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NSSCAS Mathematics (8227) study materials
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NSSCAS Mathematics (8227) practice questions

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Question 1Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Simplify √50 + √18.

  1. 8√2 Correct answer.
  2. 16√2
  3. 4√13
  4. 8√8

Explanation

√50 = √(25×2)=5√2 and √18=√(9×2)=3√2. The surds are now like terms, so add coefficients: 5√2+3√2=8√2.

Question 2Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Solve 2^(3x - 1) = 16 for x.

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

Explanation

Write 16 as 2^4, giving 2^(3x-1)=2^4. Since the bases are equal, equate exponents: 3x-1=4, so 3x=5 and x=5/3.

Question 3Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Evaluate log_2 32 + log_3(1/9).

  1. 3 Correct answer.
  2. -3
  3. 7
  4. 0

Explanation

log_2 32 = 5 because 2^5=32. Also log_3(1/9) = -2 because 3^(-2)=1/9, so the sum is 5 + (-2)=3.

Question 4Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

The quadratic equation x^2 + kx + 9 = 0 has two distinct real roots. Which condition on k must hold?

  1. -6 < k < 6
  2. |k| > 6 Correct answer.
  3. k > 6
  4. |k| < 6

Explanation

A quadratic has two distinct real roots when its discriminant b^2 - 4ac is positive. Here b=k and a=1, c=9, so k^2 - 36 > 0. This is equivalent to |k| > 6.

Question 5Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

When f(x)=2x^3 - 5x^2 + ax - 1 is divided by x - 1, the remainder is 3. Find a.

  1. 11
  2. 7 Correct answer.
  3. 3
  4. -1

Explanation

By the remainder theorem, the remainder on division by x-1 is f(1). Compute f(1)=2-5+a-1=a-4. Setting a-4=3 gives a=7.

Question 6Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Solve the inequality x^2 - 5x + 6 < 0.

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

Explanation

Factor: x^2 -5x+6=(x-2)(x-3). The parabola opens upward and is below the x-axis between its roots, so the solution is 2 < x < 3.

Question 7Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Find the coefficient of x^3 in the expansion of (2 + x)^5.

  1. 10
  2. 20
  3. 40 Correct answer.
  4. 80

Explanation

The general term is C(5,r) 2^(5-r) x^r. For x^3, r=3, so the coefficient is C(5,3)×2^2=10×4=40.

Question 8Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Solve log_2(x + 3) + log_2(x - 1) = 3.

  1. 1 + 2√3
  2. -1 ± 2√3
  3. 4
  4. 2√3 - 1 Correct answer.

Explanation

The equation requires x>1. Combining logs gives log_2((x+3)(x-1))=3, so (x+3)(x-1)=8. This simplifies to x^2+2x-11=0, whose roots are -1±2√3; only -1+2√3 = 2√3-1 satisfies x>1.

Question 9Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

The polynomial f(x)=x^3 + px^2 - 4x + q is divisible by x - 2 and leaves remainder -6 when divided by x + 1. Find p and q.

  1. p = -3, q = 12
  2. p = 3, q = -12 Correct answer.
  3. p = 3, q = 12
  4. p = -12, q = 3

Explanation

Divisibility by x-2 gives f(2)=0: 8+4p-8+q=4p+q=0. Remainder -6 on division by x+1 gives f(-1)=-6: -1+p+4+q=p+q+3=-6, so p+q=-9. Solving q=-4p with p+q=-9 gives p=3 and q=-12.

Question 10Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Express 2x^2 - 12x + 7 in the form a(x - b)^2 + c.

  1. (2x - 3)^2 - 2
  2. 2(x-3)^2 - 11 Correct answer.
  3. 2(x - 3)^2 + 7
  4. 2(x + 3)^2 - 11

Explanation

Factor 2 from the x^2 and x terms: 2(x^2 -6x)+7. Complete the square: x^2 -6x=(x-3)^2-9, so the expression becomes 2[(x-3)^2-9]+7=2(x-3)^2-11.

