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Practical Mathematics Skills Test (Suken) study materials
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Practical Mathematics Skills Test (Suken) practice questions

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Question 1Number and algebra

Expand and simplify the algebraic expression: (2x - 3)(3x + 4) - (x - 2)^2. 次の式を展開して整理しなさい：(2x - 3)(3x + 4) - (x - 2)^2

  1. 5x^2 + 3x - 8
  2. 5x^2 - 5x - 8
  3. 5x^2 + 3x - 16 Correct answer.
  4. 7x^2 - x - 16

Explanation

Expand the first product: (2x - 3)(3x + 4) = 6x^2 + 8x - 9x - 12 = 6x^2 - x - 12. Expand the second squared term: (x - 2)^2 = x^2 - 4x + 4. Subtract the two expressions: (6x^2 - x - 12) - (x^2 - 4x + 4) = 6x^2 - x^2 - x + 4x - 12 - 4 = 5x^2 + 3x - 16.

Question 2Number and algebra

Completely factor the polynomial expression: 2x^2 + 7xy + 3y^2 - 5x - 5y + 2. 次の式を因数分解しなさい：2x^2 + 7xy + 3y^2 - 5x - 5y + 2

  1. (2x - y - 1)(x - 3y - 2)
  2. (2x + 3y - 1)(x + y - 2)
  3. (2x + y + 1)(x + 3y - 2)
  4. (2x + y - 1)(x + 3y - 2) Correct answer.

Explanation

Group by powers of x: 2x^2 + (7y - 5)x + (3y^2 - 5y + 2). Factor the y quadratic: 3y^2 - 5y + 2 = (3y - 2)(y - 1). Using cross-multiplication for the quadratic in x: (2)(y - 1) + (1)(3y - 2) = 2y - 2 + 3y - 2 = 5y - 4, but we need 7y - 5. Pairing 2 with (3y - 2) and 1 with (y - 1): 2(3y - 2) + 1(y - 1) = 6y - 4 + y - 1 = 7y - 5. Thus, the factored form is (2x + y - 1)(x + 3y - 2).

Question 3Number and algebra

Simplify the radical fraction by rationalizing the denominator: (3 - √5) / (3 + √5). 次の式の分母を有理化して簡約しなさい：(3 - √5) / (3 + √5)

  1. 7 - 3√5
  2. (7 + 3√5) / 2
  3. (14 - 6√5) / 4
  4. (7 - 3√5) / 2 Correct answer.

Explanation

Multiply numerator and denominator by the conjugate (3 - √5): (3 - √5)^2 / [(3 + √5)(3 - √5)] = (9 - 6√5 + 5) / (9 - 5) = (14 - 6√5) / 4. Dividing numerator and denominator by 2 gives (7 - 3√5) / 2.

Question 4Number and algebra

Solve the quadratic equation for real roots: 3x^2 - 5x - 2 = 0. 2次方程式 3x^2 - 5x - 2 = 0 を解きなさい。

  1. x = 3, -2/3
  2. x = 2, -1/3 Correct answer.
  3. x = 1, -2/3
  4. x = -2, 1/3

Explanation

Factor the quadratic: (3x + 1)(x - 2) = 0. Therefore, 3x + 1 = 0 => x = -1/3, or x - 2 = 0 => x = 2. The solutions are x = 2 and x = -1/3.

Question 5Number and algebra

For what values of m does the quadratic equation x^2 + 2(m - 1)x + (m^2 - 3) = 0 have two distinct real roots? 2次方程式 x^2 + 2(m - 1)x + (m^2 - 3) = 0 が異なる2つの実数解を持つような定数 m の値の範囲を求めなさい。

  1. m < 2 Correct answer.
  2. m > -2
  3. m > 2
  4. m < -2

Explanation

For the equation to have two distinct real roots, the discriminant D must satisfy D/4 > 0. Here, D/4 = (m - 1)^2 - 1*(m^2 - 3) = (m^2 - 2m + 1) - (m^2 - 3) = -2m + 4. Setting -2m + 4 > 0 gives -2m > -4, which simplifies to m < 2.

Question 6Number and algebra

Solve the quadratic inequality: 2x^2 - 5x - 3 <= 0. 2次不等式 2x^2 - 5x - 3 <= 0 を解きなさい。

  1. -1/2 <= x <= 3 Correct answer.
  2. x <= -1/2, x >= 3
  3. x <= -3, x >= 1/2
  4. -3 <= x <= 1/2

Explanation

Factor the quadratic: (2x + 1)(x - 3) <= 0. The parabola y = 2x^2 - 5x - 3 opens upwards and intersects the x-axis at x = -1/2 and x = 3. It takes values less than or equal to zero between the roots: -1/2 <= x <= 3.

