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Elementary Function - Practice Questions (37)

Question 1: 1. If it is known that the terminal side of angle $2 \alpha$ is above the $x$ axis, then the angle $...

1. If it is known that the terminal side of angle $2 \alpha$ is above the $x$ axis, then the angle $\alpha$ is in the range ( )

  • A. A. The set of first quadrant angles
  • B. B. The set of first or second quadrant angles
  • C. C. The set of first or third quadrant angles
  • D. D. The set of first or fourth quadrant angles

Answer: C

Solution: From the question, $2 k \pi < 2 \alpha < ( 2 k + 1 ) \pi$ and $_ { k \in \mathrm { Z } }$ , then $k \pi < \alpha < k \pi + \frac { \pi } { 2 }$ , the $\therefore$ angle $\alpha$ ranges over the set of first or third quadrant angles.

Question 2: 2. Stretching the horizontal coordinates of all points on the graph of the function $f ( x )$ to twi...

2. Stretching the horizontal coordinates of all points on the graph of the function $f ( x )$ to twice their original values yields the graph of the function $g ( x ) = \cos 2 x$, which $f ( x )$ is

  • A. A. even function with period $2 \pi$
  • B. B. Odd functions with period $2 \pi$
  • C. C. even function with period $\frac { \pi } { 2 }$
  • D. D. Odd functions with period $\frac { \pi } { 2 }$

Answer: C

Solution: Stretching the horizontal coordinates of all points on the graph of the function $f ( x )$ to twice the original to obtain the graph of the function $g ( x ) = \cos 2 x$. Then $f ( x ) = \cos 4 x$ is an even function with period $\frac { 2 \pi } { 4 } = \frac { \pi } { 2 }$.

Question 3: 3. After translating the graph of function $y = \sin \left( x + \frac { \pi } { 3 } \right)$ to the ...

3. After translating the graph of function $y = \sin \left( x + \frac { \pi } { 3 } \right)$ to the right by $\frac { \pi } { 6 }$ units of length, the resulting graph corresponds to a function of

  • A. A. $y = \sin \left( x + \frac { \pi } { 6 } \right)$
  • B. B. $y = \sin \left( x - \frac { \pi } { 6 } \right)$
  • C. C. $y = \cos x$
  • D. D. $y = - \cos x$

Answer: A

Solution: After translating the graph of the function $y = \sin \left( x + \frac { \pi } { 3 } \right)$ to the right by $\frac { \pi } { 6 }$ unit lengths The resulting graph corresponds to the function $y = \sin \left[ \left( x - \frac { \pi } { 6 } \right) + \frac { \pi } { 3 } \right]$, which is $y = \sin \left( x + \frac { \pi } { 6 } \right)$.

Question 4: 6. If ${ } ^ { f ( x ) }$ is a power function and ${ } ^ { f ( x ) }$ is monotonically decreasing on...

6. If ${ } ^ { f ( x ) }$ is a power function and ${ } ^ { f ( x ) }$ is monotonically decreasing on $^ { ( 0 , + \infty ) }$, then the analytic formula of ${ } ^ { f ( x ) }$ may be

  • A. A. $f ( x ) = - x ^ { 2 }$
  • B. B. $f ( x ) = \frac { 2 } { x }$
  • C. C. $f ( x ) = x ^ { 3 }$
  • D. D. $f ( x ) = \frac { 1 } { x ^ { 2 } }$

Answer: D

Solution: Since the graphs of the power functions all pass through the point ${ } ^ { ( 1,1 ) }$, it is clear that options A , B are not satisfied, i.e., neither A, B is a power function; The function $f ( x ) = x ^ { 3 }$ is a power function, but monotonically increasing on $( 0 , + \infty )$; The function $f ( x ) = \frac { 1 } { x ^ { 2 } } = x ^ { - 2 }$ is an idempotent function and monotonically decreasing on $( 0 , + \infty )$, D is satisfied.

Question 5: 7. The value range of the function $y = 3 ^ { x }$ is ( ).

7. The value range of the function $y = 3 ^ { x }$ is ( ).

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

Answer: A

Solution: SOLUTION: $\because$ Since $3 ^ { x } > 0$ $\therefore$ function $y = 3 ^ { x }$ has a value domain of $( 0 , + \infty )$ , the

Question 6: 8. If ${ } _ { \pi < \alpha < 2 \pi } , \sin \alpha + \cos \alpha = \frac { 1 } { 5 }$ is known, the...

8. If ${ } _ { \pi < \alpha < 2 \pi } , \sin \alpha + \cos \alpha = \frac { 1 } { 5 }$ is known, then ${ } _ { \tan \alpha }$ is equal to ( ).

  • A. A. $- \frac { 3 } { 4 }$
  • B. B. $- \frac { 3 } { 4 }$ or $- \frac { 4 } { 3 }$
  • C. C. $\frac { 3 } { 4 }$ or $\frac { 4 } { 3 }$
  • D. D. $\frac { 3 } { 5 }$

Answer: A

Solution: SOLUTION: $\because _ { \pi < \alpha < 2 \pi } , \sin \alpha + \cos \alpha = \frac { 1 } { 5 }$ $\therefore$ squares to $1 + 2 \sin \alpha \cos \alpha = \frac { 1 } { 25 }$, which is $\sin \alpha \cos \alpha = - \frac { 12 } { 25 } < 0$, $\therefore \begin{array} { l l } \sin \alpha < 0 & , \cos \alpha > 0 \end{array}$, $\because \sin ^ { 2 } \alpha + \cos ^ { 2 } \alpha = 1$ to $\because \sin ^ { 2 } \alpha + \cos ^ { 2 } \alpha = 1$ yields: $\left( \frac { 1 } { 5 } - \cos \alpha \right) ^ { 2 } + \cos ^ { 2 } \alpha = 1$, which is solved by $\cos \alpha = \frac { 4 } { 5 }$, or $- \frac { 3 } { 5 }$ (rounding off). $\therefore \sin \alpha = \frac { 1 } { 5 } - \frac { 4 } { 5 } = - \frac { 3 } { 5 }$, which gives: $\tan \alpha = - \frac { 3 } { 4 }$.

Question 7: 9. If $\sin \left( \frac { \pi } { 6 } + \alpha \right) = - \frac { 4 } { 5 }$ is known, then $\cos ...

