JEE Main 2023 Jan 31 Shift 1 Question Paper has been updated here. NTA conducted JEE Main 2023 Jan 31 Shift 1 from 9 AM to 12 PM for B.E./B.Tech paper. According to the initial reaction of the students, the paper was reported to be in line with the previous year's trends. The students found the paper to be moderately challenging. Candidates can download the memory-based JEE Main 2023 Question Paper PDF with Solution and Answer Key for Jan 31 Shift 1 using the link below.
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JEE Main 2023 Questions with Solutions
Section A
Question 1:A bar magnet with a magnetic moment 5.0 Am² is placed in parallel position relative to a magnetic field of 0.4 T. The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is:
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If a source of electromagnetic radiation having power 15 kW produces \( 10^{16} \) photons per second, the radiation belongs to a part of spectrum is:
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The amplitude of \( 15 \sin(1000 \pi t) \) is modulated by \( 10 \sin(4 \pi t) \) signal. The amplitude modulated signal contains frequency(ies) of:
(A) 500 Hz
(B) 2 Hz
(C) 250 Hz
(D) 498 Hz
(E) 502 Hz
Choose the correct answer from the options given below:
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As shown in figure, a 70 kg garden roller is pushed with a force of \( F = 200 \, N \) at an angle of 30° with horizontal. The normal reaction on the roller is: (Given \( g = 10 \, m/s^2 \))
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The initial speed of a projectile fired from ground is \( u \). At the highest point during its motion, the speed of the projectile is \( \frac{\sqrt{5}}{2} u \). The time of flight of the projectile is:
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Spherical insulating ball and a spherical metallic ball of the same size and mass are dropped from the same height. Choose the correct statement out of the following (Assume negligible air friction):
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A free neutron decays into a proton but a free proton does not decay into neutron. This is because:
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The effect of increase in temperature on the number of electrons in conduction band (\( n_e \)) and resistance of a semiconductor will be:
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The maximum potential energy of a block executing simple harmonic motion is 25 J. \( A \) is amplitude of oscillation. At \( \frac{A}{2} \), the kinetic energy of the block is:
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The pressure of a gas changes linearly with volume from A to B as shown in the figure. If no heat is supplied to or extracted from the gas, then change in the internal energy of the gas will be:
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Which of the following correctly represents the variation of electric potential (\( V \)) of a charged spherical conductor of radius (\( R \)) with radial distance (\( r \)) from the centre?
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R
Assertion A: The beam of electrons shows wave nature and exhibits interference and diffraction.
Reason R: Davisson Germer experimentally verified the wave nature of electrons.
In the light of the above statements, choose the most appropriate answer from the options given below:
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The drift velocity of electrons for a conductor connected in an electrical circuit is \( V_d \). The conductor is now replaced by another conductor with the same material and same length but double the area of cross-section. The applied voltage remains the same. The new drift velocity of electrons will be:
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At a certain depth \( d \) below the surface of the Earth, the value of acceleration due to gravity becomes four times that of its value at a height 3R above the Earth's surface. Where \( R \) is the radius of Earth (Take \( R = 6400 \, km \)). The depth \( d \) is equal to:
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If 1000 droplets of water of surface tension 0.07 N/m, having same radius 1 mm each, combine to form a single drop. In the process, the released surface energy is:
(Take \( \pi = \frac{22}{7} \))
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A rod with circular cross-section area 2 cm\(^2\) and length 40 cm is wound uniformly with 400 turns of an insulated wire. If a current of 0.4 A flows in the wire windings, the total magnetic flux produced inside the windings is \( 4 \times 10^{-6} \, Wb \). The relative permeability of the rod is:
(Given: Permeability of vacuum \( \mu_0 = 4 \pi \times 10^{-7} \, N/A^2 \))
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The correct relation between \( \gamma = \frac{C_p}{C_v} \) and temperature \( T \) is:
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Two polaroide A and B are placed in such a way that the pass-axis of polaroids are perpendicular to each other. Now, another polaroid C is placed between A and B bisecting the angle between them. If intensity of unpolarised light is \( I_0 \), then intensity of transmitted light after passing through polaroid B will be:
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If \( R \), \( X_L \), and \( X_C \) represent resistance, inductive reactance, and capacitive reactance, then which of the following is dimensionless:
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100 balls each of mass \( m \) moving with speed \( v \) simultaneously strike a wall normally and are reflected back with the same speed, in time \( t \). The total force exerted by the balls on the wall is:
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A thin rod having a length of 1 m and area of cross-section \( 3 \times 10^{-6} \, m^2 \) is suspended vertically from one end. The rod is cooled from 210°C to 160°C. After cooling, a mass \( M \) is attached at the lower end of the rod such that the length of the rod again becomes 1 m. Young's modulus and the coefficient of linear expansion of the rod are \( 2 \times 10^{11} \, N/m^2 \) and \( 2 \times 10^{-5} \, K^{-1} \), respectively. The value of \( M \) is ______ kg. (Take \( g = 10 \, m/s^2 \))
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The speed of a swimmer is 4 km/h in still water. If the swimmer makes his strokes normal to the flow of the river of width 1 km, he reaches a point 750 m down the stream on the opposite bank. The speed of the river water is ........ km/h.
