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Calculus - Practice Questions (36)

Question 1: 1. The derivative of the function $y = \sqrt [ 5 ] { x ^ { 4 } }$ is

1. The derivative of the function $y = \sqrt [ 5 ] { x ^ { 4 } }$ is

  • A. A. $\frac { 1 } { 5 } x ^ { 3 }$
  • B. B. $\frac { 2 } { 5 } x ^ { 3 }$
  • C. C. $\frac { 4 } { 5 } x ^ { - \frac { 1 } { 5 } }$
  • D. D. $- \frac { 4 } { 5 } x ^ { - \frac { 1 } { 5 } }$

Answer: C

Solution: 试题分析:由求导公式 $y ^ { \prime } = \frac { 4 } { 5 } x ^ { - \frac { 1 } { 5 } }$ 故选C. Points : Exponential power derivation formula.

Question 2: 2. If $f ( x ) = \cos x$, then $f ^ { \prime } \left( \frac { \pi } { 2 } \right)$ has the value ( )

2. If $f ( x ) = \cos x$, then $f ^ { \prime } \left( \frac { \pi } { 2 } \right)$ has the value ( )

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

Answer: C

Solution: Because $f ( x ) = \cos x$, so $f ^ { \prime } ( x ) = - \sin x$, so $f ^ { \prime } \left( \frac { \pi } { 2 } \right) = - 1$

Question 3: 5. It is known that the slope of the tangent line to the graph of the function $f ( x ) = \frac { 2 ...

5. It is known that the slope of the tangent line to the graph of the function $f ( x ) = \frac { 2 } { x } - a x$ at the point $( - 1 , f ( - 1 ) )$ is ${ } _ { 1 }$, and the equation of this tangent line is

  • A. A. $x - y - 4 = 0$
  • B. B. $x - y - 6 = 0$
  • C. C. $x + y - 4 = 0$
  • D. D. $x + y - 5 = 0$

Answer: A

Solution: $f ^ { \prime } ( x ) = - \frac { 2 } { x ^ { 2 } } - a , f ^ { \prime } ( 1 ) = - 2 - a = 1 , a = - 3$, so $f ( x ) = \frac { 2 } { x } + 3 x , f ( - 1 ) = - 2 - 3 = - 5$, so the equation of the tangent line is $y + 5 = x + 1 , x - y - 4 = 0$.

Question 4: 6. If the function $f ( x ) = \sin x$, then $f \left( \frac { \pi } { 4 } \right) + f ^ { \prime } \...

6. If the function $f ( x ) = \sin x$, then $f \left( \frac { \pi } { 4 } \right) + f ^ { \prime } \left( \frac { \pi } { 4 } \right) = ( )$

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

Answer: B

Solution: According to the question, $f ^ { \prime } ( x ) = \cos x$ , then $f \left( \frac { \pi } { 4 } \right) + f ^ { \prime } \left( \frac { \pi } { 4 } \right) = \sin \frac { \pi } { 4 } + \cos \frac { \pi } { 4 } = \sqrt { 2 }$ Therefore: B.

Question 5: 7. "$a \leq 0$" is the "Existence of extremes in function $f ( x ) = a x + \ln x$". is neither s...

7. "$a \leq 0$" is the "Existence of extremes in function $f ( x ) = a x + \ln x$". is neither sufficient nor necessary

  • A. A. sufficient and unnecessary condition (math.)
  • B. B. necessary but insufficient condition (math.)
  • C. C. necessary and sufficient condition
  • D. D. both... (and...)

Answer: B

Solution: According to the question, the function $f ( x )$ has extreme values, i.e., its derivative function has positive and negative values. $f ^ { \prime } ( x ) = a + \frac { 1 } { x } ( x > 0 )$, to lead function $a < 0$, so $a \leq 0$ is a necessary and insufficient condition for $a \leq 0$ to be positive and negative. Point: This question mainly examines two knowledge points, one is the derivative and the method of finding the extreme value, and one is the judgment of the sufficient conditions. Extreme value is defined as the left increase and right decrease of the maximum value, left decrease and right increase of the minimum value, so the need for the derivative of the function has a positive and negative. In this question, the derivative of the function has $\frac { 1 } { x } \Rightarrow a + \frac { 1 } { x }$, that is, to $\frac { 1 } { x }$ upward, or downward translation, according to the needs of the extreme value, must be downward translation, so $a < 0$. Since the range of $a \leq 0$ is relatively large, a large range is a necessary and insufficient condition for a small range.

Question 6: 8. The following formula is correct

8. The following formula is correct

  • A. A. $[ \ln ( 2 x + 1 ) ] ^ { \prime } = \frac { 1 } { 2 x + 1 }$
  • B. B. $\left( x \cdot 2 ^ { x } \right) ^ { \prime } = 2 ^ { x } ( 1 + x \ln 2 )$
  • C. C. $\left( \frac { \cos x } { x } \right) ^ { \prime } = \frac { x \sin x + \cos x } { x ^ { 2 } }$
  • D. D. $( \sqrt { x } ) ^ { \prime } = - \frac { 1 } { 2 \sqrt { x } }$

Answer: B

Solution: Option A: $[ \ln ( 2 x + 1 ) ] ^ { \prime } = \frac { 2 } { 2 x + 1 }$, option A is wrong; Option B: $\left( x \cdot 2 ^ { x } \right) ^ { \prime } = 1 \cdot 2 ^ { x } + x \cdot 2 ^ { x } \ln 2 = 2 ^ { x } ( 1 + x \ln 2 )$, option B is correct; Option C: $\left( \frac { \cos x } { x } \right) ^ { \prime } = \frac { - \sin x \cdot x - \cos x } { x ^ { 2 } } = - \frac { x \sin x + \cos x } { x ^ { 2 } }$, option C is wrong; C: $\left( \frac { \cos x } { x } \right) ^ { \prime } = \frac { - \sin x \cdot x - \cos x } { x ^ { 2 } } = - \frac { x \sin x + \cos x } { x ^ { 2 } }$, C option is wrong; D: $( \sqrt { x } ) ^ { \prime } = \left( x ^ { \frac { 1 } { 2 } } \right) ^ { \prime } = \frac { 1 } { 2 } x ^ { - \frac { 1 } { 2 } } = \frac { 1 } { 2 \sqrt { x } }$, D option is wrong;

Question 7: 9. The derivative of the function $y = f ( x )$ at $x = x _ { 0 }$ is known to be $f ^ { \prime } \l...

