JEE Main 1 February 2023 Shift 1 question paper with solutions

JEE Main 1 February 2023 Shift 1: all 90 questions in paper order (Physics, Chemistry, Maths) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.

Physics

Question 1

Physics · Mechanical Properties of Solids · Single correct

( P + $\frac{a}{V^2}$) (V - b) = RT represents the equation of state of some gases. Where P is the pressure, V is the volume, T is the temperature and a, b, R are the constants. The physical quantity, which has dimensional formula as that of $\frac{b^2}{a}$, will be:

Bulk modulus
Modulus of rigidity
Compressibility
Energy density

Answer: (c)

Solution

Given $[b] = [V]$ and $\left[ \frac{a}{b^2} \right] = [P]$. Therefore, $\left[ \frac{b^2}{a} \right] = \frac{1}{[P]} = \frac{1}{[B]} = [K]$.

Question 2

Physics · Motion in a Straight Line · Single correct

An object moves with speed $v_1$, $v_2$, and $v_3$ along a line segment $AB$, $BC$ and $CD$ respectively as shown in figure. Where $AB = BC$ and $AD = 3 \, AB$, then average speed of the object will be:

$\frac{(v_1 + v_2 + v_3)}{3}$
$\frac{v_1 v_2 v_3}{3(v_1 v_2 + v_2 v_3 + v_3 v_1)}$
$\frac{3 v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}$
$\frac{(v_1 + v_2 + v_3)}{3 v_1 v_2 v_3}$

Answer: (c)

Solution

Given $AB = x$, $BC = x$, and $2x + CD = 3x$. Therefore, $CD = x$. The average velocity $ $ is given by: $$ = \frac{3x}{\frac{x}{v_1} + \frac{x}{v_2} + \frac{x}{v_3}} = \frac{3v_1 v_2 v_3}{v_2 v_3 + v_1 v_3 + v_1 v_2}.$$

Question 3

Physics · Motion in a Plane · Single correct

A child stands on the edge of the cliff 10 m above the ground and throws a stone horizontally with an initial speed of 5 $\mathrm{ms^{-1}}$. Neglecting the air resistance, the speed with which the stone hits the ground will be _____ $\mathrm{ms^{-1}}$ (given, g = 10 $\mathrm{ms^{-2}}$).

Answer: (b)

Solution

Given $v_y = \sqrt{2gh} = \sqrt{200}$. $v_{net} = \sqrt{25 + 200} = 15 \, \mathrm{m/s}$.

Question 4

Physics · Laws of Motion · Single correct

A block of mass $5 \, \mathrm{kg}$ is placed at rest on a table of rough surface. Now, if a force of $30 \, \mathrm{N}$ is applied in the direction parallel to surface of the table, the block slides through a distance of $50 \, \mathrm{m}$ in an interval of time $10 \, \mathrm{s}$. Coefficient of kinetic friction is (given, $g = 10 \, \mathrm{ms^{-2}}$):

0.60
0.75
0.50
0.25

Answer: (c)

Solution

Given the equation for displacement: $$S = ut + \frac{1}{2} at^2$$ Substitute the values: $$50 = 0 + \frac{1}{2} \times a \times 100$$ Solve for $a$: $$a = 1 \, \mathrm{m/s^2}$$ Using the equation of motion: $$F - \mu mg = ma$$ Substitute the values: $$30 - \mu \times 50 = 5 \times 1$$ Solve for $\mu$: $$50\mu = 25$$ Therefore, $$\mu = \frac{1}{2}$$

Question 5

Physics · Gravitation · Single correct

If earth has a mass nine times and radius twice to the of a planet P. Then $\frac{v_e}{3} \sqrt{x} \, \mathrm{ms^{-1}}$ will be the minimum velocity required by a rocket to pull out of gravitational force of P, where $v_e$ is escape velocity on earth. The value of $x$ is

Answer: (a)

Solution

The escape velocity for the planet is given by $$v_{(escape) plant} = \sqrt{\frac{2GM_P}{R_P}}$$ Substituting the given values, we have $$= \sqrt{\frac{2G \left( \frac{M_e}{9} \right)}{\left( \frac{R_e}{2} \right)}} = \frac{v_e \sqrt{2}}{3} \therefore x = 2$$

Question 6

Physics · Gravitation · Single correct

Given below are two statements: Statement-I: Acceleration due to gravity is different at different places on the surface of earth. Statement-II: Acceleration due to gravity increases as we go down below the earth's surface. In the light of the above statements, choose the correct answer from the options given below

Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Statement I is false but Statement II is true

Answer: (c)

Solution

The effective gravity $g_{eff}$ is given by the equation: $$g_{eff} = g - \omega^2 R_e \sin^2 \theta,$$ where $\theta$ is the co-latitude angle. Another expression for effective gravity is: $$g_{eff} = g \left( 1 - \frac{d}{R_e} \right),$$ where $d$ is the depth.

