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SPM Add Math · 2025 · Pahang · Percubaan Kertas 1 (Pahang)

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  1. 1

    Diagram 1 shows a rhombus ABCDABCD. The equation of the straight lines ABAB and BCBC are 2yx8=02y-x-8=0 and 2y+x4=02y+x-4=0 respectively. The straight line DBDB is extended to a point NN such that DB:BN=2:1DB:BN=2:1. Find

    图 1: 坐标平面上的菱形 ABCD,B 在上方,A 在左,C 在右,D 在下方且坐标为 (2,1)(-2,-1);对角线 DB 延长至点 N。

    (a)

    the coordinates of NN,

    [4 分]
    (b)

    the equation of the straight line ADAD.

    [2 分]

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  2. 2
    (a)

    Given that h(x)=(12x)3h(x)=(1-2x)^{3}, evaluate h(1)h'(1).

    [3 分]
    (b)

    A curve has an equation y=3x+4xy=3x+\frac{4}{x}. If xx is decreasing at a constant rate of 3 units per second, find the rate of change of yy when x=2x=2.

    [4 分]

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  3. 3

    The variable xx and yy are related by the equation xy=ax2+bx\frac{x}{y}=ax^{2}+bx such that aa and bb are constants. Two straight line graphs are obtained by plotting the relations from the equation. Table 1.1 and Table 1.2 show the respective coordinates that lie on the straight lines. Express aa in terms of bb.

    Table 1.1:

    xx0
    1y\frac{1}{y}2k-2k

    Table 1.2:

    1x\frac{1}{x}0
    1xy\frac{1}{xy}k4k-4
    ()

    Express aa in terms of bb.

    [3 分]

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  4. 4

    Solve the following system of linear equations using the elimination method and / or substitution. 2x3y+z=162x-3y+z=16 3xy+2z=193x-y+2z=19 4x+3y+3z=184x+3y+3z=18

    ()

    Solve the system of linear equations.

    [5 分]

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  5. 5
    (a)

    Sketch the graph of y=2+tan2xy=2+\tan 2x for 0xπ0\leq x\leq\pi.

    [3 分]
    (b)

    Diagram 2 shows an angle of 109π-\frac{10}{9}\pi rad on a Cartesian plane. State the reference angle in term of π\pi rad.

    图 2: 角 109π-\frac{10}{9}\pi rad,终边在第二象限。

    [1 分]

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  6. 6
    (a)

    Diagram 3.1 shows a sector of a circle with centre OO and an arc of ADAD which subtends an angle of θ\theta rad at OO. It is given that the radius of the sector of circle is rr unit and the area of sector AODAOD is AA unit2^{2}. Based on Diagram 3.1, derive A=12r2θA=\frac{1}{2}r^{2}\theta.

    图 3.1: 圆心为 O、半径为 r、圆心角为 θ\theta 的扇形 AOD。

    [2 分]
    (b)

    Diagram 3.2 shows another arc EFEF with the same centre OO and radius of 16 cm drawn on Diagram 3.1. Given that OBEOBE and OCFOCF are two straight lines such that BB and CC lie on arc ADAD. Arc lengths of ABAB, BCBC and CDCD are equal. Lines AEAE and DFDF are the tangents to arc ADAD at points AA and DD respectively. [Use π=3.142\pi=3.142] Given length of OA=12OA=12 cm, calculate the area, in cm2^{2}, of the whole Diagram 3.2.

    图 3.2: 在图 3.1 上叠加半径 16 cm 的弧 EF;OBE、OCF 为直线,AE、DF 为切线。

    [3 分]

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  7. 7

    The function hh is defined by h(x)=2x5h(x)=|2x-5|.

    (a)

    State h(1)h(1).

    [1 分]
    (b(i))

    Find the domain for h(x)5h(x)\leq 5.

    [2 分]
    (b(ii))

    Hence, sketch the graph of the function for h(x)h(x).

    [2 分]

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  8. 8

    (a) refers to Diagram 4 (a standard normal distribution curve). (b) refers to the mass of students in a school which has a normal distribution with mean of 55 kg and standard deviation of 10 kg.

    (a)

    Diagram 4 shows a standard normal distribution curve. Given P(0<z<k)=0.3125P(0<z<k)=0.3125. State P(z>k)P(z>k).

    图 4: 标准正态分布曲线,阴影区域由 00kk

    [1 分]
    (b(i))

    Find the mass of the student which gives a standard score of 0.50.5,

    [2 分]
    (b(ii))

    Find the percentage of students with mass greater than 4848 kg.

    [3 分]

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  9. 9
    (a)

    The amount of caffeine in a person's body after drinking coffee is given by C=C0e0.15tC=C_{0}e^{-0.15t} such that C0C_{0} is the initial caffeine content, in mg, and CC is the caffeine content, in mg, after tt hours. Find the time needed for the caffeine level to reduce to half of its original value. Round off your answer correct to the nearest whole number.

    [3 分]
    (b)

    Diagram 5 shows a triangular metal ABCABC designed by an engineer to support a building structure. Determine whether triangle ABCABC is a right-angled triangle at CC or not. Show your calculations.

