Page 2 =====
In the determination of Young's modulus (Y = 4MLg / πd²) by using Searle's method, a wire of length L = 2m and diameter d = 0.5mm is used. For a load M = 2.5kg, an extension ℓ = 0.25mm in the length of the wire is observed. Quantities d and ℓ are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement
(A) due to the errors in the measurements of d and ℓ are the same.
(B) due to the error in the measurement of d is twice that due to the error in the measurement of ℓ.
(C) due to the error in the measurement of ℓ is twice that due to the error in the measurement of d.
(D) due to the error in the measurement of d is four times that due to the error in the measurement of ℓ.
Sol.
L.C. = 0.5 / 100 = 0.005mm
ΔY / Y = Δℓ / ℓ + 2Δ(d) / d
Δℓ / ℓ = (0.005 × 10⁻³) / (0.25 × 10⁻³) = 1 / 50
2Δ(d) / d = (2 × 0.005 × 10⁻³) / (0.5 × 10⁻³) = 1 / 50
A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in the x-y plane with centre at O and constant angular speed ω. If the angular momentum of the system, calculated about O and P are denoted by L_O and L_P respectively, then
(A) L_O and L_P do not vary with time.
(B) L_O varies with time while L_P remains constant.
(C) L_O remains constant while L_P varies with time.
(D) L_O and L_P both vary with time.
Sol.
===== Page 3 =====
A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surface are of the same radius of curvature R = 14cm. For this bi-convex lens, for an object distance of 40cm, the image distance will be
A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surface are of the same radius of curvature R = 14cm. For this bi-convex lens, for an object distance of 40cm, the image distance will be
(A) -280.0 cm
Sol.
(B)
P_T = (1.5 - 1)(1/14 - 0) + (1.2 - 1)(0 - 1/(-14)) = 0.5/14 + 0.2/14 = 1/20
f = +20cm
1/v - 1/(-40) = 1/20
1/v = 1/20 - 1/40 = 1/40
∴ v = 40cm
A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angular speed ω as shown in the figure. At time t = 0, a small insect starts from O and moves with constant speed v, with respect to the rod towards the other end. It reaches the end of the rod at t = T and stops. The angular speed of the system remains ω throughout. The magnitude of the torque (|τ|) about O, as a function of time is best represented by which plot?
Sol. (B)
τ = ω dI/dt = ω d/dt (C + mv²t²)
= mωv²2t.
===== Page 4 =====
A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300K in a container. The ratio of the rms speeds (v_rms(helium) / v_rms(argon)) is
(A) 0.32
(B) 0.45
(C) 2.24
(D) 3.16
Sol. (D)
v_rms = √(3RT/M)
Required ratio = √(M_Ar / M_He) = √(40/4) = √10 = 3.16
Two large vertical and parallel metal plates having a separation of 1cm are connected to a DC voltage source of potential difference X. A proton is released at rest midway between the two plates. It is found to move at 45° to the vertical JUST after release. Then X is nearly
(A) 1 × 10⁻⁵ V
(B) 1 × 10⁻⁷ V
(C) 1 × 10⁻⁹ V
(D) 1 × 10⁻¹⁰ V
Sol. (C)
qE = mg
q(V/d) = mg
V = mgd / q
= (1.67 × 10⁻²⁷ × 10 × 10⁻²) / (1.6 × 10⁻¹⁹)
= 10⁻²⁸ / 10⁻¹⁹ = 10⁻⁹ V
Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at temperatures 2T and 3T respectively. The temperature of the middle (i.e. second) plate under steady state condition is
(A) (65/2)^(1/4) T
(B) (97/4)^(1/4) T
(C) (97/2)^(1/4) T
(D) (97)^(1/4) T
Sol. (C)
σA(2T)⁴ + σA(3T)⁴ = σ2A(T')⁴
16T⁴ + 81T⁴ = 2(T')⁴
97T⁴ = 2(T')⁴
(T')⁴ = 97/2 T⁴
∴ T' = (97/2)^(1/4) T
===== Page 5 =====