Question 7Number and algebra

Express the complex number fraction in standard form a + bi: (3 + 4i) / (1 - 2i), where i^2 = -1. 複素数 (3 + 4i) / (1 - 2i) を a + bi の形で表しなさい。

  1. -1 + 2i Correct answer.
  2. -1 - 2i
  3. 1 + 2i
  4. 11/5 + (10/5)i

Explanation

Multiply numerator and denominator by the complex conjugate (1 + 2i): (3 + 4i)(1 + 2i) / [(1 - 2i)(1 + 2i)] = (3 + 6i + 4i + 8i^2) / (1 - 4i^2) = (3 + 10i - 8) / (1 + 4) = (-5 + 10i) / 5 = -1 + 2i.

Question 8Number and algebra

If alpha and beta are the roots of 2x^2 - 6x + 3 = 0, find the value of alpha^2 + beta^2. 2次方程式 2x^2 - 6x + 3 = 0 の2つの解を α, β とするとき、α^2 + β^2 の値を求めなさい。

  1. 6 Correct answer.
  2. 15/2
  3. 9/2
  4. 3

Explanation

By Vieta's formulas, alpha + beta = -(-6)/2 = 3, and alpha * beta = 3/2. We know alpha^2 + beta^2 = (alpha + beta)^2 - 2*alpha*beta. Substituting the values: 3^2 - 2*(3/2) = 9 - 3 = 6.

Question 9Number and algebra

Find the remainder when the polynomial P(x) = 2x^3 - 5x^2 + 4x - 7 is divided by (x - 2). 多項式 P(x) = 2x^3 - 5x^2 + 4x - 7 を x - 2 で割ったときの余りを求めなさい。

  1. -7
  2. -3 Correct answer.
  3. 5
  4. 3

Explanation

By the Remainder Theorem, the remainder when P(x) is divided by (x - 2) is equal to P(2). Calculate P(2) = 2(2)^3 - 5(2)^2 + 4(2) - 7 = 2(8) - 5(4) + 8 - 7 = 16 - 20 + 8 - 7 = -4 + 8 - 7 = 4 - 7 = -3.

Question 10Number and algebra

In an arithmetic progression, the 4th term is 17 and the 10th term is 41. Find the first term a_1 and common difference d. 等差数列において、第4項が17、第10項が41であるとき、初項 a と公差 d を求めなさい。

  1. a_1 = 1, d = 4
  2. a_1 = 4, d = 5
  3. a_1 = 5, d = 4 Correct answer.
  4. a_1 = 5, d = 3

Explanation

The n-th term is a_n = a_1 + (n - 1)d. Thus: a_4 = a_1 + 3d = 17, and a_10 = a_1 + 9d = 41. Subtracting the first equation from the second: (a_1 + 9d) - (a_1 + 3d) = 41 - 17 => 6d = 24 => d = 4. Substitute d = 4 back: a_1 + 3(4) = 17 => a_1 + 12 = 17 => a_1 = 5.

Question 11Number and algebra

Calculate the sum of the first 20 terms of an arithmetic progression with first term a_1 = 3 and common difference d = 4. 初項 3、公差 4 の等差数列の初項から第20項までの和 S_20 を求めなさい。

  1. 1640
  2. 840
  3. 800
  4. 820 Correct answer.

Explanation

The sum formula is S_n = (n/2) * [2a_1 + (n - 1)d]. For n = 20, a_1 = 3, d = 4: S_20 = (20/2) * [2(3) + (19)(4)] = 10 * [6 + 76] = 10 * 82 = 820.

Question 12Number and algebra

Find the sum of the infinite geometric series: 6 + 2 + 2/3 + 2/9 + ... 無限等比級数 6 + 2 + 2/3 + 2/9 + ... の和を求めなさい。

  1. 9 Correct answer.
  2. 18
  3. 27/2
  4. 8

Explanation

The first term is a = 6, and the common ratio is r = 2/6 = 1/3. Since |r| = 1/3 < 1, the series converges to S = a / (1 - r) = 6 / (1 - 1/3) = 6 / (2/3) = 6 * (3/2) = 9.

Question 13Number and algebra

Find the coefficient of x^3 * y^2 in the binomial expansion of (2x - 3y)^5. (2x - 3y)^5 の展開式における x^3 y^2 の係数を求めなさい。

  1. 240
  2. 1080
  3. 720 Correct answer.
  4. -720

Explanation

The general term in the binomial expansion is 5_C_r * (2x)^(5 - r) * (-3y)^r. For x^3 * y^2, we set r = 2: Term = 5_C_2 * (2x)^3 * (-3y)^2 = 10 * (8x^3) * (9y^2) = 10 * 8 * 9 * x^3 * y^2 = 720 * x^3 * y^2. The coefficient is 720.