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In the figure given below, a block of mass \( M = 490 \, g \) is placed on a frictionless table is connected with two springs having the same spring constant (\( K = 2 \, N/m \)). If the block is horizontally displaced through \( X \) m, then the number of complete oscillations it will make in 14t seconds will be:
\begin{figure[h]
\centering
\end{figure
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In a medium, the speed of light wave decreases to 0.2 times its speed in free space. The ratio of relative permittivity to the refractive index of the medium is \( x : 1 \). The value of \( x \) is .......
(Given speed of light in free space \( c = 3 \times 10^8 \, m/s \) and for the given medium \( \mu_r = 1 \))
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A solid sphere of mass 1 kg rolls without slipping on a plane surface. Its kinetic energy is \( 7 \times 10^3 \) J. The speed of the centre of mass of the sphere is ....... cm/s.
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An inductor of 0.5 mH, a capacitor of 20 μF, and resistance of 20 Ω are connected in series with a 220 V AC source. If the current is in phase with the emf, the amplitude of current of the circuit is \( \sqrt{K} \) A. The value of \( x \) is:
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Expression for an electric field is given by \( \vec{E} = 4000 \hat{i} \, V/m \). The electric flux through the cube of side 20 cm when placed in the electric field (as shown in the figure) is ....... Vcm.
\begin{figure[h]
\centering
\end{figure
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A lift of mass \( M = 500 \, kg \) is descending with a speed of \( 2 \, m/s \). Its supporting cable begins to slip, thus allowing it to fall with a constant acceleration of \( 2 \, m/s^2 \). The kinetic energy of the lift at the end of fall through a distance of 6 m will be ....... kJ.
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For hydrogen atom, \( \lambda_1 \) and \( \lambda_2 \) are the wavelengths corresponding to the transitions 1 and 2, respectively, as shown in the figure. The ratio of \( \lambda_1 \) and \( \lambda_2 \) is \( \frac{x}{32} \). The value of \( x \) is:
\begin{figure[h]
\centering
\end{figure
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Two identical cells, when connected either in parallel or in series, give the same current in an external resistance of 5Ω. The internal resistance of each cell will be ........ \Omega.
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Chemistry
SECTION A
Question 31:
Nd\(^{2+}\) = ..........
View Solution
The electronic configuration of Nd (atomic number 60) is: \[ Nd(60) = [Xe] 4f^4 5d^0 6s^2 \]
For \( Nd^{2+} \), two electrons are removed (typically from the 6s orbital first): \[ Nd^{2+} = [Xe] 4f^4 5d^0 5s^0 \]
Thus, the correct answer is \( 4f^4 \). Quick Tip: The electron configuration for transition metal ions is derived by removing electrons from the highest energy orbitals, typically starting from the outermost \( s \)-orbital.
The methods NOT involved in concentration of ore are
(A) Liquation
(B) Leaching
(C) Electrolysis
(D) Hydraulic washing
(E) Froth flotation
Choose the correct answer from the options given below:
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Consider the following reaction \[ Propanal + Methanal \xrightarrow{(i) dil. NaOH} \xrightarrow{(ii) A} \xrightarrow{(iii) NaCN} \xrightarrow{(iv) H_2O} Product B \, (C_5H_8O_3) \]
The correct statement for product B is:
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The reaction proceeds as follows: \[ CH_3CH_2CHO + HCHO \xrightarrow{(i) NaOH, \(\Delta\)} CH_3CH_2CH(OH)CH_2CN \]
Following the steps in the given reaction:
1. Propanal reacts with methanol in the presence of NaOH to give an aldol product.
2. The aldol product undergoes nucleophilic substitution with NaCN.
3. The product forms a carboxylic acid on hydrolysis.
The resulting product undergoes decarboxylation to give a racemic mixture. The carboxylic acid thus formed will give \(CO_2\) gas when reacted with \(NaHCO_3\) solution.
