Charles University Hradec Králové Entrance Exam Sample Test 1, with Answers (2027)

Answer: A

A prion is a misfolded protein that converts normal copies of the same protein into its abnormal shape; it carries no DNA or RNA at all. A capsid around an RNA genome describes a virus, not a prion. Prions have no ribosomes and no metabolism, so antibiotics have nothing to act on, and they are not cells.

Answer: C

Every cell needs ribosomes to make proteins, so prokaryotes have them too (70S, smaller than the 80S ribosomes of the eukaryotic cytosol). The nucleus, mitochondria, Golgi apparatus and endoplasmic reticulum are membrane-bound organelles, which only eukaryotic cells possess.

Answer: E

The mitotic spindle is built from microtubules (tubulin), not from actin, so this is the one false statement. Actin filaments do form the contractile ring of cytokinesis, slide against myosin in muscle, push out the leading edge of a crawling cell and shape the pseudopodia that engulf a particle in phagocytosis.

Answer: B

A hypotonic solution has a lower solute concentration than the cell, so water moves into the cell by osmosis; a red cell has no wall to resist it and may burst (haemolysis). Shrinking and crenation happen in a hypertonic solution, and no net movement only in an isotonic one. Osmosis is the movement of water, not of salt, and cells have no pump that moves water.

Answer: D

The carrier moves glucose against its gradient using the energy of sodium flowing down its own gradient, which the Na+/K+ ATPase keeps up on the other side of the cell: that is secondary active transport. It is not primary active transport, because the carrier itself does not split ATP. Neither form of diffusion can move a substance against its concentration gradient, and pinocytosis takes in fluid in vesicles rather than through a carrier.

Answer: B

mRNA is complementary and antiparallel to the template, with uracil in place of thymine, so 3'-TACGGA-5' gives 5'-AUGCCU-3'. The first option contains thymine, so it is DNA, not RNA. The fourth has the right bases with the ends reversed (read 5' to 3' it is UCCGUA), and the third and fifth copy the template instead of pairing with it.

Answer: C

HIV is a retrovirus: the virion carries two copies of single-stranded RNA, not double-stranded DNA, so this is the false statement. The DNA appears only inside the infected cell, made by reverse transcriptase and then integrated into the host genome. HIV binds CD4 on helper T cells, and it buds through the host's plasma membrane, which becomes its envelope.

Answer: A

DNA is replicated in S phase, so in G2 each of the 46 chromosomes consists of two sister chromatids joined at the centromere: 46 chromosomes, 92 chromatids. The chromosome number counts centromeres and does not double until the sister chromatids separate in anaphase. 46 chromatids is the G1 state, and 23 chromosomes is the number in a gamete.

Answer: D

Sister chromatids stay joined through meiosis I and are separated only in anaphase II, as in mitosis. Pairing of homologues and crossing over happen in prophase I, the random orientation of bivalents at metaphase I gives independent assortment, and anaphase I pulls the homologues apart. That is why meiosis I is the reductional division.

Answer: E

A son receives his Y chromosome from his father and one of his mother's two X chromosomes, so he has a one in two chance of inheriting her affected X, and with no second X he will then have the disease. 1/4 is the probability that any child, of either sex, is an affected son, which is a different question. Daughters receive the father's normal X and so cannot be affected here.

Answer: D

Affected people are q², so q² = 1/2,500 and q = 1/50; p is then 49/50, close to 1. Carriers are 2pq = 2 × 49/50 × 1/50, about 1/25. 1 in 50 is the allele frequency q, not the carrier frequency, and 1 in 1,250 comes from doubling q² instead of calculating 2pq.

Answer: C

In the sliding filament model the thin filaments slide over the thick ones towards the centre of the sarcomere, so the regions where they do not overlap (the I band and the H zone) shrink. The A band is the length of the thick filaments and stays constant. Neither type of filament changes its own length, which is the whole point of the sliding model.

Answer: B

The pulmonary veins bring blood that has just been oxygenated in the lungs back to the left atrium. The pulmonary artery is the artery that carries deoxygenated blood, from the right ventricle to the lungs. The venae cavae and the coronary sinus return deoxygenated blood from the body and the heart wall to the right atrium.