Question 14Number and algebra

Solve the fractional inequality: (x - 1) / (x + 2) < 0. 分数不等式 (x - 1) / (x + 2) < 0 を解きなさい。

  1. -2 < x < 1 Correct answer.
  2. x < 1
  3. x < -2, x > 1
  4. x > -2

Explanation

A fraction A/B is negative when the product A*B < 0 (with B != 0). Here, (x - 1)(x + 2) < 0. The roots are -2 and 1. The product is strictly negative strictly between the roots: -2 < x < 1.

Question 15Number and algebra

Solve the absolute value inequality: |2x - 5| <= 7. 絶対値不等式 |2x - 5| <= 7 を解きなさい。

  1. x <= -1, x >= 6
  2. 1 <= x <= 6
  3. -1 <= x <= 6 Correct answer.
  4. -6 <= x <= 1

Explanation

|2x - 5| <= 7 unwraps into: -7 <= 2x - 5 <= 7. Add 5 to all parts: -2 <= 2x <= 12. Divide by 2: -1 <= x <= 6.

Question 16Number and algebra

Find the sum of the series: sum_{k=1}^{n} (3k^2 - k). 数列の和 sum_{k=1}^{n} (3k^2 - k) を n を用いて表しなさい。

  1. (3n^3 + 2n) / 2
  2. n^2 * (n + 1) Correct answer.
  3. n * (n + 1)^2
  4. (n * (n + 1) * (2n + 1)) / 2

Explanation

Recall sum(k^2) = n(n + 1)(2n + 1)/6 and sum(k) = n(n + 1)/2. Then: 3 * [n(n + 1)(2n + 1)/6] - [n(n + 1)/2] = [n(n + 1)(2n + 1)/2] - [n(n + 1)/2] = [n(n + 1)/2] * [(2n + 1) - 1] = [n(n + 1)/2] * (2n) = n^2 * (n + 1).

Question 17Number and algebra

Find the values of real numbers a and b such that the identity holds for all x: (2x + 1)/(x^2 - x - 2) = a/(x - 2) + b/(x + 1). 恒等式 (2x + 1)/(x^2 - x - 2) = a/(x - 2) + b/(x + 1) が成り立つとき、定数 a, b の値を求めなさい。

  1. a = 3/2, b = 1/2
  2. a = 5/3, b = 1/3 Correct answer.
  3. a = 1, b = 1
  4. a = 2, b = -1

Explanation

Combine the right side over a common denominator: [a(x + 1) + b(x - 2)] / [(x - 2)(x + 1)] = [(a + b)x + (a - 2b)] / (x^2 - x - 2). Equating coefficients with the numerator 2x + 1: 1) a + b = 2, and 2) a - 2b = 1. Subtracting equation 2 from equation 1 gives 3b = 1 => b = 1/3. Then a = 2 - 1/3 = 5/3.

Question 18Number and algebra

If x + 1/x = 4, what is the value of x^3 + 1/x^3? x + 1/x = 4 のとき、x^3 + 1/x^3 の値を求めなさい。

  1. 48
  2. 56
  3. 64
  4. 52 Correct answer.

Explanation

Recall the identity: x^3 + 1/x^3 = (x + 1/x)^3 - 3*(x)*(1/x)*(x + 1/x) = (x + 1/x)^3 - 3(x + 1/x). Substituting x + 1/x = 4: 4^3 - 3(4) = 64 - 12 = 52.

Question 19Number and algebra

Find the sum of all positive integers less than 100 that are divisible by 3. 100未満の正の整数のうち、3の倍数であるものの総和を求めなさい。

  1. 1716
  2. 1650
  3. 1683 Correct answer.
  4. 3300

Explanation

The positive multiples of 3 less than 100 form an arithmetic progression: 3, 6, 9, ..., 99. The number of terms is n = 99 / 3 = 33. The sum of this AP is S = (n/2) * (first + last) = (33/2) * (3 + 99) = (33/2) * 102 = 33 * 51 = 1683.

Question 20Number and algebra

Solve the system of equations for real values (x, y): x + y = 5 and x^2 + y^2 = 13. 連立方程式 x + y = 5, x^2 + y^2 = 13 の実数解の組 (x, y) を求めなさい。

  1. (3, 2) and (2, 3) Correct answer.
  2. (4, 1) and (1, 4)
  3. (-3, 8) and (8, -3)
  4. (5, 0) and (0, 5)

Explanation

From (x + y)^2 = x^2 + 2xy + y^2, we have 5^2 = 13 + 2xy => 25 - 13 = 2xy => 2xy = 12 => xy = 6. Since x and y have sum 5 and product 6, they are roots of t^2 - 5t + 6 = 0 => (t - 2)(t - 3) = 0. Thus, {x, y} = {2, 3}, giving pairs (3, 2) and (2, 3).