Question 7

Physics · Mechanical Properties of Fluids · Single correct

A mercury drop of radius $10^{-3} \, \mathrm{m}$ is broken into 125 equal size droplets. Surface tension of mercury is $0.45 \, \mathrm{Nm}^{-1}$. The gain in surface energy is:

$2.26 \times 10^{-5} \, \mathrm{J}$
$28 \times 10^{-5} \, \mathrm{J}$
$17.5 \times 10^{-5} \, \mathrm{J}$
$5 \times 10^{-5} \, \mathrm{J}$

Answer: (a)

Solution

Initial surface energy = 0.45 $\times$ 4 $\pi$ (10^{-3})^2 $$\frac{4}{3} \pi (10^{-3})^3 = 125 \times \frac{4 \pi}{3} R_{new}^3$$ Therefore, $10^{-3} = 5 R_{new}$ Thus, $$R_{new} = \frac{10^{-3}}{5} m$$ So, final surface energy = 0.45 $\times$ 125 $\times$ 4 $\pi$ $\left$( $\frac{10^{-3}}{5}$ $\right$)^2 Increase in energy = 0.45 $\times$ 4 $\pi$ $\times$ (10^{-3})^2 $\left$[ $\frac{125}{25}$ - 1 $\right$] = 4 $\times$ 0.45 $\times$ 4 $\pi$ $\times$ 10^{-6} = 2.26 $\times$ 10^{-5} J

Question 8

Physics · Thermodynamics · Single correct

A sample of gas at temperature $T$ is adiabatically expanded to double its volume. The work done by the gas in the process is ( given, $\gamma$ = $\frac{3}{2}$):

$W = TR [\sqrt{2} - 2]$
$W = \frac{T}{R} [\sqrt{2} - 2]$
$W = \frac{R}{T} [2 - \sqrt{2}]$
$W = RT [2 - \sqrt{2}]$

Answer: (d)

Solution

Given $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$. $TV^{1/2} = T_2 (2V)^{1/2}$. $T_2 = \frac{T}{\sqrt{2}}$. $W = \frac{R(T_1 - T_2)}{\gamma - 1} = \frac{R \left( T - \frac{T}{\sqrt{2}} \right)}{1} = RT \left( 2 - \sqrt{2} \right)$.

Question 9

Physics · Kinetic Theory · Single correct

The average kinetic energy of a molecule of the gas is

proportional to absolute temperature
proportional to volume
proportional to pressure
dependent on the nature of the gas

Answer: (a)

Solution

Basic theory Translational K.E on average of a molecule is $\frac{3}{2} KT$ which is independent of nature, pressure and volume.

Question 10

Physics · Waves · Single correct

A steel wire with mass per unit length $7.0 \times 10^{-3} \, \mathrm{kg \, m^{-1}}$ is under tension of $70 \, \mathrm{N}$. The speed of transverse waves in the wire will be:

$200 \, \pi \, \mathrm{m/s}$
$100 \, \mathrm{m/s}$
$10 \, \mathrm{m/s}$
$50 \, \mathrm{m/s}$

Answer: (b)

Solution

The velocity $v$ is given by the formula $$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{70}{70 \times 10^{-3}}} = 100 \, \mathrm{m/s}$$

Question 11

Physics · Electric Charges and Fields · Single correct

Let $\sigma$ be the uniform surface charge density of two infinite thin plane sheets shown in figure. Then the electric fields in three different region $E_I$, $E_{II}$ and $E_{III}$ are:

$\vec{E}_I = \frac{2\sigma}{\varepsilon_0} \hat{n}, \vec{E}_{II} = 0, \vec{E}_{III} = \frac{2\sigma}{\varepsilon_0} \hat{n}$
$\vec{E}_I = 0, \vec{E}_{II} = \frac{\sigma}{\varepsilon_0} \hat{n}, \vec{E}_{III} = 0$
$\vec{E}_I = \frac{\sigma}{2\varepsilon_0} \hat{n}, \vec{E}_{II} = 0, \vec{E}_{III} = \frac{\sigma}{2\varepsilon_0} \hat{n}$
$\vec{E}_I = -\frac{\sigma}{\varepsilon_0} \hat{n}, \vec{E}_{II} = 0, \vec{E}_{III} = \frac{\sigma}{\varepsilon_0} \hat{n}$

Answer: (d)

Solution

Assuming RHS to be $\hat{n}$ $$\vec{E}_I = \frac{\sigma}{2 \epsilon_0} (-\hat{n}) + \frac{\sigma}{2 \epsilon_0} (-\hat{n}) = -\frac{\sigma}{\epsilon_0} \hat{n}$$ $$\vec{E}_{II} = 0,$$ $$\vec{E}_{III} = \frac{\sigma}{2 \epsilon_0} (\hat{n}) + \frac{\sigma}{2 \epsilon_0} (\hat{n}) = \frac{\sigma}{\epsilon_0} (\hat{n})$$

Question 12

Physics · Current Electricity · Numerical

The equivalent resistance between $A$ and $B$ of the network shown in figure:

Answer: 4

Solution

Wheatstone bridge is in balanced condition. $$\frac{1}{R_{eq}} = \frac{1}{4R} + \frac{1}{8R}$$ $$R_{eq} = \frac{8R}{3}$$

Question 13

Physics · Moving Charges and Magnetism · Single correct

Find the magnetic field at the point P in figure. The curved portion is a semicircle connected to two long straight wires.