Question 11Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Rationalise the denominator and simplify 6/(√5 − √2).

  1. 6(√5 + √2)
  2. 2(√5 + √2) Correct answer.
  3. 2(√5 − √2)
  4. (√5 + √2)/2

Explanation

Multiply numerator and denominator by the conjugate √5 + √2. The denominator becomes 5 − 2 = 3, so the expression is 6(√5 + √2)/3 = 2(√5 + √2).

Question 12Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Evaluate 16^(−3/4).

  1. 1/2
  2. −8
  3. 1/8 Correct answer.
  4. 8

Explanation

Write 16 as 2^4, so 16^(−3/4) = (2^4)^(−3/4) = 2^(−3) = 1/8. The negative exponent gives a reciprocal, not a negative value.

Question 13Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Given that log_2 3 = p, express log_2 12 in terms of p.

  1. p/2
  2. p + 4
  3. 2p
  4. p + 2 Correct answer.

Explanation

Since 12 = 4 × 3, the product law gives log_2 12 = log_2 4 + log_2 3. Now log_2 4 = 2, so the result is p + 2.

Question 14Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

The quadratic equation 2x^2 − 8x + 5 = 0 has roots α and β. Find the value of α + β and αβ.

  1. α + β = −4, αβ = 5/2
  2. α + β = 4, αβ = −5/2
  3. α + β = 4, αβ = 5/2 Correct answer.
  4. α + β = 8, αβ = 5

Explanation

For ax^2 + bx + c = 0, the sum of roots is −b/a and the product is c/a. Here a = 2, b = −8, c = 5, so α + β = 8/2 = 4 and αβ = 5/2.

Question 15Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Solve 4^(x + 1) = 8^(2x − 1).

  1. x = 5/4 Correct answer.
  2. x = 1
  3. x = 4/5
  4. x = −5/4

Explanation

Express both sides with base 2: 4^(x + 1) = 2^(2x + 2) and 8^(2x − 1) = 2^(6x − 3). Equating exponents gives 2x + 2 = 6x − 3, so 4x = 5 and x = 5/4.

Question 16Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Solve log_2 x + log_4 x = 3.

  1. x = 8
  2. x = 4 Correct answer.
  3. x = 16
  4. x = 2

Explanation

Use change of base: log_4 x = (log_2 x)/(log_2 4) = (1/2)log_2 x. Thus (3/2)log_2 x = 3, so log_2 x = 2 and x = 4.

Question 17Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

The polynomial f(x) = x^3 − 3x^2 + ax + b has (x − 1)^2 as a factor. Find a and b.

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

Explanation

A repeated factor (x − 1)^2 means both f(1) = 0 and f′(1) = 0. From f(1) = 1 − 3 + a + b = 0 we get a + b = 2, and f′(x) = 3x^2 − 6x + a gives f′(1) = a − 3 = 0, so a = 3 and b = −1.

Question 18Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Solve the inequality (x − 1)/(x + 3) ≥ 2.

  1. −3 < x ≤ −7
  2. −7 ≤ x < −3 Correct answer.
  3. −7 < x < −3
  4. x ≤ −7 or x > −3

Explanation

Subtract 2 from both sides and combine over the common denominator: (x − 1)/(x + 3) − 2 = (−x − 7)/(x + 3) = −(x + 7)/(x + 3). The inequality becomes (x + 7)/(x + 3) ≤ 0. The critical values are −7 and −3; testing intervals gives −7 ≤ x < −3, including −7 but excluding −3.

Question 19Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Find the term independent of x in the expansion of (2x − 3/x)^6.

  1. −2160
  2. 2160
  3. 4320
  4. −4320 Correct answer.

Explanation

The general term is C(6,r)(2x)^(6−r)(−3/x)^r = C(6,r)2^(6−r)(−3)^r x^(6−2r). The term is independent of x when 6 − 2r = 0, so r = 3. Its coefficient is C(6,3) × 2^3 × (−3)^3 = 20 × 8 × (−27) = −4320.