Question 21Number and algebra

Determine the range of k for which the system has no real solutions: x^2 + y^2 = 4 and y = x + k. 円 x^2 + y^2 = 4 と直線 y = x + k が共有点を持たないような定数 k の値の範囲を求めなさい。

  1. -4 < k < 4
  2. -2√2 < k < 2√2
  3. k < -2√2, k > 2√2 Correct answer.
  4. k < -4, k > 4

Explanation

Substitute y = x + k into the circle equation: x^2 + (x + k)^2 = 4 => 2x^2 + 2kx + (k^2 - 4) = 0. For no intersection, the discriminant D/4 < 0: D/4 = k^2 - 2(k^2 - 4) = k^2 - 2k^2 + 8 = -k^2 + 8 < 0. Thus k^2 > 8, which means k < -2√2 or k > 2√2. (Equivalently, the distance from origin to line |k|/√2 > 2 => |k| > 2√2).

Question 22Number and algebra

Compute (1 + i)^8 and give its value in the form a + bi. 複素数 (1 + i)^8 の値を計算しなさい。

  1. 16i
  2. -16
  3. 16 + 16i
  4. 16 Correct answer.

Explanation

Compute (1 + i)^2 = 1 + 2i + i^2 = 2i. Then (1 + i)^8 = [(1 + i)^2]^4 = (2i)^4 = 2^4 * i^4 = 16 * 1 = 16.

Question 23Number and algebra

By mathematical induction, what is the value of the product: P_n = (1 - 1/4)(1 - 1/9)(1 - 1/16)...(1 - 1/n^2) for n >= 2? 積 P_n = (1 - 1/4)(1 - 1/9)...(1 - 1/n^2) (n >= 2) の値を n を用いて表しなさい。

  1. n / (2(n + 1))
  2. (n + 1) / n
  3. (n - 1) / (2n)
  4. (n + 1) / (2n) Correct answer.

Explanation

Each factor is (1 - 1/k^2) = (k^2 - 1)/k^2 = [(k - 1)(k + 1)] / (k * k). Writing out the product telescopes: [(1*3)/(2*2)] * [(2*4)/(3*3)] * [(3*5)/(4*4)] * ... * [((n - 1)(n + 1))/(n*n)]. All internal factors cancel, leaving the first denominator component 1/2 and the final numerator component (n + 1)/n: (1/2) * [(n + 1)/n] = (n + 1) / (2n).

Question 24Number and algebra

Find the constant term (independent of x) in the expansion of (x^2 - 2/x)^6. (x^2 - 2/x)^6 の展開式において、x を含まない項（定数項）を求めなさい。

  1. 240 Correct answer.
  2. 60
  3. -240
  4. 160

Explanation

The general term is 6_C_r * (x^2)^(6 - r) * (-2/x)^r = 6_C_r * (-2)^r * x^(12 - 2r - r) = 6_C_r * (-2)^r * x^(12 - 3r). For the constant term, set 12 - 3r = 0 => 3r = 12 => r = 4. Term = 6_C_4 * (-2)^4 = 15 * 16 = 240.

Question 25Number and algebra

Find the sum of all solutions to the equation: 2^(2x) - 5 * 2^x + 4 = 0. 方程式 2^(2x) - 5 * 2^x + 4 = 0 のすべての実数解の和を求めなさい。

  1. 4
  2. 3
  3. 2 Correct answer.
  4. 5

Explanation

Let t = 2^x with t > 0. The equation becomes t^2 - 5t + 4 = 0 => (t - 1)(t - 4) = 0 => t = 1 or t = 4. 1) 2^x = 1 => x = 0. 2) 2^x = 4 => x = 2. The sum of the solutions is 0 + 2 = 2.

Question 26Functions

Find the coordinates of the vertex of the parabola: y = -2x^2 + 8x - 3. 放物線 y = -2x^2 + 8x - 3 の頂点の座標を求めなさい。

  1. (-2, -27)
  2. (2, 1)
  3. (4, -3)
  4. (2, 5) Correct answer.

Explanation

Complete the square: y = -2(x^2 - 4x) - 3 = -2[(x - 2)^2 - 4] - 3 = -2(x - 2)^2 + 8 - 3 = -2(x - 2)^2 + 5. The vertex is at (2, 5).

Question 27Functions

Find the maximum and minimum values of the function y = x^2 - 4x + 1 on the closed interval 0 <= x <= 3. 定義域 0 <= x <= 3 における2次関数 y = x^2 - 4x + 1 の最大値と最小値を求めなさい。

  1. Maximum: 1, Minimum: -2
  2. Maximum: -2, Minimum: -3
  3. Maximum: 1, Minimum: -3 Correct answer.
  4. Maximum: 5, Minimum: -3

Explanation

Complete the square: y = (x - 2)^2 - 3. The vertex is at x = 2, which lies inside the interval [0, 3]. Thus, the minimum is y(2) = -3. Evaluate the boundaries: y(0) = 0 - 0 + 1 = 1, and y(3) = 3^2 - 4(3) + 1 = 9 - 12 + 1 = -2. Comparing values: the maximum is y(0) = 1, and the minimum is y(2) = -3.

