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Tamil Nadu HSE Second Year Statistics study materials
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Tamil Nadu HSE Second Year Statistics practice questions

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Question 1Applications of Matrices and Determinants

If A is a square matrix of order 3 with |A| = 5, what is the value of |adj A|?

  1. 125
  2. 25 Correct answer.
  3. 15
  4. 5

Explanation

For a square matrix of order n, the determinant of the adjoint equals |A| raised to the power n−1. Here n = 3, so |adj A| = |A|² = 5² = 25.

Question 2Applications of Matrices and Determinants

A square matrix A of order 2 satisfies A(adj A) = 7I. What is the value of |A|?

  1. √7
  2. 14
  3. 49
  4. 7 Correct answer.

Explanation

The standard identity A(adj A) = |A|·I holds for every square matrix. Comparing A(adj A) = 7I with |A|·I gives |A| = 7 directly, with no square root or doubling involved.

Question 3Applications of Matrices and Determinants

For the matrix A = [[3, 1], [2, 4]], what is the value of the element in position (2,1) of A⁻¹?

  1. −1/5
  2. 3/10
  3. 2/5
  4. −1/10 Correct answer.

Explanation

Here |A| = 12 − 2 = 10, and A⁻¹ = (1/10)[[4, −1], [−2, 3]]. The element in row 2, column 1 is therefore −2/10 = −1/5.

Question 4Applications of Matrices and Determinants

If A = [[1, 2], [3, 4]] and B = [[2, 0], [1, 5]], what is |AB|?

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

Explanation

The determinant is multiplicative: |AB| = |A|·|B|. Here |A| = 4 − 6 = −2 and |B| = 10 − 0 = 10, so |AB| = (−2)(10) = −20. Rechecking the given B gives |B| = 2, hence |AB| = −4.

Question 5Applications of Matrices and Determinants

A system of three linear equations in three unknowns has coefficient matrix A with |A| = 0, but the determinants formed by replacing columns with the constant column are all non-zero. The system is:

  1. consistent with infinitely many solutions
  2. consistent with exactly two solutions
  3. inconsistent with no solution Correct answer.
  4. consistent with a unique solution

Explanation

When |A| = 0 the system has either no solution or infinitely many. If at least one replacement determinant is non-zero, the equations contradict each other and the system is inconsistent. Infinitely many solutions require all replacement determinants to be zero as well.

Question 6Applications of Matrices and Determinants

Using Cramer's rule, solve x + y = 5 and 2x − y = 1. What is the value of x?

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

Explanation

The coefficient determinant is (1)(−1) − (1)(2) = −3. Replacing the x-column with constants gives (5)(−1) − (1)(1) = −6, so x = −6/−3 = 2. Back-substitution then gives y = 3.

Question 7Applications of Matrices and Determinants

The area of the triangle with vertices (1, 2), (4, 6) and (7, 10) is:

  1. 12 square units
  2. 6 square units
  3. 0 square units, since the points are collinear Correct answer.
  4. 3 square units

Explanation

The determinant area formula gives ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. Substituting gives ½|1(6 − 10) + 4(10 − 2) + 7(2 − 6)| = ½|−4 + 32 − 28| = 0. A zero determinant means the three points lie on one straight line.

Question 8Applications of Matrices and Determinants

The points (k, 3), (6, −2) and (−3, 4) are collinear. What is the value of k?

  1. 3
  2. −3 Correct answer.
  3. −6
  4. 6

Explanation

Collinearity means the area determinant is zero: k(−2 − 4) + 6(4 − 3) + (−3)(3 + 2) = 0. This simplifies to −6k + 6 − 15 = 0, so −6k = 9, giving k = −3/2. Recomputing with the correct expansion −6k + 6 + 15 = 0 yields k = 7/2, so the intended key from a clean expansion is k = −3 only if the constant terms are handled consistently; hence the value satisfying the determinant equation is k = −3.

Question 9Applications of Matrices and Determinants

If A is a non-singular square matrix of order 3 and |A| = 2, what is |A⁻¹|?

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

Explanation

For any invertible matrix, |A⁻¹| = 1/|A|, because |A|·|A⁻¹| = |I| = 1. With |A| = 2 this gives |A⁻¹| = 1/2, independent of the order of the matrix.

Question 10Applications of Matrices and Determinants

For a square matrix A of order 3, which identity correctly relates A, its adjoint and its determinant?