A small block is connected to one end of a massless spring of un-stretched length 4.9m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2m and released from rest at t = 0. It then executes simple harmonic motion with angular frequency ω = π/3 rad/s. Simultaneously at t = 0, a small pebble is projected with speed v from point P at an angle of 45° as shown in the figure. Point P is at a horizontal distance of 10m from O. If the pebble hits the block at t = 1s, the value of v is (take g = 10m/s²)
(A) √50 m/s
(B) √51 m/s
(C) √52 m/s
(D) √53 m/s
Sol. (A)
Young's double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are β_G, β_R and β_B, respectively. Then,
(A) β_G > β_B > β_R
(B) β_B > β_G > β_R
(C) β_R > β_B > β_G
(D) β_R > β_G > β_B
Sol. (D)
λ_R > λ_G > λ_B
∴ β_R > β_G > β_B
Consider a thin spherical shell of radius R with centre at the origin, carrying uniform positive surface charge density. The variation of the magnitude of the electric field |E(r)| and the electric potential V(r) with the distance r from the centre, is best represented by which graph?
Sol. (D)
===== Page 6 =====
Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields E = E₀ ĵ and B = B₀ ĵ. At time t = 0, this charge has velocity v in the x-y plane, making an angle θ with the x-axis. Which of the following option(s) is (are) correct for time t > 0?
(A) If θ = 0°, the charge moves in a circular path in the x-z plane.
(B) If θ = 0°, the charge undergoes helical motion with constant pitch along the y-axis.
(C) If θ = 10°, the charge undergoes helical motion with its pitch increasing with time, along the y-axis.
(D) If θ = 90°, the charge undergoes linear but accelerated motion along the y-axis.
(D) If θ = 90°, the charge undergoes linear but accelerated motion along the y-axis.
Sol. (C, D)
If θ = 90°, B exerts no force on q.
If θ = 0°, 10°, the charge particle moves in helix with increasing pitch due to E along y-axis.
A cubical region of side a has its centre at the origin. It encloses three fixed point charges, -q at (0, -a/4, 0), +3q at (0,0,0) and -q at (0, +a/4, 0). Choose the correct options(s)
(A) The net electric flux crossing the plane x = +a/2 is equal to the net electric flux crossing the plane x = -a/2
(B) The net electric flux crossing the plane y = +a/2 is more than the net electric flux crossing the plane y = -a/2
(C) The net electric flux crossing the entire region is q/ε₀
(D) The net electric flux crossing the plane z = +a/2 is equal to the net electric flux crossing the plane x = +a/2
Sol. (A, C, D)
Net flux through the cubical region = (-q + 3q - q)/ε₀ = q/ε₀
The flux passing through the faces x = -a/2, x = +a/2 and z = +a/2 are same due to symmetry.
A person blows into open-end of a long pipe. As a result, a high pressure pulse of air travels down the pipe. When this pulse reaches the other end of the pipe,
(A) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is open.
(B) a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is open.
(C) a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed.
(D) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed.
Sol. (B, D)
At the open end, the phase of a pressure wave changes by π radian due to reflection. At the closed end, there is no change in the phase of a pressure wave due to reflection.
===== Page 7 =====
A small block of mass of 0.1kg lies on a fixed inclined plane PQ which makes an angle θ with the horizontal. A horizontal force of 1N acts on the block through its centre of mass as shown in the figure. The block remains stationary if (take g = 10m/s²)
(A) θ = 45°
(B) θ > 45° and a frictional force acts on the block towards P.
(C) θ > 45° and a frictional force acts on the block towards Q.
(D) θ < 45° and a frictional force acts on the block towards Q.
