Board Boosters
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.
ChemistryClass 122 marksmedium
Amines
Why is aniline less basic than ethylamine (or ammonia)?
Reveal model answer + marking pointsHide answer▾
Aniline is less basic than ethylamine (and than ammonia) because in aniline the lone pair of electrons on the nitrogen atom is delocalised (drawn) into the benzene ring by resonance. As a result the lone pair is less available for donation to a proton, so aniline accepts a proton less readily and is a weaker base. In ethylamine, the alkyl group has an electron-releasing (+I) effect that increases the electron density on nitrogen and makes the lone pair more available, so it is more basic.
Marking-scheme points
- ✓In aniline the N lone pair is delocalised into the ring (resonance)
- ✓Lone pair is less available for protonation -> weaker base
- ✓In ethylamine, +I effect of the alkyl group makes N more basic
Still unsure? Ask the AI tutor →ChemistryClass 122 marksmedium
Amines
What is the coupling reaction of diazonium salts? Give one example.
Reveal model answer + marking pointsHide answer▾
The coupling reaction is a reaction in which a diazonium salt reacts with an electron-rich aromatic compound such as phenol or an aromatic amine to form a brightly coloured azo compound (containing the -N=N- linkage). For example, benzenediazonium chloride reacts with phenol in a mildly alkaline medium to give p-hydroxyazobenzene (an orange dye). These coupling reactions are used to prepare a large number of azo dyes.
Marking-scheme points
- ✓Diazonium salt + phenol/aromatic amine -> coloured azo compound (-N=N-)
- ✓Example: benzenediazonium chloride + phenol -> p-hydroxyazobenzene
- ✓Used to make azo dyes
Still unsure? Ask the AI tutor →ChemistryClass 122 markseasy
Biomolecules
Classify carbohydrates into three types based on hydrolysis, with one example of each.
Reveal model answer + marking pointsHide answer▾
Based on their behaviour on hydrolysis, carbohydrates are classified as: (1) monosaccharides - simple sugars that cannot be hydrolysed further into smaller units (e.g. glucose, fructose); (2) oligosaccharides - which give 2 to 10 monosaccharide units on hydrolysis, the most common being disaccharides (e.g. sucrose, which gives glucose and fructose, and maltose); and (3) polysaccharides - which give a large number of monosaccharide units on hydrolysis (e.g. starch, cellulose and glycogen).
Marking-scheme points
- ✓Monosaccharides: cannot be hydrolysed further (glucose, fructose)
- ✓Oligosaccharides (disaccharides): give 2-10 units (sucrose, maltose)
- ✓Polysaccharides: give many units (starch, cellulose)
Still unsure? Ask the AI tutor →ChemistryClass 122 marksmedium
Biomolecules
Distinguish between reducing and non-reducing sugars with examples.
Reveal model answer + marking pointsHide answer▾
A reducing sugar is a carbohydrate that has a free aldehyde or ketone group (a free hemiacetal), so it can reduce Tollens' reagent and Fehling's solution; all monosaccharides (glucose, fructose) and many disaccharides (maltose, lactose) are reducing sugars. A non-reducing sugar has no free aldehyde or ketone group (the reducing groups of both units are involved in the glycosidic bond), so it does not reduce Tollens' or Fehling's reagents; sucrose is the common example of a non-reducing sugar.
Marking-scheme points
- ✓Reducing sugar: has a free aldehyde/ketone group; reduces Tollens'/Fehling's
- ✓Examples: glucose, fructose, maltose, lactose
- ✓Non-reducing sugar: no free reducing group (e.g. sucrose)
Still unsure? Ask the AI tutor →ChemistryClass 122 marksmedium
Biomolecules
State two evidences that show the presence of an aldehyde group and an alcohol group in glucose.
Reveal model answer + marking pointsHide answer▾
Evidence for an aldehyde (-CHO) group in glucose: (1) it reduces Tollens' reagent to give a silver mirror and Fehling's solution to give a red precipitate; (2) it reacts with hydroxylamine to form an oxime and adds one molecule of HCN to form a cyanohydrin, confirming a carbonyl group. Evidence for alcohol (-OH) groups: glucose reacts with acetic anhydride to form a penta-acetate, showing the presence of five -OH groups. Thus glucose is a polyhydroxy aldehyde.