    图 5: 三角形 ABC,AC=6AC=\sqrt{6} m,BC=23+2BC=\frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}} m,AB=1646AB=\sqrt{16-4\sqrt{6}} m。

    [3 分]

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  10. 10
    (a)

    Find the equation of the curve that has the gradient function dydx=4x+9\frac{dy}{dx}=4x+9 and passes through the point A(2,3)A(2,-3).

    [3 分]
    (b(i))

    Given 13f(x)dx=4\int_{1}^{3}f(x)\,dx=4, 36f(x)dx=5\int_{3}^{6}f(x)\,dx=5 and 16f(x)dx=k\int_{1}^{6}f(x)\,dx=k. Find the value of kk,

    [1 分]
    (b(ii))

    Find 1372f(x)dx\int_{1}^{3}7-2f(x)\,dx.

    [3 分]

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  11. 11

    The graph of a quadratic function f(x)=px28x+qf(x)=px^{2}-8x+q, where pp and qq are constants, has a maximum point.

    (a)

    Given pp is an integer such that 2<p<2-2<p<2, state the value of pp.

    [1 分]
    (b)

    Using the answer from 11(a), find the value of qq when the graph touches the xx-axis at one point.

    [2 分]

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  12. 12

    A teacher plans to save money over a period of 2 years as preparation for further studies. He considers two different saving plans:

    Plan A: The teacher saves RMxx in the first month and increases the amount by RM20 every month thereafter.

    Plan B: The teacher saves RM2100 in the first year and increases the savings in the following year by RM400.

    (a)

    For Plan A, if he makes a saving of RM240 in the 3rd month, calculate his savings in the 6th month.

    [3 分]
    (b)

    Determine which plan allows the teacher to reach the savings target of RM10320 earlier. Justify your answer.

    [4 分]

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  13. 13

    【Bahagian B:第 13、14、15 题任选两题作答 / Answer any two of Questions 13–15】

    (a)

    Given the graph of quadratic function f(x)=rx2+2x+rf(x)=rx^{2}+2x+r does not intersect the xx-axis. Find the range of values of rr by using number line method.

    [3 分]
    (b(i))

    Diagram 6 shows a graph of f(x)=2x24mx+2m2+2nf(x)=2x^{2}-4mx+2m^{2}+2n with a=2a=2, b=4mb=-4m, and c=2m2+2nc=2m^{2}+2n. On Diagram 6, sketch a graph of y=g(x)y=g(x) when the variable of aa becomes 1.

    图 6: 抛物线 y=f(x)y=f(x),与 xx 轴交于 x=1x=1x=5x=5,最低点在 y=8y=-8

    [1 分]
    (b(ii))

    Express f(x)=2x24mx+2m2+2nf(x)=2x^{2}-4mx+2m^{2}+2n in the form a(xh)2+ka(x-h)^{2}+k, such that hh and kk are constants. Hence, find the value of mm and of nn.

    [4 分]

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  14. 14

    【Bahagian B:第 13、14、15 题任选两题作答 / Answer any two of Questions 13–15】

    (a)

    Diagram 7 shows a parallelogram DEFGDEFG. Point PP lies on straight line DEDE and point YY lies on straight line GFGF. PTPT is a straight line passing through the point FF. Given that DY=a+6b\vec{DY}=\underset{\sim}{a}+6\underset{\sim}{b}, and DT=32a+9b\vec{DT}=\frac{3}{2}\underset{\sim}{a}+9\underset{\sim}{b}. A straight line is drawn by connecting point DD to point TT. Using the vector method, determine whether the line passes through point YY or not.

    图 7: 平行四边形 DEFG,Y 在 GF 上,P 在 DE 上,T 在 PF 延长线外侧。

    [4 分]
    (b(i))

    Diagram 8 shows a regular hexagon PQRSTUPQRSTU drawn on a Cartesian plane with centre at origin OO. Express PQ+TS\vec{PQ}+\vec{TS} as a single vector.

    图 8: 以原点 O 为中心的正六边形 PQRSTU,P、Q 在上,U、R 在左右,T、S 在下。

    [1 分]
    (b(ii))

    Given OR=5i\vec{OR}=5\underset{\sim}{i}, OT=52i532j\vec{OT}=-\frac{5}{2}\underset{\sim}{i}-\frac{5\sqrt{3}}{2}\underset{\sim}{j}, find the unit vector in the direction of RT\vec{RT}, in terms of i\underset{\sim}{i} and j\underset{\sim}{j}.

    [3 分]

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  15. 15

    【Bahagian B:第 13、14、15 题任选两题作答 / Answer any two of Questions 13–15】

    It is given that f1(x)=k+x3xf^{-1}(x)=\frac{k+x}{3x}, x0x\neq 0 and fg(x)=23x2+2fg(x)=\frac{2}{3x^{2}+2}, find

    (a)

    f(x)f(x) in terms of kk,

    [2 分]
    (b)

    the value of kk such that f(1)=4fg(2)f(1)=4fg(\sqrt{2}),

    [2 分]
    (c)

    hence, find g(x)g(x) and determine whether the inverse function of g(x)g(x) exists or not by using a horizontal line test. Give your justification.

    [4 分]

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