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CSCA Math December 2025 Past Paper

All 48 questions from CSCA Math December 2025 Past Paper, each with the correct answer and a worked solution — free to read.

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CSCA Math December 2025 Past Paper · #1· Topic: 判断两个集合的包含关系
If A={1,2}A=\{1,2\} and B={1,2,3}B=\{1,2,3\}, then ( ).
  • A．B⊆AB \subseteq A
  • B．A∈BA \in B
  • Correct answer: C．A⊆BA \subseteq B
  • D．A=BA=B
Show solution(Answer: C)
Answer: C All elements of AA are contained in BB, so A⊆BA\subseteq B. Choose C.
CSCA Math December 2025 Past Paper · #2· Topic: 交集的概念及运算
If A=(−1,3)A=(-1,3) and B=(2,4)B=(2,4), then A∩B=A\cap B= ( ).
  • A．[2,3)[2,3)
  • B．(2,3](2,3]
  • C．{2,3}\{2,3\}
  • Correct answer: D．(2,3)(2,3)
Show solution(Answer: D)
Answer: D The common part of (−1,3)(-1,3) and (2,4)(2,4) is (2,3)(2,3); neither endpoint is included. Choose D.
CSCA Math December 2025 Past Paper · #3· Topic: 解不含参数的一元二次不等式
The solution set of x2−5x+4<0x^2-5x+4<0 is ( ).
  • Correct answer: A．(1,4)(1,4)
  • B．(−∞,−4)∪(−1,+∞)(-\infty,-4)\cup(-1,+\infty)
  • C．(−4,−1)(-4,-1)
  • D．(−∞,1)∪(4,+∞)(-\infty,1)\cup(4,+\infty)
Show solution(Answer: A)
Answer: A x2−5x+4=(x−1)(x−4)<0x^2-5x+4=(x-1)(x-4)<0, so 1<x<41<x<4. Choose A.
CSCA Math December 2025 Past Paper · #4· Topic: 具体函数的定义域
The domain of y=1x+1y=\frac{1}{\sqrt{x+1}} is ( ).
  • A．(0,+∞)(0,+\infty)
  • B．[−1,+∞)[-1,+\infty)
  • Correct answer: C．(−1,+∞)(-1,+\infty)
  • D．(−∞,−1)(-\infty,-1)
Show solution(Answer: C)
Answer: C The denominator cannot be zero and the radicand must be positive, so x+1>0x+1>0. Choose C.
CSCA Math December 2025 Past Paper · #5· Topic: 等差数列通项公式的基本量计算
If {an}\{a_n\} is an arithmetic sequence with a1=2a_1=2 and common difference d=3d=3, then a100=a_{100}= ( ).
  • Correct answer: A．299
  • B．305
  • C．302
  • D．298
Show solution(Answer: A)
Answer: A a100=a1+99d=2+99×3=299a_{100}=a_1+99d=2+99\times3=299. Choose A.
CSCA Math December 2025 Past Paper · #6· Topic: 判断点所在的象限
The point (−1,3−π)(-1,3-\pi) lies in which quadrant?
  • A．Quadrant I
  • B．Quadrant II
  • C．Quadrant IV
  • Correct answer: D．Quadrant III
Show solution(Answer: D)
Answer: D Since 3−π<03-\pi<0, both coordinates are negative. The point is in Quadrant III. Choose D.
CSCA Math December 2025 Past Paper · #7· Topic: 特殊角的三角函数值
(sin⁡π4+cos⁡π4)tan⁡π4=\left(\sin\frac{\pi}{4}+\cos\frac{\pi}{4}\right)\tan\frac{\pi}{4}= ( ).
  • A．222\sqrt{2}
  • B．1
  • Correct answer: C．2\sqrt{2}
  • D．2
Show solution(Answer: C)
Answer: C sin⁡π4=cos⁡π4=22\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2} and tan⁡π4=1\tan\frac{\pi}{4}=1, so the value is 2\sqrt{2}. Choose C.
CSCA Math December 2025 Past Paper · #8· Topic: 函数奇偶性的定义与判断
Which of the following functions is odd?