9. The derivative of the function $y = f ( x )$ at $x = x _ { 0 }$ is known to be $f ^ { \prime } \left( x _ { 0 } \right) = - 1$, then $\lim _ { \Delta x \rightarrow 0 } \frac { f \left( x _ { 0 } + 2 \Delta x \right) - f \left( x _ { 0 } \right) } { \Delta x } =$ $\_\_\_\_$.

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

Answer: D

Solution: According to the question, the derivative of the function $y = f ( x )$ at ${ } ^ { x = x _ { 0 } }$ is $f ^ { \prime } \left( x _ { 0 } \right) = - 1$ , . and $\lim _ { \Delta x \rightarrow 0 } \frac { f \left( x _ { 0 } + 2 \Delta x \right) - f \left( x _ { 0 } \right) } { \Delta x } = 2 \lim _ { \Delta x \rightarrow 0 } \frac { f \left( x _ { 0 } + 2 \Delta x \right) - f \left( x _ { 0 } \right) } { 2 \Delta x } = 2 f ^ { \prime } \left( x _ { 0 } \right) = - 2$ , .

Question 8: 10. It is known that the derivative of the function $f ( x )$ at $x = 1$ is 1, then $\lim _ { \Delta...

10. It is known that the derivative of the function $f ( x )$ at $x = 1$ is 1, then $\lim _ { \Delta x \rightarrow 0 } \frac { f ( 1 + \Delta x ) - f ( 1 ) } { 3 \Delta x } =$

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

Answer: C

Solution: $\lim _ { \Delta x \rightarrow 0 } \frac { f ( 1 + \Delta x ) - f ( 1 ) } { 3 \Delta x } = \frac { 1 } { 3 } \lim _ { \Delta x \rightarrow 0 } \frac { f ( 1 + \Delta x ) - f ( 1 ) } { \Delta x } = \frac { 1 } { 3 } f ^ { \prime } ( 1 ) = \frac { 1 } { 3 }$.

Question 9: 11. It is known that the derivative function of $f ( x ) = \frac { 2 x + 1 } { x ^ { 2 } }$ is $f ^ ...

11. It is known that the derivative function of $f ( x ) = \frac { 2 x + 1 } { x ^ { 2 } }$ is $f ^ { \prime } ( x )$ and $f ^ { \prime } ( i ) = ( i$ is in imaginary units $)$.

  • A. A. $- 1 - 2 i$
  • B. B. $- 2 - 2 i$
  • C. C. $- 2 + 2 i$
  • D. D. $2 - 2 i$

Answer: D

Solution: $\because f ^ { \prime } ( x ) = \frac { 2 x ^ { 2 } - 2 x ( 2 x + 1 ) } { x ^ { 4 } } = \frac { - 2 x ^ { 2 } - 2 x } { x ^ { 4 } } \quad \therefore f ^ { \prime } ( i ) = 2 - 2 i$, so choose D.

Question 10: 12. $\int _ { 0 } ^ { 3 } \left( x ^ { 2 } + 4 \right) d x =$

12. $\int _ { 0 } ^ { 3 } \left( x ^ { 2 } + 4 \right) d x =$

  • A. A. 9
  • B. B. 12
  • C. C. 21
  • D. D. 25

Answer: C

Solution: $\int _ { 0 } ^ { 3 } \left( x ^ { 2 } + 4 \right) d x = \left. \left( \frac { 1 } { 3 } x ^ { 3 } + 4 x \right) \right| _ { 0 } ^ { 3 } = \frac { 1 } { 3 } \times 3 ^ { 3 } + 4 \times 3 - \frac { 1 } { 3 } \times 0 ^ { 3 } + 4 \times 0 = 21$

Question 11: 13. If the function $y = f ( x )$ is derivable on R and satisfies $x f ^ { \prime } ( x ) + f ( x ) ...

13. If the function $y = f ( x )$ is derivable on R and satisfies $x f ^ { \prime } ( x ) + f ( x ) > 0$ with constants ${ } ^ { a }$ , $b$ $\_\_\_\_$ ), then The following inequality must hold

  • A. A. $\quad a f ( a ) > b f ( b )$
  • B. B. $a f ( b ) > b f ( a )$
  • C. C. $a f ( a ) < b f ( b )$
  • D. D. $a f ( b ) < b f ( a )$

Answer: A

Solution: TEST ANALYSIS: Let $g ( x ) = x f ( x ) , \therefore g ^ { \prime } ( x ) = x f ^ { \prime } ( x ) + f ( x ) > 0$ be constant and $\therefore g ( x )$ be monotonically increasing on $R$. $\because a > b , \therefore g ( a ) > g ( b )$. That is $a f ( a ) > b f ( b )$. Therefore, A is correct. Point: Use the derivative to study the monotonicity of a function.

Question 12: 14. The function $f ( x )$ is known to be an even function with domain R. When $x > 0$ is $f ^ { \pr...

14. The function $f ( x )$ is known to be an even function with domain R. When $x > 0$ is $f ^ { \prime } ( x ) < 0$, then the solution set of inequality $f \left( x ^ { 2 } - x \right) - f ( x ) > 0$ is

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

Answer: B

Solution: Since $f ^ { \prime } ( x ) < 0$ when $x > 0$ is $f ^ { \prime } ( x ) < 0$, the even function ${ } ^ { f ( x ) }$ is monotonically decreasing on $^ { ( 0 , + \infty ) }$. Therefore $f \left( x ^ { 2 } - x \right) - f ( x ) > 0$ is transformed into: $f \left( \left| x ^ { 2 } - x \right| \right) > f ( | x | )$ , which is a function of $f \left( \left| x ^ { 2 } - x \right| \right) > f ( | x | )$. So $\left| x ^ { 2 } - x \right| < | x |$, obviously $x = 0$ does not satisfy the inequality. Solve: $| x - 1 | < 1$, so $x \in ( 0,2 )$.

Question 13: The figure is a round table filled with water, if a hole is opened at the bottom, and the volume of ...

