x - y | = 0$, CD is wrong.

Question 34: 34. Set the function $f ( x ) = \frac { x } { x + 1 }$, the following options are wrong ( )

34. Set the function $f ( x ) = \frac { x } { x + 1 }$, the following options are wrong ( )

  • A. A. Arbitrary $x \neq 0 , x \neq - 1 , f ( x ) + f \left( \frac { 1 } { x } \right) = 1$
  • B. B. $f ( x )$ is an increasing function on $^ { ( - \infty , - 1 ) }$ and $^ { ( - 1 , + \infty ) }$.
  • C. C. For any $a \in ( 0,1 )$, the graph of the exponential function $y = a ^ { x }$ has exactly 2 common points with the graph of $f ( x )$.
  • D. D. Arbitrarily $a < 0$, the graph of the quadratic function $y = a x ^ { 2 }$ has only one common point $f ( x )$ with the graph of $( 0,0 )$.

Answer: D

Solution: For Option A: Any $\underset { x \neq 0 , x \neq - 1 } { } , f ( x ) + f \left( \frac { 1 } { x } \right) = \frac { x } { x + 1 } + \frac { \frac { 1 } { x } } { \frac { 1 } { x } + 1 } = \frac { x } { x + 1 } + \frac { 1 } { x + 1 } = 1$, thus A is correct; For Option B: Since $f ( x ) = \frac { x } { x + 1 }$ is defined by $( - \infty , - 1 ) \cup ( - 1 , + \infty )$, and $( - \infty , - 1 ) \cup ( - 1 , + \infty )$ and $f ( x ) = 1 - \frac { 1 } { x + 1 }$, it can be seen that $f ( x )$ is an increasing function on $( - \infty , - 1 )$ and $( - 1 , + \infty )$, so B is correct; For option C: If ${ } ^ { a \in ( 0,1 ) }$, the graph of the exponential function $y ^ { x = a ^ { x } }$ is plotted against $f ( x ) ^ { \text {的图象,} }$ respectively. ![](/images/questions/function/image-003.jpg) The graphs of the exponential function $y = a ^ { x }$ and $f ( x )$ have exactly 2 common points, so C is correct; For option D: For example, if $a = - 4$, then $g ( x ) = - 4 x ^ { 2 }$, then $g \left( - \frac { 1 } { 2 } \right) = f \left( - \frac { 1 } { 2 } \right) = - 1$ It can be seen that $\left( - \frac { 1 } { 2 } , - 1 \right)$ is the intersection of the graph of the quadratic function $y = - 4 x ^ { 2 }$ and the graph of $f ( x )$, so D is wrong;

Question 35: 35. The Euler function $\varphi ( n ) \left( n \in \mathbf { N } ^ { * } \right)$ has a value equal ...

35. The Euler function $\varphi ( n ) \left( n \in \mathbf { N } ^ { * } \right)$ has a value equal to the number of all positive integers not exceeding the positive integer $n$ that are mutually prime with $n$. For example, $\varphi ( 1 ) = 1 , \varphi ( 4 ) = 2$. If $m \in \mathbf { N } ^ { * }$ and $\sum _ { i = 1 } ^ { m } \varphi ( 2 i ) = 13$, then $\varphi ( m ) = ( \quad )$

  • A. A. 3
  • B. B. 4
  • C. C. 5
  • D. D. 6

Answer: B

Solution:

Question 36: 36. The power function $f ( x ) = \left( 2 m ^ { 2 } - 5 m - 2 \right) x ^ { m }$ is known to be an ...

36. The power function $f ( x ) = \left( 2 m ^ { 2 } - 5 m - 2 \right) x ^ { m }$ is known to be an odd function in the domain of definition, then $m = ( )$

  • A. A. $- \frac { 1 } { 2 }$
  • B. B. $- \frac { 1 } { 2 }$ or 3
  • C. C. $\frac { 1 } { 2 }$
  • D. D. 3

Answer: D

Solution: By the function is a power function can be $2 m ^ { 2 } - 5 m - 2 = 1$, the solution $m = 3 ^ { \text {或 } } m = - \frac { 1 } { 2 }$, the When $m = 3$, $f ( x ) = x ^ { 3 }$ is an odd function on R, which is consistent with the question; When $m = - \frac { 1 } { 2 }$, $f ( x ) = x ^ { - \frac { 1 } { 2 } }$, with a domain of definition of $( 0 , + \infty )$, is a non-singular and non-even function, which does not satisfy the question.

Question 37: 37. Known function $f ( x ) = \left\{ \begin{array} { l } 2 \sin x , \sin x \geq \cos x \\ - \cos x ...