Question 28Functions

Find the distance from the point (2, -1) to the line 3x - 4y + 5 = 0 in the Cartesian plane. 座標平面において、点 (2, -1) と直線 3x - 4y + 5 = 0 の距離を求めなさい。

  1. 2.5
  2. 5
  3. 15
  4. 3 Correct answer.

Explanation

The distance formula from point (x_0, y_0) to line ax + by + c = 0 is d = |a*x_0 + b*y_0 + c| / sqrt(a^2 + b^2). Substituting a = 3, b = -4, c = 5, x_0 = 2, y_0 = -1: d = |3(2) - 4(-1) + 5| / sqrt(3^2 + (-4)^2) = |6 + 4 + 5| / sqrt(9 + 16) = |15| / sqrt(25) = 15 / 5 = 3.

Question 29Functions

Find the center and radius of the circle defined by: x^2 + y^2 - 6x + 4y - 12 = 0. 円の方程式 x^2 + y^2 - 6x + 4y - 12 = 0 の中心の座標と半径を求めなさい。

  1. Center: (6, -4), Radius: 5
  2. Center: (-3, 2), Radius: 5
  3. Center: (3, -2), Radius: 5 Correct answer.
  4. Center: (3, -2), Radius: 25

Explanation

Complete the squares for x and y: (x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4 => (x - 3)^2 + (y + 2)^2 = 25. Thus, the center is (3, -2) and the radius is r = sqrt(25) = 5.

Question 30Functions

Find the equations of the tangent lines to the circle x^2 + y^2 = 5 passing through the external point (3, 1). 点 (3, 1) を通る円 x^2 + y^2 = 5 の接線の方程式を求めなさい。

  1. 2x - y - 5 = 0 and x + 2y - 5 = 0 Correct answer.
  2. 2x + y - 5 = 0 and x - 2y - 5 = 0
  3. 3x + y - 5 = 0 and x + 3y - 5 = 0
  4. 2x - y + 5 = 0 and x + 2y + 5 = 0

Explanation

Let the point of tangency on the circle be (x_1, y_1). The equation of the tangent line is x_1*x + y_1*y = 5. Since it passes through (3, 1): 3x_1 + y_1 = 5 => y_1 = 5 - 3x_1. Since (x_1, y_1) lies on the circle: x_1^2 + (5 - 3x_1)^2 = 5 => x_1^2 + 25 - 30x_1 + 9x_1^2 = 5 => 10x_1^2 - 30x_1 + 20 = 0 => x_1^2 - 3x_1 + 2 = 0 => (x_1 - 1)(x_1 - 2) = 0. Case 1: x_1 = 2 => y_1 = -1, yielding tangent line 2x - y = 5. Case 2: x_1 = 1 => y_1 = 2, yielding tangent line x + 2y = 5.

Question 31Functions

Solve for x: log_2(x) + log_2(x - 2) = 3. 方程式 log_2(x) + log_2(x - 2) = 3 を解きなさい。

  1. x = -2, 4
  2. x = 2
  3. x = 8
  4. x = 4 Correct answer.

Explanation

By domain constraints for logarithms, x > 0 and x - 2 > 0 => x > 2. Combine the logarithms: log_2[x(x - 2)] = 3 => x(x - 2) = 2^3 = 8 => x^2 - 2x - 8 = 0 => (x - 4)(x + 2) = 0 => x = 4 or x = -2. Since x > 2, x = -2 is an extraneous solution. Thus, x = 4 is the only valid solution.

Question 32Functions

Solve the logarithmic inequality: log_{0.5}(x - 1) > -2. 対数不等式 log_{0.5}(x - 1) > -2 を解きなさい。

  1. x > 5
  2. 1 < x < 5 Correct answer.
  3. 1 < x < 3
  4. x < 5

Explanation

Domain constraint: x - 1 > 0 => x > 1. Since the base 0.5 is strictly between 0 and 1, the inequality direction reverses when removing the logarithm: x - 1 < (0.5)^(-2). Since (0.5)^(-2) = (1/2)^(-2) = 4, we have x - 1 < 4 => x < 5. Combining with the domain constraint x > 1 yields 1 < x < 5.

Question 33Functions

Solve the trigonometric equation on 0 <= theta < 2*pi: 2*cos^2(theta) + sin(theta) - 1 = 0. 0 <= theta < 2*pi のとき、方程式 2*cos^2(theta) + sin(theta) - 1 = 0 を解きなさい。

  1. theta = pi/2, 7pi/6, 11pi/6 Correct answer.
  2. theta = pi/2, pi/6, 5pi/6
  3. theta = 7pi/6, 11pi/6
  4. theta = pi/6, 5pi/6, 3pi/2

Explanation

Use cos^2(theta) = 1 - sin^2(theta): 2(1 - sin^2(theta)) + sin(theta) - 1 = 0 => 2 - 2*sin^2(theta) + sin(theta) - 1 = 0 => -2*sin^2(theta) + sin(theta) + 1 = 0 => 2*sin^2(theta) - sin(theta) - 1 = 0 => (2*sin(theta) + 1)(sin(theta) - 1) = 0. Case 1: sin(theta) = 1 => theta = pi/2. Case 2: sin(theta) = -1/2 => theta = 7pi/6 or 11pi/6. Thus, theta = pi/2, 7pi/6, 11pi/6.