Answer: E

A person makes antibodies only against the ABO antigens they lack, and someone with group AB has both A and B antigens, so their plasma has neither anti-A nor anti-B. Both antibodies are found in group O. RhD negative means the D antigen is absent. They can receive red cells from O, A or B RhD-negative donors as well as AB, and their own red cells would be attacked by the antibodies of O, A and B recipients.

Answer: A

Cooling increases haemoglobin's affinity for oxygen and shifts the curve to the left, so less oxygen is released at a given partial pressure. More carbon dioxide, a lower pH (the Bohr effect), more 2,3-bisphosphoglycerate and a higher temperature all do the opposite, shifting the curve to the right, which helps actively working tissues take up oxygen.

Answer: D

Pepsin is secreted by the chief cells of the stomach as pepsinogen and activated by stomach acid, and it starts breaking proteins into shorter peptides. Trypsin also digests protein, but it comes from the pancreas and works in the small intestine. Amylase and lactase act on carbohydrates and lipase on fats.

Answer: A

Insulin is secreted by the beta cells of the islets of Langerhans in the pancreas, not by the liver, although the liver is one of its main targets. The liver does make albumin and most clotting factors, produces bile that is stored in the gallbladder, stores glucose as glycogen and turns ammonia from deaminated amino acids into urea in the urea cycle.

Answer: C

ADH causes water channels (aquaporin-2) to be inserted into the cells of the collecting duct, so water leaves the urine into the concentrated medulla. The proximal tubule and the descending limb are permeable to water whether or not ADH is present. The glomerulus and Bowman's capsule form the filtrate and do not reabsorb water.

Answer: B

Glucose is always freely filtered, and in health all of it is reabsorbed by sodium-glucose carriers in the proximal tubule. When the blood level is high the filtered load exceeds the maximum rate those carriers can work at (the transport maximum), and the excess stays in the urine. Filtration of glucose does not depend on its concentration, the tubule does not secrete glucose, and its reabsorption needs neither insulin nor ADH.

Answer: E

When the membrane reaches threshold, voltage-gated sodium channels open and sodium rushes in down its electrochemical gradient, making the inside positive. The outflow of potassium causes repolarisation, which follows. The sodium-potassium pump moves sodium out of the cell, not in, and an inflow of chloride would make the inside more negative, not less.

Answer: D

The human retina is inverted: light passes through the ganglion and bipolar cell layers first and reaches the rods and cones, at the back of the retina, last. The other statements are true. Rods are more sensitive and serve night vision, the fovea is packed with cones, the optic disc has no photoreceptors, and contraction of the ciliary muscle slackens the suspensory ligaments so the lens rounds up for near vision.

Answer: A

After ovulation the corpus luteum secretes progesterone, which keeps the endometrium thick and ready for implantation; when the corpus luteum regresses, progesterone falls and menstruation follows. FSH and LH come from the anterior pituitary and act on the ovary, LH triggering ovulation. Oxytocin and prolactin act in labour and lactation.

Answer: B

Thyroid hormones normally inhibit the release of TRH and TSH, so when the thyroid fails this negative feedback is lost and the anterior pituitary secretes more TSH. TSH is made in the pituitary, not the thyroid. The last option gets the direction of the effect right but the reason wrong: thyroid hormone inhibits the pituitary, it does not stimulate it.

Answer: E

Plasma cells are activated B lymphocytes that have differentiated to secrete large amounts of antibody. Helper T cells help to activate B cells but make no antibody themselves. Neutrophils and macrophages are phagocytes, and natural killer cells kill infected or abnormal cells directly.

Answer: C

Bacteria are prokaryotes: their circular chromosome lies free in the cytoplasm in a region called the nucleoid, with no nuclear envelope. The other statements are true. Bacteria have 70S ribosomes, often carry plasmids (for example with antibiotic-resistance genes), most have a peptidoglycan wall, and penicillin kills susceptible bacteria by blocking the building of that wall.

