- a$B-A$
- b$A\cap B$
- c$A\cup B$
- d
Answer (b): $A-B$ is the part of $A$ outside $B$. Removing it from $A$ leaves the part of $A$ inside $B$, which is $A\cap B$.
Revise: Algebra (Class 11 notes)
TU IOST | Entrance paper
Answer (b): $A-B$ is the part of $A$ outside $B$. Removing it from $A$ leaves the part of $A$ inside $B$, which is $A\cap B$.
Revise: Algebra (Class 11 notes)
Answer (c): $y=\sqrt{a^2-x^2}$ is defined for $-a\le x\le a$ and takes values from $0$ (at $x=\pm a$) up to $a$ (at $x=0$), so the range is $[0,a]$.
Revise: Algebra (Class 11 notes)
Answer (c): $\sqrt{a^2}=a$, so the expression is $\sqrt{a^4}=a^2$. Hence $\log_a a^2=2$.
Revise: Algebra (Class 11 notes)
Answer (a): $A^2=-I$, so $A^4=I$ and $A^{100}=(A^4)^{25}=I$, the unit matrix, which the option '1' stands for. The printed 1 is read as the identity matrix.
Revise: Algebra (Class 11 notes)
Answer (c): Let each equal $k$: $\ln a=\frac{\ln k}{x}$ etc. $b^2=ac$ gives $\frac{2}{y}=\frac1x+\frac1z$, so $x,y,z$ are in H.P.
Revise: Algebra (Class 11 notes)
Answer (c): Split as $\frac{a^x-1}{x}-\frac{b^x-1}{x}\to\ln a-\ln b=\ln(a/b)$.
Revise: Calculus (Class 12 notes)
Answer (a): $\frac{d}{dx}a^x=a^x\ln a$, so $\frac{d}{dx}5^x=5^x\ln5$.
Revise: Calculus (Class 12 notes)
Answer (a): The only restriction is $x\neq0$ (division by zero), so the domain is all reals except $0$.
Revise: Algebra (Class 11 notes)
Answer (a): Put $u=e^x+1$, $du=e^x dx$. The integral becomes $\int\frac{du}{u}=\ln(e^x+1)+C$.
Revise: Calculus (Class 12 notes)
Answer (c): Equal roots need $D=0$: $k^2-4(5)(45)=0$, so $k^2=900$ and $k=\pm30$.
Revise: Algebra (Class 11 notes)
Answer (b): The conjugate root is $2-\sqrt3$. Sum $=4$, product $=4-3=1$, so $x^2-4x+1=0$.
Revise: Algebra (Class 11 notes)
Answer (c): Write the second as $2x-y-\frac53=0$ and the first as $2x-y+4=0$. Distance $=\frac{|4+5/3|}{\sqrt5}=\frac{17}{3\sqrt5}=\frac{17}{\sqrt{45}}$.
Revise: Analytic geometry (Class 11 notes)
Answer (a): $r^2=g^2+f^2-c=\frac94+1+\frac34=4$, so $r=2$.
Revise: Analytic geometry (Class 11 notes)
Answer (c): $|(1+i)^{-1}|=\frac{1}{|1+i|}=\frac{1}{\sqrt2}$.
Revise: Algebra (Class 11 notes)
Answer (b): By De Moivre, $z^n=\cos n\theta+i\sin n\theta$ and $z^{-n}=\cos n\theta-i\sin n\theta$. The sum is $2\cos n\theta$.
Revise: Algebra (Class 11 notes)
Answer (a): The terms are $1\cdot2,\ 2\cdot3,\ 3\cdot4,\ 4\cdot5$, so $t_n=n(n+1)$.
Revise: Algebra (Class 11 notes)
Answer (d): The infinite G.P. gives $y=\frac{x}{1-x}$. Then $y-xy=x$, so $x(1+y)=y$ and $x=\frac{y}{1+y}$.
Revise: Algebra (Class 11 notes)
Answer (a): Chain rule: $y'=-\sin(\cos\sqrt x)\cdot(-\sin\sqrt x)\cdot\frac{1}{2\sqrt x}=\frac{\sin(\cos\sqrt x)\sin\sqrt x}{2\sqrt x}$.
Revise: Calculus (Class 12 notes)
Answer (b): Put $x=\cos\theta$. Then $\cos3\theta=4\cos^3\theta-3\cos\theta$, so $3\theta=\cos^{-1}(4x^3-3x)$.