  • A．y=cos⁡xy=\cos x
  • Correct answer: B．y=x+x3y=x+x^3
  • C．y=x4y=x^4
  • D．y=xsin⁡xy=x\sin x
Show solution(Answer: B)
Answer: B For f(x)=x+x3f(x)=x+x^3, f(−x)=−f(x)f(-x)=-f(x), so it is odd. Choose B.
CSCA Math December 2025 Past Paper · #9· Topic: 求反函数
The inverse function of y=x3y=x^3 is ( ).
  • Correct answer: A．y=x3y=\sqrt[3]{x}
  • B．y=3xy=3^x
  • C．y=x/3y=x/3
  • D．y=x3y=x^3
Show solution(Answer: A)
Answer: A From y=x3y=x^3, interchange xx and yy: x=y3x=y^3, hence y=x3y=\sqrt[3]{x}. Choose A.
CSCA Math December 2025 Past Paper · #10· Topic: 根据解析式直接判断函数的单调性
Which of the following functions is increasing?
  • A．y=log⁡0.5xy=\log_{0.5}x
  • B．y=(13)xy=\left(\frac{1}{3}\right)^x
  • Correct answer: C．y=2x+1y=2x+1
  • D．y=sin⁡xy=\sin x
Show solution(Answer: C)
Answer: C y=2x+1y=2x+1 is a linear function with positive slope, so it is increasing. Choose C.
CSCA Math December 2025 Past Paper · #11· Topic: 判断点所在的象限
Which point lies in the third quadrant?
  • A．(−1,2)(-1,2)
  • B．(1,2)(1,2)
  • C．(1,−2)(1,-2)
  • Correct answer: D．(−1,−2)(-1,-2)
Show solution(Answer: D)
Answer: D A point in Quadrant III has both coordinates negative; (−1,−2)(-1,-2) fits. Choose D.
CSCA Math December 2025 Past Paper · #12· Topic: 分式不等式
The solution set of x−1x−2≤1\frac{x-1}{x-2}\le1 is ( ).
  • A．(−∞,1](-\infty,1]
  • B．(2,+∞)(2,+\infty)
  • Correct answer: C．(−∞,2)(-\infty,2)
  • D．[1,2][1,2]
Show solution(Answer: C)
Answer: C x−1x−2≤1\frac{x-1}{x-2}\le1 is equivalent to 1x−2≤0\frac{1}{x-2}\le0 with x≠2x\ne2, giving x<2x<2. Choose C.
CSCA Math December 2025 Past Paper · #13· Topic: 等差中项的应用
If a,b,ca,b,c form an arithmetic sequence and a+c=20a+c=20, then b=b= ( ).
  • A．20
  • Correct answer: B．10
  • C．5
  • D．0
Show solution(Answer: B)
Answer: B In an arithmetic sequence, b=a+c2=202=10b=\frac{a+c}{2}=\frac{20}{2}=10. Choose B.
CSCA Math December 2025 Past Paper · #14· Topic: 直线的倾斜角
The inclination angle of the line y=3x+10y=\sqrt{3}x+10 is ( ).
  • A．45∘45^\circ
  • B．30∘30^\circ
  • C．120∘120^\circ
  • Correct answer: D．60∘60^\circ
Show solution(Answer: D)
Answer: D The slope is 3\sqrt{3}. Thus tan⁡θ=3\tan\theta=\sqrt{3}, so θ=60∘\theta=60^\circ. Choose D.
CSCA Math December 2025 Past Paper · #15· Topic: 比较指数幂的大小
Which inequality is correct?
  • A．2.1−2>1.2−22.1^{-2}>1.2^{-2}
  • Correct answer: B．2.13>1.232.1^3>1.2^3
  • C．32.3>333^{2.3}>3^3
  • D．0.75−0.2>0.75−0.40.75^{-0.2}>0.75^{-0.4}
Show solution(Answer: B)
Answer: B Because 2.1>1.2>02.1>1.2>0, cubing preserves the inequality: 2.13>1.232.1^3>1.2^3. Choose B.
CSCA Math December 2025 Past Paper · #16· Topic: 由终边或终边上的点求三角函数值
If P(1,y)P(1,y) lies on the terminal side of angle α\alpha and tan⁡α=2\tan\alpha=2, then y=y= ( ).