Answer: E

The atomic number, 17, is the number of protons, and in a neutral atom also the number of electrons. The mass number, 37, is the number of nucleons, so there are 37 - 17 = 20 neutrons. Statements A to D are therefore correct. Chlorine-35 and chlorine-37 are isotopes of the same element: they have the same number of protons (17) and differ only in the number of neutrons (18 and 20), so statement E is the one that is not correct.

Answer: C

Each C=O bond in carbon dioxide is polar, but the molecule is linear (O=C=O), so the two bond dipoles are equal and point in opposite directions and cancel: the molecule has no overall dipole. Water is bent and ammonia is trigonal pyramidal, so their bond dipoles do not cancel. HCl has a single polar bond, so it must be polar. In CH3Cl the C-Cl bond is much more polar than the C-H bonds, so the dipoles do not cancel either.

Answer: A

Down group 1 each element adds an electron shell, so the atoms get larger (B is correct) and the outer electron is further from the nucleus and better shielded. It is therefore removed more easily, so the first ionisation energy falls (C is correct), and it is attracted less strongly in a bond, so electronegativity decreases down the group, not increases: A is the statement that is not correct. All the alkali metals lose their single outer electron to form M+ ions (D) and react with water, for example 2Na + 2H2O -> 2NaOH + H2 (E).

Answer: B

n = m / M = 9.0 g / 18 g/mol = 0.50 mol. The number of molecules is n x N_A = 0.50 x 6.0 x 10^23 = 3.0 x 10^23. 6.0 x 10^23 would be one whole mole (18 g), 1.2 x 10^24 comes from dividing 18 by 9 instead of 9 by 18, and 5.4 x 10^24 comes from multiplying the mass by the Avogadro constant without converting to moles.

Answer: C

Dilution does not change the amount of solute: n = c x V = 2.0 mol/L x 0.050 L = 0.10 mol. In 0.250 L this gives c = 0.10 mol / 0.250 L = 0.40 mol/L (equivalently c1V1 = c2V2, so c2 = 2.0 x 50 / 250). 0.10 mol/L confuses the amount (0.10 mol) with the concentration, and 10 mol/L comes from inverting the volume ratio; a diluted solution cannot be more concentrated than the original.

Answer: D

n(HCl) = 2.0 mol/L x 0.100 L = 0.20 mol. The equation needs 2 mol HCl per 1 mol CaCO3, so n(CaCO3) = 0.20 / 2 = 0.10 mol, and m = 0.10 mol x 100 g/mol = 10 g. 20 g ignores the 1 : 2 ratio (taking 0.20 mol of CaCO3), 40 g applies the ratio the wrong way round, and 2 g uses the concentration without the volume.

Answer: B

NaOH is a strong base and dissociates completely, so [OH-] = 0.001 = 10^-3 mol/L and pOH = 3. At 25 °C pH + pOH = 14, so pH = 14 - 3 = 11. pH 3 is the pOH mistaken for the pH (it would make the solution acidic), and pH 13 would be the value for a 0.1 mol/L solution.

Answer: D

Sodium acetate is the salt of a strong base (NaOH) and a weak acid (acetic acid). The acetate ion is a weak base and takes a proton from water, CH3COO- + H2O <-> CH3COOH + OH-, so the solution is alkaline. NaCl, KNO3 and KBr come from a strong acid and a strong base, so their solutions are neutral. NH4Cl comes from a weak base and a strong acid: the ammonium ion gives up a proton to water and the solution is acidic.

Answer: E

n(H+) = 0.10 x 0.100 = 0.010 mol and n(OH-) = 0.080 x 0.100 = 0.0080 mol. They neutralise 1 : 1, leaving 0.010 - 0.0080 = 0.0020 mol of H+ in excess. The total volume is 200 mL = 0.200 L, so [H+] = 0.0020 / 0.200 = 0.010 mol/L = 10^-2 mol/L and pH = 2. pH 7 assumes the acid and base cancel exactly, which they do not because the amounts differ. pH 1 is the pH of the acid before mixing, and pH 12 would need the base to be in excess.