Sol. (A,C)
At θ = 45°, mg sinθ = 1 × cosθ
At θ > 45°, mg sinθ > 1 × cosθ (friction acts upward)
At θ < 45°, mg sinθ < 1 × cosθ (friction acts downward)
For the resistance network shown in the figure, choose the correct option(s)
(A) The current through PQ is zero.
(C) The potential at S is less than that at Q.
(B) I₁ = 3A
(D) I₂ = 2A
Sol. (A,B,C,D)
Nodes P and Q are equipotential and nodes S and T are equipotential from wheastone bridge, no current passes through PQ and ST.
I₁ = 12/4 = 3A
I₂ = I₁(12/(6+12)) = 2A
===== Page 8 =====
A circular wire loop of radius R is placed in the x-y plane centered at the origin O. A square loop of side a(a<
Sol. (7)
M = Nφ/I = [2(μ₀IR² / (2(8R)²))² cos45°] / I = μ₀a² / (8R 2^(1/2)) = μ₀a² / (R 2^(?))
p = 7
An infinitely long solid cylinder of radius R has a uniform volume charge density ρ. It has a spherical cavity of radius R/2 with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point P, which is at a distance 2R from the axis of the cylinder, is given by the expression 23ρR / (16kε₀). The value of k is
Sol. (6)
E = λ(ĵ) / (2πε₀(2R)) + K(ρ(4/3)π(R³/8))(-ĵ) / (4R²)
E = ρπR²(ĵ) / (4πε₀R) + KπρR(-ĵ) / 24
E = KρπR(ĵ) + (K/24)ρπR(-ĵ)
E = KρπR(23/24)(ĵ) = (23/(96ε₀))ρR(ĵ)
⇒ k = 6
===== Page 9 =====
A proton is fired from very far away towards a nucleus with charge Q = 120e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is: (take the proton mass, m_p = (5/3) × 10⁻²⁷ kg; h/e = 4.2 × 10⁻¹⁵ J.s/C; 1/(4πε₀) = 9 × 10⁹ m/F; 1 fm = 10⁻¹⁵ m)
Sol. (7)
0 + (1/2)mv² = K(Q)e / (10 × 10⁻¹⁵) = K(120e)e / (10 × 10⁻¹⁵)
(1/2) × (5/3) × 10⁻²⁷ v² = (9 × 10⁹ × 120 × (1.6 × 10⁻¹⁹)²) / (10 × 10⁻¹⁵)
v = √(331.776 × 10¹⁵)
λ = h / mv
λ = (4.2 × 10⁻¹⁵ × 1.6 × 10⁻¹⁹) / ((5/3) × 10⁻²⁷ × √(331.776 × 10¹⁵)) = 7 × 10⁻¹⁵ = 7 fm
A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in the figure. The moment of inertia of this lamina about axes passing though O and P is I_O and I_P respectively. Both these axes are perpendicular to the plane of the lamina. The ratio I_P / I_O to the nearest integer is
Sol.