Thus, the correct statement for product B is that it is a racemic mixture and gives a gas with saturated \(NaHCO_3\) solution. Quick Tip: A racemic mixture is a mixture containing equal amounts of enantiomers, and a carboxylic acid can release \(CO_2\) when reacted with \(NaHCO_3\).
The correct order of basicity of oxides of vanadium is:
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When \( Cu^{2+} \) ion is treated with KI, a white precipitate, X appears in solution. The solution is titrated with sodium thiosulphate, the compound Y is formed. X and Y respectively are:
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The reaction sequence is:
Cobalt chloride when dissolved in water forms pink colored complex X which has octahedral geometry. This solution on treating with cone HCl forms deep blue complex Y which has a Z geometry. X, Y and Z, respectively, are:
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Identify X, Y, and Z in the following reaction. (Equation not balanced) \[ ClO + NO_2 \rightarrow X \xrightarrow{H_2O} Y + Z \]
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The correct order of melting point of dichlorobenzenes is:
A protein 'X' with molecular weight of 70,000 u, on hydrolysis gives amino acids. One of these amino acids is
Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from \( n = 4 \) to \( n = 2 \) of He\(^{+}\) spectrum?
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Match items of column I and II
Column I (Mixture of compounds) Column II (Separation Technique)
A. H₂O/CH₃Cl i. Crystallization
B. ii. Differential solvent extraction
C. Kerosene/Naphthalene iii. Column chromatography
D. C₆H₅O/NaCl iv. Fractional Distillation
Correct match is:
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The correct increasing order of the ionic radii is:
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H\(_2\)O\(_2\) acts as a reducing agent in:
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Which of the following artificial sweeteners has the highest sweetness value in comparison to cane sugar?
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Match List I with List II
List I List II
A. \(XeF_4\) I. See-saw
B. \(SF_4\) II. Square planar
C. \(NH_4^+\) III. Bent T-shaped
D. \(BrF_3\) IV. Tetrahedral
Choose the correct answer from the options given below:
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Choose the correct set of reagents for the following conversion \[ trans (Ph-CH=CH-CH_3) \rightarrow cis (Ph-CH=CH-CH_3) \]
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Adding surfactants in non-polar solvent, the micelles structure will look like (Surfactant structure):
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An organic compound 'A' with empirical formula \(C_6H_6\)O gives sooty flame on burning. Its reaction with bromine solution in low polarity solvent results in high yield of B. B is:
Which one of the following statements is correct for electrolysis of brine solution?
View Solution
SECTION B
Question 51:
The logarithm of equilibrium constant for the reaction \[ Pd^{2+} (aq) + 4Cl^- (aq) \rightleftharpoons PdCl_4^{2-} (aq) \]
is:
Given: \[ 2.303RT = 0.06\, V, \quad F = 96,485\, C/mol, \quad E^\circ = 0.83\, V, \quad E^\circ = 0.65\, V \]
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\[ \textbf{A} \rightarrow \textbf{B} \]
The rate constants of the above reaction at 200 K and 300 K are 0.03 min\(^{-1}\) and 0.05 min\(^{-1}\) respectively.
The activation energy for the reaction is \hspace{2cm} J (Nearest integer)
Given: \[ \text{ln} \, 10 = 2.3, \quad R = 8.3 \, \text{J K}^{-1} \text{mol}^{-1}, \quad \log 5 = 0.70, \quad \log 3 = 0.48, \quad \log 2 = 0.30 \]
View Solution
We can use the Arrhenius equation to solve for the activation energy: \[ \log \left( \frac{K_{300}}{K_{200}} \right) = \frac{E_a}{2.3 \times 8.314} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
Substituting the given values: \[ \log \left( \frac{0.05}{0.03} \right) = \frac{E_a}{2.3 \times 8.314} \left( \frac{1}{200} - \frac{1}{300} \right) \]
Solving for \( E_a \): \[ \log \left( \frac{0.05}{0.03} \right) = \log 5 - \log 3 = 0.70 - 0.48 = 0.22 \] \[ 0.22 = \frac{E_a}{2.3 \times 8.314} \times \left( \frac{1}{200} - \frac{1}{300} \right) \] \[ 0.22 = \frac{E_a}{2.3 \times 8.314} \times \left( \frac{1}{60000} \right) \]
Solving for \( E_a \): \[ E_a = 2519.88 \, J \quad \Rightarrow \quad E_a \approx 2520 \, J \] Quick Tip: The activation energy can be calculated using the Arrhenius equation, where the ratio of rate constants at two different temperatures is related to the exponential factor involving the activation energy.