9. If $\sin \left( \frac { \pi } { 6 } + \alpha \right) = - \frac { 4 } { 5 }$ is known, then $\cos \left( \frac { \pi } { 3 } - \alpha \right) =$ is known.

  • A. A. $\frac { 4 } { 5 }$
  • B. B. $\frac { 3 } { 5 }$
  • C. C. $- \frac { 4 } { 5 }$
  • D. D. $- \frac { 3 } { 5 }$

Answer: C

Solution: SOLUTION: $\because \sin \left( \frac { \pi } { 6 } + \alpha \right) = - \frac { 4 } { 5 }$, $\therefore \cos \left( \frac { \pi } { 3 } - \alpha \right) = \cos \left[ \frac { \pi } { 2 } - \left( \frac { \pi } { 6 } + \alpha \right) \right] = \sin \left( \frac { \pi } { 6 } + \alpha \right) = - \frac { 4 } { 5 }$ $\therefore \cos \left( \frac { \pi } { 3 } - \alpha \right) = \cos \left[ \frac { \pi } { 2 } - \left( \frac { \pi } { 6 } + \alpha \right) \right] = \sin \left( \frac { \pi } { 6 } + \alpha \right) = - \frac { 4 } { 5 }$, $\therefore \cos \left( \frac { \pi } { 3 } - \alpha \right) = \cos \left[ \frac { \pi } { 2 } - \left( \frac { \pi } { 6 } + \alpha \right) \right] = \sin \left( \frac { \pi } { 6 } + \alpha \right) = - \frac { 4 } { 5 }$, $\therefore \cos \left( \frac { \pi } { 3 } - \alpha \right) = \cos \left[ \frac { \pi } { 2 } - \left( \frac { \pi } { 6 } + \alpha \right) \right] = \sin \left( \frac { \pi } { 6 } + \alpha \right) = - \frac { 4 } { 5 }$

Question 8: 10. If $\sqrt [ 3 ] { a ^ { 4 } } \cdot a ^ { \frac { 1 } { 4 } } \cdot a ^ { \frac { 5 } { 12 } } =...

10. If $\sqrt [ 3 ] { a ^ { 4 } } \cdot a ^ { \frac { 1 } { 4 } } \cdot a ^ { \frac { 5 } { 12 } } = 4$ is known, then $a =$ is known.

  • A. A. 2
  • B. B. - 2
  • C. C. 4
  • D. D. 2 or -2

Answer: A

Solution: Since $\sqrt [ 3 ] { a ^ { 4 } } \cdot a ^ { \frac { 1 } { 4 } } \cdot a ^ { \frac { 5 } { 12 } } = a ^ { \frac { 4 } { 3 } } \cdot a ^ { \frac { 1 } { 4 } } \cdot a ^ { \frac { 5 } { 12 } } = a ^ { 2 }$, $a ^ { 2 } = 4$, solve for $a = \pm 2$; To make sense of the equation, $a > 0$ , so $a = 2$ ;

Question 9: 11. The following angles have the same final side as $985 ^ { \circ }$.

11. The following angles have the same final side as $985 ^ { \circ }$.

  • A. A. $165 ^ { \circ }$
  • B. B. $265 ^ { \circ }$
  • C. C. ${ } ^ { 85 ^ { \circ } }$
  • D. D. $- 105 ^ { \circ }$

Answer: B

Solution: Using the concept of angles with the same endpoints, we know that the angle with the same endpoint as $985 ^ { \circ }$ is $985 ^ { \circ } + k 360 ^ { \circ } ( k \in Z )$ then it is when $k = 2,985 ^ { \circ } - 360 ^ { \circ } \times 2 = 265 ^ { \circ }$ is $k = 2,985 ^ { \circ } - 360 ^ { \circ } \times 2 = 265 ^ { \circ }$.

Question 10: 12. If the central angle of a sector is $108 ^ { \circ }$ and the radius is 10 cm, then the area of ...

12. If the central angle of a sector is $108 ^ { \circ }$ and the radius is 10 cm, then the area of the sector is

  • A. A. $30 \pi \mathrm {~cm} ^ { 2 }$
  • B. B. $60 \pi \mathrm {~cm} ^ { 2 }$
  • C. C. $5400 \pi \mathrm {~cm} ^ { 2 }$
  • D. D. $10800 \pi \mathrm {~cm} ^ { 2 }$

Answer: A

Solution: The fan has a central angle of $108 ^ { \circ } = \frac { 3 \pi } { 5 }$ and a radius of 10 cm. Then the area of the sector is $\frac { 1 } { 2 } \times \frac { 3 \pi } { 5 } \times 10 ^ { 2 } = 30 \pi \mathrm {~cm} ^ { 2 }$.

Question 11: 13. If $a = \ln 2 , b = 5 ^ { - \frac { 1 } { 2 } } , c = \int ^ { 1 } x d x$, then the magnitude re...

13. If $a = \ln 2 , b = 5 ^ { - \frac { 1 } { 2 } } , c = \int ^ { 1 } x d x$, then the magnitude relation of $a , b , c$ is

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

Answer: C

Solution: $\because \frac { 1 } { 2 } = \ln \sqrt { e } < \ln 2 < \ln e = 1$ $\therefore \frac { 1 } { 2 } < a < 1$ $\because b = 5 ^ { - \frac { 1 } { 2 } } = \frac { 1 } { \sqrt { 5 } } < \frac { 1 } { 2 } , \quad c = \int _ { 0 } ^ { 1 } x \mathrm {~d} x = \left. \frac { 1 } { 2 } x ^ { 2 } \right| _ { 0 } ^ { 1 } = \frac { 1 } { 2 }$ $\therefore b < c < a$

Question 12: 14. A function that also has the following properties: "(1) the least positive period is $\pi$ and (...