I_p = [4mR²/2 + m(4R²)] - [mR²/4 + (m/2)5R²]
I_p = mR²[(2 + 4) - (1/8 + 5/4)]
I_p = mR²(6 - 11/8) = 37/8 mR²
I_O = (4mR²/2) - (mR²/4 + (m/2)R²)
I_O = mR²[2 - (1/8 + 1/4)] = mR²[2 - 3/8] = 13/8 mR²
I_P / I_O = (37/8) / (13/8) ≈ 3 (Nearest Integer)
===== Page 10 =====
A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a as shown in the figure. Both the cylinder and the cavity are infinity long. A uniform current density J flows along the length. If the magnitude of the magnetic field at the point P is given by (N/12) μ₀aJ, then the value of N is
Sol. (5)
B = μ₀(Jπa²) / (2πa) - μ₀(Jπa²/4) / (2π(3a/2))
B = 5μ₀Ja / 12 = μ₀NJa / 12
So N = 5
===== Page 11 =====
A compound M_pX_q has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below. The empirical formula of the compound is
Sol. (B)
X = 8 × 1/8 + 6 × 1/2 = 4
M = 4 × 1/4 + 1 = 2
So, unit cell formula of the compound is M₂X₄ and the empirical formula of the compound is MX₂
The carboxyl functional group (COOH) is present in
(A) picric acid
(B) barbituric acid
(C) ascorbic acid
(D) aspirin
Sol. (D)
As per IUPAC nomenclature, the name of the complex [Co(H₂O)₄(NH₃)₂]Cl₃ is
(A) Tetraquadiamimecobalt (III) chloride
(B) Tetraquadiamimecobalt (III) chloride
(C) Diaminetetraaquacobalt (III) chloride
(D) Diaminetetraaquacobalt (III) chloride
Sol. (D)
[Co(H₂O)₄(NH₃)₂]Cl₃
Diaminetetraaquacobalt (III) chloride
===== Page 12 =====
In allene (C₃H₄), the type(s) of hybridization of the carbon atoms is (are)
(A) sp and sp
(B) sp and sp
(C) only sp
(D) sp and sp
Sol. (B)
H₂C = C = CH₂
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a₀ is Bohr radius]
(A) h² / (4π²ma₀²)
(B) h² / (16π²ma₀²)
(C) h² / (32π²ma₀²)
(D) h² / (64π²ma₀²)
Sol. (C)
As per Bohr's postulate, mvr = nh / 2π
So, v = nh / 2πmr
KE = (1/2)mv²
So, KE = (1/2)m(nh / 2πmr)²
Since, r = a₀ × n² / z
So, for 2nd Bohr orbit r = a₀ × 2² / 1 = 4a₀
KE = (1/2)m(2²h² / (4π²m² × (4a₀)²))
KE = h² / (32π²ma₀²)
===== Page 13 =====
Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen?
(A) HNO₃, NO, NH₄Cl, N₂
(B) HNO₃, NO, N₂, NH₄Cl
(C) HNO₃, NH₄Cl, NO, N₂
(D) NO, HNO₃, NH₄Cl, N₂
Sol. (B)
HNO₃, NO, N₂, NH₄Cl
For one mole of a van der Waals gas when b = 0 and T = 300 K, the PV vs. 1/V plot is shown below. The value of the van der Waals constant a (atm.litre².mol⁻²) is
(A) 1.0
(B) 4.5
(C) 1.5
(D) 3.0
Sol. (C)
van der Waals equation for 1 mole of real gas is,
(P + a/V²)(V - b) = RT
but, b = 0 (given)
⇒ (P + a/V²)(V) = RT
∴ PV = -a × 1/V + RT ...(i)
y = mx + c
Slope = tan(π - θ) = -a
So, tanθ = a = (21.6 - 20.1)/(3 - 2) = 1.5
or, tanθ = (24.6 - 20.1)/(3 - 0) = 1.5
The number of aldol reaction(s) that occurs in the given transformation is
===== Page 14 =====
The colour of light absorbed by an aqueous solution of CuSO₄ is
(A) orange-red
(B) blue-green
(C) yellow
(D) violet
Sol. (A)
Aqueous solution of copper sulphate absorbs orange red light and appears blue (complementary colour).
The number of optically active products obtained from the complete ozonolysis of the given compound is
(A) 0
(B) 1
(C) 2
(D) 4
Sol. (A)
===== Page 15 =====
Which of the following hydrogen halides react(s) with AgNO₃(aq) to give a precipitate that dissolves in Na₂S₂O₃(aq)?
(A) HCl
(B) HF
(C) HBr
(D) HI
Sol. (A,C,D)
HX + AgNO₃ → AgX↓ + HNO₃ (X = Cl, Br, I)
AgX + 2Na₂S₂O₃ → Na₃[Ag(S₂O₃)₂] + NaX
Identify the binary mixture(s) that can be separated into individual compounds, by differential extraction, as shown in the given scheme.