The figure is a round table filled with water, if a hole is opened at the bottom, and the volume of water flowing out of any equal time interval is equal, the height of the water in the container $h$ with the time $t$ as a function of $h = f ( t )$, the definition of domain is $D$, and let $t _ { 0 } \in D , t _ { 0 } \pm \Delta t \in D , k _ { 1 } , k _ { 2 }$ denote the average rate of change of $^ { f ( t ) }$ over the interval $^ { \left[ t _ { 0 } - \Delta t , t _ { 0 } \right] , \left[ t _ { 0 } , t _ { 0 } + \Delta t \right] ( \Delta t > 0 ) }$, respectively, and then ![](/images/questions/calculus/image-001.jpg) size relations

  • A. A. $k _ { 1 } > k _ { 2 }$
  • B. B. $k _ { 1 } < k _ { 2 }$
  • C. C. $k _ { 1 } = k _ { 2 }$
  • D. D. It is not possible to determine the ${ } ^ { k _ { 1 } , k _ { 2 } }$ of the ${ } ^ { k _ { 1 } , k _ { 2 } }$.

Answer: A

Solution: From the shape of the container, the decrease of height is getting bigger and bigger in the same change time, and the rate of change of height $h$ is less than 0, so the average rate of change of ${ } ^ { f ( t ) }$ on the interval $\left[ t _ { 0 } - \Delta t , t _ { 0 } \right] , \left[ t _ { 0 } , t _ { 0 } + \Delta t \right] ( \Delta t > 0 )$ changes from big to small, i.e., the rate of change of

Question 14: 16. The slope of the tangent line to a point $P$ on the curve $y = \frac { 1 } { x }$ is - 4, and th...

16. The slope of the tangent line to a point $P$ on the curve $y = \frac { 1 } { x }$ is - 4, and the coordinates of point $P$ are

  • A. A. $\left( \frac { 1 } { 2 } , 2 \right)$
  • B. B. $\left( \frac { 1 } { 2 } , 2 \right)$ or $\left( - \frac { 1 } { 2 } , - 2 \right)$
  • C. C. $\left( - \frac { 1 } { 2 } , - 2 \right)$
  • D. D. $\left( \frac { 1 } { 2 } , - 2 \right)$

Answer: B

Solution: $\because$ curve $\mathrm { y } = \frac { 1 } { x } , \therefore \mathrm { y } ^ { \prime } = - \frac { 1 } { x ^ { 2 } }$, the Let $\mathrm { P } \left( \mathrm { x } _ { 0 } , \frac { 1 } { x _ { 0 } } \right)$ The slope of the tangent line to $\because$ at point P is $- 4 , \therefore - \frac { 1 } { x _ { 0 } { } ^ { 2 } } = - 4$, and solve for $x _ { 0 } = \frac { 1 } { 2 }$ or $x _ { 0 } = - \frac { 1 } { 2 }$, $x _ { 0 } = \frac { 1 } { 2 }$ or $x _ { 0 } = - \frac { 1 } { 2 }$. $\therefore$ The coordinates of point P are $\left( \frac { 1 } { 2 } , 2 \right)$ or $\left( - \frac { 1 } { 2 } , - 2 \right)$.

Question 15: 17. The monotonically decreasing interval of the function $y = \frac { 1 } { 2 } x ^ { 2 } - \ln x$ ...

17. The monotonically decreasing interval of the function $y = \frac { 1 } { 2 } x ^ { 2 } - \ln x$ is ( ).

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

Answer: B

Solution: The definition domain of the function is $( 0 , + \infty )$. From $y = \frac { 1 } { 2 } x ^ { 2 } - \ln x$ , we have $y ^ { \prime } = x - \frac { 1 } { x } = \frac { x ^ { 2 } - 1 } { x }$ . From $y ^ { \prime } = x - \frac { 1 } { x } = \frac { x ^ { 2 } - 1 } { x } < 0$ , we have $x \left( x ^ { 2 } - 1 \right) < 0$ . Since $x > 0$, we have $0 < x < 1$. So the monotonically decreasing interval of the function is $( 0,1 ]$.

Question 16: 18. The function $f ( x ) = x ^ { 3 } + a x ^ { 2 } + b x + c$, where $a , b , c$ is a real number, ...

18. The function $f ( x ) = x ^ { 3 } + a x ^ { 2 } + b x + c$, where $a , b , c$ is a real number, and when $a ^ { 2 } - 3 b < 0$ is $f ( x )$ is a function of $a ^ { 2 } - 3 b < 0$ on R

  • A. A. incremental function
  • B. B. reduced function (math.)
  • C. C. a constant (math.)
  • D. D. Unable to determine the monotonicity of the function

Answer: A

Solution: SOLUTION: Since $f ( x ) = x ^ { 3 } + a x ^ { 2 } + b x + c$, so $f ^ { \prime } ( x ) = 3 x ^ { 2 } + 2 a x + b$ then $\Delta = 4 a ^ { 2 } - 12 b = 4 \left( a ^ { 2 } - 3 b \right) < 0$ , so $f ^ { \prime } ( x ) > 0$ is constant on R. So $f ( x )$ is an increasing function on R ;

Question 17: 19. If the function $f ( x ) = k x + \mathrm { e } ^ { 2 x }$ is monotonically increasing on the int...

19. If the function $f ( x ) = k x + \mathrm { e } ^ { 2 x }$ is monotonically increasing on the interval ${ } ^ { ( - 1,1 ) }$, then the range of the real number $k$ is ( )

  • A. A. $\left[ - 2 \mathrm { e } ^ { 2 } , + \infty \right)$
  • B. B. $\left[ - 2 \mathrm { e } ^ { - 2 } , + \infty \right)$
  • C. C. $\left[ - \mathrm { e } ^ { 2 } , + \infty \right)$
  • D. D. $\left[ - \mathrm { e } ^ { - 2 } , + \infty \right)$

Answer: B

Solution: As the function $f ( x ) = k x + \mathrm { e } ^ { 2 x }$, then $f ^ { \prime } ( x ) = k + 2 \mathrm { e } ^ { 2 x }$, and Since the function $f ^ { ( x ) }$ is monotonically increasing on the interval $^ { ( - 1,1 ) }$, then $f ^ { \prime } ( x ) \geq 0$ holds on the interval $( - 1,1 )$. i.e. $k \geq - 2 \mathrm { e } ^ { 2 x }$ is constant on the interval $( - 1,1 )$. Since $y = - 2 \mathrm { e } ^ { 2 x }$ is monotonically decreasing on the interval $( - 1,1 )$, then $y < - 2 \mathrm { e } ^ { - 2 }$ , the Therefore, $k \geq - 2 \mathrm { e } ^ { - 2 }$, i.e., the real number $k$ is in the range $\left[ - 2 \mathrm { e } ^ { - 2 } , + \infty \right)$.