Revise: Trigonometry (Class 11 notes)
Answer (c): $\tan ax=\tan(\frac\pi2-bx)$ gives $ax=n\pi+\frac\pi2-bx$, so $x=\frac{(2n+1)}{a+b}\cdot\frac\pi2$.
Revise: Trigonometry (Class 11 notes)
Answer (b): Add all columns to the first: each row sum is $1+\omega+\omega^2=0$, so the first column becomes zero and the determinant is $0$.
Revise: Algebra (Class 11 notes)
Answer (a): Multiply $(y-x)(y+x)=0$ to get $y^2-x^2=0$.
Revise: Analytic geometry (Class 11 notes)
Answer (b): $A-B$ removes the point $2$ (the only common point) from $A$, leaving $[-3,2)$.
Revise: Algebra (Class 11 notes)
Answer (c): Both one-sided limits exist but differ, which is a jump discontinuity.
Revise: Calculus (Class 12 notes)
Answer (d): Area $=\int_0^4 3x,dx=\frac32x^2\Big|_0^4=24$.
Revise: Calculus (Class 12 notes)
Answer (a): An epitaph is the text written on a tombstone in memory of the dead.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (b): With a pronoun object, a phrasal verb is split: 'handed it over'. 'Handed over it' is wrong.
Revise: Grammar and Usage (Class 11 notes)
Answer (b): 'Call on' means to visit a person, which fits visiting her sister in the village.
Revise: Grammar and Usage (Class 11 notes)
Answer (c): 'With' shows possession of a feature: a coat with five pockets.
Revise: Grammar and Usage (Class 11 notes)
Answer (b): The idiom means to attack unfairly, against the rules.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (a): The verb 'disapprove' is followed by the preposition 'of'.
Revise: Grammar and Usage (Class 11 notes)
Answer (d): 'Switched off' is the standard phrase for turning a light off. 'Take off' is wrong in form and 'left off' means stopped or omitted.
Revise: Grammar and Usage (Class 11 notes)
Answer (d): 'Mercy' is a noun. Thicken and impoverish are verbs and mercantile is an adjective.
Revise: Grammar and Usage (Class 11 notes)
Answer (a): 'Good' describes 'swimming' and is a clear adjective. 'Lawn' is a noun; 'knitting' is a gerund used as a noun modifier.
Revise: Grammar and Usage (Class 11 notes)
Answer (c): In 'I know him' the subject performs the action. The others are passive.
Revise: Grammar and Usage (Class 11 notes)
Answer (a): 'Memoranda' is the plural of memorandum, so the plural verb 'have been' is used.
Revise: Grammar and Usage (Class 11 notes)
Answer (b): 'Furniture' is an uncountable noun and takes a singular verb. Police is plural and the others are plural forms.
Revise: Grammar and Usage (Class 11 notes)
Answer (b): With 'neither...nor' the verb agrees with the nearer subject, 'they', so 'are'.
Revise: Grammar and Usage (Class 11 notes)
Answer (c): The collective noun for sheep is a flock, not a herd. A herd of cattle, a pile of books and a pack of wolves are all correct.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (d): 'Daren't' is a negative modal-like verb, so the positive question tag is 'dare they?'.
Revise: Grammar and Usage (Class 11 notes)
Answer (d): 'They are too' completes the sentence meaning 'and they are hungry too'. The other options clash in subject or construction.
Revise: Grammar and Usage (Class 11 notes)
Answer (a): LIKE-li-hood is stressed on the first syllable. Religious, fantastic and orthography are stressed on the second.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (a): e-co-nom-i-cal has five syllables.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (b): 'So' and 'sow' (to plant seeds) are both pronounced /soʊ/. The other pairs sound different.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (d): 'They' and 'that' both begin with the voiced sound /ð/.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (a): A jockey is a professional rider in horse races.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (c): 'Appropriate' means suitable, so its opposite is unsuitable.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (a): 'Sporadic' means happening at irregular, scattered times, i.e. occasional.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (b): Option b is wrongly punctuated: it needs a comma after 'said to me' and the question mark belongs inside the quotation marks.
Revise: Grammar and Usage (Class 11 notes)
Answer (c): Father + ly gives the adjective 'fatherly'.
Revise: Vocabulary and Word Formation (Class 11 notes)
Answer (a): $V=\frac{W}{q}=\frac{ML^2T^{-2}}{IT}=ML^2T^{-3}I^{-1}$.
Revise: Physical Quantities (Class 11 notes)
Answer (b): Taking downward positive, $v^2=u^2+2gh$ gives $9u^2=u^2+2gh$, so $h=\frac{4u^2}{g}$.