$\frac{\mu_0 i}{2r} \left( 1 + \frac{2}{\pi} \right)$
$\frac{\mu_0 i}{2r} \left( 1 + \frac{1}{\pi} \right)$
$\frac{\mu_0 i}{2r} \left( \frac{1}{2} + \frac{1}{2\pi} \right)$
$\frac{\mu_0 i}{2r} \left( \frac{1}{2} + \frac{1}{\pi} \right)$

Answer: (c)

Solution

The expression for the magnetic field at point P is given by: $$B_P = \left( \frac{\mu_0 i}{4r} + \frac{\mu_0 i}{4\pi r} \right) = \frac{\mu_0 i}{2r} \left( \frac{1}{2} + \frac{1}{2\pi} \right)$$

Question 14

Physics · Current Electricity · Single correct

Match the List-I with List-II \begin{tabular}{|c|l|c|l|} \hline & \text{List I} & & \text{List II}\\ \hline \text{A.} & \text{AC generator} & \text{I.} & \text{Presence of both L and C}\\ \hline \text{B.} & \text{Transformer} & \text{II.} & \text{Electromagnetic Induction}\\ \hline \text{C.} & \text{Resonance phenomenon to occur} & \text{III.} & \text{Quality factor}\\ \hline \text{D.} & \text{Sharpness of resonance} & \text{IV.} & \text{Mutual Inductance}\\ \hline \end{tabular}

A-IV, B-II, C-I, D-III
A-II, B-I, C-III, D-IV
A-II, B-IV, C-I, D-III
A-IV, B-III, C-I, D-II

Answer: (c)

Solution

Based on theory

Question 15

Physics · Current Electricity · Single correct

Match the List-I with List-II: Choose the correct answer from the options given below: \begin{tabular}{|c|l|c|l|} \hline & \text{List I} & & \text{List II}\\ \hline \text{A.} & \text{Microwaves} & \text{I.} & \text{Radio active decay of the nucleus}\\ \hline \text{B.} & \text{Gamma rays} & \text{II.} & \text{Rapid acceleration and deceleration of electron in aerials}\\ \hline \text{C.} & \text{Radio waves} & \text{III.} & \text{Inner shell electrons}\\ \hline \text{D.} & \text{X-rays} & \text{IV.} & \text{Klystron valve}\\ \hline \end{tabular}

A-I, B-II, C-III, D-IV
A-IV, B-I, C-II, D-III
A-I, B-III, C-IV, D-II
A-IV, B-III, C-II, D-I

Answer: (b)

Solution

Based on theory

Question 16

Physics · Wave Optics · Single correct

'n' polarizing sheets are arranged such that each makes an angle $45^\circ$ with the proceeding sheet. An unpolarized light of intensity I is incident into this arrangement. The output intensity is found to be $\frac{I}{64}$. The value of n will be:

Answer: (b)

Solution

After passing through first sheet $$I_1 = \frac{I}{2}$$ After passing through second sheet $$I_2 = I_1 \cos^2(45^\circ) = \frac{I}{4}$$ After passing through $n^{th}$ sheet $$I_n = \frac{I}{2^n} = \frac{I}{64}$$ $$n = 6$$

Question 17

Physics · Dual Nature of Radiation and Matter · Single correct

A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of $\lambda$. An alpha particle having certain kinetic energy has the same de-Broglie wavelength $\lambda$. The ratio of kinetic energy of proton and that of alpha particle is:

2 : 1
4 : 1
1 : 2
1 : 4

Answer: (b)

Solution

The kinetic energy is given by $$KE = \frac{p^2}{2m} = \frac{h^2}{2m \lambda^2}$$. The ratio of kinetic energies is $$\frac{KE_p}{KE_\alpha} = \frac{m_\alpha}{m_p} = 4:1$$.

Question 18

Physics · Nuclei · Single correct

The mass of proton, neutron and helium nucleus are respectively $1.0073$ $\mathrm{u}$, 1.0087 $\mathrm{u}$ and 4.0015 $\mathrm{u}$. The binding energy of helium nucleus is:

14.2 $\mathrm{MeV}$
28.4 $\mathrm{MeV}$
56.8 $\mathrm{MeV}$
7.1 $\mathrm{MeV}$

Answer: (b)

Solution

The binding energy (B.E) of Helium is given by the equation: $$B.E of Helium = (2m_p + 2m_N - m_{He}) c^2$$ This evaluates to: $$= 28.4 \, MeV$$

Question 19

Physics · Physical World, Units and Measurements · Single correct

Match the List I with List II Choose the correct answer from the options given below: \begin{tabular}{|c|l|c|l|} \hline & \text{List I} & & \text{List II}\\ \hline \text{A.} & \text{Intrinsic Semiconductor} & \text{I.} & \text{Fermi-level near conduction band}\\ \hline \text{B.} & \text{n-type semiconductor} & \text{II.} & \text{Fermi-level at middle}\\ \hline \text{C.} & \text{p-type semiconductor} & \text{III.} & \text{Fermi-level near valence band}\\ \hline \text{D.} & \text{Metals} & \text{IV.} & \text{Fermi-level inside conduction band}\\ \hline \end{tabular}

(A) $\rightarrow$ I, (B) $\rightarrow$ II, (C) $\rightarrow$ III, (D) $\rightarrow$ IV
(A) $\rightarrow$ II, (B) $\rightarrow$ I, (C) $\rightarrow$ III, (D) $\rightarrow$ IV
(A) $\rightarrow$ II, (B) $\rightarrow$ III, (C) $\rightarrow$ I, (D) $\rightarrow$ IV
(A) $\rightarrow$ III, (B) $\rightarrow$ I, (C) $\rightarrow$ II, (D) $\rightarrow$ IV

Answer: (c)

Solution

Based on theory

Question 20

Physics · Communication Systems · Single correct

Which of the following frequencies does not belong to FM broadcast.