The enthalpy change for the conversion of \[ \frac{1}{2} Cl_2(g) \rightarrow Cl^-(aq) \, is \, (-) \, ...... kJ mol^{-1} \, (Nearest integer) \]
Given: \[ \Delta_{f}H^\circ(Cl_2(g)) = 240 \, kJ mol^{-1}, \quad \Delta_{g}H^\circ(Cl_2(g)) = -350 \, kJ mol^{-1}, \quad \Delta_{hyd}H^\circ(Cl^-) = -380 \, kJ mol^{-1} \]
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On complete combustion, 0.492 g of an organic compound gave 0.792 g of \(CO_2\). The % of carbon in the organic compound is \underline{\hspace{2cm}} (Nearest integer).
Given: \[ \text{Weight of carbon in 0.792 g CO}_2 \]
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At 27°C, a solution containing 2.5 g of solute in 250.0 mL of solution exerts an osmotic pressure of 400 Pa. The molar mass of the solute is \hspace{2cm} g mol\(^{-1}\) (Nearest integer).
Given: \[ R = 0.083 \, \text{L bar K}^{-1} \text{mol}^{-1}, \quad \pi = \text{osmotic pressure}, \quad T = 300 \, \text{K}, \quad \text{osmotic pressure} = 400 \, \text{Pa}, \quad 1 \, \text{bar} = 10^5 \, \text{Pa} \]
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Zinc reacts with hydrochloric acid to give hydrogen and zinc chloride. The volume of hydrogen gas produced at STP from the reaction of 11.5 g of zinc with excess HCl is \hspace{2cm} L (Nearest integer).
Given: \[ \text{Molar mass of Zn} = 65.4 \, \text{g mol}^{-1}, \quad \text{Molar volume of H}_2 \, \text{at STP} = 22.7 \, \text{L/mol} \]
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How many of the transformations given below would result in aromatic amines?
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For reaction: \[ \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \rightleftharpoons \text{SO}_3(g) \] \( K_p = 2 \times 10^{12} \) at 27°C and 1 atm pressure. The \( K_c \) for the same reaction is \hspace{2cm} (Nearest integer).
Given: \[ R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \]
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The oxidation state of phosphorus in hypophosphoric acid is ________________.
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The total pressure of a mixture of non-reacting gases X (0.6 g) and Y (0.45 g) in a vessel is 740 mm of Hg. The partial pressure of the gas X is \hspace{2cm} mm of Hg (Nearest Integer).
Given: \[ \textbf{Molar mass of X} = 20 \, \textbf{g mol}^{-1}, \quad \textbf{Molar mass of Y} = 45 \, \textbf{g mol}^{-1} \]
View Solution
Mathematics
Section – A
Question 61:
If the maximum distance of normal to the ellipse \[ \frac{x^2}{4} + \frac{y^2}{b^2} = 1, \, b < 2, from the origin is 1, then the eccentricity of the ellipse is: \]
View Solution
Step 1:
Equation of the normal is given by \[ 2x \sec \theta - b \cos \theta = 4 - b^2 \]
Step 2:
The distance from the origin to the normal is \[ Distance = \frac{4 - b^2}{4 \sec \theta + b \cos \theta} \]
Step 3:
The distance will be maximum when \[ 4 \sec \theta + b \cos \theta is minimum \]
This occurs when \[ \tan \theta = \frac{b}{2} \]
Step 4:
Substituting into the equation, we get \[ 4 - b^2 = b^2 + b^2 = \sqrt{3} \] Quick Tip: For such problems, we solve by simplifying the given equations step by step and applying geometrical concepts like normal distances and eccentricity.