14. A function that also has the following properties: "(1) the least positive period is $\pi$ and (2) it is an increasing function on the interval $\left[ - \frac { \pi } { 6 } , \frac { \pi } { 6 } \right]$" is

  • A. A. $y = \sin \left( 2 x - \frac { \pi } { 6 } \right)$
  • B. B. $y = \sin \left( \frac { x } { 2 } + \frac { \pi } { 6 } \right)$
  • C. C. $y = \cos \left( \frac { 1 } { 2 } x + \frac { \pi } { 3 } \right)$
  • D. D. $y = \cos \left( 2 x - \frac { \pi } { 6 } \right)$

Answer: A

Solution: For A, the period $y = \sin \left( 2 x - \frac { \pi } { 6 } \right)$ of $T = \frac { 2 \pi } { 2 } = \pi$. When $- \frac { \pi } { 6 } \leq x \leq \frac { \pi } { 6 }$, $- \frac { \pi } { 2 } \leq 2 x - \frac { \pi } { 6 } \leq \frac { \pi } { 6 }$ , so the function is increasing on the interval $\left[ - \frac { \pi } { 6 } , \frac { \pi } { 6 } \right]$, correct; For B, $y = \sin \left( \frac { x } { 2 } + \frac { \pi } { 6 } \right)$ has period $T = \frac { 2 \pi } { \frac { 1 } { 2 } } = 4 \pi$ , which is not true; For C, the period $T = \frac { 2 \pi } { \frac { 1 } { 2 } } = 4 \pi$ of $y = \cos \left( \frac { 1 } { 2 } x + \frac { \pi } { 3 } \right)$, not true; For D, the period $y = \cos \left( 2 x - \frac { \pi } { 6 } \right)$ of $T = \frac { 2 \pi } { 2 } = \pi$. When $- \frac { \pi } { 6 } \leq x \leq \frac { \pi } { 6 }$ is $- \frac { \pi } { 2 } \leq 2 x - \frac { \pi } { 6 } \leq \frac { \pi } { 6 }$, so the function increases and then decreases on the interval $\left[ - \frac { \pi } { 6 } , \frac { \pi } { 6 } \right]$, which is not true;

Question 13: 15. Known angle $\theta ^ { \text {的终边过点 } P ( 3,2 ) \text { ,则 } \frac { \sin 2 \theta } { 2 \cos ^...

15. Known angle $\theta ^ { \text {的终边过点 } P ( 3,2 ) \text { ,则 } \frac { \sin 2 \theta } { 2 \cos ^ { 2 } \theta - 3 \sin ^ { 2 } \theta } = }$

  • A. A. 2
  • B. B. 2
  • C. C. $\frac { 5 } { 2 }$
  • D. D. 3

Answer: B

Solution: Since the terminal side of angle $\theta$ passes through the point $P ( 3,2 )$, $\tan \theta = \frac { 2 } { 3 }$, so $\frac { \sin 2 \theta } { 2 \cos ^ { 2 } \theta - 3 \sin ^ { 2 } \theta } = \frac { 2 \sin \theta \cos \theta } { 2 \cos ^ { 2 } \theta - 3 \sin ^ { 2 } \theta } = \frac { 2 \tan \theta } { 2 - 3 \tan ^ { 2 } \theta } = \frac { 2 \times \frac { 2 } { 3 } } { 2 - 3 \times \left( \frac { 2 } { 3 } \right) ^ { 2 } } = 2$.

Question 14: 16. The function $f ( x ) _ { \text {满足 } } f ( x ) - f ( \pi - x ) = 0$ is known to be monotonicall...

16. The function $f ( x ) _ { \text {满足 } } f ( x ) - f ( \pi - x ) = 0$ is known to be monotonically decreasing on the interval $\left( \frac { \pi } { 4 } , \frac { \pi } { 2 } \right)$, and $f ( x ) _ { \text {的解析式 } }$ is known to be monotonically decreasing on the interval $f ( x ) _ { \text {的解析式 } }$. may be

  • A. A. $f ( x ) = \sin x$
  • B. B. $f ( x ) = \sin 2 x$
  • C. C. $f ( x ) = \cos x$
  • D. D. $f ( x ) = \cos 2 x$

Answer: D

Solution: By the meaning of $f ( x ) = f ( \pi - x )$, the line $x = \frac { \pi } { 2 }$ is the symmetry axis of the graph of $f ( x )$, so we can exclude options B and C. Because $f ( x )$ monotonically decreases in the interval [[INLINE_FORMULA_4 ]] is monotonically decreasing on the interval $\left( \frac { \pi } { 4 } , \frac { \pi } { 2 } \right)$, we can exclude A.

Question 15: 17. It is known that the function $f ( x ) = \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } + \sin ...

17. It is known that the function $f ( x ) = \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } + \sin x + a$, if $f ( \ln m ) = 1 , f \left( \ln \frac { 1 } { m } \right) = 3$, then $a =$

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

Answer: B

Solution: Because of the function $f ( x ) = \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } + \sin x + a$ , we have $f ( - x ) + f ( x ) = 2 a$ . Since $f \left( \ln \frac { 1 } { m } \right) = f ( - \ln m ) = 3 , f ( \ln m ) = 1$, it follows that $f ( \ln m ) + f \left( \ln \frac { 1 } { m } \right) = 1 + 3 = 4 = 2 a$. So $a = 2$ .

Question 16: 18. If $\tan \alpha = - 2$ is known, then the value of $\frac { \sin ( \pi + 2 \alpha ) } { \cos ^ {...

18. If $\tan \alpha = - 2$ is known, then the value of $\frac { \sin ( \pi + 2 \alpha ) } { \cos ^ { 2 } \alpha }$ is

  • A. A. 4
  • B. B. 2
  • C. C. - 2
  • D. D. - 4

Answer: A

Solution: $\because \tan \alpha = - 2$ $\therefore \frac { \sin ( \pi + 2 \alpha ) } { \cos ^ { 2 } \alpha } = \frac { - \sin 2 \alpha } { \cos ^ { 2 } \alpha } = \frac { - 2 \sin \alpha \cos \alpha } { \cos ^ { 2 } \alpha } = - 2 \tan \alpha = 4$ .

Question 17: 19. If $2 ^ { m } = 3 ^ { n } = 36$ is known, then $\frac { 1 } { m } + \frac { 1 } { n } =$ is know...

19. If $2 ^ { m } = 3 ^ { n } = 36$ is known, then $\frac { 1 } { m } + \frac { 1 } { n } =$ is known.

  • A. A. 1
  • B. B. 2
  • C. C. $\frac { 1 } { 6 }$
  • D. D. $\frac { 1 } { 2 }$

Answer: D

Solution: From the question, since $2 ^ { m } = 3 ^ { n } = 36$ , then $m = \log _ { 2 } 36 , n = \log _ { 3 } 36$ , the so $\frac { 1 } { m } + \frac { 1 } { n } = \log _ { 36 } 2 + \log _ { 36 } 3 = \log _ { 36 } ( 2 \times 3 ) = \log _ { 36 } 6 = \frac { \log _ { 6 } 6 } { \log _ { 6 } 36 } = \frac { 1 } { 2 }$ , .