106 $\mathrm{MHz}$
64 $\mathrm{MHz}$
99 $\mathrm{MHz}$
89 $\mathrm{MHz}$

Answer: (b)

Solution

FM broadcast range is $88 \, \mathrm{MHz}$ to $108 \, \mathrm{MHz}$.

Question 21

Physics · Work, Energy and Power · Numerical

A small particle moves to position $5\hat{i} - 2\hat{j} + \hat{k}$ from its initial position $2\hat{i} + 3\hat{j} - 4\hat{k}$ under the action of force $5\hat{i} + 2\hat{j} + 7\hat{k} \, \mathrm{N}$. The value of work done will be

Answer: 40

Solution

Given $W = \mathbf{F} \cdot (\mathbf{r}_f - \mathbf{r}_i)$. $$W = (5\hat{i} + 2\hat{j} + 7\hat{k}) \cdot ((5\hat{i} - 2\hat{j} + \hat{k}) - (2\hat{i} + 3\hat{j} - 4\hat{k}))$$ $W = 40 \, \mathrm{J}$

Question 22

Physics · System of Particles and Rotational Motion · Numerical

A solid cylinder is released from rest from the top of an inclined plane of inclination $30^\circ$ and length $60 \, \mathrm{cm}$. If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is _________ $\mathrm{ms^{-1}}$. (Given $g = 10 \, \mathrm{ms^{-2}}$)

Answer: 2

Solution

The velocity is given by the equation $$v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$$ where $h = 60 \sin 30^\circ = 30 \, \mathrm{cm}$. Also, $$k^2 = \frac{R^2}{2}.$$

Question 23

Physics · Mechanical Properties of Solids · Numerical

A certain pressure $P$ is applied to 1 litre of water and 2 litre of a liquid separately. Water gets compressed to 0.01$\%$ whereas the liquid gets compressed to 0.03$\%$. The ratio of Bulk modulus of water to that of the liquid is $\frac{3}{x}$. The value of $x$ is

Answer: 1

Solution

Given $$B_{water} = \frac{-\Delta P}{\left(\frac{\Delta V}{V}\right)} = \frac{-\Delta P}{\frac{0.01}{100}}$$ $$B_{liquid} = \frac{-\Delta P}{\frac{0.03}{100}}$$ $$\frac{B_{water}}{B_{liquid}} = 3$$ Therefore, $$x = 1$$

Question 24

Physics · Oscillations · Numerical

The amplitude of a particle executing SHM is 3 cm. The displacement at which its kinetic energy will be 25$\%$ more than the potential energy is:

Answer: 2

Solution

Given $KE = PE + \frac{PE}{4}$. Therefore, $KE = \frac{5}{4} PE$. Now, $\frac{1}{2} m \omega^2 \left(A^2 - x^2\right) = \frac{5}{4} \times \frac{1}{2} m \omega^2 x^2$. This implies $A^2 - x^2 = \frac{5}{4} x^2$. Rearranging gives $\frac{9x^2}{4} = A^2$. Thus, $x = \frac{2}{3} A$. Therefore, $x = \frac{2}{3} \times 3 cm$, which gives $x = 2 cm$.

Question 25

Physics · Electric Charges and Fields · Numerical

Two equal positive point charges are separated by a distance $2a$. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge $q_0$ becomes maximum is $\frac{a}{\sqrt{x}}$. The value of $x$ is __________.

Answer: 2

Solution

The force is given by the equation $$F = \frac{2Kqq_0x}{(x^2 + a^2)^{3/2}}.$$ For $F$ to be maximum, $$\frac{dF}{dx} = 0.$$ Solving for $x$, we find $$x = \frac{a}{\sqrt{2}}.$$

Question 26

Physics · Experimental Physics · Numerical

In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf 1.5 V is found to be 60 cm. If this cell is replaced by another cell of emf E, the length of null point increases by 40 cm. The value of E is $\frac{x}{10} \, \mathrm{V}$. The value of x is .

Answer: 25

Solution

$\dfrac{E_1}{E_2}=\dfrac{l_1}{l_2}$ $\dfrac{1.5}{E_2}=\dfrac{60}{60+40}=\dfrac{6}{10}=\dfrac{3}{5}$ $E_2=\dfrac{5}{2}=\dfrac{x}{10}$ $\therefore\ x=25$

Question 27

Physics · Electrostatic Potential and Capacitance · Numerical

A charge particle of 2 $\mu \mathrm{C}$ accelerated by a potential difference of 100 $\mathrm{V}$ enters a region of uniform magnetic field of magnitude 4 $\mathrm{mT}$ at right angle to the direction of field. The charge particle completes semicircle of radius 3 $\mathrm{cm}$ inside magnetic field. The mass of the charge particle is _____ $\times 10^{-18} \mathrm{kg}$.