37. Known function $f ( x ) = \left\{ \begin{array} { l } 2 \sin x , \sin x \geq \cos x \\ - \cos x , \sin x < \cos x \text { ,给出下列结论:(1)} f ( x ) \text { 是周期函数 ;(2)} f ( x ) \end{array} \right.$ has a minimum value of - 1; (3) $f ( x ) _ { \text {在区间 } } \left( \frac { \pi } { 2 } , 2 \pi \right)$ is monotonically decreasing on $f ( x ) _ { \text {在区间 } } \left( \frac { \pi } { 2 } , 2 \pi \right)$. The number of correct conclusions is ( )

  • A. A. 0
  • B. B. 1
  • C. C. 2
  • D. D. 3

Answer: B

Solution: From the known $f ( x ) = \left\{ \begin{array} { l } 2 \sin x , \frac { \pi } { 4 } + 2 k \pi \leq x \leq \frac { 5 \pi } { 4 } + 2 k \pi \\ - \cos x , - \frac { 3 \pi } { 4 } + 2 k \pi < x < \frac { \pi } { 4 } + 2 k \pi \end{array} , k \in \mathrm { Z } \right.$, it is easy to know $f ( x + 2 \pi ) = f ( x )$. Therefore ${ } ^ { f ( x ) }$ is a periodic function, so (1) is correct; When $x \in \left[ 2 k \pi + \frac { \pi } { 4 } , 2 k \pi + \frac { 5 \pi } { 4 } \right] ( k \in \mathbf { Z } )$, $f ( x ) = 2 \sin x \in [ - \sqrt { 2 } , 2 ] , - \sqrt { 2 } < - 1$ , so (2) is wrong; Combining the analytic equations, we know that $f ( x ) _ { \text {在 } } \left( \frac { \pi } { 2 } , \frac { 5 \pi } { 4 } \right)$ is monotonically decreasing on $\left( \frac { 5 \pi } { 4 } , 2 \pi \right)$, while $2 \sin \frac { 5 \pi } { 4 } = - \sqrt { 2 } < \frac { \sqrt { 2 } } { 2 } = - \cos \frac { 5 \pi } { 4 }$, so (3) is wrong).

Question 38: 38. If the real number $a = \log _ { 2 } 3 , b = \log _ { \frac { 1 } { 3 } } \frac { 1 } { 2 } , c ...

38. If the real number $a = \log _ { 2 } 3 , b = \log _ { \frac { 1 } { 3 } } \frac { 1 } { 2 } , c = \cos 2$ is known, then ( )

  • A. A. $a > b > c$
  • B. B. $a > c > b$
  • C. C. $b > c > a$
  • D. D. $c > b > a$

Answer: A

Solution: Because $a = \log _ { 2 } 3 > 1 , \quad b = \log _ { \frac { 1 } { 3 } } \frac { 1 } { 2 } = \log _ { 3 } 2 \in ( 0,1 )$. and $\frac { \pi } { 2 } < 2 < \pi$ , therefore ${ } _ { C } = \cos 2 < 0$ , and and so $a > b > c$ .

Question 39: 39. It is known that the function $\varphi ( x )$ defined on $\mathbf { R }$ satisfies: when $x _ { ...

39. It is known that the function $\varphi ( x )$ defined on $\mathbf { R }$ satisfies: when $x _ { 1 } \neq x _ { 2 }$, there is always $\frac { \varphi \left( x _ { 1 } \right) - \varphi \left( x _ { 2 } \right) } { x _ { 1 } - x _ { 2 } } > 0$, and if for any $x \in \mathbf { R } , \varphi \left( \mathrm { e } ^ { x } - b \right) \geq \varphi ( a x )$, it holds. Then the maximum value of $a b$ is ( ).

  • A. A. $\sqrt { \mathrm { e } }$
  • B. B. $\frac { \mathrm { e } } { 2 }$
  • C. C. ${ } _ { e }$
  • D. D. $\mathrm { e } ^ { 2 }$