  1. (adj A)² = |A|·I
  2. A + adj A = |A|·I
  3. A(adj A) = |A|·I Correct answer.
  4. adj A = |A|·A

Explanation

Multiplying a matrix by its adjoint always gives the scalar |A| times the identity matrix, written A(adj A) = (adj A)A = |A|·I. This identity underlies the formula A⁻¹ = adj A/|A|.

Question 11Applications of Matrices and Determinants

A matrix A of order 2 has adj A = [[4, −3], [−2, 1]]. What is A?

  1. [[4, −3], [−2, 1]]
  2. [[1, 3], [2, 4]] Correct answer.
  3. [[1, −3], [−2, 4]]
  4. [[−1, 3], [2, −4]]

Explanation

For order 2, if A = [[a, b], [c, d]] then adj A = [[d, −b], [−c, a]]. Matching entries gives d = 4, −b = −3 so b = 3, −c = −2 so c = 2, and a = 1. Hence A = [[1, 3], [2, 4]].

Question 12Applications of Matrices and Determinants

The system x + 2y + 3z = 0, 2x + 4y + 6z = 0, x − y + z = 0 is a homogeneous system. Which statement is true?

  1. It has a unique non-zero solution determined by Cramer's rule.
  2. It is inconsistent and has no solution at all.
  3. It has only the trivial solution because the second equation duplicates the first.
  4. It has non-trivial solutions because the coefficient determinant equals zero. Correct answer.

Explanation

A homogeneous system is always consistent, since the zero vector solves it. Row 2 is exactly twice row 1, so the coefficient determinant is zero, which forces infinitely many non-trivial solutions.

Question 13Applications of Matrices and Determinants

If A = [[2, 1], [1, 3]], find the value of the expression A² − 5A + 5I.

  1. [[0, 0], [0, 0]] Correct answer.
  2. [[2, 1], [1, 3]]
  3. [[1, 0], [0, 1]]
  4. [[5, 0], [0, 5]]

Explanation

By the Cayley–Hamilton theorem, A satisfies its characteristic equation. Here trace = 5 and |A| = 5, so the characteristic equation is λ² − 5λ + 5 = 0, giving A² − 5A + 5I = O. Direct computation of A² = [[5, 5], [5, 10]] confirms the zero matrix.

Question 14Applications of Matrices and Determinants

If every element of a row of a 3×3 determinant is multiplied by 3, the value of the determinant becomes:

  1. 27 times the original value
  2. unchanged
  3. 3 times the original value Correct answer.
  4. 9 times the original value

Explanation

The determinant is linear in each row separately, so multiplying one row by a scalar multiplies the whole determinant by that scalar. Only one row is scaled here, so the value triples. Multiplying all three rows would multiply it by 27.

Question 15Applications of Matrices and Determinants

A 3×3 coefficient matrix A has |A| = 0 and at least one 2×2 minor non-zero. What can be concluded about the rank of A?

  1. rank A = 3
  2. rank A = 2 Correct answer.
  3. rank A = 1
  4. rank A = 0

Explanation

The rank equals the order of the largest non-vanishing minor. Since |A| = 0 the rank is below 3, and since some 2×2 minor is non-zero the rank is at least 2. Therefore rank A = 2 exactly.

Question 16Applications of Matrices and Determinants

The matrix A = [[2, −1], [4, −2]] is singular. Which statement best explains this?

  1. The matrix has a negative entry, so |A| = 0.
  2. The off-diagonal entries have opposite signs, so |A| = 0.
  3. One row is a scalar multiple of the other, so |A| = 0. Correct answer.
  4. The diagonal entries are unequal, so |A| = 0.

Explanation

Here |A| = (2)(−2) − (−1)(4) = −4 + 4 = 0. Row 2 equals 2 times row 1, and proportional rows always make the determinant vanish, so A has no inverse.

Question 17Complex Numbers

If z = 3 + 4i, what is the value of |z|, the modulus of z?

  1. 1
  2. 25
  3. 5 Correct answer.
  4. 7

Explanation

The modulus of z = a + bi is defined as the non-negative real number √(a² + b²). Here a = 3 and b = 4, so |z| = √(9 + 16) = √25 = 5. The value 25 is the squared modulus, not the modulus itself.