Answer: 144

Solution

Given $$r = \frac{mv}{qB} = \frac{\sqrt{2km}}{qB}, m = \frac{r^2 q^2 B^2}{2k}$$ Substituting the values: $$m = \frac{\frac{1}{100} \times \frac{3}{100} \times 2 \times 2 \times 4 \times 10^{-3} \times 4 \times 10^{-3} \times 10^{-12}}{2 \times (100) \times 10^{-6}}$$ Simplifying gives: $$= 144 \times 10^{-18} \, \mathrm{kg}$$

Question 28

Physics · Alternating Current · Numerical

A series LCR circuit is connected to an ac source of 220V, 50Hz. The circuit contain a resistance $R = 100\Omega$ and an inductor of inductive reactance $X_L = 79.6\Omega$. The capacitance of the capacitor needed to maximize the average rate at which energy is supplied will be _________ $\mu$ F.

Answer: 40

Solution

To maximize the average rate at which energy supplied i.e. power will be maximum. So in LCR circuit power will be maximum at the condition of resonance and in resonance condition $$X_L = X_C$$ $$79.6 = \frac{1}{\omega C}$$ $$\therefore C = \frac{1}{2\pi \times 50 \times 79.6}$$ $$\therefore C = 40 \mu \mathrm{F}$$

Question 29

Physics · Ray Optics and Optical Instruments · Numerical

A thin cylindrical rod of length 10 cm is placed horizontally on the principle axis of a concave mirror of focal length 20 cm. The rod is placed in a such a way that mid point of the rod is at 40 cm from the pole of mirror. The length of the image formed by the mirror will be $\frac{x}{3}$ cm. The value of $x$ is

Answer: 32

Solution

Given $U_A = -45 \, \mathrm{cm}$, $f = -20 \, \mathrm{cm}$. $$V_A = \frac{-45 \times (-20)}{-45 - (-20)} = \frac{-900}{25} = -36 \, \mathrm{cm}$$ And $U_B = -35 \, \mathrm{cm}$. $$V_B = \frac{-35 \times (-20)}{-35 - (-20)} = \frac{700}{-15}$$ Therefore, $V_A - V_B = length of image$ $$= \left(-36 + \frac{140}{3}\right) \, \mathrm{cm}$$ $$= \frac{-108 + 140}{3} \, \mathrm{cm}$$ $$= \frac{32}{3} \, \mathrm{cm}$$ Therefore, $x = 32$

Question 30

Physics · Atoms · Numerical

A light of energy 12.75 eV is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is $\frac{x}{\pi} \times 10^{-17}$ eVs. The value of $x$ is _______ (use $h = 4.14 \times 10^{-15}$ eVs, $c = 3 \times 10^{8}$ ms$^{-1}$).

Answer: 828

Solution

In the ground state energy $= -13.6 \, \mathrm{eV}$. So energy $$\frac{-13.6 \, \mathrm{eV}}{n^2} = -13.6 + 12.75$$ $$\frac{-13.6 \, \mathrm{eV}}{n^2} = -0.85$$ $$n = \sqrt{16}$$ $$n = 4$$ Angular momentum $$= \frac{nh}{2\pi} = \frac{4h}{2\pi} = \frac{2h}{\pi}$$ Angular momentum $$= \frac{2}{\pi} \times 4.14 \times 10^{-15}$$ $$= \frac{828 \times 10^{-17}}{\pi} \, \mathrm{eVs}$$

Chemistry

Question 31

Chemistry · Hydrogen · Single correct

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: Hydrogen is an environment friendly fuel. Reason R: Atomic number of hydrogen is 1 and it is a very light element. In the light of the above statements, choose the correct answer from the options given below

A is true but R is false
Both A and R are true but R is NOT the correct explanation of A
A is false but R is true
Both A and R are true and R is the correct explanation of A

Answer: (b)

Solution

No pollution occurs by combustion of hydrogen and very low density of hydrogen.

Question 32

Chemistry · The s-Block Elements · Single correct

Match List I with List II \begin{tabular}{|c|l|c|c|} \hline \text{List-I} & & \text{List-II} & \\ \hline \text{(A)} & \text{Slaked lime} & \text{(I)} & $\mathrm{NaOH}$ \\ \hline \text{(B)} & \text{Dead burnt plaster} & \text{(II)} & $\mathrm{Ca(OH)_2}$ \\ \hline \text{(C)} & \text{Caustic soda} & \text{(III)} & $\mathrm{Na_2CO_3\cdot10H_2O}$ \\ \hline \text{(D)} & \text{Washing soda} & \text{(IV)} & $\mathrm{CaSO_4}$ \\ \hline \end{tabular} Choose the correct answer form the options given below:

(A) – I, (B) – IV, (C ) – II, (D) – III
(A) – III, (B) – I, (C ) – II, (D) – IV
(A) – II, (B) – IV, (C ) – I, (D) – III
(A) – III, (B) – II, (C ) – IV, (D) – I

Answer: (c)

Solution

From S-block NCERT

Question 33

Chemistry · The s-Block Elements · Multiple correct

Choose the correct statement(s): A. Beryllium oxide is purely acidic in nature. B. Beryllium carbonate is kept in the atmosphere of $\mathrm{CO}_2$. C. Beryllium sulphate is readily soluble in water. D. Beryllium shows anomalous behavior. Choose the correct answer from the options given below:

A, B and C only
B, C and D only
A and B only
A only

Answer: (b)

Solution

A. Beryllium oxide is amphoteric in nature. B. Beryllium carbonate is kept in the atmosphere of $\mathrm{CO_2}$ because it is thermally less stable. C. Beryllium sulphate is readily soluble in water due to high degree of hydration. D. Beryllium shows anomalous behaviour due to small size, high ionization energy and high value of $\phi$ (polarising power).