For all \( z \in \mathbb{C} \) on the curve \( C \) such that \( | z | = 1 \), let the locus of the point \( z + \frac{1}{z} \) be the curve \( C_1 \). Then:
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A wire of length 20 m is to be cut into two pieces. A piece of length \( \ell_1 \) is bent to make a square of area \( A_1 \), and the other piece of length \( \ell_2 \) is made into a circle of area \( A_2 \). If \( 2A_1 + 3A_2 \) is minimum, then \( \frac{\ell_1}{\ell_2} \) is equal to:
View Solution
For the system of linear equations:
\( x + y + z = 6 \)
\( \alpha x + \beta y + 7z = 3 \)
\( x + 2y + 3z = 14 \)
Which of the following is NOT true?
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Let the shortest distance between the lines \[ L: \frac{x - 5}{2} = \frac{y - \lambda}{0} = \frac{z + 1}{1}, \quad \lambda \geq 0 \quad and \quad L_1: x + 1 = y - 1 = 4 - z = 2\sqrt{6} \]
If \( (\alpha, \beta, \gamma) \) lies on \( L \), then which of the following is NOT possible?
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Let \( y = f(x) \) represent a parabola with focus \( \left( -\frac{1}{2}, 0 \right) \) and directrix \( y = -\frac{1}{2} \).
Then \[ S = \left\{ x \in \mathbb{R} : \tan^{-1} \left( \sqrt{f(x)} + \sin \left( \sqrt{f(x) + 1} \right) \right) = \frac{\pi}{2} \right\} \]
contains:
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Let \[ A = \begin{pmatrix} 1 & 0 & 0
0 & 4 & -1
0 & 12 & -3 \end{pmatrix} \]
Then the sum of the diagonal elements of the matrix \( (A + I)^{11} \) is equal to:
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Let \( R \) be a relation on \( \mathbb{N} \times \mathbb{N} \) defined by \[ (a, b) \, R \, (c, d) \quad if and only if \quad ad(b - c) = bc(a - d). \]
Then \( R \) is:
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Let \[ y = f(x) = \sin^3 \left( \frac{\pi}{3} \cos \left( \frac{\pi}{3\sqrt{2}} \left( -4x^3 + 5x^2 + 1 \right)^{\frac{3}{2}} \right) \right) \]
Then, at \( x = 1 \),
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If the sum and product of four positive consecutive terms of a G.P. are 126 and 1296, respectively, then the sum of common ratios of all such GPs is:
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The number of real roots of the equation \[ \sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6} \]
is:
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Let a differentiable function \(f\) satisfy \[ f(x) + \int_3^x f(t) \, dt = \sqrt{x+1}, \quad x \geq 3. \]
Then \(f(8)\) is equal to:
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If the domain of the function \[ f(x) = \frac{\lfloor x \rfloor}{1 + x^2}, \textbf{ where } \lfloor x \rfloor \textbf{ is the greatest integer less than or equal to } x, \textbf{ is } [2, 6), \textbf{ then its range is:} \]
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Let \( \mathbf{a} = 2\hat{i} + \hat{j} + \hat{k} \), and \( \mathbf{b} \) and \( \mathbf{c} \) be two nonzero vectors such that \[ \left| \mathbf{a + \mathbf{b} + \mathbf{c}} \right| = \left| \mathbf{a + \mathbf{b} - \mathbf{c}} \right| \quad \text{and} \quad \mathbf{b} \cdot \mathbf{c} = 0. \]
Consider the following two statements:
(A) The magnitude of the vector a plus λ times the vector c is greater than or equal to the magnitude of vector a, for all real numbers λ.(B) \( \mathbf{a} \) and \( \mathbf{c} \) are always parallel.
View Solution
Step 1: Start with the given equation: \[ \left| \mathbf{a} + \mathbf{b} + \mathbf{c} \right| = \left| \mathbf{a} + \mathbf{b} - \mathbf{c} \right|. \]
Square both sides of the equation: \[ \left( \mathbf{a} + \mathbf{b} + \mathbf{c} \right)^2 = \left( \mathbf{a} + \mathbf{b} - \mathbf{c} \right)^2. \]
Expand both sides: \[ \mathbf{a}^2 + 2\mathbf{a} \cdot \mathbf{b} + 2\mathbf{a} \cdot \mathbf{c} + \mathbf{b}^2 + 2\mathbf{b} \cdot \mathbf{c} + \mathbf{c}^2 = \mathbf{a}^2 + 2\mathbf{a} \cdot \mathbf{b} - 2\mathbf{a} \cdot \mathbf{c} + \mathbf{b}^2 - 2\mathbf{b} \cdot \mathbf{c} + \mathbf{c}^2. \]
Simplifying the equation: \[ 2 \mathbf{a} \cdot \mathbf{c} + 2 \mathbf{b} \cdot \mathbf{c} = -2 \mathbf{a} \cdot \mathbf{c} - 2 \mathbf{b} \cdot \mathbf{c}. \]
Since \( \mathbf{b} \cdot \mathbf{c} = 0 \), we have: \[ 4 \mathbf{a} \cdot \mathbf{c} = 0 \quad \Rightarrow \quad \mathbf{a} \cdot \mathbf{c} = 0. \]
Step 2: Therefore, \( \mathbf{a} \) and \( \mathbf{c} \) are perpendicular, not parallel. Hence, statement (B) is incorrect.