Answer: C

By Le Chatelier's principle, raising the temperature favours the endothermic direction, which here is the reverse reaction, so the equilibrium yield of ammonia falls: C is not correct. Higher pressure favours the side with fewer gas molecules (2 on the right against 4 on the left), so A is correct. Removing a product or adding a reactant both shift the equilibrium to the right (B and E). A catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached sooner but its position does not change (D).

Answer: B

An enzyme, like any catalyst, provides a pathway with a lower activation energy and so speeds up the forward and reverse reactions equally. It shortens the time needed to reach equilibrium but does not change the equilibrium constant, so B is not correct. Almost all enzymes are proteins (a few catalytic RNAs exist), their activity has an optimum temperature and pH, and a competitive inhibitor competes with the substrate for the active site.

Answer: C

The rate is second order in A and first order in B. Doubling [A] multiplies the rate by 2^2 = 4; halving [B] multiplies it by 1/2. Overall the factor is 4 x 1/2 = 2, so the rate doubles. Saying it stays the same treats both changes as first order (2 x 1/2), quadrupling ignores the change in B, and eightfold multiplies by 2 for B instead of dividing.

Answer: E

The oxidation numbers in a neutral compound add up to zero. Potassium is +1 and each oxygen is -2, so +1 + x + 4(-2) = 0, which gives x = +7. +2 is manganese in Mn2+, the product when permanganate is reduced in acid, +4 is manganese in MnO2, and +6 is manganese in the manganate ion MnO4^2-. -1 is the charge of the permanganate ion, not the oxidation number of manganese.

Answer: B

Zinc goes from 0 to +2, losing two electrons: it is oxidised (A and C are correct). Copper goes from +2 to 0, gaining two electrons: Cu2+ is reduced (D and E are correct). The species that is reduced is the oxidising agent, because it takes the electrons; the reducing agent is zinc. So B, which calls Cu2+ the reducing agent, is not correct.

Answer: E

n(MnO4-) = 0.020 mol/L x 0.0200 L = 4.0 x 10^-4 mol. The equation gives 5 mol Fe2+ per 1 mol MnO4-, so n(Fe2+) = 5 x 4.0 x 10^-4 = 2.0 x 10^-3 mol. In 0.0250 L, c(Fe2+) = 2.0 x 10^-3 / 0.0250 = 0.080 mol/L. 0.016 mol/L forgets the factor of 5, 0.0032 mol/L divides by 5 instead of multiplying, 0.032 mol/L uses a 1 : 2 ratio, and 0.10 mol/L divides by the permanganate volume (20 mL) instead of the sample volume.

Answer: D

Sodium chloride is an ionic compound: a lattice of Na+ and Cl- ions held by strong electrostatic attraction, which is why it melts only at a high temperature (about 800 °C). It conducts electricity only when the ions are free to move, that is, when molten or dissolved. In the solid the ions are fixed in the lattice, so solid NaCl is an insulator and D is not correct.

Answer: C

Oxides of non-metals are usually acidic: sulfur trioxide reacts with water to form sulfuric acid, SO3 + H2O -> H2SO4. The other four are oxides of metals and are basic: Na2O and K2O give NaOH and KOH, CaO gives Ca(OH)2, and MgO, which dissolves only slightly, gives a weakly alkaline solution of Mg(OH)2.

Answer: A

The oxidising power of the halogens falls down group 17: F2 > Cl2 > Br2 > I2. A halogen displaces a halide only of a halogen below it, because it is the stronger oxidising agent. Chlorine oxidises bromide to bromine, so A happens. In B, C, D and E a weaker oxidising halogen would have to oxidise the halide of a stronger one (iodine against chloride or bromide, bromine against chloride or fluoride), which does not happen.

Answer: D

Four carbon atoms can be joined either in an unbranched chain, butane (CH3CH2CH2CH3), or with a branch, 2-methylpropane (CH3CH(CH3)CH3). Any other drawing, such as a bent chain, is one of these two again, so there are 2 isomers. Three isomers is the answer for C5H12 (pentane, 2-methylbutane and 2,2-dimethylpropane).