Question 18: 20. If the function $f ( x ) = \ln x + a x ^ { 2 } - 2 x$ is monotonically increasing in the interva...

20. If the function $f ( x ) = \ln x + a x ^ { 2 } - 2 x$ is monotonically increasing in the interval $^ { ( 1,2 ) }$, then the real number $a$ is in the range ( )

  • A. A. $\left( - \infty , \frac { 3 } { 8 } \right]$
  • B. B. $\left( \frac { 3 } { 8 } , \frac { 1 } { 2 } \right)$
  • C. C. $\left( \frac { 1 } { 2 } , + \infty \right)$
  • D. D. $\left[ \frac { 1 } { 2 } , + \infty \right)$

Answer: D

Solution: From the function $f ( x ) = \ln x + a x ^ { 2 } - 2 x$ we obtain $f ^ { \prime } ( x ) = \frac { 1 } { x } + 2 a x - 2$ if ${ } ^ { f ( x ) }$ is monotonically increasing in the interval $^ { ( 1,2 ) }$. then $f ^ { \prime } ( x ) \geq 0$ is constant in $x \in { } ^ { ( 1,2 ) }$. i.e. $a \geq \frac { 1 } { x } - \frac { 1 } { 2 x ^ { 2 } }$ is constant in $x \in ( 1,2 )$. such that $g ( x ) = \frac { 1 } { x } - \frac { 1 } { 2 x ^ { 2 } } = - \frac { 1 } { 2 } \left( \frac { 1 } { x } - 1 \right) ^ { 2 } + \frac { 1 } { 2 }$, the by $\frac { 1 } { x } \in \left( \frac { 1 } { 2 } , 1 \right)$. $\therefore g ( x ) < g ( 1 ) = \frac { 1 } { 2 }$, $\therefore g ( x ) < g ( 1 ) = \frac { 1 } { 2 }$, $\therefore g ( x ) < g ( 1 ) = \frac { 1 } { 2 }$ Therefore $a \geq \frac { 1 } { 2 }$ , . That is, the real number $a$ is in the range $\left[ \frac { 1 } { 2 } , + \infty \right)$.

Question 19: 21. In the sport of high diving, the height of an athlete's center of gravity relative to the water'...

21. In the sport of high diving, the height of an athlete's center of gravity relative to the water's surface at $t$ (in seconds) $h$ (in: meters) satisfies the relation ${ } ^ { h ( t ) = a t ^ { 2 } + 5 t + 11 }$, and the average rate of change of $h$ when $1 \leq t \leq 2$ is - 10 m/s, then when $t = 3$ when $h$'s instantaneous rate of change is ( )

  • A. A. - 15 m/s
  • B. B. 15 m/s
  • C. C. - 25 m/s
  • D. D. 25 m/s

Answer: C

Solution: From the question $\frac { h ( 2 ) - h ( 1 ) } { 2 - 1 } = 3 a + 5 = - 10$ , solve $a = - 5$ , then $h ( t ) = - 5 t ^ { 2 } + 5 t + 11$ , and and thus $h ^ { \prime } ( t ) = - 10 t + 5$, so $h ^ { \prime } ( 3 ) = - 10 \times 3 + 5 = - 25$.

Question 20: 23. The collinear line of the parabola ${ } ^ { 2 } = 4 y$ intersects the ${ } ^ { y }$ axis at ${ }...

23. The collinear line of the parabola ${ } ^ { 2 } = 4 y$ intersects the ${ } ^ { y }$ axis at ${ } ^ { M }$, and the equation of the line that is tangent to the parabola at the point ${ } ^ { M }$ is ( ).

  • A. A. $y = 2 x - 1$ or $y = - 2 x - 1$
  • B. B. $y = 3 x - 1$ or $y = - 3 x - 1$
  • C. C. $y = 4 x - 1$ or $y = - 4 x - 1$
  • D. D. $y = x - 1$ or $y = - x - 1$

Answer: D

Solution: SOLUTION: From the question, the parabola $x ^ { 2 } = 4 y$ is collinear to $y = - 1$ , then $M ( 0 , - 1 )$ , and $M ( 0 , - 1 )$ From $x ^ { 2 } = 4 y$ we get $y = \frac { x ^ { 2 } } { 4 }$, which is derived from $y ^ { \prime } = \frac { x } { 2 }$, and $y ^ { \prime } = \frac { x } { 2 }$, which is derived from $y ^ { \prime } = \frac { x } { 2 }$. Let the coordinates of the tangent point be $\left( x _ { 0 } , \frac { x _ { 0 } ^ { 2 } } { 4 } \right)$, then $\frac { \frac { x _ { 0 } ^ { 2 } } { 4 } + 1 } { x _ { 0 } } = \frac { x _ { 0 } } { 2 }$, solve for $x _ { 0 } = \pm 2$, $\therefore$ $\therefore$ the slope of the tangent line is $k = \pm 1$ , the $k = \pm 1$ $\therefore$ The equation of the tangent line is $y + 1 = \pm x$, i.e., $y = x - 1$ or $y = - x - 1$.

Question 21: 24. If $f ( x ) = - \frac { 1 } { 2 } x ^ { 2 } + 2 x f ^ { \prime } ( 2019 ) - 2019 \ln x$ is known...

24. If $f ( x ) = - \frac { 1 } { 2 } x ^ { 2 } + 2 x f ^ { \prime } ( 2019 ) - 2019 \ln x$ is known, $f ^ { \prime } ( 1 ) =$ is known.

  • A. A. 2017
  • B. B. 2018
  • C. C. 2019
  • D. D. 2020

Answer: D

Solution: $f ^ { \prime } ( x ) = - x + 2 f ^ { \prime } ( 2019 ) - \frac { 2019 } { x }$ Let $\mathrm { x } = 2019$ , get $f ^ { \prime } ( 2019 ) = 2020$ , then $f ^ { \prime } ( 1 ) = - 1 + 4040 - 2019 = 2020$

Question 22: 25. The function $f ( x ) = \frac { 1 } { 3 } x ^ { 3 } + x + \cos x$ is known, and if $f ( 1 ) > f ...