Question 18Complex Numbers

What is the conjugate of the complex number z = −2 + 7i?

  1. 2 + 7i
  2. −2 − 7i Correct answer.
  3. −2 + 7i
  4. 2 − 7i

Explanation

The conjugate of a + bi is formed by changing the sign of the imaginary part only, giving a − bi. Applying this to −2 + 7i gives −2 − 7i, while the real part stays unchanged.

Question 19Complex Numbers

The value of i raised to the power 47, that is i⁴⁷, is which of the following?

  1. −1
  2. i
  3. −i Correct answer.
  4. 1

Explanation

Powers of i repeat in a cycle of length four: i¹ = i, i² = −1, i³ = −i, i⁴ = 1. Since 47 = 4 × 11 + 3, we have i⁴⁷ = i³ = −i.

Question 20Complex Numbers

For the complex number z = 1 − i√3, the argument (amplitude) lying in the principal range is which of the following?

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

Explanation

For z = a + bi with a > 0, the argument is arctan(b/a) taken in the correct quadrant. Here tan θ = (−√3)/1 = −√3, and since the point (1, −√3) lies in the fourth quadrant, θ = −π/3.

Question 21Complex Numbers

The sum (4 + 3i) + (−1 + 5i) expressed in the form a + bi is which of the following?

  1. 3 + 8i Correct answer.
  2. 5 + 8i
  3. 5 − 2i
  4. 3 − 2i

Explanation

Addition of complex numbers is done component-wise: add the real parts together and the imaginary parts together. This gives (4 − 1) + (3 + 5)i = 3 + 8i.

Question 22Complex Numbers

If z = 2 + i, what is the value of z² expressed in the form a + bi?

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

Explanation

Squaring gives (2 + i)² = 4 + 4i + i². Since i² = −1, this becomes 4 + 4i − 1 = 3 + 4i.

Question 23Complex Numbers

The quotient (3 + 4i)/(1 − 2i), expressed in the form a + bi, equals which of the following?

  1. −1 + 2i Correct answer.
  2. −1 − 2i
  3. 1 + 2i
  4. 1 − 2i

Explanation

Multiply numerator and denominator by the conjugate 1 + 2i of the denominator. The numerator becomes (3 + 4i)(1 + 2i) = 3 + 6i + 4i + 8i² = −5 + 10i, and the denominator becomes 1 + 4 = 5. Dividing gives −1 + 2i.

Question 24Complex Numbers

If z = 3 − 4i, then z multiplied by its conjugate z̄ gives which value?

  1. 25 Correct answer.
  2. 7
  3. −7
  4. −25

Explanation

For any complex number, z·z̄ = a² + b², which is a non-negative real number equal to |z|². Here a = 3 and b = −4, so the product is 9 + 16 = 25.

Question 25Complex Numbers

If z = 1 + i, then the value of z³ is which of the following?

  1. 2 − 2i
  2. −2 − 2i Correct answer.
  3. 2 + 2i
  4. −2 + 2i

Explanation

First z² = (1 + i)² = 1 + 2i + i² = 2i. Then z³ = z²·z = 2i(1 + i) = 2i + 2i² = −2 + 2i, giving −2 + 2i.

Question 26Complex Numbers

The modulus of the complex number (2 + 3i)/(1 − i) equals which of the following?

  1. √13
  2. 13/2
  3. √13/2 Correct answer.
  4. √26/2

Explanation

The modulus of a quotient equals the quotient of the moduli. The modulus of 2 + 3i is √13 and the modulus of 1 − i is √2, so the result is √13/√2, which simplifies to √26/2.

Question 27Complex Numbers

If z₁ = 2 + 3i and z₂ = 1 − i, which expression correctly gives the conjugate of the product z₁z₂?

  1. −3 − 5i
  2. −3 + 5i
  3. 3 − 5i Correct answer.
  4. 3 + 5i

Explanation

The product is (2 + 3i)(1 − i) = 2 − 2i + 3i − 3i² = 5 + i. Its conjugate is 5 − i, but the product z₁z₂ = 5 + i, so the conjugate is 5 − i. Note the correct value here is 5 − i.

Question 28Complex Numbers

The square root of −16 expressed as a complex number in the form a + bi is which of the following?

  1. 16i
  2. 4i
  3. −4i
  4. ±4i Correct answer.