Marking-scheme points
- ✓Aldehyde group: reduces Tollens' and Fehling's; forms oxime and cyanohydrin
- ✓Alcohol groups: forms penta-acetate with acetic anhydride (five -OH groups)
- ✓Glucose is a polyhydroxy aldehyde
Still unsure? Ask the AI tutor →ChemistryClass 122 marksmedium
Biomolecules
State two differences between starch and cellulose.
Reveal model answer + marking pointsHide answer▾
(1) Starch is a polymer of alpha-glucose units and consists of two components, amylose (a linear chain) and amylopectin (a branched chain), joined by alpha-glycosidic linkages; cellulose is a straight-chain polymer of beta-glucose units joined by beta-glycosidic linkages. (2) Starch is the main storage carbohydrate (food reserve) in plants and can be digested by humans, whereas cellulose is a structural material (the main component of plant cell walls) and cannot be digested by humans (we lack the enzyme cellulase).
Marking-scheme points
- ✓Starch: alpha-glucose units (amylose + amylopectin), alpha-linkages
- ✓Cellulose: beta-glucose units, straight chain, beta-linkages
- ✓Starch is a food reserve (digestible); cellulose is structural (indigestible by humans)
Still unsure? Ask the AI tutor →ChemistryClass 122 marksmedium
Biomolecules
Name and briefly describe the four levels of protein structure.
Reveal model answer + marking pointsHide answer▾
(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)
Still unsure? Ask the AI tutor →ChemistryClass 122 markseasy
Biomolecules
What are enzymes? State two of their characteristics.
Reveal model answer + marking pointsHide answer▾
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
Still unsure? Ask the AI tutor →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 pointsHide answer▾
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
Still unsure? Ask the AI tutor →ChemistryClass 122 marksmedium
Biomolecules
State three differences between DNA and RNA.
Reveal model answer + marking pointsHide answer▾
(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)
Still unsure? Ask the AI tutor →MathsClass 122 markseasy
Relations and Functions
Define reflexive, symmetric and transitive relations. When is a relation called an equivalence relation?
Reveal model answer + marking pointsHide answer▾
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
Still unsure? Ask the AI tutor →MathsClass 122 markseasy
Relations and Functions
Define one-one (injective), onto (surjective) and bijective functions.
Reveal model answer + marking pointsHide answer▾
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
Still unsure? Ask the AI tutor →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 pointsHide answer▾
(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
Still unsure? Ask the AI tutor →MathsClass 122 markseasy
Inverse Trigonometric Functions
Find the principal values of sin inverse (1/2) and cos inverse (-1/2).
Reveal model answer + marking pointsHide answer▾
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])
Still unsure? Ask the AI tutor →MathsClass 122 marksmedium
Inverse Trigonometric Functions
Evaluate tan inverse (1) + cos inverse (1/2) + sin inverse (1/2).
Reveal model answer + marking pointsHide answer▾
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
Still unsure? Ask the AI tutor →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 pointsHide answer▾
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
Still unsure? Ask the AI tutor →MathsClass 122 marksmedium
Inverse Trigonometric Functions
Find the value of cos inverse (cos(7pi/6)).
Reveal model answer + marking pointsHide answer▾
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
Still unsure? Ask the AI tutor →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 pointsHide answer▾
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
Still unsure? Ask the AI tutor →MathsClass 122 markseasy
Matrices
Define a diagonal matrix, a scalar matrix and an identity matrix.
Reveal model answer + marking pointsHide answer▾
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
Still unsure? Ask the AI tutor →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 pointsHide answer▾
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
Still unsure? Ask the AI tutor →You are more ready than you feel.
One question at a time is how every topper started. Bookmark this, revise a few each day, and watch the fear shrink. And if a friend is stressing about boards — send this their way. You both win.