Question 34

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

Resonance in carbonate ion ($\mathrm{CO}_3^{2-}$) is Which of the following is true?

It is possible to identify each structure individually by some physical or chemical method.
All these structures are in dynamic equilibrium with each other.
Each structure exists for equal amount of time.
$\mathrm{CO}_3^{2-}$ has a single structure i.e., resonance hybrid of the above three structures.

Answer: (d)

Solution

Resonating structures are hypothetical and resonance hybrid is a real structure which is the weighted average of all the resonating structures.

Question 35

Chemistry · Hydrocarbons · Multiple correct

But-2-yne is reacted separately with one mole of Hydrogen as shown below: Identify the incorrect statements from the options given below: \text{A. A is more soluble than B.} \text{B. The boiling point \& melting point of A are higher and lower than B respectively.} \text{C. A is more polar than B because dipole moment of A is zero.} \text{D. }$Br_2$ \text{ adds easily to B than A.}

B and C only
B, C and D only
A, C and D only
A and B only

Answer: (b)

Solution

Incorrect statements are C and D only, correct choice is not available.

Question 36

Chemistry · Environmental Chemistry · Single correct

How can photochemical smog be controlled?

By using tall chimneys
By complete combustion of fuel
By using catalytic converters in the automobiles/industry
By using catalyst

Answer: (c)

Solution

NCERT (Environmental chemistry)

Question 37

Chemistry · The Solid State · Single correct

Which of the following represents the lattice structure of $A_{0.95}O$ containing $A^{2+}$, $A^{3+}$ and $O^{2-}$ ions? $A^{2+}$ $A^{3+}$ $O^{2-}$

B and C only
B only
A and B only
A only

Answer: (d)

Solution

Applying electrical neutrality principle in metal deficiency defect. $3A^{2+}$ are replaced by $2A^{3+}$; thus one vacant site per pair of $A^{3+}$ is created.

Question 38

Chemistry · Surface Chemistry · Single correct

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Amongst He, Ne, Ar and Kr; 1 g of activated charcoal adsorbs more of Kr. Reason R : The critical volume $V_c$ (cm$^3$ mol$^{-1}$) and critical pressure $P_c$ (atm) is highest for Krypton but the compressibility factor at critical point $Z_c$ is lowest for Krypton. In the light of the above statements, choose the correct answer from the options given below.

A is true but R is false
A is false but R is true
Both A and R are true but R is NOT the correct explanation of A
Both A and R are true and R is the correct explanation A

Answer: (a)

Solution

Adsorption is proportional to van der Waals attraction forces. $$Z_c = \frac{3}{8}$$ for all real gases.

Question 39

Chemistry · General Principles and Processes of Isolation of Elements · Single correct

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: In an Ellingham diagram, the oxidation of carbon to carbon monoxide shows a negative slope with respect to temperature. Reason R: CO tends to get decomposed at higher temperature. In the light of the above statements, choose the correct answer from the options given below

Both A and R are correct and R is the correct explanation of A
A is not correct but R is correct
Both A and R are correct but R is NOT the correct explanation of A
A is correct but R is not correct

Answer: (d)

Solution

The reaction is given by: $$2\mathrm{C}(s) + \mathrm{O_2}(g) \rightarrow 2\mathrm{CO}(g)$$ $\Delta_r S^\circ$ is positive, $\Delta_r G^\circ = \Delta_r H^\circ - T \Delta_r S^\circ$; thus the slope is negative. As temperature increases, $\Delta_r G^\circ$ becomes more negative, thus it has a lower tendency to get decomposed.

Question 40

Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct

Given below are two statements: Statement I: Chlorine can easily combine with oxygen to form oxides; and the product has a tendency to explode. Statement II: Chemical reactivity of an element can be determined by its reaction with oxygen and halogens. In the light of the above statements, choose the correct answer from the options given below.

Both the statements I and II are true
Statement I is true but Statement II is false
Statement I is false but Statement II is true
Both the Statements I and II are false

Answer: (a)

Solution

Chlorine oxides, $\mathrm{Cl_2O}$, $\mathrm{ClO_2}$, $\mathrm{Cl_2O_6}$, and $\mathrm{Cl_2O_7}$ are

Question 41

Chemistry · Co-ordination Compounds · Single correct

A solution of $\mathrm{FeCl_3}$, when treated with $\mathrm{K_4[Fe(CN)_6]}$, gives a Prussian blue precipitate due to the formation of:

K[Fe_2(CN)_6]
$\mathrm{Fe[Fe(CN)_6]}$
Fe_3[Fe(CN)_6]_2
Fe_4[Fe(CN)_6]_3

Answer: (d)

Solution

Formation of Prussian blue complex takes place.