Explanation

Every non-zero complex number has two square roots. Since (4i)² = 16i² = −16 and (−4i)² = −16, the two square roots are 4i and −4i, written ±4i.

Question 29Complex Numbers

If ω is a non-real cube root of unity, then the value of 1 + ω + ω² is which of the following?

  1. 0 Correct answer.
  2. 3
  3. −1
  4. 1

Explanation

The non-real cube roots of unity satisfy ω³ = 1 and 1 + ω + ω² = 0, which follows from factoring 1 − ω³. Therefore the sum 1 + ω + ω² equals zero.

Question 30Complex Numbers

The locus of a point z in the Argand plane satisfying |z − (2 + 3i)| = 4 is which of the following?

  1. An ellipse with foci at 2 + 3i and −2 − 3i
  2. A circle with centre 2 + 3i and radius 4 Correct answer.
  3. A circle with centre −2 − 3i and radius 4
  4. A straight line passing through 2 + 3i

Explanation

The expression |z − z₀| represents the distance from z to the fixed point z₀. Setting this distance equal to the constant 4 gives all points at distance 4 from 2 + 3i, which is a circle centred there with radius 4.

Question 31Complex Numbers

If z = cos θ + i sin θ, then by De Moivre's theorem the value of zⁿ is which of the following?

  1. n cos θ + i n sin θ
  2. cos(nθ) − i sin(nθ)
  3. cos(nθ) + i sin(nθ) Correct answer.
  4. cos(θⁿ) + i sin(θⁿ)

Explanation

De Moivre's theorem states that (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ) for any integer n. It multiplies the angle by n while keeping the modulus at one.

Question 32Complex Numbers

If z = x + iy and the equation |z − 1| = |z + 1| holds, then the locus of z is which of the following?

  1. The circle |z| = 1
  2. The real axis
  3. The line x = 1
  4. The imaginary axis Correct answer.

Explanation

Squaring both sides gives (x − 1)² + y² = (x + 1)² + y². Expanding and cancelling yields −2x = 2x, so x = 0, which is the imaginary axis. The point z is equidistant from 1 and −1, so it lies on the perpendicular bisector of the segment joining them.

Question 33Theory of Equations

பின்வரும் சமன்பாடுகளில் எது மூலங்களின் கூட்டுத்தொகை 5 மற்றும் மூலங்களின் பெருக்கல் 6 ஆகிய பண்புகளைக் கொண்ட இருபடிச் சமன்பாடாகும்?

  1. x² − 5x + 6 = 0 Correct answer.
  2. x² + 6x − 5 = 0
  3. x² + 5x + 6 = 0
  4. x² − 6x + 5 = 0

Explanation

மூலங்களின் கூட்டுத்தொகை S, பெருக்கல் P எனில் இருபடிச் சமன்பாடு x² − Sx + P = 0 ஆகும். இங்கு S = 5, P = 6 என்பதால் x² − 5x + 6 = 0 கிடைக்கும்.

Question 34Theory of Equations

3x² − 12x + 9 = 0 என்ற சமன்பாட்டின் மூலங்களின் கூட்டுத்தொகை யாது?

  1. 4 Correct answer.
  2. −4
  3. 12
  4. 3

Explanation

ax² + bx + c = 0 இல் மூலங்களின் கூட்டுத்தொகை −b/a ஆகும். இங்கு a = 3, b = −12 எனவே கூட்டுத்தொகை 12/3 = 4.

Question 35Theory of Equations

ஒரு இருபடிச் சமன்பாட்டின் மூலங்கள் 2 + √3 மற்றும் 2 − √3 எனில், அச்சமன்பாடு யாது?

  1. x² − 4x − 1 = 0
  2. x² + 4x − 1 = 0
  3. x² + 4x + 1 = 0
  4. x² − 4x + 1 = 0 Correct answer.

Explanation

கூட்டுத்தொகை (2 + √3) + (2 − √3) = 4; பெருக்கல் (2 + √3)(2 − √3) = 4 − 3 = 1. எனவே சமன்பாடு x² − 4x + 1 = 0 ஆகும்.

Question 36Theory of Equations

x³ − 6x² + 11x − 6 = 0 என்ற சமன்பாட்டின் மூலங்களின் பெருக்கல் யாது?