Answer: E

In benzene the six p electrons are delocalised around the planar ring of sp2 carbons, so all six C-C bonds are the same length, between that of a single and a double bond. E describes the Kekulé structure with alternating bond lengths, which does not match the real molecule, so it is not correct. Because addition would destroy the stable delocalised system, benzene typically undergoes electrophilic substitution instead (C is correct).

Answer: B

Carbons 2 and 3 are each bonded to H, Cl, CH3 and the rest of the molecule, so there are two chiral centres and at most 2^2 = 4 stereoisomers: (2R,3R), (2S,3S), (2R,3S) and (2S,3R). But the two halves of the molecule are identical, so (2R,3S) and (2S,3R) are the same compound, a meso form with an internal plane of symmetry that is not optically active. That leaves a pair of enantiomers plus the meso form: 3 stereoisomers. 4 is the 2^n maximum without checking for symmetry, and 2 counts only the enantiomers.

Answer: A

Propan-1-ol, CH3CH2CH2OH, is a primary alcohol. Its first oxidation product is the aldehyde propanal, CH3CH2CHO; distilling it off stops it being oxidised further to propanoic acid, which is what forms under reflux. Propanone is the ketone formed by oxidising the secondary alcohol propan-2-ol, propene would come from dehydration rather than oxidation, and propan-2-ol is an isomer of the starting alcohol, not an oxidation product.

Answer: A

An ester has no O-H group, so its molecules cannot hydrogen-bond to one another. Carboxylic acids can (they even form dimers), so esters boil at lower temperatures than acids of similar molar mass, and A is not correct. Esterification is a condensation of an acid and an alcohol that releases water (B), ethyl ethanoate, CH3COOCH2CH3, comes from ethanol and ethanoic acid (C), and alkaline hydrolysis (saponification) gives the carboxylate salt and the alcohol (D).

Answer: D

Sucrose is made of glucose and fructose joined by a glycosidic bond; glucose and galactose make lactose. So D is not correct. Glucose has six carbons and an aldehyde group in its open-chain form (an aldohexose, C6H12O6), and that aldehyde group makes it a reducing sugar. Cellulose is a linear polymer of beta-glucose units joined by beta-1,4 bonds.

Answer: A

A linear chain of n amino acids has n - 1 peptide bonds, so a tripeptide has 2 (and releasing 2 water molecules as it forms). The order matters because a peptide has a distinct N-terminus and C-terminus, so the number of sequences using each amino acid once is 3! = 3 x 2 x 1 = 6 (for example Gly-Ala-Ser and Ser-Ala-Gly are different). 3 bonds counts the residues rather than the links between them, and 27 = 3^3 would allow each amino acid to be used more than once.

Answer: B

1 km/h = 1000 m / 3600 s = 1/3.6 m/s, so 72 km/h = 72 / 3.6 = 20 m/s. 259.2 m/s comes from multiplying by 3.6 instead of dividing, and 7.2 m/s and 200 m/s come from shifting the decimal point by a power of ten as if the conversion factor were 10.

Answer: D

Work = force x distance, so 1 J = 1 N x 1 m. A newton is kg m s^-2 (from F = ma), so 1 J = kg m s^-2 x m = kg m^2 s^-2. The same result follows from kinetic energy, (1/2)mv^2: kg x (m/s)^2. kg m s^-2 is the newton, kg m^2 s^-3 is the watt (a joule per second), kg m^-1 s^-2 is the pascal and kg m s^-1 is the unit of momentum.

Answer: C

At constant speed the lifting force equals the weight, mg = 2.0 x 10 = 20 N, and the work done is the gain in potential energy, mgh = 20 N x 5.0 m = 100 J. Power = work / time = 100 J / 4.0 s = 25 W. 100 W gives the work in joules as if it were the power, 2.5 W leaves out g (m x h / t), 20 W is the weight in newtons, not a power, and 10 W divides the work by 10 instead of by the time.