Question 18: 20. The function $f ( x ) = \sin x \cos x + \cos ^ { 2 } x$ has a maximum value of

20. The function $f ( x ) = \sin x \cos x + \cos ^ { 2 } x$ has a maximum value of

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

Answer: A

Solution: $f ( x ) = \sin x \cos x + \cos ^ { 2 } x = \frac { 1 } { 2 } \sin 2 x + \frac { 1 + \cos 2 x } { 2 } = \frac { 1 } { 2 } \sin 2 x + \frac { 1 } { 2 } \cos 2 x + \frac { 1 } { 2 }$ $= \frac { \sqrt { 2 } } { 2 } \left( \frac { \sqrt { 2 } } { 2 } \sin 2 x + \frac { \sqrt { 2 } } { 2 } \cos 2 x \right) + \frac { 1 } { 2 } = \frac { \sqrt { 2 } } { 2 } \left( \sin 2 x \cos \frac { \pi } { 4 } + \frac { \sqrt { 2 } } { 2 } \cos 2 x \sin \frac { \pi } { 4 } \right) + \frac { 1 } { 2 }$ $= \frac { \sqrt { 2 } } { 2 } \sin \left( 2 x + \frac { \pi } { 4 } \right) + \frac { 1 } { 2 }$. Because $x \in \mathrm { R }$, so $2 x + \frac { \pi } { 4 } \in \mathrm { R }$ When $2 x + \frac { \pi } { 4 } = \frac { \pi } { 2 } + 2 k \pi , k \in \mathrm { Z }$ is $f ( x ) _ { \text {取得最大值,即 } } f ( x ) = \frac { \sqrt { 2 } + 1 } { 2 }$.

Question 19: 21. function (math.) $$ y = \log _ { a } ( 4 x - 1 ) \quad , \quad ( a > 0 \text { 且 } a \neq 1 ) \...

21. function (math.) $$ y = \log _ { a } ( 4 x - 1 ) \quad , \quad ( a > 0 \text { 且 } a \neq 1 ) \quad \text { 图象必过的定点是 } $$

  • A. A. $\left( \frac { 1 } { 4 } , 1 \right)$
  • B. B. $( 1,0 )$
  • C. C. $( 0,1 )$
  • D. D. $\left( \frac { 1 } { 2 } , 0 \right)$

Answer: D

Solution: Test analysis: Let $4 x - 1 = 1 , x = \frac { 1 } { 2 }$, then $y = 0$, so the graph of the function passes through the fixed point $\left( \frac { 1 } { 2 } , 0 \right)$, so option D is correct. The graph of a logarithmic function.

Question 20: 22. Let ${ } ^ { a = \log _ { 3 } \pi } , b = \ln 2 , c = \cos 2$, then

22. Let ${ } ^ { a = \log _ { 3 } \pi } , b = \ln 2 , c = \cos 2$, then

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

Answer: C

Solution: From the question, according to the monotonicity of the logarithmic function, we can get $a = \log _ { 3 } \pi > \log _ { 3 } 3 = 1$ $b = \ln 2 > \ln 1 = 0 , b = \ln 2 < \ln e = 1$, which is $b \in ( 0,1 )$. And by $c = \cos 2 < 0$, so $a > b > c$. [点睛]本题主要考查了对数函数的单调性的应用,以及余弦函数的性质的应用,其中解解 According to the monotonicity of the logarithmic function and the nature of the cosine function, the range of values of $a , b , c$ is the key to the answer, which focuses on the reasoning and arithmetic ability, and belongs to the basic questions.

Question 21: 23. The following functions are monotonically increasing on $\mathbf { R }$ ( )

23. The following functions are monotonically increasing on $\mathbf { R }$ ( )

  • A. A. $f ( x ) = \tan x$
  • B. B. $f ( x ) = \left( \frac { 1 } { 2 } \right) ^ { x }$
  • C. C. $f ( x ) = x ^ { \frac { 1 } { 2 } }$
  • D. D. $f ( x ) = \left\{ \begin{array} { l } x - 1 , x \leq 1 \\ \ln x , x > 1 \end{array} \right.$

Answer: D

Solution: For A, $f ( x ) = \tan x$ is monotonically increasing on $\left( - \frac { \pi } { 2 } + k \pi , \frac { \pi } { 2 } + k \pi \right) , k \in \mathbf { Z }$, so A is wrong; For B, $f ( x ) = \left( \frac { 1 } { 2 } \right) ^ { x }$ is monotonically decreasing on $\mathbf { R }$, so B is wrong; For C, $f ( x ) = x ^ { \frac { 1 } { 2 } }$ is defined by $[ 0 , + \infty )$ and monotonically increases on $[ 0 , + \infty )$, so C is wrong; For D, $f ( x ) = \left\{ \begin{array} { l } x - 1 , x \leq 1 \\ \ln x , x > 1 , \end{array} \right.$ is monotonically increasing on $f ( x ) = \left\{ \begin{array} { l } x - 1 , x \leq 1 \\ \ln x , x > 1 , \end{array} \right.$. The function $y = x - 1$ is monotonically increasing when $x \leq 1$ and $y = x - 1 \leq 0$; When $x > 1$, $y = \ln x$ is monotonically increasing and ${ } ^ { y = \ln x > 0 }$; So the function $f ( x )$ is monotonically increasing on $\mathbf { R }$, so D is correct.

Question 22: 24. In the isoperimetric series $\left\{ a _ { n } \right\}$ with all positive numbers in each term ...