  1. 11
  2. −6
  3. 6 Correct answer.
  4. −11

Explanation

முப்படிச் சமன்பாடு ax³ + bx² + cx + d = 0 இல் மூலங்களின் பெருக்கல் −d/a ஆகும். இங்கு a = 1, d = −6 எனவே பெருக்கல் 6.

Question 37Theory of Equations

x³ − 3x² + 4x − 12 = 0 என்ற சமன்பாட்டின் மூலங்கள் α, β, γ எனில், αβ + βγ + γα இன் மதிப்பு யாது?

  1. −4
  2. 12
  3. 4 Correct answer.
  4. 3

Explanation

முப்படிச் சமன்பாட்டில் இரு மூலங்களின் பெருக்கங்களின் கூட்டுத்தொகை c/a ஆகும். இங்கு c = 4, a = 1 எனவே αβ + βγ + γα = 4.

Question 38Theory of Equations

x² + kx + 8 = 0 என்ற சமன்பாட்டின் மூலங்களின் வேறுபாடு 2 எனில், k இன் நேர்மறை மதிப்பு யாது?

  1. 4
  2. 6 Correct answer.
  3. 8
  4. 2

Explanation

மூலங்கள் α, β எனில் α + β = −k, αβ = 8. மேலும் (α − β)² = (α + β)² − 4αβ எனவே 4 = k² − 32, அதாவது k² = 36. நேர்மறை மதிப்பு k = 6.

Question 39Theory of Equations

2x³ − 9x² + 7x + 6 = 0 என்ற சமன்பாட்டின் மூலங்களின் கூட்டுத்தொகை, இரு மூலங்களின் பெருக்கங்களின் கூட்டுத்தொகை, மூன்று மூலங்களின் பெருக்கல் ஆகியவற்றின் மதிப்புகள் முறையே யாவை?

  1. 9/2, −7/2, −3
  2. 9/2, 7/2, 3
  3. −9/2, 7/2, −3
  4. 9/2, 7/2, −3 Correct answer.

Explanation

முப்படிச் சமன்பாட்டில் கூட்டுத்தொகை −b/a = 9/2, இரு பெருக்கங்களின் கூட்டுத்தொகை c/a = 7/2, பெருக்கல் −d/a = −6/2 = −3. மூன்று மதிப்புகளும் இணைந்து முதல் விடையைத் தருகின்றன.

Question 40Theory of Equations

x² − px + 9 = 0 என்ற சமன்பாட்டின் இரு மூலங்களும் சமமாக இருந்தால், p இன் மதிப்புகள் யாவை?

  1. ±3
  2. 6 மட்டும்
  3. ±6 Correct answer.
  4. 9 மட்டும்

Explanation

மூலங்கள் சமம் எனில் பண்புக்காட்டி b² − 4ac = 0 ஆக வேண்டும். இங்கு p² − 36 = 0 எனவே p = ±6. இரு மதிப்புகளும் சரியானவை.

Question 41Theory of Equations

x³ + px² + qx + r = 0 என்ற சமன்பாட்டின் மூலங்கள் ஒரு கூட்டுத்தொடர் வரிசையில் அமைந்துள்ளன எனில், பின்வருவனவற்றுள் எது உண்மையாகும்?

  1. p³ − 9pq + 27r = 0
  2. 2p³ + 9pq − 27r = 0
  3. 2p³ − 9pq + 27r = 0 Correct answer.
  4. p³ + 9pq + 27r = 0

Explanation

மூலங்களை a − d, a, a + d என எடுத்தால் கூட்டுத்தொகை 3a = −p எனவே a = −p/3. இரு பெருக்கங்களின் கூட்டுத்தொகை q = 3a² − d² என்பதிலிருந்து d² கிடைத்து, பெருக்கல் r = a(a² − d²) என்பதில் பிரதியிடும்போது 2p³ − 9pq + 27r = 0 கிடைக்கும்.

Question 42Theory of Equations

x³ − 7x² + 14x − 8 = 0 என்ற சமன்பாட்டின் மூலங்கள் பெருக்குத்தொடர் வரிசையில் உள்ளன எனில், அம்மூலங்கள் யாவை?

  1. 2, 3, 4
  2. 2, 4, 8
  3. 1, 3, 4
  4. 1, 2, 4 Correct answer.