Step 3: Now, consider statement (A): \[ \left| \mathbf{a} + \lambda \mathbf{c} \right| \geq \left| \mathbf{a} \right|. \]
This is always true for any value of \( \lambda \in \mathbb{R} \), because the magnitude of a vector added to a scalar multiple of another vector is always greater than or equal to the magnitude of the original vector. Thus, statement (A) is correct. Quick Tip: When dealing with vector magnitudes and dot products, remember that the square of the magnitude of a vector is always non-negative. Use the dot product property to simplify equations involving vector magnitudes.
Let \( \alpha \in (0, 1) \) and \( \beta = \log(1 - \alpha) \). Let \[ P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots + \frac{x^n}{n}, \quad x \in (0, 1). \]
Then the integral
\[ \int_0^\alpha \frac{1}{1 - t} \, dt \] is equal to:
View Solution
Step 1: Start with the given integral: \[ \int_0^\alpha \frac{1}{1 - t} \, dt. \]
This can be rewritten as: \[ \int_0^\alpha \frac{1}{1 - t} \, dt = -\int_0^\alpha \frac{d}{1 - t}. \]
Step 2: Now, express the series expansion for \( P_n(x) \): \[ P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots + \frac{x^n}{n}. \]
Step 3: After integrating the series term-by-term, we get: \[ -\int_0^\alpha \frac{d}{1 - t} = -P_0(\alpha) - \beta. \]
Step 4: Hence, the value of the integral is: \[ \int_0^\alpha \frac{1}{1 - t} \, dt = -(\beta + P_0(\alpha)). \] Quick Tip: When solving integrals involving logarithmic expressions, consider series expansions for functions like \( P_n(x) \) and integrate term-by-term.
If \( \sin^{-1} \left( \frac{\alpha}{17} \right) + \cos^{-1} \left( \frac{4}{5} \right) - \tan^{-1} \left( \frac{77}{36} \right) = 0, \quad 0 < \alpha < 13, \)
then \( \sin^{-1} (\sin \alpha) + \cos^{-1} (\cos \alpha) \) is equal to:
View Solution
Let a circle \( C_1 \) be obtained on rolling the circle \[ x^2 + y^2 - 4x - 6y + 11 = 0 \] upwards 4 units on the tangent \( T \) to it at the point (3, 2). Let \( C_2 \) be the image of \( C_1 \) in \( T \).
Let A and B be the centers of circles \( C_1 \) and \( C_2 \) respectively, and M and N be respectively the feet of perpendiculars drawn from A and B on the x-axis. Then the area of the trapezium AMNB is:
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(S1) \( (p \Rightarrow q) \vee (p \land \neg q) \) is a tautology
(S2) \( (\neg p) \Rightarrow (\neg q) \) \land \( ((\neg p) \vee q) \) is a contradiction. Then:
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The value of \[ \int \frac{(2 + 3 \sin x)}{\sin x (1 + \cos x)} \, dx \]
is equal to:
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A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is:
View Solution
Section – B
Question 81:
Let 5 digit numbers be constructed using the digits 0, 2, 3, 4, 7, 9 with repetition allowed, and are arranged in ascending order with serial numbers.
Then the serial number of the number 42923 is:
View Solution
Let \( a_1, a_2, \dots, a_n \) be in A.P. If \( a_5 = 2a_1 \text{ and } a_1 = 18, \) then
\[ 12 \left( \frac{1}{\sqrt{a_0} + \sqrt{a_1}} + \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \cdots + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}} \right) \]
is equal to:
View Solution
Step 1: Given that \( a_5 = 2a_1 \) and \( a_1 = 18 \), we know that \( a_5 = 2 \times 18 = 36 \).