  • Correct answer: A．2
  • B．3\sqrt{3}
  • C．12\frac{1}{2}
  • D．1
Show solution(Answer: A)
Answer: A tan⁡α=y1=2\tan\alpha=\frac{y}{1}=2, so y=2y=2. Choose A.
CSCA Math December 2025 Past Paper · #17· Topic: 等差数列通项公式的基本量计算
If {an}\{a_n\} is arithmetic with a1=1,a2=2a_1=1,a_2=2, then a2025=a_{2025}= ( ).
  • A．2024
  • Correct answer: B．2025
  • C．20
  • D．2
Show solution(Answer: B)
Answer: B d=a2−a1=1d=a_2-a_1=1, so a2025=1+2024=2025a_{2025}=1+2024=2025. Choose B.
CSCA Math December 2025 Past Paper · #18· Topic: 由圆心（或半径）求圆的方程
A circle has center (−3,2)(-3,2) and radius 4. Its equation is ( ).
  • Correct answer: A．(x+3)2+(y−2)2=16(x+3)^2+(y-2)^2=16
  • B．(x−3)2+(y+2)2=16(x-3)^2+(y+2)^2=16
  • C．(x−3)2+(y+2)2=4(x-3)^2+(y+2)^2=4
  • D．(x+3)2+(y−2)2=4(x+3)^2+(y-2)^2=4
Show solution(Answer: A)
Answer: A A circle with center (−3,2)(-3,2) and radius 4 has equation (x+3)2+(y−2)2=16(x+3)^2+(y-2)^2=16. Choose A.
CSCA Math December 2025 Past Paper · #19· Topic: 求平面两点间的距离
The distance from P(−1,−1)P(-1,-1) to Q(1,1)Q(1,1) is ( ).
  • Correct answer: A．222\sqrt{2}
  • B．2
  • C．2\sqrt{2}
  • D．4
Show solution(Answer: A)
Answer: A The distance is (1+1)2+(1+1)2=22\sqrt{(1+1)^2+(1+1)^2}=2\sqrt{2}. Choose A.
CSCA Math December 2025 Past Paper · #20· Topic: 直线的点斜式方程及辨析
The equation of the line through (1,2)(1,2) and (2,4)(2,4) is ( ).
  • Correct answer: A．y=2xy=2x
  • B．y=2x+2y=2x+2
  • C．y=x+1y=x+1
  • D．y=x+2y=x+2
Show solution(Answer: A)
Answer: A The slope is 4−22−1=2\frac{4-2}{2-1}=2, and the line through (1,2)(1,2) is y=2xy=2x. Choose A.
CSCA Math December 2025 Past Paper · #21· Topic: 由圆的一般方程确定圆心和半径
Given x2+y2+2x−2y=0x^2+y^2+2x-2y=0, the radius of the circle is ( ).
  • Correct answer: A．2\sqrt{2}
  • B．1
  • C．2
  • D．12\frac{1}{2}
Show solution(Answer: A)
Answer: A Completing the square gives (x+1)2+(y−1)2=2(x+1)^2+(y-1)^2=2, so the radius is 2\sqrt{2}. Choose A.
CSCA Math December 2025 Past Paper · #22· Topic: 特殊角的三角函数值
If α\alpha is acute and sin⁡α=32\sin\alpha=\frac{\sqrt{3}}{2}, then tan⁡α=\tan\alpha= ( ).
  • A．14\frac{1}{4}
  • Correct answer: B．3\sqrt{3}
  • C．12\frac{1}{2}
  • D．33\frac{\sqrt{3}}{3}
Show solution(Answer: B)
Answer: B Since α\alpha is acute and sin⁡α=32\sin\alpha=\frac{\sqrt{3}}{2}, α=60∘\alpha=60^\circ and tan⁡α=3\tan\alpha=\sqrt{3}. Choose B.
CSCA Math December 2025 Past Paper · #23· Topic: 等比数列通项公式的基本量计算
The first three terms of a geometric sequence are 1, 2, 4. Its general term is ( ).
  • A．an=2na_n=2^n
  • Correct answer: B．an=2n−1a_n=2^{n-1}
  • C．an=8a_n=8
  • D．an=2na_n=2n
Show solution(Answer: B)
Answer: B The sequence has first term 1 and common ratio 2, so an=2n−1a_n=2^{n-1}. Choose B.