Question 20Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

The roots of x^2 + (k − 2)x + (k + 1) = 0 differ by 3. Find the possible values of k.

  1. k = −1 only
  2. k = 9 or k = −1 Correct answer.
  3. k = 9 only
  4. k = 3 or k = −3

Explanation

If the roots are α and β, then (α − β)^2 = (α + β)^2 − 4αβ. Here α + β = 2 − k and αβ = k + 1, so (2 − k)^2 − 4(k + 1) = 9. Expanding gives k^2 − 8k − 9 = 0, hence (k − 9)(k + 1) = 0, so k = 9 or k = −1.

Question 21Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Expand and simplify (√5 + 2)(√5 − 2).

  1. 5 − 4√5
  2. 4√5
  3. 1 Correct answer.
  4. 9

Explanation

This is a difference of two squares: (a+b)(a−b)=a²−b², with a=√5 and b=2. Hence the product is 5−4=1. The cross terms cancel, so no surd remains.

Question 22Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Evaluate 27^(2/3).

  1. 81
  2. 9 Correct answer.
  3. 18
  4. 6

Explanation

Write 27^(2/3) as (27^(1/3))^2. Since 27^(1/3)=3, squaring gives 3²=9. The denominator 3 denotes a cube root, and the numerator 2 denotes a square.

Question 23Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Which logarithmic equation is equivalent to 4^3 = 64?

  1. log_3 64 = 4
  2. log_4 3 = 64
  3. log_4 64 = 3 Correct answer.
  4. log_64 4 = 3

Explanation

By definition, log_a b = c is equivalent to a^c = b. Here a=4, b=64 and c=3, so the equivalent statement is log_4 64 = 3. The base of the logarithm is the base that is raised to the power.

Question 24Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Solve log_3(x + 2) − log_3 x = 1.

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

Explanation

Combine the logarithms using log_a M − log_a N = log_a(M/N). This gives log_3((x+2)/x)=1, so (x+2)/x=3. Solving gives x=1, which satisfies x>0 for the original logarithms.

Question 25Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

The equation x^2 − 6x + 4 = 0 has roots α and β. Find the value of α² + β².

  1. 28 Correct answer.
  2. 32
  3. 44
  4. 20

Explanation

For x²−6x+4=0, the sum of roots is α+β=6 and the product is αβ=4. Use α²+β²=(α+β)²−2αβ. Substituting gives 36−8=28.

Question 26Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Find the remainder when f(x) = x^4 − 3x^2 + 2x − 5 is divided by x + 2.

  1. 5
  2. −13
  3. 3
  4. −5 Correct answer.

Explanation

The remainder theorem states that the remainder on division by x+2 is f(−2). Substitute: (−2)^4−3(−2)^2+2(−2)−5=16−12−4−5. This simplifies to −5, so the remainder is −5.

Question 27Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Solve the inequality |2x − 5| < 7.

  1. x > −1
  2. x < −1 or x > 6
  3. x < 6
  4. −1 < x < 6 Correct answer.

Explanation

For |u|<a with a>0, the inequality is equivalent to −a<u<a. Thus −7<2x−5<7. Adding 5 gives −2<2x<12, and dividing by 2 gives −1<x<6.

Question 28Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Find the coefficient of x^4 in the expansion of (1 + 2x)^7.

  1. 35
  2. 1120
  3. 280
  4. 560 Correct answer.

Explanation

The general term in (1+2x)^7 is C(7,r)(2x)^r. For x^4 take r=4, giving C(7,4)×2^4 x^4. Since C(7,4)=35 and 2^4=16, the coefficient is 35×16=560.

Question 29Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Solve 2^(2x) − 5(2^x) + 4 = 0.

  1. x = 0 or x = 2 Correct answer.
  2. x = 0 only
  3. x = 2 only
  4. x = 1 or x = 4

Explanation

Let y=2^x, so 2^(2x)=y². The equation becomes y²−5y+4=0, which factorises as (y−1)(y−4)=0. Hence 2^x=1 or 2^x=4, giving x=0 or x=2.