25. The function $f ( x ) = \frac { 1 } { 3 } x ^ { 3 } + x + \cos x$ is known, and if $f ( 1 ) > f \left( \log _ { 2 } x \right)$, then the range of values of the real number ${ } _ { x }$ is ( ).

  • A. A. $( 0,2 )$
  • B. B. $( 0,1 )$
  • C. C. $( 2 , + \infty )$
  • D. D. $( 1 , + \infty )$

Answer: A

Solution: Because $f ( x ) = \frac { 1 } { 3 } x ^ { 3 } + x + \cos x$ , . Therefore $f ^ { \prime } ( x ) = x ^ { 2 } + 1 - \sin x > 0$ , . so the function $y = f ( x )$ is monotonically increasing on R, and so $f ( 1 ) > f \left( \log _ { 2 } x \right)$ , . is equivalent to $\left\{ \begin{array} { l } \log _ { 2 } x < 1 \\ x > 0 \end{array} \right.$ and solves for $0 < x < 2$.

Question 23: 26. It is known that the function $f ( x ) = e ^ { x } + f ^ { \prime } ( 0 ) x ^ { 2 } + 3 x + 2$, ...

26. It is known that the function $f ( x ) = e ^ { x } + f ^ { \prime } ( 0 ) x ^ { 2 } + 3 x + 2$, then $f ^ { \prime } ( 1 ) = ( )$

  • A. A. $e + 5$
  • B. B. $e + 8$
  • C. C. $e + 11$
  • D. D. $e + 12$

Answer: C

Solution: From the question $f ^ { \prime } ( x ) = e ^ { x } + 2 f ^ { \prime } ( 0 ) x + 3$ , so $f ^ { \prime } ( 0 ) = e ^ { 0 } + 0 + 3 = 4$ , the $f ^ { \prime } ( 1 ) = e + 2 \times 4 + 3 = e + 11$

Question 24: 27. The graph of the function $f ( x ) = a e ^ { x } + b x - 1$ is known to be tangent to the $x$ ax...

27. The graph of the function $f ( x ) = a e ^ { x } + b x - 1$ is known to be tangent to the $x$ axis at the origin of the coordinates, then the values of $a$ and $b$ are ( ) respectively.

  • A. A. $a = - 1 , b = 1$
  • B. B. $a = 1 , b = - 1$
  • C. C. $a = - 1 , b = 0$
  • D. D. $a = 0 , b = - 1$

Answer: B

Solution: Since $f ( x ) = a e ^ { x } + b x - 1$, then $f ^ { \prime } ( x ) = a e ^ { x } + b$ , and From the known conditions, we have $\left\{ \begin{array} { l } f ( 0 ) = a - 1 = 0 \\ f ^ { \prime } ( 0 ) = a + b = 0 \end{array} \right.$, which solves for $\left\{ \begin{array} { l } a = 1 \\ b = - 1 \end{array} \right.$.

Question 25: 28. If $x = \ln 3$ is known to be a minima of the function $f ( x ) = \mathrm { e } ^ { x } + a x$, ...

28. If $x = \ln 3$ is known to be a minima of the function $f ( x ) = \mathrm { e } ^ { x } + a x$, then $a =$ ( )

  • A. A. $\ln 3$
  • B. B. $- \ln 3$
  • C. C. 3
  • D. D. - 3

Answer: D

Solution: Because $f ( x ) = \mathrm { e } ^ { x } + a x$, therefore $f ^ { \prime } ( x ) = \mathrm { e } ^ { x } + a$. Since $x = \ln 3$ is a minima of $f ( x )$, $f ^ { \prime } ( \ln 3 ) = 3 + a = 0$ , solving for $a = - 3$. When $a = - 3$ is $f ^ { \prime } ( x ) = \mathrm { e } ^ { x } - 3$, $f ^ { \prime } ( x ) = \mathrm { e } ^ { x } - 3$ , the When $x > \ln 3$, $f ^ { \prime } ( x ) > 0 , f ( x )$ is monotonically increasing; when $x < \ln 3$, $f ^ { \prime } ( x ) < 0 , f ( x )$ is monotonically decreasing. Therefore, $a = - 3$, $x = \ln 3$ is the minima of $f ( x )$, and therefore $a = - 3$.

Question 26: 29. The function $f ( x ) = \frac { 1 } { x }$ is known to be $\lim _ { \Delta x \rightarrow 0 } \fr...

29. The function $f ( x ) = \frac { 1 } { x }$ is known to be $\lim _ { \Delta x \rightarrow 0 } \frac { f ( 1 + \Delta x ) - f ( 1 ) } { \Delta x } = ( )$.

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

Answer: A

Solution: Since $f ( 1 + \Delta x ) = \frac { 1 } { 1 + \Delta x } , f ( 1 ) = \frac { 1 } { 1 } = 1$, then $f ( 1 + \Delta x ) - f ( 1 ) = \frac { 1 } { 1 + \Delta x } - 1 = \frac { - \Delta x } { 1 + \Delta x }$, therefore $\lim _ { \Delta x \rightarrow 0 } \frac { f ( 1 + \Delta x ) - f ( 1 ) } { \Delta x } = \lim _ { \Delta x \rightarrow 0 } \frac { - 1 } { 1 + \Delta x } = - 1$.

Question 27: 31. The following derivation operations are correct

31. The following derivation operations are correct

  • A. A. $\left( x + \frac { 1 } { x } \right) ^ { \prime } = 1 + \frac { 1 } { x ^ { 2 } }$
  • B. B. $\left( \log _ { 2 } x \right) ^ { \prime } = \frac { 1 } { x \ln 2 }$
  • C. C. $\left( 3 ^ { x } \right) ^ { \prime } = 3 ^ { x } \cdot \log _ { 3 } \mathrm { e }$
  • D. D. $\left( \frac { x ^ { 2 } } { \mathrm { e } ^ { x } } \right) ^ { \prime } = \frac { 2 x + x ^ { 2 } } { \mathrm { e } ^ { x } }$

Answer: B

Solution: A. $\left( x + \frac { 1 } { x } \right) ^ { \prime } = 1 - \frac { 1 } { x ^ { 2 } }$, A is wrong; B. $\left( \log _ { 2 } x \right) ^ { \prime } = \frac { 1 } { x \ln 2 }$, B is correct; C. $\left( 3 ^ { x } \right) ^ { \prime } = 3 ^ { x } \cdot \ln 3 = 3 ^ { x } \cdot \frac { 1 } { \log _ { 3 } \mathrm { e } }$, C is wrong; D. $\left( \frac { x ^ { 2 } } { \mathrm { e } ^ { x } } \right) ^ { \prime } = \frac { 2 x - x ^ { 2 } } { \mathrm { e } ^ { x } }$, so D is wrong.