(A) C₆H₅OH and C₆H₅COOH
(B) C₆H₅COOH and C₆H₅CH₂OH
(C) C₆H₅CH₂OH and C₆H₅OH
(D) C₆H₅CH₂OH and C₆H₅CH₂COOH
Sol. (B,D)
(A) Both are soluble in NaOH, hence inseparable.
(B) Only benzoic acid (C₆H₅COOH) is soluble in NaOH and NaHCO₃, while benzyl alcohol (C₆H₅CH₂OH) is not. Hence, separable.
(C) Although NaOH can enable separation between benzyl alcohol (C₆H₅CH₂OH) and phenol (C₆H₅OH) as only the later is soluble in NaOH. However, in NaHCO₃, both are insoluble. Hence, inseparable.
(D) α-phenyl acetic acid (C₆H₅CH₂COOH) is soluble in NaOH and NaHCO₃. While benzyl alcohol (C₆H₅CH₂OH) is not. Hence, separable.
For an ideal gas, consider only P-V work in going from an initial state X to the final state Z. The final state Z can be reached by either of the two paths shown in the figure. Which of the following choice(s) is(are) correct? [Take ΔS as change in entropy and w as work done]
===== Page 16 =====
Which of the following molecules, in pure form, is (are) unstable at room temperature?
Sol. (B,C)
Compound cyclobutadiene and cyclopentadienone being antiaromatic are unstable at room temperature.
Choose the correct reason(s) for the stability of the lyophobic colloidal particles.
(A) Preferential adsorption of ions on their surface from the solution
(B) Preferential adsorption of solvent on their surface from the solution
(C) Attraction between different particles having opposite charges on their surface
(D) Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles
Sol. (A,D)
Lyophobic colloids are stable due to preferential adsorption of ions on their surface from solution and potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles that makes lyophobic sol stable.
SECTION III: Integer Answer Type
This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive).
29.2% (w/w) HCl stock solution has density of 1.25 g mL⁻¹. The molecular weight of HCl is 36.5 g mol⁻¹. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is
Sol. (8)
===== Page 17 =====
The substituents R₁ and R₂ for nine peptides are listed in the table given below. How many of these peptides are positively charged at pH = 7.0?
H₃N-CH-CO-NH-CH-CO-NH-CH-CO-NH-CH-CO
H R₁ R₂ H H H H H
H H H H CH₃ H H H
Table:
Peptide | R₁ | R₂
I | H | H
II | H | CH₃
III | CH₃ | COOH
IV | CH₃ | CONH₂
V | CH₃ | CONH₂
VI | (CH₂)₂NH₂ | (CH₂)₂NH₂
VII | CH₃ | COOH
VIII | CH₃ | OH
IX | (CH₂)₂NH₂ | CH₃
Sol.
Peptides with isoelectric point (pI) > 7 would exist as cation in neutral solution (pH = 7) IV, VI, VIII and IX
An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are t_{1/8} and t_{1/10} respectively. What is the value of [t_{1/8} / t_{1/10}] × 10? (take log₁₀2 = 0.3)
Sol.
When the following aldohexose exists in its D-configuration, the total number of stereoisomers in its pyranose form is
===== Page 18 =====
The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below. To which group, element X belongs in the periodic table?
⁶³₂₉Cu + ¹H → 6₀n + α + 2¹H + X
Sol.
⁶³₂₉Cu + ¹H → 6₀n + ⁴₂He + 2¹H + ᴬ_Z X
Mass number: 63 + 1 = 1 × 6 + 4 + 1 × 2 + A
A = 64 - 12 = 52
Atomic number: 29 + 1 = 6 × 0 + 2 + 2 × 1 + Z
Z = 30 - 4 = 26
ᴬ_Z X = ⁵²₂₆Fe
Hence X is in group '8' in the periodic table.