Question 30Algebra (surds, indices, logarithms, quadratics, polynomials, inequalities, binomial expansion)

Solve √(2x + 3) − √(x + 1) = 1.

  1. x = 3 only
  2. x = −1 only
  3. There is no real solution
  4. x = −1 or x = 3 Correct answer.

Explanation

Isolate one square root and square both sides: √(2x+3)=1+√(x+1), leading to x+1=2√(x+1). Squaring again gives (x+1)²=4(x+1), so (x+1)(x−3)=0. Checking x=−1 and x=3 in the original equation shows both are valid.

Question 31Functions and graphs (domain and range, composite and inverse functions, transformations)

The function f is defined by f(x) = 1/(x^2 − 9). What is the domain of f?

  1. x ∈ ℝ, x ≠ ±3 Correct answer.
  2. x ∈ ℝ, x > 3
  3. x ∈ ℝ, x ≠ 3
  4. x ∈ ℝ

Explanation

The denominator must be non-zero, so x^2 − 9 ≠ 0. Solving x^2 = 9 gives x = 3 or x = −3, so both values must be excluded. Therefore the domain is all real numbers except ±3.

Question 32Functions and graphs (domain and range, composite and inverse functions, transformations)

The graph of y = f(x) is transformed to the graph of y = f(x − 2) + 3. Which description of the transformation is correct?

  1. A translation of 2 units to the right and 3 units down
  2. A translation of 2 units to the right and 3 units up Correct answer.
  3. A translation of 2 units to the left and 3 units down
  4. A translation of 2 units to the left and 3 units up

Explanation

Replacing x by x − 2 shifts the graph horizontally in the positive x-direction by 2 units. Adding 3 outside the function shifts the graph vertically upwards by 3 units. The combined transformation is therefore 2 right and 3 up.

Question 33Functions and graphs (domain and range, composite and inverse functions, transformations)

What is the range of the function f(x) = x^2 − 4x + 7 for x ∈ ℝ?

  1. (−∞, 3]
  2. [3, ∞) Correct answer.
  3. [7, ∞)
  4. (−∞, 7]

Explanation

Complete the square: x^2 − 4x + 7 = (x − 2)^2 + 3. Since (x − 2)^2 ≥ 0, the smallest value is 3, attained at x = 2. The quadratic opens upwards, so the range is [3, ∞).

Question 34Functions and graphs (domain and range, composite and inverse functions, transformations)

The functions f and g are defined by f(x) = 2x + 1 and g(x) = x^2 − 3. What is fg(x), meaning f(g(x))?

  1. 2x^2 − 2
  2. 4x^2 + 4x − 2
  3. 2x^2 − 5 Correct answer.
  4. x^2 − 2

Explanation

Apply g first: g(x) = x^2 − 3. Then substitute this into f: f(x^2 − 3) = 2(x^2 − 3) + 1 = 2x^2 − 5.

Question 35Functions and graphs (domain and range, composite and inverse functions, transformations)

The function f is defined by f(x) = (3x − 2)/(x + 1) for x ≠ −1. What is f^{-1}(x)?

  1. (x + 2)/(3 − x), x ≠ 3 Correct answer.
  2. (x + 1)/(3x − 2), x ≠ 2/3
  3. (x − 2)/(x + 3), x ≠ −3
  4. (3x + 2)/(x − 1), x ≠ 1

Explanation

Let y = (3x − 2)/(x + 1). Rearranging gives y(x + 1) = 3x − 2, so x(y − 3) = −2 − y. Hence x = (y + 2)/(3 − y), and swapping x and y gives f^{-1}(x) = (x + 2)/(3 − x), with x ≠ 3.

Question 36Functions and graphs (domain and range, composite and inverse functions, transformations)

What is the domain of the function f(x) = √(5 − 2x)?

  1. x ≥ 2/5
  2. x ≤ 5/2 Correct answer.
  3. x ≤ 2/5
  4. x ≥ 5/2

Explanation

For a real square root, the expression under the root must be non-negative: 5 − 2x ≥ 0. Solving gives 2x ≤ 5, so x ≤ 5/2. The domain is therefore x ≤ 5/2.