24. In the isoperimetric series $\left\{ a _ { n } \right\}$ with all positive numbers in each term and common ratio $q \neq 1$, let $P = \frac { 1 } { 2 } \left( \log _ { 0.5 } a _ { 5 } + \log _ { 0.5 } a _ { 7 } \right)$, $Q = \log _ { 0.5 } \frac { a _ { 3 } + a _ { 9 } } { 2 }$, then the relationship between ${ } _ { P }$ and are related in size by ( )

  • A. A. $P \geq Q$
  • B. B. $P < Q$
  • C. C. $P \leq Q$
  • D. D. $P > Q$

Answer: D

Solution: $P = \frac { 1 } { 2 } \left( \log _ { 0.5 } a _ { 5 } + \log _ { 0.5 } a _ { 7 } \right) = \frac { 1 } { 2 } \log _ { 0.5 } a _ { 5 } a _ { 7 } = \log _ { 0.5 } \sqrt { a _ { 5 } a _ { 7 } } = \log _ { 0.5 } a _ { 6 }$, $a _ { 3 } = a _ { 9 }$ $Q = \log _ { 0.5 } \frac { a _ { 3 } + a _ { 9 } } { 2 } \leq \log _ { 0.5 } \sqrt { a _ { 3 } a _ { 9 } } = \log _ { 0.5 } a _ { 6 }$ (taking the equal sign if and only if $a _ { 3 } = a _ { 9 }$), $a _ { 3 } = a _ { 9 }$ $\because \left\{ a _ { n } \right\}$ the terms are all positive and $q \neq 1 , \therefore a _ { 3 } \neq a _ { 9 } , \therefore Q < \log _ { 0.5 } a _ { 6 }$ $\therefore P > Q$

Question 23: 25. Regardless of the value of ${ } ^ { a }$, the graph of the function $f ( x ) = \log _ { a } x - ...

25. Regardless of the value of ${ } ^ { a }$, the graph of the function $f ( x ) = \log _ { a } x - 2$ must pass through the point

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

Answer: C

Solution: Test analysis: When $x = 1$, the value of the function is constant - 2, so the fixed point is $( 1 , - 2 )$. Points : Exponential function graphs over the fixed point.

Question 24: 26. It is known that the function $f ( x )$ defined on $\mathbf { R }$ satisfies $f ( - x ) + f ( x ...

26. It is known that the function $f ( x )$ defined on $\mathbf { R }$ satisfies $f ( - x ) + f ( x ) = 0$ and when $x \leq 0$ is $f ( x ) = \frac { - 2 } { 2 ^ { x } } + 2$ then $f ( 1 ) =$ is $f ( 1 ) =$. FORMULA_5]]

  • A. A. 2
  • B. B. 4
  • C. C. - 2
  • D. D. - 4

Answer: A

Solution: From the question: the function $f ( x )$ is odd when $x \leq 0$, $f ( x ) = \frac { - 2 } { 2 ^ { x } } + 2$, so $f ( 1 ) = - f ( - 1 ) = - \left[ \left( \frac { - 2 } { 2 ^ { - 1 } } \right) + 2 \right] = 2$.

Question 25: If $a > 0 , b > 0 , a \neq 1 , b \neq 1 , a b \neq 1 , \log _ { a } m = 4 , \log _ { b } m = 6$ is k...

If $a > 0 , b > 0 , a \neq 1 , b \neq 1 , a b \neq 1 , \log _ { a } m = 4 , \log _ { b } m = 6$ is known, then $\log _ { a b } m =$ is known.

  • A. A. $\frac { 1 } { 5 }$
  • B. B. $\frac { 1 } { 24 }$
  • C. C. $\frac { 5 } { 12 }$
  • D. D. $\frac { 12 } { 5 }$

Answer: D

Solution: $\because \log _ { a } m = 4 , \log _ { b } m = 6 , \therefore \log _ { m } a = \frac { 1 } { \log _ { a } m } = \frac { 1 } { 4 } , \log _ { m } b = \frac { 1 } { \log _ { b } m } = \frac { 1 } { 6 }$ $\therefore \log _ { m } a + \log _ { m } b = \frac { 1 } { 4 } + \frac { 1 } { 6 } = \frac { 5 } { 12 } = \log _ { m } a b , \quad \therefore \log _ { a b } m = \frac { 12 } { 5 }$ .

Question 26: If it is known that $\sin \varphi = - \frac { 3 } { 5 }$ and $| \varphi | < \frac { \pi } { 2 }$ , t...

If it is known that $\sin \varphi = - \frac { 3 } { 5 }$ and $| \varphi | < \frac { \pi } { 2 }$ , then $\tan \varphi =$

  • A. A. $- \frac { 4 } { 3 }$
  • B. B. $\frac { 4 } { 3 }$
  • C. C. $- \frac { 3 } { 4 }$
  • D. D. $\frac { 3 } { 4 }$

Answer: C

Solution: Since $| \varphi | < \frac { \pi } { 2 }$, then $\cos \varphi > 0$, and $\sin \varphi = - \frac { 3 } { 5 }$. So $\cos \varphi = \sqrt { 1 - \sin ^ { 2 } \varphi } = \frac { 4 } { 5 }$, then $\tan \varphi = \frac { \sin \varphi } { \cos \varphi } = - \frac { 3 } { 4 }$.

Question 27: 30. If $\sqrt [ n ] { a ^ { n } } + ( \sqrt [ n + 1 ] { a } ) ^ { n + 1 } = 0 , a \neq 0$ and $n \in...

30. If $\sqrt [ n ] { a ^ { n } } + ( \sqrt [ n + 1 ] { a } ) ^ { n + 1 } = 0 , a \neq 0$ and $n \in \mathbf { N } ^ { * } , n \geq 2$, then

  • A. A. $a > 0$ and $n$ is even
  • B. B. $a < 0$ and $n$ is even
  • C. C. $a > 0$ and $n$ is an odd number
  • D. D. $a < 0$ and $n$ is an odd number

Answer: B

Solution:

Question 28: 31. If $M = \{ x \mid - 2 \leq x \leq 2 \} , N = \left\{ x \mid y = \log _ { 2 } ( x - 1 ) \right\}$...

31. If $M = \{ x \mid - 2 \leq x \leq 2 \} , N = \left\{ x \mid y = \log _ { 2 } ( x - 1 ) \right\}$, then $M \cap N =$

  • A. A. $\{ x \mid - 2 \leq x < 0 \}$
  • B. B. $\{ x \mid - 1 < x < 0 \}$
  • C. C. $\{ x \mid - 2 < x < 0 \}$
  • D. D. $\{ x \mid 1 < x \leq 2 \}$

Answer: D

Solution: SOLUTION: Since $N = \left\{ x \mid y = \log _ { 2 } ( x - 1 ) \right\}$, $N = \{ x \mid x > 1 \} , ~ \because ^ { M } = \{ x \mid - 2 \leq x \leq 2 \}$, $\therefore M \cap N = \{ x \mid 1 < x \leq 2 \}$, the

Question 29: 32. If the solution set of the inequality ${ } _ { x }$ with respect to $\frac { \cos x - 2 } { x ^ ...