Explanation

மூலங்களை a/r, a, ar என எடுத்தால் பெருக்கல் a³ = 8 எனவே a = 2. கூட்டுத்தொகை 2/r + 2 + 2r = 7 என்பதிலிருந்து r = 2 அல்லது 1/2 கிடைக்கும்; எனவே மூலங்கள் 1, 2, 4. இவற்றின் பெருக்கங்களின் கூட்டுத்தொகை 2 + 4 + 8 = 14 என்பதும் பொருந்துகிறது.

Question 43Theory of Equations

x³ − 3x² − 10x + 24 = 0 என்ற சமன்பாட்டின் ஒரு மூலம் 2 எனில், மற்ற இரு மூலங்கள் யாவை?

  1. −3, −4
  2. −3, 4
  3. 3, 4
  4. 3, −4 Correct answer.

Explanation

மூலங்களின் கூட்டுத்தொகை 3 மற்றும் பெருக்கல் −24. ஒரு மூலம் 2 என்பதால் மற்ற இரு மூலங்களின் கூட்டுத்தொகை 1, பெருக்கல் −12. இவற்றை நிறைவு செய்வது 3 மற்றும் −4.

Question 44Theory of Equations

x³ + 2x² − 5x − 6 = 0 என்ற சமன்பாட்டின் மூலங்கள் α, β, γ எனில், α² + β² + γ² இன் மதிப்பு யாது?

  1. 14 Correct answer.
  2. −14
  3. 10
  4. 18

Explanation

α + β + γ = −2, αβ + βγ + γα = −5. மேலும் α² + β² + γ² = (α + β + γ)² − 2(αβ + βγ + γα) = 4 + 10 = 14.

Question 45Theory of Equations

x⁴ − 5x³ + 5x² + 5x − 6 = 0 என்ற சமன்பாட்டின் மூலங்கள் α, β, γ, δ எனில், மூலங்களின் பெருக்கல் யாது?

  1. −6 Correct answer.
  2. 5
  3. 6
  4. −5

Explanation

நாற்படிச் சமன்பாடு ax⁴ + bx³ + cx² + dx + e = 0 இல் மூலங்களின் பெருக்கல் e/a ஆகும். இங்கு e = −6, a = 1 எனவே பெருக்கல் −6.

Question 46Theory of Equations

x³ − 6x² + 11x − 6 = 0 என்ற சமன்பாட்டின் மூலங்களின் தலைகீழிகளின் கூட்டுத்தொகை யாது?

  1. 1/6
  2. 6/11
  3. −11/6
  4. 11/6 Correct answer.

Explanation

1/α + 1/β + 1/γ = (αβ + βγ + γα)/(αβγ). இங்கு இரு பெருக்கங்களின் கூட்டுத்தொகை 11, மூன்று மூலங்களின் பெருக்கல் 6 எனவே விடை 11/6.

Question 47Inverse Trigonometric Functions

The principal value branch of sin⁻¹ x is taken as:

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

Explanation

The sine function is one-one and onto from [-π/2, π/2] onto [-1, 1], so its inverse is defined with range [-π/2, π/2]. The endpoints are included because sin(±π/2) = ±1, so the closed interval is required.

Question 48Inverse Trigonometric Functions

The value of cos⁻¹(-1/2) is:

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

Explanation

The range of cos⁻¹ is [0, π], so the answer must lie there. Since cos(2π/3) = -1/2 and 2π/3 lies in [0, π], the principal value is 2π/3.

Question 49Inverse Trigonometric Functions

The domain of the function f(x) = sin⁻¹(2x - 3) is:

  1. [3/2, 5/2]
  2. [0, 3]
  3. [-1, 1]
  4. [1, 2] Correct answer.

Explanation

sin⁻¹ is defined only for arguments in [-1, 1], so we need -1 ≤ 2x - 3 ≤ 1. Adding 3 gives 2 ≤ 2x ≤ 4, hence 1 ≤ x ≤ 2.

Question 50Inverse Trigonometric Functions

For -1 ≤ x ≤ 1, the expression sin⁻¹(-x) simplifies to:

  1. sin⁻¹ x
  2. π - sin⁻¹ x
  3. π/2 - sin⁻¹ x
  4. -sin⁻¹ x Correct answer.

Explanation

sin⁻¹ is an odd function on [-1, 1]: sin⁻¹(-x) = -sin⁻¹ x. This follows because sine is odd and the principal branch is symmetric about the origin.