Question 28: 32. A function $y = f ( x )$ is said to have the property $T$ if there exist two points on the image...

32. A function $y = f ( x )$ is said to have the property $T$ if there exist two points on the image of the function $y = f ( x )$ such that the tangent lines to the image of the function at these points are perpendicular to each other. The following functions have the property $T$.

  • A. A. $y = \cos x$
  • B. B. $y = \ln x$
  • C. C. $y = \mathrm { e } ^ { x }$
  • D. D. $y = x ^ { 3 }$

Answer: A

Solution: From the question, ${ } ^ { y = f ( x ) }$ has the property $T$, i.e. there exists ${ } ^ { x _ { 1 } , x _ { 2 } }$ such that $f ^ { \prime } \left( x _ { 1 } \right) \cdot f ^ { \prime } \left( x _ { 2 } \right) = - 1$; For $\mathrm { A } , ~ y ^ { \prime } = - \sin x$, there exists $x _ { 1 } = \frac { \pi } { 2 } , ~ x _ { 2 } = - \frac { \pi } { 2 }$ such that $f ^ { \prime } \left( x _ { 1 } \right) \cdot f ^ { \prime } \left( x _ { 2 } \right) = \left( - \sin \frac { \pi } { 2 } \right) \left[ - \sin \left( - \frac { \pi } { 2 } \right) \right] = - 1$, A is correct; For B, the domain of definition of $y = \ln x$ is $( 0 , + \infty ) , y ^ { \prime } = \frac { 1 } { x } > 0$ . Therefore, there is no ${ } ^ { x _ { 1 } , x _ { 2 } }$, which makes $f ^ { \prime } \left( x _ { 1 } \right) \cdot f ^ { \prime } \left( x _ { 2 } \right) = - 1$, B is wrong; For C, $y ^ { \prime } = \mathrm { e } ^ { x } > 0$, so there is no ${ } ^ { x _ { 1 } , x _ { 2 } }$ such that $f ^ { \prime } \left( x _ { 1 } \right) \cdot f ^ { \prime } \left( x _ { 2 } \right) = - 1$, C is wrong; For D, $y ^ { \prime } = 3 x ^ { 2 } \geq 0$, so there is no ${ } ^ { x _ { 1 } , x _ { 2 } }$, making $f ^ { \prime } \left( x _ { 1 } \right) \cdot f ^ { \prime } \left( x _ { 2 } \right) = - 1$, D wrong;

Question 29: 33. Given that the function $f ( x ) = \ln x - \frac { 1 } { 3 } a x ^ { 3 }$ is monotonically decre...

33. Given that the function $f ( x ) = \ln x - \frac { 1 } { 3 } a x ^ { 3 }$ is monotonically decreasing on $( 1 , + \infty )$, the range of the real number $a$ is

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

Answer: B

Solution: Since $f ( x ) = \ln x - \frac { 1 } { 3 } a x ^ { 3 }$, then $f ^ { \prime } ( x ) = \frac { 1 } { x } - a x ^ { 2 }$, $f ^ { \prime } ( x ) \leq 0$ holds for any $x > 1$. From the question $f ^ { \prime } ( x ) \leq 0$ is constant for any $x > 1$, then $a \geq \frac { 1 } { x ^ { 3 } }$ is constant for any $x > 1$. When $x > 1$ is $\frac { 1 } { x ^ { 3 } } \in ( 0,1 ) , \therefore a \geq 1$.

Question 30: 34. The area of the plane formed by the line $x = - 2 , x = 2 , y = 0$ and the curve $y = x ^ { 2 } ...

34. The area of the plane formed by the line $x = - 2 , x = 2 , y = 0$ and the curve $y = x ^ { 2 } - x$ is ( )

  • A. A. $\frac { 16 } { 3 }$
  • B. B. $\frac { 17 } { 3 }$
  • C. C. $\frac { 8 } { 3 }$
  • D. D. $\frac { 5 } { 3 }$

Answer: B

Solution: Analysis: Make the graph of the function $y = x ^ { 2 } - x$ and the line $x = - 2 , x = 2$ to determine the upper and lower limits of the integral. Explanation: As shown in the figure. ![](/images/questions/calculus/image-002.jpg) $S = \int _ { - 2 } ^ { 0 } \left( x ^ { 2 } - x \right) d x + \int _ { 0 } ^ { 1 } \left( - x ^ { 2 } + x \right) d x + \int _ { 1 } ^ { 2 } \left( x ^ { 2 } - x \right) d x$ $= \left. \left( \frac { 1 } { 3 } x ^ { 3 } - \frac { 1 } { 2 } x ^ { 2 } \right) \right| _ { - 2 } ^ { 0 } + \left. \left( - \frac { 1 } { 3 } x ^ { 3 } + \frac { 1 } { 2 } x ^ { 2 } \right) \right| _ { 0 } ^ { 1 } + \left. \left( \frac { 1 } { 3 } x ^ { 3 } - \frac { 1 } { 2 } x ^ { 2 } \right) \right| _ { 1 } ^ { 2 } = \frac { 17 } { 3 }$.

Question 31: 35. The function $f ( x ) = \sin 2 x - k \sin x + x$ is monotonically decreasing on the interval $\l...

35. The function $f ( x ) = \sin 2 x - k \sin x + x$ is monotonically decreasing on the interval $\left( 0 , \frac { \pi } { 2 } \right)$, then the value of $k$ is in the range ( ).

  • A. A. $[ - 3 , + \infty )$
  • B. B. $[ 3 , + \infty )$
  • C. C. $[ 0 , + \infty )$
  • D. D. $[ - 3,3 ]$

Answer: B

Solution:

Question 32: 36. If the function $f ( x ) = x ^ { 2 } + 2 \cos x$ is known, the solution set of the inequality $f...

36. If the function $f ( x ) = x ^ { 2 } + 2 \cos x$ is known, the solution set of the inequality $f ( 2 x - 1 ) < f ( 3 x )$ is ( ).