Question 42

Chemistry · The d-and f-Block Elements · Single correct

Highest oxidation state of Mn is exhibited in $\mathrm{Mn_2O_7}$. The correct statements about $\mathrm{Mn_2O_7}$ are

A and C only
A and D only
B and D only
B and C only

Answer: (a)

Solution

Question 43

Chemistry · Co-ordination Compounds · Single correct

Which of the following complex will show largest splitting of d-orbitals?

$[\mathrm{Fe(C_2O_4)_3}]^{3-}$
$[\mathrm{FeF_6}]^{3-}$
$[\mathrm{Fe(CN)_6}]^{3-}$
$[\mathrm{Fe(NH_3)_6}]^{3+}$

Answer: (c)

Solution

$\overline{\mathrm{CN}}$ is a strong field ligand so maximum splitting in $d$ orbitals take place.

Question 44

Chemistry · Co-ordination Compounds · Single correct

Which of the following are examples of double salts? (A) $\mathrm{FeSO_4\cdot(NH_4)_2SO_4\cdot6H_2O}$ (B) $\mathrm{CuSO_4\cdot4NH_3\cdot H_2O}$ (C) $\mathrm{K_2SO_4\cdot Al_2(SO_4)_3\cdot24H_2O}$ (D) $\mathrm{Fe(CN)_2\cdot4KCN}$ Choose the correct answer.

A and C only
A and B only
A, B and D only
B and D only

Answer: (a)

Solution

Double salt contains two or more types of salts. $\mathrm{CuSO_4.4NH_3.H_2O}$ and $\mathrm{Fe(CN)_2.4KCN}$ are complex compounds.

Question 45

Chemistry · Analytical Chemistry · Single correct

Identify the incorrect option from the following:

Answer: (b)

Solution

In alcoholic KOH, elimination reaction takes place

Question 46

Chemistry · Alcohols, Phenols and Ethers · Single correct

Decreasing order of dehydration of the following alcohols is

a > d > b > c
b > d > c > a
b > a > d > c
d > b > c > a

Answer: (b)

Solution

Dehydration of alcohol is directly proportional to the stability of carbocation.

Question 47

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

In the following reaction, 'A' is 'A' Major product.

Answer: (b)

Solution

Initially lone pair electron of $-\mathrm{NH_2}$ attack on electrophilic carbon, after then lone pair electron of oxygen attacks leading to formation of cyclic compound.

Question 48

Chemistry · Co-ordination Compounds · Single correct

Match List I with List II \begin{tabular}{|c|l|c|l|} \hline \text{List-I} & & \text{List-II} & \\ \hline \text{(A) Tranquilizers} & & \text{(I) Anti blood clotting} & \\ \hline \text{(B) Aspirin} & & \text{(II) Salvarsan} & \\ \hline \text{(C) Antibiotic} & & \text{(III) Antidepressant drugs} & \\ \hline \text{(D) Antiseptic} & & \text{(IV) Soframicine} & \\ \hline \end{tabular} Choose the correct answer from the options given below:

$(A) - IV, (B) - II, (C) - I, (D) - III$
$(A) - II, (B) - I, (C) - III, (D) - IV$
$(A) - II, (B) - I, (C) - IV, (D) - IV$
$(A) - II, (B) - IV, (C) - I, (D) - III$

Answer: (c)

Solution

NCERT (Chemistry in every day life)

Question 49

Chemistry · Biomolecules · Single correct

The correct representation in six-membered pyranose form for the following sugar [X] is

Answer: (b)

Solution

By Howorth structure of mannose

Question 50

Chemistry · Co-ordination Compounds · Single correct

Match List I and List II \begin{tabular}{|c|l|c|l|} \hline \text{List I} & & \text{List II} & \\ \hline \text{Test} & & \text{Functional group / Class of Compound} & \\ \hline \text{(A) Molisch's Test} & & \text{(I) Peptide} & \\ \hline \text{(B) Biuret Test} & & \text{(II) Carbohydrate} & \\ \hline \text{(C) Carbylamine Test} & & \text{(III) Primary amine} & \\ \hline \text{(D) Schiff's Test} & & \text{(IV) Aldehyde} & \\ \hline \end{tabular} Choose the correct answer from the options given below:

(A) - I, (B) - II, (C ) - III, (D) - IV
(A) - III, (B) - IV, (C ) -I, (D) - II
(A) - II, (B) - I, (C ) - III, (D) - IV
(A) - III, (B) - IV, (C ) -II, (D) - I

Answer: (c)

Solution

Match the tests in List I with the functional groups or class of compounds in List II. Molisch's Test is used for detecting carbohydrates. Biuret Test is used for detecting peptides. Carbylamine Test is used for detecting primary amines. Schiff's Test is used for detecting aldehydes.

Question 51

Chemistry · Solutions · Numerical

The density of $3 \, \mathrm{M}$ solution of NaCl is $1.0 \, \mathrm{g \, mL^{-1}}$. Molality of the solution is _______ $\times 10^{-2} \, \mathrm{m}$. (Nearest integer). Given: Molar mass of Na and Cl is $23$ and $35.5 \, \mathrm{g \, mol^{-1}}$ respectively.