===== Page 19 =====
[Repeated 1s, page mostly blank/artifact]
===== Page 20 =====
⇒ αx + ((4/5)α - 4)y = 9 ...(ii)
Comparing (i) and (ii)
h/α = k/((4/5)α - 4) = (h² + k²)/9
α = 20h / (4h - 5k)
Now, h(4h - 5k)/20h = (h² + k²)/9
20(h² + k²) = 9(4h - 5k)
20(x² + y²) - 36x + 45y = 0.
The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is
(A) 75
(B) 150
(C) 210
(D) 243
Sol. (B)
Number of ways
= 3⁵ - ³C₁2⁵ + ³C₂1⁵
= 243 - 96 + 3 = 150.
The integral ∫ sec²x / (sec x + tan x)^(9/2) dx equals (for some arbitrary constant K)
(A) -1/(sec x + tan x)^(11/2) [1/11 - 1/7(sec x + tan x)²] + K
(B) 1/(sec x + tan x)^(11/2) [1/11 - 1/7(sec x + tan x)²] + K
(C) -1/(sec x + tan x)^(11/2) [1/11 + 1/7(sec x + tan x)²] + K
(D) 1/(sec x + tan x)^(11/2) [1/11 + 1/7(sec x + tan x)²] + K
Sol. (C)
I = ∫ sec²x / (sec x + tan x)^(9/2) dx
Let sec x + tan x = t
⇒ sec x - tan x = 1/t
Now (sec x tan x + sec²x) dx = dt
sec x (sec x + tan x) dx = dt
sec x dx = dt/t, (1/2)(t + 1/t) = sec x
I = (1/2)∫ (t + 1/t) / t^(9/2) dt
= (1/2)∫ (t^(-7/2) + t^(-11/2))dt
= (1/2)[t^(-5/2)/(-5/2) + t^(-9/2)/(-9/2)]
= -1/5 t^(-5/2) - 1/9 t^(-9/2)
= -1/7 t^(-7/2) - 1/11 t^(-11/2)
= -1/(sec x + tan x)^(11/2) [1/11 + 1/7(sec x + tan x)²] + k
===== Page 21 =====
= 1/2 [t^(-7/2)/(-7/2) + t^(-11/2)/(-11/2)]
= -1/7 t^(-7/2) - 1/11 t^(-11/2)
= -1/7 1/t^(7/2) - 1/11 1/t^(11/2)
= -1/t^(11/2) (1/11 + t²/7) = -1/(sec x + tan x)^(11/2) [1/11 + 1/7(sec x + tan x)²] + k
The point P is the intersection of the straight line joining the points Q(2, 3, 5) and R(1, -1, 4) with the plane 5x - 4y - z = 1. If S is the foot of the perpendicular drawn from the point T(2, 1, 4) to QR, then the length of the line segment PS is
(A) 1/√2
(B) √2
(C) 2
(D) 2√2
Sol. (A)
D. R. of QR is 1, 4, 1
Coordinate of P = (4/3, 1/3, 13/3)
D. R. of PT is 2, 2, -1
Angle between QR and PT is 45°
And PT = 1
⇒ PS = TS = 1/√2
Let f(x) = { x² |cos π/x|, x ≠ 0; 0, x = 0 }, x ∈ R, then f is
(A) differentiable both at x = 0 and at x = 2
(B) differentiable at x = 0 but not differentiable at x = 2
(C) not differentiable at x = 0 but differentiable at x = 2
(D) differentiable neither at x = 0 nor at x = 2
Sol. (B)
f'(0) = lim_{h→0} (f(0+h) - f(0))/h
= lim_{h→0} (h² |cos π/h| - 0)/h
= lim_{h→0} h |cos π/h| = 0
so, f(x) is differentiable at x = 0
f'(2⁺) = lim_{h→0} (f(2+h) - f(2))/h
= lim_{h→0} ((2+h)² |cos π/(2+h)| - 0)/h
===== Page 22 =====
Let z be a complex number such that the imaginary part of z is nonzero and a = z² + z + 1 is real. Then a cannot take the value
(A) -1
(B) 1/3
(C) 1/2
(D) 3/4
Sol.