  • A. A. $\left( - 1 , \frac { 1 } { 5 } \right)$
  • B. B. $\left( - \frac { 1 } { 5 } , 1 \right)$
  • C. C. $( - \infty , - 1 ) \cup \left( \frac { 1 } { 5 } , + \infty \right)$
  • D. D. $\left( - \infty , - \frac { 1 } { 5 } \right) \cup ( 1 , + \infty )$

Answer: C

Solution: Since the function $y = f ( x )$ has domain $R , f ( - x ) = ( - x ) ^ { 2 } + 2 \cos ( - x ) = x ^ { 2 } + 2 \cos x = f ( x )$, the function $y = f ( x )$ is even. When $x \geq 0$ is $f ^ { \prime } ( x ) = 2 x - 2 \sin x = g ( x ) , g ^ { \prime } ( x ) = 2 - 2 \cos x \geq 0$, the function $x \geq 0$ is even. So $f ^ { \prime } ( x ) = 2 x - 2 \sin x$ is monotonically increasing in $[ 0 , + \infty )$, so $f ^ { \prime } ( x ) \geq f ^ { \prime } ( 0 ) = 0$ So the function $y = f ( x )$ is monotonically increasing in $[ 0 , + \infty )$. From $f ( 2 x - 1 ) < f ( 3 x )$, we can get $f ( | 2 x - 1 | ) < f ( | 3 x | )$, then $| 2 x - 1 | < | 3 x |$. Squaring both sides of the inequality yields $9 x ^ { 2 } > ( 2 x - 1 ) ^ { 2 }$ , and This yields $( x + 1 ) ( 5 x - 1 ) > 0$, which solves for $x < - 1$ or $x > \frac { 1 } { 5 }$. Therefore, the solution set of the inequality $f ( 2 x - 1 ) < f ( 3 x )$ is $( - \infty , - 1 ) \cup \left( \frac { 1 } { 5 } , + \infty \right)$.

Question 33: 37. It is known that the parabola $C : x ^ { 2 } = 4 y$, the line through the point $M ( 0,4 )$ and ...

37. It is known that the parabola $C : x ^ { 2 } = 4 y$, the line through the point $M ( 0,4 )$ and $C$ intersects at the points $A , B$, and the tangent line at the points $C _ { \text {在 } } A , B$ intersects at the point $N$. Of the following four points, the one that can be the midpoint of the line segment $M N$ is ( )

  • A. A. $( 0,1 )$
  • B. B. $\left( \frac { 1 } { 2 } , 0 \right)$
  • C. C. $\left( \frac { 1 } { 2 } , 1 \right)$
  • D. D. $\left( 1 , - \frac { 1 } { 2 } \right)$

Answer: B

Solution: It may be useful to set $A \left( x _ { 1 } , y _ { 1 } \right) , B \left( x _ { 2 } , y _ { 2 } \right) , N \left( x _ { 0 } , y _ { 0 } \right)$, and from $x ^ { 2 } = 4 y$, $y = \frac { 1 } { 4 } x ^ { 2 }$, then $y ^ { \prime } = \frac { 1 } { 2 } x$. The equation of the tangent to the parabola at the point $A$ is then $y - y _ { 1 } = \frac { 1 } { 2 } x _ { 1 } \left( x - x _ { 1 } \right)$, and Since $x _ { 1 } ^ { 2 } = 4 y _ { 1 }$, the equation is simplified to $x _ { 1 } x = 2 \left( y + y _ { 1 } \right)$ and Similarly, the equation of the tangent to the parabola at the point ${ } ^ { B }$ is $x _ { 2 } x = 2 \left( y + y _ { 2 } \right)$, and And the tangent lines intersect at the point $N \left( x _ { 0 } , y _ { 0 } \right)$, so we get $\left\{ \begin{array} { l } x _ { 1 } x _ { 0 } = 2 \left( y _ { 0 } + y _ { 1 } \right) \\ x _ { 2 } x _ { 0 } = 2 \left( y _ { 0 } + y _ { 2 } \right) \end{array} \right.$. i.e., the points $A , B$ are on the line $x _ { 0 } x - 2 y - 2 y _ { 0 } = 0$, i.e., the equation of the line $A B$ is $x _ { 0 } x - 2 y - 2 y _ { 0 } = 0$, and $x _ { 0 } x - 2 y - 2 y _ { 0 } = 0$ is $x _ { 0 } x - 2 y - 2 y _ { 0 } = 0$. Since the point $M ( 0,4 )$ is on the line $A B$, substituting solves for $y _ { 0 } = - 4$, which is $N \left( x _ { 0 } , - 4 \right)$. Therefore, the midpoint of segment $M N$ is $N \left( \frac { x _ { 0 } } { 2 } , 0 \right)$, so the option that can be the midpoint of segment $M N$ is $\left( \frac { 1 } { 2 } , 0 \right)$. ![](/images/questions/calculus/image-003.jpg)

Question 34: 38. The function $f ( x ) = 2 ( x - 1 ) \mathrm { e } ^ { x } - x ^ { 2 } - a x$ is known to be mono...

38. The function $f ( x ) = 2 ( x - 1 ) \mathrm { e } ^ { x } - x ^ { 2 } - a x$ is known to be monotonically increasing on R. The maximum value of $a$ is ( ).

  • A. A. 0
  • B. B. $\frac { 1 } { 6 }$
  • C. C. e
  • D. D. 3

Answer: A

Solution: From the function $f ( x ) = 2 ( x - 1 ) \mathrm { e } ^ { x } - x ^ { 2 } - a x$, it follows that $f ^ { \prime } ( x ) = 2 x \mathrm { e } ^ { x } - 2 x - a$, the Since $f ( x )$ is monotonically increasing on R, $f ^ { \prime } ( x ) = 2 x \mathrm { e } ^ { x } - 2 x - a \geq 0$ is constant and i.e. $a \leq 2 x \mathrm { e } ^ { x } - 2 x$ is constant. Let $g ( x ) = 2 x \mathrm { e } ^ { x } - 2 x$ be $g ^ { \prime } ( x ) = ( 2 x + 2 ) \mathrm { e } ^ { x } - 2$, then $g ^ { \prime } ( x ) = ( 2 x + 2 ) \mathrm { e } ^ { x } - 2$, and When $x < 0$ is $g ^ { \prime } ( x ) < 0$; when $x > 0$ is $g ^ { \prime } ( x ) > 0$, then $g ^ { \prime } ( x ) > 0$ then $g ^ { ( x ) }$ is monotonically decreasing on $^ { ( - \infty , 0 ) }$ and monotonically increasing on $( 0 , + \infty )$. So $g ( x ) _ { \text {min } } = g ( 0 ) = 0$, i.e., $a \leq 0$, so the real number $a$ has a maximum value of zero.