Question 37Functions and graphs (domain and range, composite and inverse functions, transformations)

What is the range of the function f(x) = 2^(x + 1) − 3?

  1. (−∞, −3)
  2. [−3, ∞)
  3. (3, ∞)
  4. (−3, ∞) Correct answer.

Explanation

For every real x, 2^(x + 1) > 0, and it can take every positive value. Subtracting 3 therefore gives values strictly greater than −3, and every such value is attained. Hence the range is (−3, ∞).

Question 38Functions and graphs (domain and range, composite and inverse functions, transformations)

The graph of y = f(x) is transformed to the graph of y = f(−x) + 2. Which transformation is described?

  1. A reflection in the y-axis followed by a translation of 2 units up Correct answer.
  2. A reflection in the x-axis followed by a translation of 2 units up
  3. A reflection in the x-axis followed by a translation of 2 units down
  4. A reflection in the y-axis followed by a translation of 2 units down

Explanation

Replacing x by −x reflects the graph in the y-axis, because the sign of the input is changed. Adding 2 outside the function moves the graph vertically upwards by 2 units. Thus the transformation is a y-axis reflection followed by a translation 2 up.

Question 39Functions and graphs (domain and range, composite and inverse functions, transformations)

The functions f and g are defined by f(x) = √x and g(x) = 1/(x − 2). What is the domain of the composite function fg(x) = f(g(x))?

  1. x > 2 Correct answer.
  2. x ≥ 2
  3. x ≠ 2
  4. x > 0

Explanation

First g(x) requires x ≠ 2. Then f(g(x)) = √(1/(x − 2)) requires the input to f to be non-negative, so 1/(x − 2) ≥ 0. This fraction is positive only when x − 2 > 0, giving x > 2; the value x = 2 is already excluded.

Question 40Functions and graphs (domain and range, composite and inverse functions, transformations)

The function f is defined by f(x) = x^2 + 4x + 1 for x ≥ −2. What is f^{-1}(x)?

  1. −2 + √(x + 3), x ≥ −3 Correct answer.
  2. −2 − √(x + 3), x ≥ −3
  3. −2 + √(x − 3), x ≥ 3
  4. 2 + √(x + 3), x ≥ −3

Explanation

Complete the square: f(x) = (x + 2)^2 − 3. For x ≥ −2, the range is [−3, ∞) and x + 2 ≥ 0. Solving y = (x + 2)^2 − 3 gives x + 2 = √(y + 3), so f^{-1}(x) = −2 + √(x + 3) for x ≥ −3.

Question 41Functions and graphs (domain and range, composite and inverse functions, transformations)

The function f is defined by f(x) = 1/√(x − 3). What is the domain of f?

  1. x > 3 Correct answer.
  2. x < 3
  3. x ≥ 3
  4. x ≠ 3

Explanation

For the square root to be defined, x − 3 must be positive, and it cannot be zero because it is in the denominator. Hence x − 3 > 0, giving x > 3.

Question 42Functions and graphs (domain and range, composite and inverse functions, transformations)

What is the range of the function f(x) = |x − 5| + 1 for x ∈ ℝ?

  1. [5, ∞)
  2. [0, ∞)
  3. [1, ∞) Correct answer.
  4. (1, ∞)

Explanation

The absolute value |x − 5| is never negative and reaches 0 when x = 5. Therefore f(x) is at least 1 and can take every value from 1 upward.

Question 43Functions and graphs (domain and range, composite and inverse functions, transformations)

The functions f and g are defined by f(x) = 3x − 2 and g(x) = x². What is the value of f(g(−2))?

  1. 4
  2. 64
  3. 10 Correct answer.
  4. −8

Explanation

Work from the inside outward: g(−2) = (−2)² = 4, then f(4) = 3(4) − 2 = 10.

Question 44Functions and graphs (domain and range, composite and inverse functions, transformations)

The function f is defined by f(x) = (2x − 7)/3. What is f⁻¹(x)?