32. If the solution set of the inequality ${ } _ { x }$ with respect to $\frac { \cos x - 2 } { x ^ { 2 } - m x - n } > 0$ is $( - 2,3 )$, then $m n =$

  • A. A. 5
  • B. B. - 5
  • C. C. 6
  • D. D. - 6

Answer: C

Solution:

Question 30: 33. If the function $f ( x ) = 2 \sin ^ { 2 } \left( x - \frac { \pi } { 4 } \right) + \sqrt { 3 } \...

33. If the function $f ( x ) = 2 \sin ^ { 2 } \left( x - \frac { \pi } { 4 } \right) + \sqrt { 3 } \sin \left( 2 x - \frac { \pi } { 6 } \right) - 1$ , then the following conclusion is incorrect ( )

  • A. A. The least positive period of the function $f ( x )$ is $\frac { \pi } { 2 }$
  • B. B. The function $f ( x )$ is monotonically increasing on the interval $\left[ - \frac { \pi } { 12 } , \frac { 5 \pi } { 12 } \right]$.
  • C. C. Function $f ( x )$ image about ${ } ^ { x = - \frac { \pi } { 12 } }$ symmetric
  • D. D. The graph of the function $f ( x )$ is symmetric about the point $\left( \frac { 2 \pi } { 3 } , 0 \right)$.

Answer: A

Solution: According to the dihedral and induction formulas, $2 \sin ^ { 2 } \left( x - \frac { \pi } { 4 } \right) - 1 = \cos \left( 2 x - \frac { \pi } { 2 } \right) = \sin 2 x$, thus $f ( x ) = \sqrt { 3 } \sin \left( 2 x - \frac { \pi } { 6 } \right) - \sin 2 x = \frac { 1 } { 2 } \sin 2 x - \frac { \sqrt { 3 } } { 2 } \cos 2 x = \sin \left( 2 x - \frac { \pi } { 3 } \right)$. A option, according to the trigonometric period formula, $T = \frac { 2 \pi } { 2 } = \pi$ , A option is wrong; B option, let $2 x - \frac { \pi } { 3 } \in \left[ 2 k \pi - \frac { \pi } { 2 } , 2 k \pi + \frac { \pi } { 2 } \right] , k \in \mathbf { Z } \quad$, when solving $x \in \left[ k \pi - \frac { \pi } { 12 } , k \pi + \frac { 5 \pi } { 12 } \right] , k \in \mathbf { Z } \quad , k = 0$ can be INLINE_FORMULA_5]] is monotonically increasing on the interval $\left[ - \frac { \pi } { 12 } , \frac { 5 \pi } { 12 } \right]$, option B is correct; In option C, the graph of $2 x - \frac { \pi } { 3 } = k \pi + \frac { \pi } { 2 } , k \in \mathbf { Z }$ is symmetric with respect to $x = - \frac { \pi } { 12 }$ when $2 x - \frac { \pi } { 3 } = k \pi + \frac { \pi } { 2 } , k \in \mathbf { Z }$ is solved by $x = \frac { k \pi } { 2 } + \frac { 5 \pi } { 12 } , k \in \mathbf { Z } , ~ k = - 1$, and option C is correct; In option D, $2 x - \frac { \pi } { 3 } = k \pi , k \in \mathbf { Z }$ is the horizontal coordinate of the center of symmetry, and $x = \frac { k \pi } { 2 } + \frac { \pi } { 6 } , k \in \mathbf { Z }$ is the horizontal coordinate of the center of symmetry, and $x = \frac { k \pi } { 2 } + \frac { \pi } { 6 } , k \in \mathbf { Z }$ is the horizontal coordinate of the center of symmetry. D, $x = \frac { k \pi } { 2 } + \frac { \pi } { 6 } , k \in \mathbf { Z }$ is the horizontal coordinate of the center of symmetry, so that $x = \frac { k \pi } { 2 } + \frac { \pi } { 6 } = \frac { 2 \pi } { 3 }$ solves $k = 1$ and the image of $f ( x )$ is symmetric about the point $\left( \frac { 2 \pi } { 3 } , 0 \right)$, and option D is correct.

Question 31: 34. It is known that the function $f ( x )$ defined on $R$ satisfies: for any $x \in R$, there is an...

34. It is known that the function $f ( x )$ defined on $R$ satisfies: for any $x \in R$, there is an $f ( x + 1 ) = f ( 1 - x )$, and when $x \in ( - \infty , 1 )$, $( x - 1 ) \cdot f ^ { \prime } ( x ) > 0$ (where $( x - 1 ) \cdot f ^ { \prime } ( x ) > 0$) is the same as $x \in ( - \infty , 1 )$ when $( x - 1 ) \cdot f ^ { \prime } ( x ) > 0$ (where $f ^ { \prime } ( x )$ is the derivative of $f ( x )$). Let $a = f \left( \log _ { 2 } 3 \right)$, $b = f \left( \log _ { 3 } 2 \right) , c = f \left( 2 ^ { 1.5 } \right)$, then the magnitude relation of $a , b , c$ is ( )