Step 2: In an arithmetic progression, the general form for the \( n \)-th term is: \[ a_n = a_1 + (n-1)d \]
where \( a_1 \) is the first term and \( d \) is the common difference.
Step 3: From the condition \( a_5 = 36 \), we can write: \[ a_5 = a_1 + 4d \]
Substituting \( a_1 = 18 \) and \( a_5 = 36 \), we get: \[ 36 = 18 + 4d \] \[ 18 = 4d \quad \Rightarrow \quad d = \frac{18}{4} = 4.5. \]
Step 4: Now, let's calculate \( a_{18} \). Using the formula for the general term: \[ a_{18} = a_1 + 17d \]
Substituting \( a_1 = 18 \) and \( d = 4.5 \): \[ a_{18} = 18 + 17 \times 4.5 = 18 + 76.5 = 94.5. \]
Step 5: Now, calculate the sum of the terms in the given series: \[ 12 \left( \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}} \right) \]
This is a sum involving the terms of the form \( \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} \). We can use the following approximation: \[ \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} \approx \frac{1}{\sqrt{a_k} + \sqrt{a_k + d}}. \]
Step 6: We simplify the series and calculate the sum: \[ 12 \times \left( \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} \right) \quad for each \( k \). \]
The final result is: \[ 12 \times 9 = 108. \] Quick Tip: In problems involving arithmetic progressions and sums of series, make sure to use the general term formula and simplify terms systematically to calculate the total sum efficiently.
Let \( \theta \) be the angle between the planes
\[ P_1: \vec{r} \cdot ( \hat{i} + \hat{j} + 2 \hat{k}) = 9 \quad \text{and} \quad P_2: \vec{r} \cdot (2 \hat{i} - \hat{j} + \hat{k}) = 15. \]
Let \( L \) be the line that meets \( P_2 \) at the point (4, -2, 5) and makes an angle \( \theta \) with the normal of \( P_2 \). If \( \alpha \) is the angle between \( L \) and \( P_2 \), then
\[ (\tan^2 \theta)(\cot^2 \alpha) \] is equal to:
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Question 84:
Let α > 0, be the smallest number such that the expansion of (3/x3 + 2/x)30 has a term βx-α, β ∈ ℕ. Then α is equal to:
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Let \( \vec{a} \) and \( \vec{b} \) be two vectors such that \[ |\vec{a}| = \sqrt{14}, \quad |\vec{b}| = \sqrt{6}, \quad |\vec{a} \times \vec{b}| = \sqrt{48}. \]
Then \( (\vec{a} \cdot \vec{b})^2 \textbf{ is equal to:} \)
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Let the line \( L: \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1} \) intersect the plane \[ 2x + y + 3z = 16 \textbf{ at the point } P. \]
Let the point Q be the foot of perpendicular from the point R(1, -1, -3) \text{ on the line L.
If \alpha is the area of triangle PQR, then \alpha^2 is equal to:
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The remainder on dividing \( 5^{99} \) by 11 is:
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If the variance of the frequency distribution
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline x_i & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline f_i & 3 & 6 & 16 & \alpha & 9 & 5 & 6 \\ \hline \end{array} \]
is given, find the value of \( \alpha \).
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Let for \( x \in \mathbb{R} \) \[ f(x) = \frac{x + |x|}{2} \quad for \quad x \geq 0 \quad and \quad f(x) = \frac{x}{2} \quad for \quad x < 0 \] \[ g(x) = \begin{cases} x^2 & for \quad x \geq 0,
x & for \quad x < 0 \end{cases} \]
\text{Then the area bounded by the curve \( y = f \circ g(x) \) \text{ and the lines \( y = 0, 2y - x = 15 \) \text{ is equal to:
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Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11, is equal to:
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Also Check:
JEE Main 2023 Jan 31 Shift 1 Question Paper by Coaching Institute
| Coaching Institutes | Question Paper with Answer Key PDF |
|---|---|
| Vedantu | Check Here |
JEE Main 2023 Paper Analysis Jan 31 Shift 1
JEE Main 2023 Paper Analysis for the exam scheduled on January 31 Shift 1 has been updated here. Candidates can check subject-wise paper analysis for the exam scheduled on January 31 Shift 1 here along with the topics with the highest weightage.









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