CSCA Math December 2025 Past Paper · #24· Topic: 由不等式的性质比较数（式）大小
Given a>ba>b, which statement is true?
  • Correct answer: A．5a>5b5a>5b
  • B．1a<1b\frac{1}{a}<\frac{1}{b}
  • C．a2>b2a^2>b^2
  • D．sin⁡a>sin⁡b\sin a>\sin b
Show solution(Answer: A)
Answer: A Multiplying both sides of a>ba>b by the positive number 5 preserves the inequality, so 5a>5b5a>5b. Choose A.
CSCA Math December 2025 Past Paper · #25· Topic: 求平面两点间的距离
The distance from P(−1,2)P(-1,2) to Q(3,1)Q(3,1) is ( ).
  • A．13\sqrt{13}
  • Correct answer: B．17\sqrt{17}
  • C．5
  • D．5\sqrt{5}
Show solution(Answer: B)
Answer: B The distance is (3+1)2+(1−2)2=17\sqrt{(3+1)^2+(1-2)^2}=\sqrt{17}. Choose B.
CSCA Math December 2025 Past Paper · #26· Topic: 判断指数函数的单调性
Which statement about y=(12)xy=\left(\frac{1}{2}\right)^x is correct?
  • A．It is increasing.
  • Correct answer: B．It is decreasing.
  • C．It is symmetric about the origin.
  • D．It is symmetric about the xx-axis.
Show solution(Answer: B)
Answer: B The base 12\frac{1}{2} is between 0 and 1, so the exponential function is decreasing. Choose B.
CSCA Math December 2025 Past Paper · #27· Topic: 二倍角的余弦公式
If sin⁡α=14\sin\alpha=\frac{1}{4}, then cos⁡2α=\cos2\alpha= ( ).
  • A．−78-\frac{7}{8}
  • B．18\frac{1}{8}
  • C．1516\frac{15}{16}
  • Correct answer: D．78\frac{7}{8}
Show solution(Answer: D)
Answer: D cos⁡2α=1−2sin⁡2α=1−2⋅116=78\cos2\alpha=1-2\sin^2\alpha=1-2\cdot\frac{1}{16}=\frac{7}{8}. Choose D.
CSCA Math December 2025 Past Paper · #28· Topic: 求直线交点坐标
The intersection of l1:y=2x+1l_1:y=2x+1 and l2:x+y+1=0l_2:x+y+1=0 is ( ).
  • A．(13,23)\left(\frac{1}{3},\frac{2}{3}\right)
  • B．(−13,−23)\left(-\frac{1}{3},-\frac{2}{3}\right)
  • C．(23,−13)\left(\frac{2}{3},-\frac{1}{3}\right)
  • Correct answer: D．(−23,−13)\left(-\frac{2}{3},-\frac{1}{3}\right)
Show solution(Answer: D)
Answer: D Substitute y=2x+1y=2x+1 into x+y+1=0x+y+1=0 to get 3x+2=03x+2=0, so the point is (−23,−13)\left(-\frac{2}{3},-\frac{1}{3}\right). Choose D.
CSCA Math December 2025 Past Paper · #29· Topic: 求双曲线的焦点坐标
The foci of x2−y2=1x^2-y^2=1 are ( ).
  • A．(0,±2)(0,\pm\sqrt{2})
  • B．(±2,0)(\pm2,0)
  • Correct answer: C．(±2,0)(\pm\sqrt{2},0)
  • D．(0,±2)(0,\pm2)
Show solution(Answer: C)
Answer: C For x2−y2=1x^2-y^2=1, a2=1a^2=1, b2=1b^2=1, so c2=2c^2=2. The foci are (±2,0)(\pm\sqrt{2},0). Choose C.
CSCA Math December 2025 Past Paper · #30· Topic: 对数的运算
log⁡232+log⁡1232=\log_2 32+\log_{\frac{1}{2}}32= ( ).
  • Correct answer: A．0
  • B．-5
  • C．5
  • D．10
Show solution(Answer: A)
Answer: A log⁡232=5\log_2 32=5 and log⁡1232=−5\log_{\frac{1}{2}}32=-5, so the sum is 0. Choose A.