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

Answer: C

Solution: From $f ( x + 1 ) = f ( 1 - x )$, the graph of $^ { y = f ( x ) }$ is symmetric about the line $^ { x = 1 }$, and $^ { x < 1 }$ is in $^ { x < 1 }$. $( x - 1 ) f ^ { \prime } ( x ) > 0$, so $f ^ { \prime } ( x ) < 0$, i.e., $f ^ { ( x ) }$ is monotonically decreasing on $^ { ( - \infty , 1 ) }$, so it is monotonically increasing on ${ } ^ { ( 1 , + \infty ) }$. $1 < \log _ { 2 } 3 < 2,2 ^ { 1.5 } > 2 , \log _ { 3 } 2 < 1 , f \left( \log _ { 3 } 2 \right) = f \left( 2 - \log _ { 3 } 2 \right) = f \left( \log _ { 3 } \frac { 9 } { 2 } \right)$, $1 < \log _ { 2 } 3 < 2,2 ^ { 1.5 } > 2 , \log _ { 3 } 2 < 1 , f \left( \log _ { 3 } 2 \right) = f \left( 2 - \log _ { 3 } 2 \right) = f \left( \log _ { 3 } \frac { 9 } { 2 } \right)$, $1 < \log _ { 2 } 3 < 2,2 ^ { 1.5 } > 2 , \log _ { 3 } 2 < 1 , f \left( \log _ { 3 } 2 \right) = f \left( 2 - \log _ { 3 } 2 \right) = f \left( \log _ { 3 } \frac { 9 } { 2 } \right)$ $\log _ { 2 } 3 > \log _ { 2 } 2 \sqrt { 2 } = \frac { 3 } { 2 } , 1 < \log _ { 3 } \frac { 9 } { 2 } < \log _ { 3 } 3 \sqrt { 3 } = \frac { 3 } { 2 }$, so $1 < \log _ { 3 } \frac { 9 } { 2 } < \log _ { 2 } 3 < 2 ^ { 1.5 }$. so $b < a < c$.

Question 32: 35. The terms of the isometric series $\left\{ a _ { n } \right\}$ are all positive and $\begin{gath...

35. The terms of the isometric series $\left\{ a _ { n } \right\}$ are all positive and $\begin{gathered} a _ { 5 } a _ { 6 } = 4 \text { ,则 } \log _ { 2 } a _ { 1 } + \log _ { 2 } a _ { 2 } + \cdots + \log _ { 2 } a _ { 10 } = ( ) \\ ( ) \end{gathered}$ is an isometric series of $\begin{gathered} a _ { 5 } a _ { 6 } = 4 \text { ,则 } \log _ { 2 } a _ { 1 } + \log _ { 2 } a _ { 2 } + \cdots + \log _ { 2 } a _ { 10 } = ( ) \\ ( ) \end{gathered}$.

  • A. A. 4
  • B. B. 6
  • C. C. 8
  • D. D. 10

Answer: D

Solution: From the algorithm of logarithmic addition and the properties of the isoperimetric series: $\log _ { 2 } a _ { 1 } + \log _ { 2 } a _ { 2 } + \cdots + \log _ { 2 } a _ { 10 } = \log _ { 2 } \left( a _ { 1 } a _ { 2 } \cdots a _ { 10 } \right)$ $= \log _ { 2 } \left( a _ { 5 } a _ { 6 } \right) ^ { 5 } = \log _ { 2 } 4 ^ { 5 } = \log _ { 2 } 2 ^ { 10 } = 10$.

Question 33: 36. In the straight triangular prism $A B C - A _ { 1 } B _ { 1 } C _ { 1 }$, $V A B C$ is a triangl...

36. In the straight triangular prism $A B C - A _ { 1 } B _ { 1 } C _ { 1 }$, $V A B C$ is a triangle with side 2, $A A _ { 1 } = 3 , N$ is the midpoint on the prism $A _ { 1 } B _ { 1 }$, $M$ is the moving point on the prism ${ } ^ { C C _ { 1 } }$, and $N$ is the plane $N$, and $N$ is the moving point on the plane $A B M$. ] is the moving point on the prism ${ } ^ { C C _ { 1 } }$, and $N$ crosses $N$ to make a perpendicular segment of the plane $A B M$, with the foot of the perpendicular segment at the point $O$, and the point $M$ is the center of the plane when the point $M$ is from the point [[$M$]. INLINE_FORMULA_10]] moves to point ${ } ^ { C _ { 1 } }$, the length of the trajectory of point $O$ is () ![](/images/questions/elementary-function/image-001.jpg)

  • A. A. $\frac { \pi } { 2 }$
  • B. B. $\pi$
  • C. C. $\frac { 3 \pi } { 2 }$
  • D. D. $\frac { 2 \sqrt { 3 } \pi } { 3 }$

Answer: B

Solution: Take $A B$ midpoint $P$ and connect $P C , C _ { 1 } N$ as shown. ![](/images/questions/elementary-function/image-002.jpg) Because $P C \perp A B , P N \perp A B$ and $P N \cap P C = P$ , so $A B \perp$ planes $P C C _ { 1 } N , A B ^ { \subset }$ planes $A B M$ , so planes $A B M \perp$ planes , plane $A B M \cap$ plane $P C C _ { 1 } N = P M$, over $N$ as $N O \perp P M , N O \subset _ { \text {plane } } ^ { P C C _ { 1 } N }$, so $N O \perp$ plane $N O \perp$ plane [[INLINE_FORMULA_14 [INLINE_FORMULA_15]]. When the point $M$ moves from the point $C$ to the point $C _ { 1 }$, the point $O$ is the circle $P N$ with diameter $P N$. INLINE_FORMULA_21]] (part), as shown in the figure. ![](/images/questions/elementary-function/image-003.jpg) When $M$ moves to the point $C _ { 1 }$, $O$ points to the highest point, when $P C = \sqrt { 3 } , C C _ { 1 } = 3 , \angle C P C _ { 1 } = \frac { \pi } { 3 }$ so $\angle O P Q = \frac { \pi } { 6 }$ and thus $\angle O Q P = \frac { 2 \pi } { 3 }$. So the arc length $I = \frac { 2 \pi } { 3 } \cdot \frac { 3 } { 2 } = \pi$, i.e., the length of the trajectory of the point $O$ is $\pi$.

Question 34: 37. The minimum value of the function $f ( x ) = \sin \left( 2 x - \frac { \pi } { 4 } \right)$ on t...

37. The minimum value of the function $f ( x ) = \sin \left( 2 x - \frac { \pi } { 4 } \right)$ on the interval $\left[ 0 , \frac { \pi } { 2 } \right]$ is

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

Answer: B

Solution: TEST ANALYSIS: $x \in \left[ 0 , \frac { \pi } { 2 } \right] \therefore 2 x \in [ 0 , \pi ] , 2 x - \frac { \pi } { 4 } \in \left[ - \frac { \pi } { 4 } , \frac { 3 \pi } { 4 } \right]$, so the minimum value is $$ \sin \left( - \frac { \pi } { 4 } \right) = - \frac { \sqrt { 2 } } { 2 } $$ ## 考点 :三角函数最值

Question 35: 38. The graph of the function $f ( x ) = \sin ( \omega x + \phi )$ (where $| \varphi | < \frac { \pi...