Question 34Functions

Express y = sin(x) + sqrt(3)*cos(x) in the composite form r*sin(x + alpha), where r > 0 and 0 <= alpha < 2*pi. 関数 y = sin(x) + sqrt(3)*cos(x) を r*sin(x + alpha) (r > 0, 0 <= alpha < 2*pi) の形に変形しなさい。

  1. 2*sin(x + pi/6)
  2. 2*sin(x + pi/3) Correct answer.
  3. sqrt(2)*sin(x + pi/3)
  4. 4*sin(x + pi/3)

Explanation

The amplitude is r = sqrt(1^2 + (sqrt(3))^2) = sqrt(1 + 3) = 2. Factor out 2: 2 * [(1/2)*sin(x) + (sqrt(3)/2)*cos(x)]. Using the angle addition formula where cos(alpha) = 1/2 and sin(alpha) = sqrt(3)/2, we find alpha = pi/3. Thus, y = 2*sin(x + pi/3).

Question 35Functions

Find the area of the region in the xy-plane defined by the system of inequalities: y >= x^2 - 2x and y <= x. 不等式 y >= x^2 - 2x かつ y <= x の表す領域の面積を求めなさい。

  1. 27/2
  2. 9
  3. 9/2 Correct answer.
  4. 3/2

Explanation

Find intersection points of y = x^2 - 2x and y = x: x^2 - 2x = x => x^2 - 3x = 0 => x(x - 3) = 0 => x = 0 and x = 3. On [0, 3], the line y = x is above the parabola. The enclosed area is given by the Archimedean 1/6 formula: Area = integral_{0}^{3} (x - (x^2 - 2x)) dx = integral_{0}^{3} (-x^2 + 3x) dx = (1/6)*(3 - 0)^3 = 27 / 6 = 9/2.

Question 36Functions

Determine the range of the function f(x) = 9^x - 4 * 3^x + 5 for all real numbers x. すべての実数 x に対して、関数 f(x) = 9^x - 4 * 3^x + 5 の取り得る値の範囲（値域）を求めなさい。

  1. All real numbers
  2. f(x) >= 1 Correct answer.
  3. f(x) >= 5
  4. f(x) > 0

Explanation

Let t = 3^x. Since x is a real number, t > 0. The function becomes g(t) = t^2 - 4t + 5 = (t - 2)^2 + 1. Since the vertex is at t = 2 (which is > 0 and corresponds to x = log_3(2)), the minimum value is achieved at t = 2, where g(2) = 1. As t -> infty, g(t) -> infty. Thus, the range is f(x) >= 1.

Question 37Functions

Find the equation of the line passing through the point (1, 3) and perpendicular to the line 2x - 5y + 7 = 0. 点 (1, 3) を通り、直線 2x - 5y + 7 = 0 に垂直な直線の方程式を求めなさい。

  1. 2x - 5y + 13 = 0
  2. 2x + 5y - 17 = 0
  3. 5x + 2y - 11 = 0 Correct answer.
  4. 5x - 2y + 1 = 0

Explanation

The given line 2x - 5y + 7 = 0 has slope m_1 = 2/5. A perpendicular line has slope m_2 = -1/m_1 = -5/2. The point-slope form through (1, 3) is y - 3 = (-5/2)(x - 1) => 2(y - 3) = -5(x - 1) => 2y - 6 = -5x + 5 => 5x + 2y - 11 = 0.

Question 38Functions

Find the length of the chord intercepted by the circle x^2 + y^2 = 25 on the line x + y - 5 = 0. 円 x^2 + y^2 = 25 が直線 x + y - 5 = 0 から切り取る線分（弦）の長さを求めなさい。

  1. 5
  2. 5√3
  3. 10
  4. 5√2 Correct answer.

Explanation

The circle has center (0, 0) and radius R = 5. The distance d from the origin to the line x + y - 5 = 0 is d = |0 + 0 - 5| / sqrt(1^2 + 1^2) = 5 / sqrt(2). By the Pythagorean theorem on half the chord: (L/2)^2 = R^2 - d^2 = 25 - (25/2) = 25/2. Thus L/2 = sqrt(25/2) = 5 / sqrt(2). The full chord length is L = 2 * (5 / sqrt(2)) = 10 / sqrt(2) = 5√2.