Given equation is z² + z + 1 - a = 0
Clearly this equation do not have real roots if D < 0
⇒ 1 - 4(1 - a) < 0
⇒ 4a < 3
a < 3/4
The ellipse E₁: x²/9 + y²/4 = 1 is inscribed in a rectangle R whose sides are parallel to the coordinate axes.
Another ellipse E₂ passing through the point (0, 4) circumscribes the rectangle R. The eccentricity of the ellipse E₂ is
(A) √2/2
(B) √3/2
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(C) 1/2
(D) 3/4
Sol. (C)
Equation of ellipse is (y + 2)(y - 2) + λ(x + 3)(x - 3) = 0
It passes through (0, 4) ⇒ λ = 4/3
Equation of ellipse is x²/12 + y²/16 = 1
e = 1/2
Alternate
Let the ellipse be x²/a² + y²/b² = 1 as it is passing through (0, 4) and (3, 2)
So, b² = 16 and 9/a² + 4/16 = 1
⇒ a² = 12
So, 12 = 16(1 - e²)
⇒ e = 1/2
The function f : [0,3] → [1,29], defined by f(x) = 2x³ - 15x² + 36x + 1, is
(A) one-one and onto
(B) onto but not one-one
(C) one-one but not onto
(D) neither one-one nor onto
Sol. (B)
f(x) = 2x³ - 15x² + 36x + 1
f'(x) = 6x² - 30x + 36
= 6(x² - 5x + 6)
= 6(x - 2)(x - 3)
f(x) is increasing in [0, 2] and decreasing in [2, 3]
f(x) is many one
f(0) = 1
f(2) = 29
f(3) = 28
Range is [1, 29]
Hence, f(x) is many-one-onto
SECTION II: Multiple Correct Answer(s) Type
This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.
Tangents are drawn to the hyperbola x²/9 - y²/4 = 1, parallel to the straight line 2x - y = 1. The points of contact of the tangents on the hyperbola are
(A) (9/(2√2), 1/√2)
(B) (-9/(2√2), -1/√2)
(C) (3√3, -2√2)
(D) (-3√3, 2√2)
Sol. (A, B)
Slope of tangent = 2
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The tangents are y = 2x ± √(9 × 4 - 4)
i.e., 2x - y = ± 4√2
⇒ x/(2√2) - y/(4√2) = 1 and x/(2√2) + y/(4√2) = 1
Comparing it with x x₁/9 - y y₁/4 = 1
We get point of contact as (9/(2√2), 1/√2) and (9/(2√2), -1/√2)
Alternate: Equation of tangent at P(θ) is (secθ/3)x - (tanθ/2)y = 1
⇒ Slope = 2secθ/(3tanθ) = 2
⇒ sinθ = 1/3
⇒ points are (9/(2√2), 1/√2) and (-9/(2√2), -1/√2)
Let θ, φ ∈ [0, 2π] be such that 2cosθ(1 - sinφ) = sin²θ(tan(θ/2) + cot(θ/2))cosφ - 1, tan(2π - θ) > 0 and -1 < sinθ < -√3/2. Then φ cannot satisfy
(A) 0 < φ < π/2
(B) π/2 < φ < 4π/3
(C) 4π/3 < φ < 3π/2
(D) 3π/2 < φ < 2π
Sol. (A,C,D)
2cosθ(1 - sinφ) = (2sin²θ/sinθ)cosφ - 1 = 2sinθcosφ - 1
2cosθ - 2cosθsinφ = 2sinθcosφ - 1
2cosθ + 1 = 2sin(θ + φ)
tan(2π - θ) > 0 ⇒ tanθ < 0 and -1 < sinθ < -√3/2
⇒ θ ∈ (3π/2, 5π/3)
1/2 < sin(θ + φ) < 1
⇒ 2π + π/6 < θ + φ < 5π/6 + 2π
2π/6 - θ_max < φ < 2π + 5π/6 - θ_min
π/2 < φ < 4π/3.