Answer: A

The work done by the brakes removes all the kinetic energy: F x s = (1/2)mv^2 = 0.5 x 1000 x 20^2 = 200000 J, so F = 200000 / 40 = 5000 N (equivalently a = v^2 / 2s = 400 / 80 = 5 m/s^2 and F = ma = 5000 N). With constant deceleration the average speed is (20 + 0) / 2 = 10 m/s, so t = 40 m / 10 m/s = 4 s (or t = v / a = 20 / 5 = 4 s). 10000 N forgets the 1/2 in the kinetic energy, and 2 s divides the distance by the initial speed instead of the average speed.

Answer: D

The volume flow rate is the same in both sections (continuity equation): A1v1 = A2v2, so v2 = A1v1 / A2 = 2.0 x 10 / 0.50 = 40 cm/s. Where the vessel is narrower the blood must move faster, which rules out 2.5 cm/s (the ratio applied the wrong way round) and 5 cm/s. 160 cm/s squares the area ratio, a mistake that comes from treating the given areas as radii.

Answer: C

At constant volume p / T is constant, with T in kelvin: T1 = 27 + 273 = 300 K and T2 = 127 + 273 = 400 K. So p2 = p1 x T2 / T1 = 300 x 400 / 300 = 400 kPa. About 1411 kPa uses Celsius temperatures (300 x 127 / 27), which is not valid, 225 kPa inverts the ratio, and 600 kPa assumes the pressure doubles. Since 400 kPa is offered, "no answer is correct" is wrong.

Answer: E

For resistors in parallel, 1/R = 1/R1 + 1/R2 = 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3, so R = 3 ohm (or R = R1R2 / (R1 + R2) = 48 / 16 = 3 ohm). The combined resistance of a parallel pair is always smaller than the smaller resistor. 16 ohm is the series total, 1/3 ohm forgets to invert 1/R, 48 ohm is the product alone and 8 ohm is the mean.

Answer: E

The lens equation gives 1/f = 1/u + 1/v, so 1/v = 1/10 - 1/15 = 3/30 - 2/30 = 1/30 and v = +30 cm: a real image 30 cm behind the lens. The magnification is v / u = 30 / 15 = 2, and a real image formed by a single lens is inverted. So the image is real, inverted, 30 cm behind the lens and twice the size of the object. A has the wrong size, B describes a virtual image (which needs the object inside the focal length), C comes from adding 1/10 and 1/15, and D calls a real image upright. None of A to D is right, so the answer is "no answer is correct".

Answer: B

32 days is 32 / 8 = 4 half-lives, and each halves the activity: 160 -> 80 -> 40 -> 20 -> 10 MBq, that is 160 / 2^4 = 160 / 16 = 10 MBq. 20 MBq stops after three half-lives, 40 MBq divides by 4 (the number of half-lives) instead of by 2^4, and 5 MBq takes one half-life too many.

Answer: A

The resistance is R = V^2 / P = 200^2 / 1000 = 40000 / 1000 = 40 ohm. At 100 V, P = V^2 / R = 100^2 / 40 = 10000 / 40 = 250 W. Because P is proportional to V^2, halving the voltage quarters the power. 500 W assumes the power is proportional to the voltage, and 1000 W assumes the power stays fixed whatever the supply.

Answer: C

Write everything as a power of 2: 2^5 x 2^3 = 2^8, and 4^3 = (2^2)^3 = 2^6. So the expression is 2^8 / 2^6 = 2^2 = 4. 16 comes from treating 4^3 as 2^4 (adding the exponents 2 + 3 instead of multiplying), 1 from treating 4^3 as 2^8, and 8 from treating it as 2^5.

Answer: B

(n + 1)! = (n + 1) x n x (n - 1)!, so (n + 1)! / (n - 1)! = (n + 1)n. Then n^2 + n - 30 = 0, which factorises as (n + 6)(n - 5) = 0, so n = 5 or n = -6; only n = 5 is a natural number. Check: 6! / 4! = 720 / 24 = 30. n = 6 gives 7! / 5! = 42 and n = 4 gives 5! / 3! = 20. Since 5 is offered, "no answer is correct" is wrong.

