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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.

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ChemistryClass 122 marksmedium

Biomolecules

Name and briefly describe the four levels of protein structure.

Reveal model answer + marking points

(1) Primary structure - the specific sequence in which the amino acids are linked in the polypeptide chain. (2) Secondary structure - the local folding of the chain into shapes such as the alpha-helix or beta-pleated sheet, held together by hydrogen bonds. (3) Tertiary structure - the overall three-dimensional folding of the whole polypeptide chain, giving a globular or fibrous shape. (4) Quaternary structure - the arrangement and association of two or more polypeptide chains (subunits), as in haemoglobin.

Marking-scheme points

  • Primary: sequence of amino acids
  • Secondary: alpha-helix or beta-pleated sheet (hydrogen bonds)
  • Tertiary: overall 3D fold; Quaternary: association of subunits (haemoglobin)
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ChemistryClass 122 markseasy

Biomolecules

What are enzymes? State two of their characteristics.

Reveal model answer + marking points

Enzymes are biological catalysts that speed up the biochemical reactions occurring in living organisms; chemically almost all enzymes are proteins (globular proteins). Characteristics: (1) they are highly specific, each enzyme usually catalysing only one particular reaction or type of reaction (lock-and-key specificity); and (2) they are highly efficient and work best under mild conditions of an optimum temperature (around body temperature) and an optimum pH; they are denatured and lose activity outside these conditions.

Marking-scheme points

  • Enzymes = biological catalysts (mostly proteins)
  • Highly specific (one enzyme, one reaction)
  • Work best at an optimum temperature and pH
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ChemistryClass 122 markseasy

Biomolecules

How are vitamins classified? Name one deficiency disease caused by a lack of vitamin C and vitamin D.

Reveal model answer + marking points

Vitamins are organic compounds required in small amounts in the diet for normal growth and health. They are classified into two groups: (1) fat-soluble vitamins (A, D, E and K), which are soluble in fats and oils and are stored in the liver and fatty tissues; and (2) water-soluble vitamins (the B group and C), which are soluble in water and must be supplied regularly in the diet as they are not stored. Deficiency of vitamin C causes scurvy, and deficiency of vitamin D causes rickets in children.

Marking-scheme points

  • Fat-soluble vitamins: A, D, E, K (stored in the body)
  • Water-soluble vitamins: B group and C (not stored)
  • Vitamin C deficiency: scurvy; vitamin D deficiency: rickets
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ChemistryClass 122 marksmedium

Biomolecules

State three differences between DNA and RNA.

Reveal model answer + marking points

(1) DNA (deoxyribonucleic acid) contains the sugar deoxyribose, whereas RNA (ribonucleic acid) contains the sugar ribose. (2) DNA contains the nitrogenous bases adenine, guanine, cytosine and thymine, whereas RNA contains adenine, guanine, cytosine and uracil (uracil replaces thymine). (3) DNA is usually a double-stranded helix and stores the genetic (hereditary) information, whereas RNA is usually single-stranded and mainly takes part in protein synthesis.

Marking-scheme points

  • Sugar: deoxyribose in DNA, ribose in RNA
  • Bases: DNA has thymine, RNA has uracil (instead of thymine)
  • DNA double-stranded (stores genes); RNA single-stranded (protein synthesis)
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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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