If y(x) satisfies the differential equation y' - ytanx = 2x secx and y(0) = 0, then
(A) y(π/4) = π²/(8√2)
(B) y(π/4) = π²/18
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[Repeated 1s, page mostly blank/artifact]
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Let S be the area of the region enclosed by y = e^(-x²), y = 0, x = 0 and x = 1. Then
(A) S ≥ 1/e
(B) S ≥ 1 - 1/e
(C) S ≤ 1/4 (1 + 1/√e)
(D) S ≤ 1/√2 + 1/√e (1 - 1/√2)
Sol. (A,B,D)
S > 1/e (As area of rectangle OCDS = 1/e)
Since e^(-x²) ≥ e^(-x) ∀ x ∈ [0,1]
⇒ S > ∫₀¹ e^(-x) dx = (1 - 1/e)
Area of rectangle OAPQ + Area of rectangle QBRS > S
S < 1/√2(1) + (1 - 1/√2)(1/√e).
Since 1/4(1 + 1/√e) < 1 - 1/e
Hence, (C) is incorrect.
SECTION III: Integer Answer Type
This section contains 5 questions. The answer to each question is single digit integer, ranging from 0 to 9 (both inclusive).
If a, b and c are unit vectors satisfying |a - b|² + |b - c|² + |c - a|² = 9, then |2a + 5b + 5c| is
Sol. (3)
As, |a - b|² + |b - c|² + |c - a|² = 3(|a|² + |b|² + |c|²) - |a + b + c|²
⇒ 3 × 3 - |a + b + c|² = 9
⇒ |a + b + c| = 0 ⇒ a + b + c = 0
⇒ b + c = -a
⇒ |2a + 5(b + c)| = |-3a| = 3|a| = 3.
Let f : R → R be defined as f(x) = |x| + |x² - 1|. The total number of points at which f attains either a local maximum or a local minimum is
Sol. (5)
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Let S be the focus of the parabola y² = 8x and let PQ be the common chord of the circle x² + y² - 2x - 4y = 0 and the given parabola. The area of the triangle PQS is
Sol.
The parabola is x = 2t², y = 4t
Solving it with the circle we get:
4t⁴ + 16t² - 4t² - 16t = 0
⇒ t⁴ + 3t² - 4t = 0 ⇒ t = 0, 1
so, the points P and Q are (0,0) and (2,4) which are also diametrically opposite points on the circle.
The focus is S = (2,0)
The area of ΔPQS = (1/2) × 2 × 4 = 4
Let p(x) be a real polynomial of least degree which has a local maximum at x = 1 and a local minimum at x = 3. If p(1) = 6 and p(3) = 2, then p'(0) is
Sol.
Let p'(x) = k(x - 1)(x - 3)
⇒ p(x) = k(x³/3 - 2x² + 3x) + c
Now, p(1) = 6 ⇒ (4/3)k + c = 6
also, p(3) = 2 ⇒ c = 2
so, k = 3
so, p'(0) = 3k = 9
The value of 6 + log_{3/2} ( (1/(3√2)) √(4 - (1/(3√2))√(4 - (1/(3√2))√(4 - ...))) ) is
Sol.
Let √(4 - (1/(3√2))√(4 - (1/(3√2))√(4 - ...))) = y
So, 4 - (1/(3√2)) y = y² (y > 0)
⇒ y² + (1/(3√2))y - 4 = 0 ⇒ y = 8/(3√2)
so, the required value is 6 + log_{3/2}( (1/(3√2)) × (8/(3√2)) )
= 6 + log_{3/2}(4/9) = 6 - 2 = 4.
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