CSCA Math December 2025 Past Paper · #31· Topic: 三角函数的图象与性质
Which statement about y=cos⁡xy=\cos x is correct?
  • A．The function is odd
  • Correct answer: B．The function has period 2π2\pi
  • C．cos⁡(x+π)=cos⁡x\cos(x+\pi)=\cos x
  • D．The function is increasing on [0,π][0,\pi]
Show solution(Answer: B)
Answer: B The function y=cos⁡xy=\cos x has period 2π2\pi. Choose B.
CSCA Math December 2025 Past Paper · #32· Topic: 根据抛物线方程求焦点或准线
Which statement about the parabola y2=4xy^2=4x is correct?
  • A．The axis of symmetry is x=0x=0
  • B．The foci are (±1,0)(\pm1,0)
  • Correct answer: C．The focus is (1,0)(1,0)
  • D．It passes through (−4,4)(-4,4)
Show solution(Answer: C)
Answer: C y2=4xy^2=4x has the form y2=4pxy^2=4px, so p=1p=1 and the focus is (1,0)(1,0). Choose C.
CSCA Math December 2025 Past Paper · #33· Topic: 由斜率判断两条直线垂直
Which line is perpendicular to l:x+y+3=0l:x+y+3=0?
  • Correct answer: A．y=x+3y=x+3
  • B．y=−2x+1y=-2x+1
  • C．y=1−xy=1-x
  • D．y=2x+1y=2x+1
Show solution(Answer: A)
Answer: A The line x+y+3=0x+y+3=0 has slope −1-1, so a perpendicular line has slope 1. y=x+3y=x+3 fits. Choose A.
CSCA Math December 2025 Past Paper · #34· Topic: 半角公式
Given cos⁡α=−12\cos\alpha=-\frac{1}{2} and α∈(π2,π)\alpha\in(\frac{\pi}{2},\pi), find sin⁡α2\sin\frac{\alpha}{2}.
  • A．−32-\frac{\sqrt{3}}{2}
  • Correct answer: B．32\frac{\sqrt{3}}{2}
  • C．12\frac{1}{2}
  • D．−12-\frac{1}{2}
Show solution(Answer: B)
Answer: B Since α/2∈(π/4,π/2)\alpha/2\in(\pi/4,\pi/2), sin⁡α2=1−cos⁡α2=32\sin\frac{\alpha}{2}=\sqrt{\frac{1-\cos\alpha}{2}}=\frac{\sqrt{3}}{2}. Choose B.
CSCA Math December 2025 Past Paper · #35· Topic: 用和、差角的余弦公式化简、求值
Using cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta, find cos⁡π12\cos\frac{\pi}{12}.
  • A．64\frac{\sqrt{6}}{4}
  • B．1+34\frac{1+\sqrt{3}}{4}
  • C．6−24\frac{\sqrt{6}-\sqrt{2}}{4}
  • Correct answer: D．6+24\frac{\sqrt{6}+\sqrt{2}}{4}
Show solution(Answer: D)
Answer: D π12=π4−π6\frac{\pi}{12}=\frac{\pi}{4}-\frac{\pi}{6}, so the difference formula gives cos⁡π12=6+24\cos\frac{\pi}{12}=\frac{\sqrt{6}+\sqrt{2}}{4}. Choose D.
CSCA Math December 2025 Past Paper · #36· Topic: 判断两个函数是否相等
Which function is identical to y=∣x∣y=|x|?
  • Correct answer: A．y=x2y=\sqrt{x^2}
  • B．y=(x)2y=(\sqrt{x})^2
  • C．y=x33y=\sqrt[3]{x^3}
  • D．y=xy=x
Show solution(Answer: A)
Answer: A For every real xx, x2=∣x∣\sqrt{x^2}=|x|. Choose A.
CSCA Math December 2025 Past Paper · #37· Topic: 特殊角的三角函数值
If α\alpha is acute and cos⁡α=12\cos\alpha=\frac{1}{2}, then sin⁡α2=\sin\frac{\alpha}{2}= ( ).