Answer: A

Write y = (2x - 3) / 5 and solve for x: 5y = 2x - 3, so 2x = 5y + 3 and x = (5y + 3) / 2. Swapping the letters, f^-1(x) = (5x + 3) / 2. Check: f((5x + 3) / 2) = (5x + 3 - 3) / 5 = x. (5x - 3) / 2 undoes the steps without reversing the sign of the 3, and 5 / (2x - 3) is the reciprocal 1 / f(x), a common confusion between the inverse function and the reciprocal.

Answer: D

Complete the square: x^2 - 6x + 5 = (x - 3)^2 - 9 + 5 = (x - 3)^2 - 4, so the vertex is (3, -4). Equivalently x = -b / 2a = 6 / 2 = 3 and y(3) = 9 - 18 + 5 = -4. (-3, -4) and (-3, 32) get the sign of the x-coordinate wrong (32 is y(-3)), and (3, 4) gets the sign of the y-coordinate wrong.

Answer: D

f'(x) = 3x^2 - 6x = 3x(x - 2), which is zero at x = 0 and x = 2. f''(x) = 6x - 6, so f''(0) = -6 < 0 (a local maximum) and f''(2) = 6 > 0 (a local minimum). f(2) = 8 - 12 + 2 = -2, so the local minimum is at (2, -2). (0, 2) is the local maximum, (1, 0) is the point of inflection, and (2, 2) has the right x but uses f(0) for the y-value.

Answer: A

The given line has gradient 2, so a perpendicular line has gradient -1/2 (the product of the gradients is -1). Through (1, 2): y - 2 = -(1/2)(x - 1), so 2y - 4 = -x + 1 and x + 2y - 5 = 0. 2x - y = 0 passes through (1, 2) but is parallel (gradient 2); 2x + y - 4 = 0 has gradient -2 and x - 2y + 3 = 0 has gradient +1/2, both through (1, 2) but not perpendicular; x + 2y + 3 = 0 has the right gradient but does not pass through (1, 2).

Answer: C

The space diagonal of the cube is a diameter of the sphere: 2 x sqrt(3) = 2 sqrt(3) cm. A cube of edge a has space diagonal a sqrt(3), so a sqrt(3) = 2 sqrt(3) and a = 2 cm. The volume is 2^3 = 8 cm^3. 3 sqrt(3) takes the radius as the edge, 24 sqrt(3) takes the diameter as the edge, and 4 sqrt(3) π is the volume of the sphere, (4/3)π(sqrt(3))^3. Since 8 cm^3 is offered, "no answer is correct" is wrong.

Answer: E

The modulus of a quotient is the quotient of the moduli: |3 + 4i| = sqrt(9 + 16) = 5 and |1 - i| = sqrt(1 + 1) = sqrt(2), so |z| = 5 / sqrt(2) = 5 sqrt(2) / 2, about 3.54. Check by dividing: z = (3 + 4i)(1 + i) / 2 = (-1 + 7i) / 2, and |z| = sqrt(1 + 49) / 2 = sqrt(50) / 2 = 5 sqrt(2) / 2. 5/2 divides by |1 - i|^2 instead of |1 - i|, 5 sqrt(2) multiplies instead of dividing, 5 ignores the denominator and 25/2 is |z|^2. None of A to D equals 5 sqrt(2) / 2, so the answer is "no answer is correct".

Answer: B

There are 6 x 6 = 36 equally likely ordered outcomes. A sum of 8 comes from (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2): 5 outcomes, so the probability is 5/36. 1/12 = 3/36 counts only the unordered pairs {2, 6}, {3, 5} and {4, 4}, and 1/6 = 6/36 is the probability of a sum of 7, the most likely total.

Answer: E

Complete the square: x^2 - 4x + 5 = (x - 2)^2 + 1, which is at least 1 for every real x. (Equivalently the discriminant is 16 - 20 = -4 < 0 and the leading coefficient is positive, so the parabola lies entirely above the x-axis.) The expression is never negative, so no real x satisfies the inequality: the solution set is empty. A, B and C come from wrongly factorising x^2 - 4x - 5 = (x - 5)(x + 1), and D describes the inequality x^2 - 4x + 5 > 0. None of A to D is right, so the answer is "no answer is correct".