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The questions your board exam loves to ask

800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.

800 board questionsModel answersMarking-scheme pointsEvery chapterCBSE · ISC · State boards

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MathsClass 122 markseasy

Relations and Functions

Define reflexive, symmetric and transitive relations. When is a relation called an equivalence relation?

Reveal model answer + marking points

A relation R on a set A is reflexive if (a, a) belongs to R for every a in A. It is symmetric if (a, b) in R implies (b, a) in R. It is transitive if (a, b) in R and (b, c) in R together imply (a, c) in R. A relation that is reflexive, symmetric and transitive at the same time is called an equivalence relation.

Marking-scheme points

  • Reflexive: (a, a) in R for all a
  • Symmetric: (a, b) in R => (b, a) in R
  • Transitive: (a, b), (b, c) in R => (a, c) in R; all three => equivalence
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MathsClass 122 markseasy

Relations and Functions

Define one-one (injective), onto (surjective) and bijective functions.

Reveal model answer + marking points

A function f from A to B is one-one (injective) if distinct elements of A have distinct images, that is, f(x1) = f(x2) implies x1 = x2. It is onto (surjective) if every element of B has at least one pre-image in A, that is, the range of f equals the codomain B. A function that is both one-one and onto is called bijective (a one-to-one correspondence).

Marking-scheme points

  • One-one: f(x1) = f(x2) => x1 = x2
  • Onto: range = codomain (every element of B has a pre-image)
  • Bijective: both one-one and onto
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MathsClass 122 marksmedium

Relations and Functions

If f(x) = x + 1 and g(x) = 2x, find (g o f)(x) and (f o g)(x).

Reveal model answer + marking points

(g o f)(x) means g(f(x)). So (g o f)(x) = g(f(x)) = g(x + 1) = 2(x + 1) = 2x + 2. (f o g)(x) means f(g(x)). So (f o g)(x) = f(g(x)) = f(2x) = 2x + 1. Note that (g o f)(x) is not equal to (f o g)(x), so composition of functions is not commutative in general.

(g o f)(x) = g(f(x))

Marking-scheme points

  • (g o f)(x) = g(f(x)) = 2(x + 1) = 2x + 2
  • (f o g)(x) = f(g(x)) = 2x + 1
  • Composition is not commutative in general
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MathsClass 122 markseasy

Inverse Trigonometric Functions

Find the principal values of sin inverse (1/2) and cos inverse (-1/2).

Reveal model answer + marking points

The principal value branch of sin inverse is [-pi/2, pi/2]. Since sin(pi/6) = 1/2, the principal value of sin inverse (1/2) = pi/6. The principal value branch of cos inverse is [0, pi]. Since cos(2pi/3) = -1/2, the principal value of cos inverse (-1/2) = 2pi/3.

Marking-scheme points

  • sin inverse principal branch [-pi/2, pi/2]
  • sin inverse (1/2) = pi/6
  • cos inverse (-1/2) = 2pi/3 (branch [0, pi])
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MathsClass 122 marksmedium

Inverse Trigonometric Functions

Evaluate tan inverse (1) + cos inverse (1/2) + sin inverse (1/2).

Reveal model answer + marking points

The principal values are: tan inverse (1) = pi/4, cos inverse (1/2) = pi/3, and sin inverse (1/2) = pi/6. Adding them: pi/4 + pi/3 + pi/6. Taking the LCD 12: 3pi/12 + 4pi/12 + 2pi/12 = 9pi/12 = 3pi/4.

Marking-scheme points

  • tan inverse 1 = pi/4, cos inverse (1/2) = pi/3, sin inverse (1/2) = pi/6
  • Sum = pi/4 + pi/3 + pi/6
  • = 9pi/12 = 3pi/4
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MathsClass 122 markseasy

Inverse Trigonometric Functions

Prove that sin inverse (x) + cos inverse (x) = pi/2 for all x in [-1, 1].

Reveal model answer + marking points

Let sin inverse (x) = theta, so sin theta = x, where theta lies in [-pi/2, pi/2]. Then x = sin theta = cos(pi/2 - theta). Since (pi/2 - theta) lies in [0, pi], we have cos inverse (x) = pi/2 - theta = pi/2 - sin inverse (x). Rearranging gives sin inverse (x) + cos inverse (x) = pi/2.

sin inverse (x) + cos inverse (x) = pi/2

Marking-scheme points

  • Let sin inverse (x) = theta, so x = sin theta = cos(pi/2 - theta)
  • cos inverse (x) = pi/2 - theta
  • Hence sin inverse (x) + cos inverse (x) = pi/2
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MathsClass 122 marksmedium

Inverse Trigonometric Functions

Find the value of cos inverse (cos(7pi/6)).