Question 39Functions

Find the maximum value of z = 2x + 3y subject to the constraints: x >= 0, y >= 0, x + 2y <= 8, and 2x + y <= 10. 条件 x >= 0, y >= 0, x + 2y <= 8, 2x + y <= 10 のもとで、z = 2x + 3y の最大値を求めなさい。

  1. 15
  2. 10
  3. 14 Correct answer.
  4. 12

Explanation

The feasible region is a polygon with vertices: 1) (0, 0) => z = 0. 2) (0, 4) from x + 2y = 8 => z = 2(0) + 3(4) = 12. 3) (5, 0) from 2x + y = 10 => z = 2(5) + 3(0) = 10. 4) Intersection of x + 2y = 8 and 2x + y = 10: multiply the second by 2: 4x + 2y = 20. Subtracting: 3x = 12 => x = 4, then y = 2. At vertex (4, 2): z = 2(4) + 3(2) = 8 + 6 = 14. The maximum value is 14.

Question 40Functions

If log_{10}(2) = 0.3010 and log_{10}(3) = 0.4771, how many digits does 6^20 have in base 10? log_{10}(2) = 0.3010, log_{10}(3) = 0.4771 とするとき、6^20 は何桁の整数か求めなさい。

  1. 15桁
  2. 20桁
  3. 16桁 Correct answer.
  4. 17桁

Explanation

Take the common logarithm: log_{10}(6^20) = 20 * log_{10}(6) = 20 * [log_{10}(2) + log_{10}(3)] = 20 * [0.3010 + 0.4771] = 20 * 0.7781 = 15.562. Since 15 <= log_{10}(6^20) < 16, it follows that 10^15 <= 6^20 < 10^16. Therefore, 6^20 has 15 + 1 = 16 digits.

Question 41Functions

Find the period and amplitude of the trigonometric function y = 3*sin(4x - pi/2). 三角関数 y = 3*sin(4x - pi/2) の振幅と周期を求めなさい。

  1. Amplitude: 3, Period: pi/2 Correct answer.
  2. Amplitude: 3, Period: pi/4
  3. Amplitude: 3, Period: 2pi
  4. Amplitude: 6, Period: pi/2

Explanation

For y = A*sin(Bx - C), the amplitude is |A| = 3. The period of sin is 2pi / B. Here B = 4, so the period is 2pi / 4 = pi/2.

Question 42Functions

What is the locus of a point P(x, y) that moves such that the ratio of its distances from A(-2, 0) and B(4, 0) is 1:2? 2点 A(-2, 0), B(4, 0) からの距離の比が 1:2 である点 P の軌跡を求めなさい。

  1. Circle: (x + 4)^2 + y^2 = 16 (Circle of Apollonius) Correct answer.
  2. Circle: (x - 2)^2 + y^2 = 9
  3. Circle: x^2 + y^2 = 8
  4. Line: x = 0

Explanation

Set up the distance ratio: AP : BP = 1 : 2 => 2*AP = BP => 4*AP^2 = BP^2. We have 4[(x + 2)^2 + y^2] = (x - 4)^2 + y^2 => 4(x^2 + 4x + 4 + y^2) = x^2 - 8x + 16 + y^2 => 4x^2 + 16x + 16 + 4y^2 = x^2 - 8x + 16 + y^2 => 3x^2 + 24x + 3y^2 = 0 => x^2 + 8x + y^2 = 0 => (x + 4)^2 + y^2 = 16. This is the Circle of Apollonius with center (-4, 0) and radius 4.

Question 43Functions

Find the range of a such that the graph of y = x^2 - 2ax + a + 2 is strictly above the x-axis for all real x. すべての実数 x に対して、放物線 y = x^2 - 2ax + a + 2 が常に x 軸より上側にあるような定数 a の値の範囲を求めなさい。

  1. a < -2, a > 1
  2. a < -1, a > 2
  3. -1 < a < 2 Correct answer.
  4. -2 < a < 1

Explanation

For the parabola (which opens upward) to remain strictly above the x-axis, it must have no real x-intercepts, meaning its discriminant D < 0. Here, D/4 = (-a)^2 - 1*(a + 2) = a^2 - a - 2 < 0. Factor: (a - 2)(a + 1) < 0. This inequality holds strictly between the roots: -1 < a < 2.

Question 44Functions

Solve the exponential inequality: (1/3)^(2x) - 4 * (1/3)^x + 3 >= 0. 指数不等式 (1/3)^(2x) - 4 * (1/3)^x + 3 >= 0 を解きなさい。

  1. x <= -1, x >= 0 Correct answer.
  2. 0 <= x <= 1
  3. -1 <= x <= 0
  4. x <= 0, x >= 1

Explanation

Let u = (1/3)^x > 0. The inequality is u^2 - 4u + 3 >= 0 => (u - 1)(u - 3) >= 0. Thus, u <= 1 or u >= 3. 1) (1/3)^x <= 1 => (1/3)^x <= (1/3)^0. Since the base 1/3 < 1, the inequality reverses: x >= 0. 2) (1/3)^x >= 3 => (1/3)^x >= (1/3)^(-1). Reversing gives x <= -1. Combining: x <= -1 or x >= 0.