  1. (3x + 7)/2 Correct answer.
  2. (3x − 7)/2
  3. (7 − 3x)/2
  4. (2x + 7)/3

Explanation

Let y = (2x − 7)/3. Rearranging gives 3y + 7 = 2x, so x = (3y + 7)/2. Replacing y by x gives f⁻¹(x) = (3x + 7)/2.

Question 45Functions and graphs (domain and range, composite and inverse functions, transformations)

The graph of y = f(x) is transformed to the graph of y = f(2x) + 1. Which description of the transformation is correct?

  1. A horizontal compression by scale factor 1/2 followed by a translation of 1 unit up Correct answer.
  2. A horizontal stretch by scale factor 1/2 followed by a translation of 1 unit right
  3. A horizontal compression by scale factor 1/2 followed by a translation of 1 unit down
  4. A horizontal stretch by scale factor 2 followed by a translation of 1 unit up

Explanation

Replacing x by 2x compresses the graph horizontally by scale factor 1/2 because each x-value is reached twice as fast. Adding 1 outside the function shifts the graph up by 1.

Question 46Functions and graphs (domain and range, composite and inverse functions, transformations)

The functions f and g are defined by f(x) = 1/(x − 4) and g(x) = √x. What is the domain of the composite function f(g(x))?

  1. x ≥ 0 only
  2. x ≥ 0 and x ≠ 16 Correct answer.
  3. x > 0 and x ≠ 16
  4. x ≥ 0 and x ≠ 4

Explanation

The outer function 1/(x − 4) is undefined when its input equals 4, and the square root requires x ≥ 0. Thus √x must not equal 4, so x ≠ 16, while x ≥ 0 remains required.

Question 47Functions and graphs (domain and range, composite and inverse functions, transformations)

What is the range of the function f(x) = 1/(x² + 2) for x ∈ ℝ?

  1. [0, 1/2]
  2. (0, 1/2] Correct answer.
  3. [1/2, ∞)
  4. (0, 1/2)

Explanation

Since x² ≥ 0, x² + 2 ≥ 2, so its reciprocal is positive and at most 1/2. As x² + 2 becomes arbitrarily large, the reciprocal approaches 0 but never reaches it.

Question 48Functions and graphs (domain and range, composite and inverse functions, transformations)

The graph of y = f(x) passes through the point (2, 5). Which point lies on the graph of y = 2f(x − 1) + 3?

  1. (2, 13)
  2. (3, 10)
  3. (3, 13) Correct answer.
  4. (1, 13)

Explanation

For y = 2f(x − 1) + 3, the old x-coordinate 2 moves to 2 + 1 = 3, and the old y-coordinate 5 becomes 2(5) + 3 = 13. Hence the corresponding point is (3, 13).

Question 49Functions and graphs (domain and range, composite and inverse functions, transformations)

The functions f and g are defined by f(x) = √(x − 2) and g(x) = x/(x − 3). What is the domain of the composite function g(f(x))?

  1. x > 2 and x ≠ 11
  2. x ≥ 2 and x ≠ 3
  3. x ≥ 2 and x ≠ 11 Correct answer.
  4. x ≥ 2 only

Explanation

For g(f(x)) we need x − 2 ≥ 0 so x ≥ 2. The outer function g has denominator f(x) − 3, so √(x − 2) cannot equal 3, giving x ≠ 11.

Question 50Functions and graphs (domain and range, composite and inverse functions, transformations)

The function f is defined by f(x) = x² + 2x + 3 for x ≤ −1. What is f⁻¹(x)?

  1. −1 − √(x − 2) Correct answer.
  2. 1 − √(x − 2)
  3. −1 + √(x − 2)
  4. 1 + √(x − 2)

Explanation

Complete the square: f(x) = (x + 1)² + 2. Since x ≤ −1, x + 1 ≤ 0, so x + 1 = −√(y − 2), giving f⁻¹(x) = −1 − √(x − 2).