Question 35: 39. If the line $y = a x + b$ is known to be tangent to the curve $\mathrm { y } = \mathrm { x } + \...

39. If the line $y = a x + b$ is known to be tangent to the curve $\mathrm { y } = \mathrm { x } + \frac { 1 } { \mathrm { x } }$, the maximum value of $2 a + b$ is ( )

  • A. A. $\frac { 1 } { 2 }$
  • B. B. 2
  • C. C. $\frac { 5 } { 2 }$
  • D. D. 5

Answer: C

Solution: Let the transverse coordinate of the tangent point be $m ( m \neq 0 )$, and derive $\mathrm { y } = \mathrm { x } + \frac { 1 } { \mathrm { x } }$ from $y ^ { \prime } = 1 - \frac { 1 } { x ^ { 2 } }$. From $\left\{ \begin{array} { l } a = 1 - \frac { 1 } { m ^ { 2 } } \\ a m + b = m + \frac { 1 } { m } \end{array} \right.$, solve for $\left\{ \begin{array} { l } a = 1 - \frac { 1 } { m ^ { 2 } } \\ b = \frac { 2 } { m } \end{array} \right.$, and $\left\{ \begin{array} { l } a = 1 - \frac { 1 } { m ^ { 2 } } \\ b = \frac { 2 } { m } \end{array} \right.$, and $\left\{ \begin{array} { l } a = 1 - \frac { 1 } { m ^ { 2 } } \\ b = \frac { 2 } { m } \end{array} \right.$. So $2 a + b = - \frac { 2 } { m ^ { 2 } } + \frac { 2 } { m } + 2 = - 2 \left( \frac { 1 } { m } - \frac { 1 } { 2 } \right) ^ { 2 } + \frac { 5 } { 2 }$ , and $2 a + b = - \frac { 2 } { m ^ { 2 } } + \frac { 2 } { m } + 2 = - 2 \left( \frac { 1 } { m } - \frac { 1 } { 2 } \right) ^ { 2 } + \frac { 5 } { 2 }$ So the maximum value of $2 a + b$ is $\frac { 5 } { 2 }$ at $m = 2$.

Question 36: 40. It is known that the function $f ( x ) = \sin \left( \omega x + \frac { \pi } { 6 } \right) ( \o...

40. It is known that the function $f ( x ) = \sin \left( \omega x + \frac { \pi } { 6 } \right) ( \omega > 0 )$ is monotonically decreasing on $( 1,2 )$ and monotonically increasing on $( 2,3 )$, and the circle $x ^ { 2 } + y ^ { 2 } = r ^ { 2 } ( r > 0 )$ contains exactly three points corresponding to the extremes of $f ( x )$. $x ^ { 2 } + y ^ { 2 } = r ^ { 2 } ( r > 0 )$ contains exactly three points corresponding to the extremes of $f ( x )$, then the range of $r$ is ( ) High School Mathematics Assignment, October 29, 2025

  • A. A. $[ 2 , \sqrt { 5 } )$
  • B. B. $\left( \sqrt { 5 } , \frac { \sqrt { 29 } } { 2 } \right]$
  • C. C. $( \sqrt { 5 } , 3 ]$
  • D. D. $\left[ \sqrt { 5 } , \frac { \sqrt { 53 } } { 2 } \right]$

Answer: B

Solution: From the known minimum value of $f ( x )$ at $x = 2$, $x = 2$ $\therefore \sin \left( 2 \omega + \frac { \pi } { 6 } \right) = - 1 , \therefore 2 \omega + \frac { \pi } { 6 } = 2 k \pi - \frac { \pi } { 2 } ( k \in \mathbf { Z } )$, solve for $\omega = k \pi - \frac { \pi } { 3 } ( k \in \mathbf { Z } )$, $\omega = k \pi - \frac { \pi } { 3 } ( k \in \mathbf { Z } )$, and The $\because$ function ${ } ^ { f ( x ) }$ is monotonically decreasing on $^ { ( 1,2 ) }$ and $\therefore \frac { T } { 2 } \geq 1$, i.e., $\frac { \pi } { \omega } \geq 1 , ~ \therefore 0 < \omega \leq \pi$. When ${ } _ { k = 1 }$, $\omega = \frac { 2 \pi } { 3 } , ~ T = 3$, the condition is satisfied that $\therefore f ( x ) = \sin \left( \frac { 2 \pi } { 3 } x + \frac { \pi } { 6 } \right)$. From the image, we know that the right side of the ${ } ^ { y }$ axis contains two points corresponding to the extremes, and the left side contains one point corresponding to the extremes. The value of ![](/images/questions/calculus/image-004.jpg) is greater than the point corresponding to the second extreme value to the right of the origin. The distance of $( 2 , - 1 ) _ { \text {到原点的距离,小于等于原点左侧第二个极值对应的点 } } \left( - \frac { 5 } { 2 } , 1 \right)$ to the origin, i.e. $r \in \left( \sqrt { 5 } , \frac { \sqrt { 29 } } { 2 } \right]$, $r \in \left( \sqrt { 5 } , \frac { \sqrt { 29 } } { 2 } \right]$, the
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Calculus

导数与微积分

36 Practice Questions

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

Topic Overview

Derivatives and calculus are core tools of mathematical analysis, focusing on the rate of change of functions, the slope of tangents, and area calculations. In the CSCA exam, this part of the content usually appears in the form of calculation and application questions, requiring candidates to master the basic rules of derivatives, the geometric significance of derivatives and simple integral operations. The questions are of moderate difficulty and focus on the application of basic formulas and computational accuracy.

Questions:36

Key Points

  • 1Derivative formulas and algorithms for elementary elementary functions
  • 2Geometric significance of derivatives (tangent slope and tangent equation)
  • 3Derivatives of Composite and Implicit Functions
  • 4Analyzing monotonicity and extreme values of functions using derivatives

Study Tips

It is recommended that the derivation formula be reinforced through categorical exercises, with special attention to the chain rule for the derivation of composite functions and the standard solution of the tangent line equation.

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