Question 45Functions

Find the coordinates of the two foci of the ellipse: x^2 / 25 + y^2 / 16 = 1. 楕円 x^2 / 25 + y^2 / 16 = 1 の焦点の座標を求めなさい。

  1. (5, 0) and (-5, 0)
  2. (sqrt(41), 0) and (-sqrt(41), 0)
  3. (0, 3) and (0, -3)
  4. (3, 0) and (-3, 0) Correct answer.

Explanation

For an ellipse x^2/a^2 + y^2/b^2 = 1 with a > b, a^2 = 25 and b^2 = 16. The major axis lies on the x-axis. The focal distance is c = sqrt(a^2 - b^2) = sqrt(25 - 16) = sqrt(9) = 3. Therefore, the foci are located at (3, 0) and (-3, 0).

Question 46Functions

Find the sum of all solutions of tan(2x) = 1 in the interval 0 <= x < pi. 0 <= x < pi の範囲において、方程式 tan(2x) = 1 のすべての解の和を求めなさい。

  1. 3pi/4 Correct answer.
  2. pi
  3. pi/2
  4. 5pi/4

Explanation

Since 0 <= x < pi, the angle 2x satisfies 0 <= 2x < 2pi. tan(2x) = 1 occurs when 2x = pi/4 or 2x = 5pi/4. Dividing by 2: x = pi/8 or x = 5pi/8. Both lie in [0, pi). Their sum is pi/8 + 5pi/8 = 6pi/8 = 3pi/4.

Question 47Functions

Find the point of intersection between the line y = 2x - 1 and the inverse function of f(x) = 3x - 5. 関数 f(x) = 3x - 5 の逆関数 f^(-1)(x) と、直線 y = 2x - 1 の交点の座標を求めなさい。

  1. (7/5, 9/5)
  2. (8/5, 11/5) Correct answer.
  3. (1, 1)
  4. (2, 3)

Explanation

To find the inverse function: y = 3x - 5 => 3x = y + 5 => x = (y + 5)/3. Swapping variables gives f^(-1)(x) = (x + 5)/3. Set equal to y = 2x - 1: (x + 5)/3 = 2x - 1 => x + 5 = 6x - 3 => 5x = 8 => x = 8/5. Substitute into y = 2(8/5) - 1 = 16/5 - 5/5 = 11/5. The intersection point is (8/5, 11/5).

Question 48Functions

What is the maximum value of f(x) = sin^2(x) + cos(x) on the real line? 実数全体において、関数 f(x) = sin^2(x) + cos(x) の最大値を求めなさい。

  1. 3/2
  2. 2
  3. 5/4 Correct answer.
  4. 1

Explanation

Substitute sin^2(x) = 1 - cos^2(x): f(x) = 1 - cos^2(x) + cos(x) = -(cos^2(x) - cos(x)) + 1 = -(cos(x) - 1/2)^2 + 1/4 + 1 = -(cos(x) - 1/2)^2 + 5/4. Since cos(x) = 1/2 is within the range [-1, 1], the function achieves its maximum value of 5/4.

Question 49Functions

For which real values of k does the circle (x - k)^2 + (y - 2)^2 = 9 meet (intersect or touch) the x-axis? 円 (x - k)^2 + (y - 2)^2 = 9 が x 軸と交わる（または接する）条件を調べなさい。

  1. The circle intersects the x-axis for all real k Correct answer.
  2. Only when |k| <= 3
  3. Never intersects for any k
  4. Only when k = 0

Explanation

The circle has radius R = 3 and center (k, 2). The distance from the center (k, 2) to the x-axis (the line y = 0) is d = |2 - 0| = 2. Since d = 2 < R = 3, the distance from the center to the x-axis is strictly less than the radius regardless of the horizontal position k. Therefore, the circle intersects the x-axis at two points for all real values of k.

Question 50Functions

Find the asymptotes of the hyperbola: x^2 / 16 - y^2 / 9 = 1. 双曲線 x^2 / 16 - y^2 / 9 = 1 の漸近線の方程式を求めなさい。

  1. y = (3/4)x and y = -(3/4)x Correct answer.
  2. y = (4/3)x and y = -(4/3)x
  3. y = 3x and y = -3x
  4. y = (9/16)x and y = -(9/16)x

Explanation

For a standard hyperbola x^2/a^2 - y^2/b^2 = 1, the equations of the asymptotes are obtained by replacing 1 with 0: x^2/a^2 - y^2/b^2 = 0 => y = +-(b/a)x. Here a^2 = 16 (so a = 4) and b^2 = 9 (so b = 3). Thus, the asymptotes are y = +-(3/4)x.