38. The graph of the function $f ( x ) = \sin ( \omega x + \phi )$ (where $| \varphi | < \frac { \pi } { 2 }$) is shown in the figure. In order to obtain the graph of $y = \sin \omega x$, it is only necessary to put all the points on the graph of $y = f ( x )$ ( ) ![](/images/questions/elementary-function/image-004.jpg)

  • A. A. Shift right $\frac { \pi } { 12 }$ units to length
  • B. B. Translate $\frac { \pi } { 12 }$ to the left by one unit.
  • C. C. Translate $\frac { \pi } { 6 }$ to the left by one unit.
  • D. D. Shift right $\frac { \pi } { 6 }$ units to length

Answer: D

Solution: Analysis: Find $\omega$ from the period, and then find $\varnothing$ from the five-point graph, so as to get the function $f ( x ) = \sin 2 ( x + \left. \frac { \pi } { 6 } \right)$, so the graph of $y = f ( x )$ can be shifted to the right $\frac { \pi } { 6 }$ unit length to get the graph of $y = \sin \omega x$, and then the conclusion can be drawn. FORMULA_4]] to the right to get the image of $y = \sin \omega x$, which leads to the conclusion. Explanation: The $\frac { 1 } { 4 } \times \frac { 2 \pi } { \omega } = \frac { 7 } { 12 } \pi - \frac { \pi } { 3 } = \frac { \pi } { 4 } \therefore \omega = 2$ can be obtained from the question. Then by the five-point method of graphing can be $2 \times \frac { \pi } { 3 } + \varnothing = \pi , \therefore \varnothing = \frac { \pi } { 3 }$, so the function $f ( x ) = \sin ( \omega x + \phi ) = \sin \left( 2 x + \frac { \pi } { 3 } \right) = \sin 2 \left( x + \frac { \pi } { 6 } \right)$. Therefore, the graph of $y = f ( x )$ is shifted to the right by $\frac { \pi } { 6 }$ units to obtain the graph of $y = \sin \omega x$, and the graph of $f ( x ) = \sin ( \omega x + \phi ) = \sin \left( 2 x + \frac { \pi } { 3 } \right) = \sin 2 \left( x + \frac { \pi } { 6 } \right)$ is shifted to the right by $y = f ( x )$.

Question 36: 39. The population model of Goose Town, a Netroots city, is approximated as $P = 320014 \mathrm { e ...

39. The population model of Goose Town, a Netroots city, is approximated as $P = 320014 \mathrm { e } ^ { 0.015 t }$, where $t = 0$ denotes the population in 2015, then the year when the population of Goose Town reaches 600012 will be approximately ( ) (refer to the data: $\ln 2 \approx 0.693 , \ln 3 \approx 1.099$, [[]], [[]]). INLINE_FORMULA_3]])

  • A. A. 2037
  • B. B. 2047
  • C. C. 2057
  • D. D. 2067

Answer: C

Solution: $P = 320014 \mathrm { e } ^ { 0.015 t } = 600012$, i.e. $\mathrm { e } ^ { 0.015 t } = \frac { 600012 } { 320014 } \approx \frac { 15 } { 8 } , ~ 0.015 t \approx \ln \frac { 15 } { 8 }$, $t \approx \frac { \ln \frac { 15 } { 8 } } { 0.015 } = \frac { \ln 3 + \ln 5 - 3 \ln 2 } { 0.015 } \approx 42 , \quad 2015 + 42 = 2057$, $t \approx \frac { \ln \frac { 15 } { 8 } } { 0.015 } = \frac { \ln 3 + \ln 5 - 3 \ln 2 } { 0.015 } \approx 42 , \quad 2015 + 42 = 2057$, $t \approx \frac { \ln \frac { 15 } { 8 } } { 0.015 } = \frac { \ln 3 + \ln 5 - 3 \ln 2 } { 0.015 } \approx 42 , \quad 2015 + 42 = 2057$

Question 37: 40. It is known that the function $f ( x ) = \left\{ \begin{array} { l } x ^ { 2 } - 6 a x + 2 , x <...

40. It is known that the function $f ( x ) = \left\{ \begin{array} { l } x ^ { 2 } - 6 a x + 2 , x < 1 \\ a ^ { x } - 2 x , x \geq 1 \end{array} ( a > 0 , a \neq 1 ) \right.$, if the function $f ( x )$ satisfies that $x _ { 1 } , x _ { 2 }$ holds for any $x _ { 1 } , x _ { 2 }$ when $x _ { 1 } \neq x _ { 2 }$ is $\frac { f \left( x _ { 1 } \right) - f \left( x _ { 2 } \right) } { x _ { 1 } - x _ { 2 } } < - 2$, the real number The range of values of $\boldsymbol { a }$ is ( ) High School Mathematics Assignment, October 29, 2025

  • A. A. $\left( 0 , \frac { 2 } { 3 } \right]$
  • B. B. $\left[ \frac { 2 } { 3 } , 1 \right)$
  • C. C. $\left[ \frac { 2 } { 3 } , \frac { 5 } { 7 } \right]$
  • D. D. $\left[ \frac { 5 } { 7 } , 1 \right)$

Answer: C

Solution:
Back to Topics

Elementary Function

基本初等函数

37 Practice Questions

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

Topic Overview

Basic elementary functions are the foundation of mathematics, mainly including power functions, exponential functions, logarithmic functions, trigonometric functions and inverse trigonometric functions. In the CSCA exam, this part of the content often appears in the form of function image transformation, property analysis, equation solving, etc. Candidates are required to be proficient in definitions, images and basic properties, and be able to apply them flexibly to problem solving.

Questions:37

Key Points

  • 1Analysis of function definition domain, value domain and basic properties
  • 2Recognition, drawing and transformation of function images (translation, telescoping)
  • 3Solve equations or inequalities using properties of functions
  • 4Trigonometric Function Induction Formula and Special Angle Valuation

Study Tips

It is recommended to memorize the characteristics of various types of functions by comparing and contrasting them by plotting the images of the functions and focusing on practicing the integrated application of image transformations and properties.

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