Answer: B

Solution: Since $x _ { 1 } \neq x _ { 2 }$ has constant $\frac { \varphi \left( x _ { 1 } \right) - \varphi \left( x _ { 2 } \right) } { x _ { 1 } - x _ { 2 } } > 0$ at $\frac { \varphi \left( x _ { 1 } \right) - \varphi \left( x _ { 2 } \right) } { x _ { 1 } - x _ { 2 } } > 0$, $\varphi ( x )$ is monotonically increasing on $\mathbf { R }$. So if $\varphi \left( \mathrm { e } ^ { x } - b \right) \geq \varphi ( a x )$, then $\mathrm { e } ^ { x } - b \geq a x$, i.e. $\mathrm { e } ^ { x } \geq a x + b$. constructor $f ( x ) = \mathrm { e } ^ { x } - a x - b ( x \in \mathbf { R } ) , ~ f ^ { \prime } ( x ) = \mathrm { e } ^ { x } - a$ , the $f ( x ) = \mathrm { e } ^ { x } - a x - b ( x \in \mathbf { R } ) , ~ f ^ { \prime } ( x ) = \mathrm { e } ^ { x } - a$ If $a = 0$, then $f ^ { \prime } ( x ) > 0$ is constant on $x \in \mathbf { R }$ and $f ( x ) \geq 0$ is constant on $b \leq 0$ when $a b = 0$ is constant. FORMULA_13]]; If $a < 0$, then $f ^ { \prime } ( x ) > 0 , f ( x )$ is monotonically increasing, at which point $f ( x ) \geq 0$ cannot be constant; If $a > 0$, $x > \ln a , f ( x )$ is monotonically increasing from $f ^ { \prime } ( x ) > 0$, then $x > \ln a , f ( x )$ is monotonically increasing. $f ^ { \prime } ( x ) < 0$ gets $x < \ln a , f ( x )$ monotonically decreasing, so $f ( x ) \geq f ( \ln a ) = a - a \ln a - b \geq 0$ That is, $b \leq a - a \ln a$, so $a b \leq a ^ { 2 } - a ^ { 2 } \ln a$, so that $g ( a ) = a ^ { 2 } - a ^ { 2 } \ln a ( a > 0 )$, so that $g ( a ) = a ^ { 2 } - a ^ { 2 } \ln a ( a > 0 )$, so that $g ( a ) = a ^ { 2 } - a ^ { 2 } \ln a ( a > 0 )$ Let $g ^ { \prime } ( a ) = a ( 1 - 2 \ln a ) = 0$ be $a = \sqrt { \mathrm { e } }$, so $a \in ( 0 , \sqrt { \mathrm { e } } ) _ { \text {时 } , ~ } g ^ { \prime } ( a ) > 0 , g ( a ) _ { \text {单调递增,} }$ is $a \in ( \sqrt { \mathrm { e } } , + \infty )$. $a \in ( 0 , \sqrt { \mathrm { e } } ) _ { \text {时 } , ~ } g ^ { \prime } ( a ) > 0 , g ( a ) _ { \text {单调递增,} }$ $a \in ( \sqrt { \mathrm { e } } , + \infty )$ when $g ^ { \prime } ( a ) < 0 , g ( a )$ monotonically decreases, so that $g ( a ) _ { \text {max } } = g ( \sqrt { \mathrm { e } } ) = \frac { \mathrm { e } } { 2 }$ Therefore, the maximum value of $a b$ is $\frac { \mathrm { e } } { 2 }$. In summary, the maximum value of $a b$ is $\frac { \mathrm { e } } { 2 }$.

Question 40: 40. Of the following functions, the minimum value is 2 ( ) High School Mathematics Assignment, ...

40. Of the following functions, the minimum value is 2 ( ) High School Mathematics Assignment, October 29, 2025

  • A. A. $y = x ^ { 2 } + 2 ( x > 0 )$
  • B. B. $y = - x ^ { 2 } + 2 x + 1$
  • C. C. $y = \frac { 9 x ^ { 2 } + 1 } { 3 x } ( x > 0 )$
  • D. D. $y = \frac { x ^ { 2 } + 2 } { 2 } + \frac { 2 } { x ^ { 2 } + 1 }$

Answer: C

Solution:
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Function

函数

40 Practice Questions

Practice with Chinese questions to prepare for the CSCA exam. You can toggle translations while practicing.

Topic Overview

Functions are a central concept in mathematics that describe correspondences between variables, and basic questions such as domain of definition, parity, and analytic form are often covered on the CSCA exam. Questions often require analyzing the properties of a function or calculating specific parameters based on its expression, and require proficiency in algebraic operations and characterization of basic function types.

Questions:40

Key Points

  • 1Find the domain of definition of the function (non-zero denominator, non-negative in the root sign, etc.)
  • 2Determining the parity of a function (using the f(-x) vs. f(x) relationship)
  • 3Solve for the values of the parameters according to the properties of the function
  • 4Simple analysis of composite or segmented functions

Study Tips

Practice by specifying the type of function, systematically checking the definitional domain constraints, and trying to simplify the analysis by substituting -x for parity judgments first.

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