Reveal model answer + marking points

The principal value branch of cos inverse is [0, pi], but 7pi/6 does not lie in this interval. First find cos(7pi/6) = cos(pi + pi/6) = -cos(pi/6) = -sqrt(3)/2. Now cos inverse (-sqrt(3)/2) = 5pi/6, since 5pi/6 lies in [0, pi] and cos(5pi/6) = -sqrt(3)/2. Therefore cos inverse (cos(7pi/6)) = 5pi/6.

Marking-scheme points

  • 7pi/6 is not in the principal branch [0, pi]
  • cos(7pi/6) = -sqrt(3)/2
  • cos inverse (-sqrt(3)/2) = 5pi/6
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MathsClass 122 markseasy

Inverse Trigonometric Functions

Write the domain and principal value range of sin inverse (x) and tan inverse (x).

Reveal model answer + marking points

For sin inverse (x): the domain is [-1, 1] and the principal value range is [-pi/2, pi/2]. For tan inverse (x): the domain is all real numbers R and the principal value range is the open interval (-pi/2, pi/2). Inverse trigonometric functions are defined by restricting the domain of the corresponding trigonometric functions so that they become one-one.

Marking-scheme points

  • sin inverse: domain [-1, 1], range [-pi/2, pi/2]
  • tan inverse: domain R, range (-pi/2, pi/2)
  • Defined by restricting the domain of the trig function
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MathsClass 122 markseasy

Matrices

Define a diagonal matrix, a scalar matrix and an identity matrix.

Reveal model answer + marking points

A diagonal matrix is a square matrix in which all the non-diagonal (off-diagonal) elements are zero. A scalar matrix is a diagonal matrix in which all the diagonal elements are equal (to the same non-zero scalar). An identity (unit) matrix is a scalar matrix in which all the diagonal elements are equal to 1; it is denoted by I and acts as the multiplicative identity, since AI = IA = A.

Marking-scheme points

  • Diagonal matrix: all off-diagonal elements are zero
  • Scalar matrix: diagonal matrix with equal diagonal elements
  • Identity matrix: scalar matrix with all diagonal elements = 1
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MathsClass 122 markseasy

Matrices

Find the values of x and y if the matrix [[x + y, 2],[5, x - y]] equals [[6, 2],[5, 2]].

Reveal model answer + marking points

Two matrices are equal when their corresponding elements are equal. Comparing: x + y = 6 and x - y = 2. Adding these two equations: 2x = 8, so x = 4. Subtracting the second from the first: 2y = 4, so y = 2. Hence x = 4 and y = 2.

Marking-scheme points

  • Equal matrices -> equate corresponding elements
  • x + y = 6 and x - y = 2
  • Solve: x = 4, y = 2
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MathsClass 122 markseasy

Matrices

What is the transpose of a matrix? Write two properties of the transpose.

Reveal model answer + marking points

The transpose of a matrix A, denoted A' (or A^T), is the matrix obtained by interchanging its rows and columns. Two properties are: (1) (A')' = A, that is, the transpose of the transpose is the original matrix; and (2) (AB)' = B' A', the transpose of a product equals the product of the transposes in reverse order. Also (A + B)' = A' + B' and (kA)' = k A'.

(AB)' = B' A'

Marking-scheme points

  • Transpose: interchange rows and columns
  • (A')' = A
  • (AB)' = B' A'
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MathsClass 122 marksmedium

Matrices

Define a symmetric and a skew-symmetric matrix.

Reveal model answer + marking points

A square matrix A is symmetric if A' = A, that is, aij = aji for all i and j (it is unchanged on taking its transpose). A square matrix A is skew-symmetric if A' = -A, that is, aij = -aji for all i and j; this forces all its diagonal elements to be zero. Every square matrix can be expressed as the sum of a symmetric and a skew-symmetric matrix.

symmetric: A' = A; skew-symmetric: A' = -A

Marking-scheme points

  • Symmetric: A' = A (aij = aji)
  • Skew-symmetric: A' = -A (aij = -aji), diagonal elements zero
  • Every square matrix = symmetric part + skew-symmetric part
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MathsClass 122 markseasy

Matrices

If a matrix has 12 elements, what are the possible orders it can have?

Reveal model answer + marking points

If a matrix has 12 elements, then the order (number of rows x number of columns) must be a pair of factors whose product is 12. The possible orders are: 1 x 12, 12 x 1, 2 x 6, 6 x 2, 3 x 4 and 4 x 3. So there are 6 possible orders. (In general, the number of possible orders equals the number of ways of writing the total number of elements as a product of two natural numbers.)

Marking-scheme points

  • Order (rows x columns) must multiply to give 12
  • Factor pairs of 12
  • Orders: 1x12, 12x1, 2x6, 6x2, 3x4, 4x3 (six orders)
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MathsClass 122 markseasy

Matrices

If A = [[1, 2],[3, 4]] and B = [[2, 1],[0, 5]], find 2A - B.