Revise: Kinematics (Class 11 notes)
Answer (c): $a=-\omega^2x$, which is largest when the displacement is largest, at the extreme position.
Revise: Periodic Motion (Class 12 notes)
Answer (a): Fractional increase is $\alpha\Delta T$ for length, $2\alpha\Delta T$ for area and $3\alpha\Delta T$ for volume, so volume increases most.
Revise: Thermal Expansion (Class 11 notes)
Answer (c): $\Delta L=\frac{FL}{AY}\propto\frac{L}{d^2}$. Ratio $=\frac{1/4}{2/1}=\frac18$.
Revise: Elasticity (Class 11 notes)
Answer (b): Continuity: $Av=$ constant, so $v\propto\frac{1}{r^2}$. $\frac{v_1}{v_2}=\frac{2^2}{3^2}=\frac49$.
Revise: Fluid Statics (Class 12 notes)
Answer (d): Constant pressure is an isobaric process.
Revise: First Law of Thermodynamics (Class 12 notes)
Answer (a): Heat to warm the ice: $0.05\times2000\times6=600$ J. Mass frozen $=\frac{600}{3.36\times10^5}\text{ kg}\approx1.785$ g.
Revise: Quantity of Heat (Class 11 notes)
Answer (b): Rays from the focus become parallel after reflection, so the image forms at infinity.
Revise: Reflection at Curved Mirrors (Class 11 notes)
Answer (d): $\eta=1-\frac{T_2}{T_1}=1-\frac{300}{400}=0.25=25%$.
Revise: Second Law of Thermodynamics (Class 12 notes)
Answer (a): ${}_a n_g=\frac32$. ${}_d n_g=\frac{{}_a n_g}{{}_a n_d}=\frac{3/2}{12/5}=\frac{15}{24}=\frac58$.
Revise: Refraction at Plane Surface (Class 11 notes)
Answer (b): $\frac1f=(\mu-1)\left(\frac2R\right)=0.5\cdot\frac2R=\frac1R$, so $f=R$.
Revise: Lenses (Class 11 notes)
Answer (b): Wavelength in glass $=\lambda/n=7200/1.5=4800$ Å.
Revise: Refraction at Plane Surface (Class 11 notes)
Answer (a): $\frac1{C_s}=\frac13+\frac19+\frac1{18}=\frac12$, so $C_s=2\mu F$. $C_p=30\mu F$. Ratio $=\frac{2}{30}=1:15$.
Revise: Capacitor (Class 11 notes)
Answer (c): Across two corners, one resistor $r$ is in parallel with two in series ($2r$): $\frac{r\cdot2r}{3r}=\frac{2r}{3}$.
Revise: DC Circuit (Class 11 notes)
Answer (b): On the same supply $P=\frac{V^2}{R}\propto\frac1R$, so the ratio is $2:1$.
Revise: DC Circuit (Class 11 notes)
Answer (b): $U=\frac12LI^2=\frac12(0.05)(16)=0.4$ J.
Revise: Electromagnetic Induction (Class 12 notes)
Answer (c): At resonance $X_L=X_C$, impedance is minimum ($=R$) and current is maximum.
Revise: Alternating Currents (Class 12 notes)
Answer (a): $\lambda=\frac{h}{mv}=\frac{6.63\times10^{-34}}{10^{-31}\times10^5}=6.63\times10^{-8}$ m.
Revise: Quantization of Energy (Class 12 notes)
Answer (a): Destructive interference needs a phase difference of an odd multiple of $\pi$; the smallest given is $\pi$.
Revise: Interference (Class 12 notes)
Answer (c): Hard X-rays have shorter wavelength and so higher frequency and penetrating power. All X-rays travel at $c$.
Revise: Electrons (Class 12 notes)
Answer (d): An alpha particle is a helium nucleus; its mass is less than its separate nucleons by the mass defect (binding energy).
Revise: Radioactivity and Nuclear Reaction (Class 12 notes)
Answer (a): $A_v=\beta\frac{R_L}{R_{in}}=50\times\frac{4000}{500}=400$.
Revise: Semiconductor Devices (Class 12 notes)
Answer (d): Closed pipe: $f_1=\frac{v}{4L}=\frac{320}{4}=80$ Hz (other resonances are odd multiples: 240, 400 Hz...).
Revise: Wave in Pipes and Strings (Class 12 notes)
Answer (d): Comparing with $y=A\sin(\omega t-kx)$ gives $\lambda=2\pi/k$. The only option consistent with the intended $k=0.025$ is $\lambda=2\pi/0.025=80\pi$, so the printed 0.0025 looks like a misprint for 0.025.