  • A．32\frac{\sqrt{3}}{2}
  • B．±12\pm\frac{1}{2}
  • C．±32\pm\frac{\sqrt{3}}{2}
  • Correct answer: D．12\frac{1}{2}
Show solution(Answer: D)
Answer: D Since α\alpha is acute and cos⁡α=12\cos\alpha=\frac{1}{2}, α=60∘\alpha=60^\circ, so sin⁡α2=sin⁡30∘=12\sin\frac{\alpha}{2}=\sin30^\circ=\frac{1}{2}. Choose D.
CSCA Math December 2025 Past Paper · #38· Topic: 诱导公式二
Which trigonometric reduction formula is incorrect?
  • Correct answer: A．tan⁡(π+α)=−tan⁡α\tan(\pi+\alpha)=-\tan\alpha
  • B．cos⁡(2π−α)=cos⁡α\cos(2\pi-\alpha)=\cos\alpha
  • C．sin⁡(π−α)=sin⁡α\sin(\pi-\alpha)=\sin\alpha
  • D．sin⁡(π2−α)=cos⁡α\sin\left(\frac{\pi}{2}-\alpha\right)=\cos\alpha
Show solution(Answer: A)
Answer: A The identity is tan⁡(π+α)=tan⁡α\tan(\pi+\alpha)=\tan\alpha, not −tan⁡α-\tan\alpha. Choose A.
CSCA Math December 2025 Past Paper · #39· Topic: 根据抛物线方程求焦点或准线
The directrix of y2=−xy^2=-x is ( ).
  • A．x=12x=\frac{1}{2}
  • Correct answer: B．x=14x=\frac{1}{4}
  • C．x=−14x=-\frac{1}{4}
  • D．x=−12x=-\frac{1}{2}
Show solution(Answer: B)
Answer: B Write y2=−xy^2=-x as y2=4pxy^2=4px, so p=−14p=-\frac{1}{4} and the directrix is x=−p=14x=-p=\frac{1}{4}. Choose B.
CSCA Math December 2025 Past Paper · #40· Topic: 求值——同角三角函数基本关系
Given cos⁡α−sin⁡αcos⁡α+sin⁡α=13\frac{\cos\alpha-\sin\alpha}{\cos\alpha+\sin\alpha}=\frac{1}{3}, then tan⁡α=\tan\alpha= ( ).
  • A．14\frac{1}{4}
  • B．13\frac{1}{3}
  • Correct answer: C．12\frac{1}{2}
  • D．23\frac{2}{3}
Show solution(Answer: C)
Answer: C Let t=tan⁡αt=\tan\alpha. Then 1−t1+t=13\frac{1-t}{1+t}=\frac{1}{3}, so t=12t=\frac{1}{2}. Choose C.
CSCA Math December 2025 Past Paper · #41· Topic: 利用等差数列通项公式求数列中的项
If a1=1a_1=1 and an=na_n=n, then a100=a_{100}= ( ).
  • A．1
  • B．1100\frac{1}{100}
  • Correct answer: C．100
  • D．50
Show solution(Answer: C)
Answer: C Since an=na_n=n, a100=100a_{100}=100. Choose C.
CSCA Math December 2025 Past Paper · #42· Topic: 研究对数函数的单调性
Which statement about y=log⁡axy=\log_a x (a>0a>0, a≠1a\ne1) is correct?
  • Correct answer: A．When 0<a<10<a<1, it is decreasing on its domain
  • B．When a>1a>1, it is decreasing on its domain
  • C．Its domain is (−∞,+∞)(-\infty,+\infty)
  • D．When a>1a>1, log⁡ax>1\log_a x>1 for every x>0x>0
Show solution(Answer: A)
Answer: A For logarithms, a>0a>0 and a≠1a\ne1. When 0<a<10<a<1, y=log⁡axy=\log_a x is decreasing on (0,+∞)(0,+\infty). Choose A.
CSCA Math December 2025 Past Paper · #43· Topic: 根据椭圆方程求 $a 、 b 、 c$
For x24+y220=1\frac{x^2}{4}+\frac{y^2}{20}=1, which statement is correct?