Reveal model answer + marking points

First find 2A = [[2, 4],[6, 8]]. Then subtract B element by element: 2A - B = [[2 - 2, 4 - 1],[6 - 0, 8 - 5]] = [[0, 3],[6, 3]].

Marking-scheme points

  • 2A = [[2, 4],[6, 8]]
  • Subtract B element by element
  • 2A - B = [[0, 3],[6, 3]]
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MathsClass 122 marksmedium

Determinants

Evaluate the determinant of the matrix [[1, 2, 3],[0, 1, 4],[5, 6, 0]].

Reveal model answer + marking points

Expand along the first row: |A| = 1 x (1 x 0 - 4 x 6) - 2 x (0 x 0 - 4 x 5) + 3 x (0 x 6 - 1 x 5) = 1 x (0 - 24) - 2 x (0 - 20) + 3 x (0 - 5) = -24 + 40 - 15 = 1. So the value of the determinant is 1.

Marking-scheme points

  • Expand along the first row
  • = 1(0 - 24) - 2(0 - 20) + 3(0 - 5)
  • = -24 + 40 - 15 = 1
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MathsClass 122 marksmedium

Determinants

State any three properties of determinants.

Reveal model answer + marking points

(1) The value of a determinant remains unchanged if its rows and columns are interchanged (that is, |A'| = |A|). (2) If any two rows (or two columns) of a determinant are interchanged, the sign of the determinant changes. (3) If any two rows (or columns) of a determinant are identical (or proportional), the value of the determinant is zero. Also, if each element of a row (or column) is multiplied by a constant k, the determinant is multiplied by k.

Marking-scheme points

  • Interchanging rows and columns does not change the value
  • Interchanging two rows/columns changes the sign
  • Two identical/proportional rows or columns give value zero
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MathsClass 122 markseasy

Determinants

Define the minor and the cofactor of an element of a determinant.

Reveal model answer + marking points

The minor of an element aij of a determinant is the value of the determinant obtained by deleting the ith row and the jth column in which that element lies; it is denoted by Mij. The cofactor of the element aij is Aij = (-1)^(i + j) x Mij, that is, the minor with a sign attached according to the position of the element. Cofactors are used to expand a determinant and to find the adjoint of a matrix.

Aij = (-1)^(i+j) Mij

Marking-scheme points

  • Minor Mij: determinant after deleting row i and column j
  • Cofactor Aij = (-1)^(i+j) Mij
  • Used in expansion of determinants and finding the adjoint
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MathsClass 122 markseasy

Determinants

What is a singular and a non-singular matrix? State the condition for a matrix to be invertible.

Reveal model answer + marking points

A square matrix A is called singular if its determinant is zero (|A| = 0), and non-singular if its determinant is non-zero (|A| is not equal to 0). The condition for a square matrix to be invertible (to have an inverse) is that it must be non-singular, that is, |A| must not be zero, because the inverse A inverse = (1/|A|) adj A is defined only when |A| is not zero.

A inverse exists if and only if |A| is not 0

Marking-scheme points

  • Singular: |A| = 0; non-singular: |A| not equal to 0
  • A matrix is invertible only if it is non-singular
  • Because A inverse = (1/|A|) adj A needs |A| not zero
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MathsClass 122 marksmedium

Determinants

If A is a 3 x 3 matrix with |A| = 5, find the values of |adj A| and |2A|.

Reveal model answer + marking points

For an n x n matrix, |adj A| = |A|^(n - 1). Here n = 3, so |adj A| = |A|^2 = 5^2 = 25. Also, for a scalar k, |kA| = k^n |A|. Here |2A| = 2^3 x |A| = 8 x 5 = 40. So |adj A| = 25 and |2A| = 40.

|adj A| = |A|^(n-1); |kA| = k^n |A|

Marking-scheme points

  • |adj A| = |A|^(n-1) = 5^2 = 25
  • |kA| = k^n |A|
  • |2A| = 2^3 x 5 = 40
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MathsClass 122 markseasy

Continuity and Differentiability

State the conditions for a function f(x) to be continuous at a point x = a.

Reveal model answer + marking points

A function f(x) is continuous at a point x = a if the following three conditions are all satisfied: (1) f(a) is defined (the function has a value at a); (2) the limit of f(x) as x approaches a exists (that is, the left-hand limit equals the right-hand limit); and (3) this limit is equal to the value of the function, that is, limit of f(x) as x approaches a = f(a).

limit(x->a) f(x) = f(a)

Marking-scheme points

  • f(a) must be defined
  • The limit as x approaches a must exist (LHL = RHL)
  • The limit must equal f(a)
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