Revise: Wave Motion (Class 12 notes)
Answer (c): $\ce{Al4C3 + 12H2O -> 4Al(OH)3 + 3CH4}$, so methane is released.
Revise: Hydrocarbons (Class 11 notes)
Answer (c): Concentrated HCl with anhydrous $\ce{ZnCl2}$ is Lucas reagent, used to distinguish 1, 2 and 3 degree alcohols.
Revise: Alcohols (Class 12 notes)
Answer (a): Valeric acid is $\ce{CH3(CH2)3COOH}$, a five carbon acid: pentanoic acid.
Revise: Carboxylic Acids (Class 12 notes)
Answer (c): Blood red $\ce{Fe(SCN)3}$ forms when both N and S are present in the sodium fusion extract (as NaSCN).
Revise: Fundamental Principles (Class 11 notes)
Answer (d): Methanol has hydrogen bonding, which gives a much higher boiling point than the other compounds.
Revise: Chemical Bonding (Class 11 notes)
Answer (c): Cannizzaro reaction needs an aldehyde with no alpha hydrogen. HCHO has none, so it qualifies.
Revise: Aldehydes and Ketones (Class 12 notes)
Answer (a): $\ce{C6H6 + CH3Cl ->[AlCl3] C6H5CH3 + HCl}$ is Friedel-Crafts alkylation.
Revise: Aromatic Hydrocarbons (Class 11 notes)
Answer (d): Only primary amines give the carbylamine (isocyanide) test; $\ce{CH3NH2}$ is primary.
Revise: Amines (Class 12 notes)
Answer (b): Hematite is $\ce{Fe2O3}$, an iron ore.
Revise: Chemistry of Metals (Class 11 notes)
Answer (c): Oxides of strongly electropositive metals like sodium give alkali with water, so they are basic.
Revise: Classification of Elements (Class 11 notes)
Answer (d): Electronegativity decreases down the halogen group, so fluorine is highest.
Revise: Classification of Elements (Class 11 notes)
Answer (a): $\ce{ZnSO4.7H2O}$ is white vitriol. Green is $\ce{FeSO4.7H2O}$, blue is $\ce{CuSO4.5H2O}$.
Revise: Chemistry of Metals (Class 11 notes)
Answer (a): Brass is copper with zinc; bronze is copper with tin.
Revise: Chemistry of Metals (Class 11 notes)
Answer (b): Dolomite is the double carbonate of calcium and magnesium.
Revise: Chemistry of Metals (Class 11 notes)
Answer (a): Zinc ($Z=30$) has a full d subshell: $[Ar]3d^{10}4s^2$.
Revise: Transition Metals (Class 12 notes)
Answer (a): In neutral medium Mn changes from $+7$ to $+4$ (3 electrons): $\frac{158}{3}=52.67$.
Revise: Volumetric Analysis (Class 12 notes)
Answer (b): Sommerfeld introduced elliptical orbits to improve Bohr's model.
Revise: Atomic Structure (Class 11 notes)
Answer (a): Moles $=\frac{8}{16}=0.5$. Atoms $=0.5\times6.022\times10^{23}=3.011\times10^{23}$.
Revise: Stoichiometry (Class 11 notes)
Answer (b): HCl is a strong acid: $[H^+]=10^{-3}$, so pH $=3$.
Revise: Ionic Equilibrium (Class 12 notes)
Answer (d): Rate depends on temperature, concentration and surface area, but molecularity is a property of the elementary step, not a rate factor.
Revise: Chemical Kinetics (Class 12 notes)
Answer (a): Specific conductance is $\frac1\rho$, with unit $ohm^{-1}cm^{-1}$, i.e. Siemen $cm^{-1}$.
Revise: Electrochemistry (Class 12 notes)
Answer (a): Hess's law says the total heat change of a reaction is the same regardless of the path taken.
Revise: Chemical Thermodynamic (Class 12 notes)
Answer (b): By Ostwald's dilution law, $\alpha$ rises as concentration falls, so the degree of dissociation increases with dilution.
Revise: Ionic Equilibrium (Class 12 notes)
Answer (a): Moles $=\frac{0.1}{22.4}=0.00446$; mass $=0.00446\times17=0.0759$ g.
Revise: Stoichiometry (Class 11 notes)
Answer (b): Dalton proposed the law of multiple proportions (1803).
Revise: Stoichiometry (Class 11 notes)