  • A．Its vertices are (±4,0)(\pm4,0) and (0,±20)(0,\pm20)
  • B．Its foci are (±4,0)(\pm4,0)
  • C．The lengths of its major and minor axes are 20 and 4
  • Correct answer: D．Its foci are (0,±4)(0,\pm4)
Show solution(Answer: D)
Answer: D Here a2=20a^2=20, b2=4b^2=4, and the major axis is vertical. Thus c2=20−4=16c^2=20-4=16, so the foci are (0,±4)(0,\pm4). Choose D.
CSCA Math December 2025 Past Paper · #44· Topic: 由两条直线平行求方程
A line ll is parallel to l1:x+2y+4=0l_1:x+2y+4=0 and passes through the intersection of l2:x+y+1=0l_2:x+y+1=0 and l3:2x+y−1=0l_3:2x+y-1=0. Which equation is ll?
  • A．x+y−7=0x+y-7=0
  • B．x+y−5=0x+y-5=0
  • C．x−2y−1=0x-2y-1=0
  • Correct answer: D．x+2y+4=0x+2y+4=0
Show solution(Answer: D)
Answer: D Solving x+y+1=0x+y+1=0 and 2x+y−1=02x+y-1=0 gives (2,−3)(2,-3). A parallel line has form x+2y+c=0x+2y+c=0; substituting gives c=4c=4, so x+2y+4=0x+2y+4=0. Choose D.
CSCA Math December 2025 Past Paper · #45· Topic: 向量的线性运算的几何应用
Let OO be the origin, and let AA and BB be two points in the plane. If PP is the midpoint of ABAB, which expression can represent OP→\overrightarrow{OP}?
  • A．OB→−OA→\overrightarrow{OB}-\overrightarrow{OA}
  • B．OA→+OB→\overrightarrow{OA}+\overrightarrow{OB}
  • Correct answer: C．12(OA→+OB→)\frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OB})
  • D．OA→−OB→\overrightarrow{OA}-\overrightarrow{OB}
Show solution(Answer: C)
Answer: C Since PP is the midpoint of ABAB, OP→=12(OA→+OB→)\overrightarrow{OP}=\frac{1}{2}(\overrightarrow{OA}+\overrightarrow{OB}). Choose C.
CSCA Math December 2025 Past Paper · #46· Topic: 复数的乘方
If z3=1z^3=1 and z≠1z\ne1, then 1+z+z2+⋯+z999=1+z+z^2+\cdots+z^{999}= ( ).
  • A．ii
  • B．1±3i2\frac{1\pm\sqrt{3} i}{2}
  • Correct answer: C．1
  • D．−1±3i2\frac{-1\pm\sqrt{3} i}{2}
Show solution(Answer: C)
Answer: C For a non-real cube root of 1, 1+z+z2=01+z+z^2=0. The first 999 terms make 333 full cycles, and z999=1z^{999}=1, so the sum is 1. Choose C.
CSCA Math December 2025 Past Paper · #47· Topic: 利用等差数列的性质计算
In an arithmetic sequence, 3(a2+a6)+2(a6+a10+a14)=243(a_2+a_6)+2(a_6+a_{10}+a_{14})=24. Find S13S_{13}.
  • A．104
  • B．39
  • C．52
  • Correct answer: D．26
Show solution(Answer: D)
Answer: D In the arithmetic sequence, 3(a2+a6)+2(a6+a10+a14)=12a7=243(a_2+a_6)+2(a_6+a_{10}+a_{14})=12a_7=24, so a7=2a_7=2 and S13=13a7=26S_{13}=13a_7=26. Choose D.
CSCA Math December 2025 Past Paper · #48· Topic: 计算古典概型问题的概率
The Chinese zodiac has 12 animals. If three people are chosen randomly, what is the probability that at least two share the same zodiac sign?
  • A．143144\frac{143}{144}
  • B．5572\frac{55}{72}
  • C．112\frac{1}{12}
  • Correct answer: D．1772\frac{17}{72}
Show solution(Answer: D)
Answer: D The probability that all three zodiac signs are different is 12⋅11⋅10123=5572\frac{12\cdot11\cdot10}{12^3}=\frac{55}{72}. Therefore the required probability is 1−5572=17721-\frac{55}{72}=\frac{17}{72}. Choose D.

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