Answer: 364

Solution

The molality $m$ is calculated as follows: $$m = \frac{1000 \times M}{1000 \times d - M \times M.W of solute}$$ Substituting the given values: $$= \frac{1000 \times 3}{1000 \times 1 - (3 \times 58.5)} = 3.64$$ Therefore, $$= 364 \times 10^{-2}$$

Question 52

Chemistry · Structure of Atom · Single correct

Electrons in a cathode ray tube have been emitted with a velocity of $1000 \, \mathrm{ms^{-1}}$. The number of following statements which is/are true about the emitted radiation is ________ . Given: $h = 6 \times 10^{-34} \, \mathrm{Js}$, $m_e = 9 \times 10^{-31} \, \mathrm{kg}$.

The deBroglie wavelength of the electron emitted is $666.67 \, \mathrm{nm}$.
The characteristic of electrons emitted depend upon the material of the electrodes of the cathode ray tube.
The cathode rays start from cathode and move towards anode.
The nature of the emitted electrons depends on the nature of the gas present in cathode ray tube.

Answer: (b)

Solution

$(A)$ $V_e = 1000 \, \mathrm{m/s}$; $h = 6 \times 10^{-34} \, \mathrm{Js}$; $m_e = 9 \times 10^{-31} \, \mathrm{kg}$ $$\lambda = \frac{h}{mv} = \frac{6 \times 10^{-34}}{9 \times 10^{-31} \times 1000} = 666.67 \times 10^{-9} \, \mathrm{m}$$ $$= 666.67 \, \mathrm{nm}$$ $(B)$ The characteristic of electrons emitted is independent of the material of the electrodes of the cathode ray tube. $(C)$ The cathode rays start from cathode and move towards anode. $(D)$ The nature of the emitted electrons is independent on the nature of the gas present in cathode ray tube.

Question 53

Chemistry · Thermodynamics · Numerical

At 25°C, the enthalpy of the following processes are given: $$\mathrm{H_2(g) + O_2(g) \rightarrow 2OH(g)} \Delta H^\circ = 78 \, \mathrm{kJ \, mol^{-1}}$$ $$\mathrm{H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g)} \Delta H^\circ = -242 \, \mathrm{kJ \, mol^{-1}}$$ $$\mathrm{H_2(g) \rightarrow 2H(g)} \Delta H^\circ = 436 \, \mathrm{kJ \, mol^{-1}}$$ $$\mathrm{\frac{1}{2}O_2(g) \rightarrow O(g)} \Delta H^\circ = 249 \, \mathrm{kJ \, mol^{-1}}$$ What would be the value of $X$ for the following reaction? (Nearest integer) $$\mathrm{H_2O(g) \rightarrow H(g) + OH(g)} \Delta H^\circ = X \, \mathrm{kJ \, mol^{-1}}$$

Answer: 499

Solution

The reaction sequence is as follows: $$\mathrm{2H_2O(g) \rightarrow 2H_2(g) + O_2(g)} +(242 \times 2) \, \mathrm{kJ \, mol^{-1}}$$ $$\mathrm{H_2(g) + O_2(g) \rightarrow 2OH} +78 \, \mathrm{kJ \, mol^{-1}}$$ $$\mathrm{H_2(g) \rightarrow 2H} +436 \, \mathrm{kJ \, mol^{-1}}$$ Adding these reactions gives: $$\mathrm{2H_2O \rightarrow 2H + 2OH} +998 \, \mathrm{kJ \, mol^{-1}}$$ For the reaction: $$\mathrm{H_2O \rightarrow H + OH}$$ The enthalpy change is: $$998 \times \frac{1}{2} = +499 \, \mathrm{kJ \, mol^{-1}}$$

Question 54

Chemistry · Equilibrium · Numerical

(i) $X(g) \rightleftharpoons Y(g) + Z(g)$ \quad $K_{p1} = 3$ (ii) $A(g) \rightleftharpoons 2B(g)$ \quad $K_{p2} = 1$ If the degree of dissociation and initial concentration of both the reactants $X(g)$ and $A(g)$ are equal, then the ratio of the total pressure at equilibrium $\left(\dfrac{P_1}{P_2}\right)$ is equal to $x : 1$. The value of $x$ is \underline{\hspace{1.5cm}} (Nearest integer)

Answer: 12

Solution

For the reaction $\mathrm{x(g) \rightleftharpoons y(g) + z(g)}$, $k_{p_1} = 3$. Initial moles at equilibrium are $n$, $n - \alpha n$, $\alpha n$, and $-\alpha n$. The expression for $k_{p_1}$ is given by: $$k_{p_1} = \frac{\left( \frac{\alpha}{1 + \alpha} \times p_1 \right)^2}{\frac{1 - \alpha}{1 + \alpha} \times p_1}$$ Simplifying, we have: $$3 = \frac{\alpha^2 \times p_1}{1 - \alpha^2}$$ For the reaction $\mathrm{A(g) \rightleftharpoons 2B(g)}$, $k_{p_2} = 1$. Initial moles at equilibrium are $n$, $x - \alpha n$, $2 \alpha n$, and $p_{total} = p_2$. The expression for $k_{p_2}$ is: $$k_{p_2} = \frac{\left( \frac{2\alpha}{1 + \alpha} \times p_2 \right)^2}{\frac{1 - \alpha}{1 + \alpha} \times p_2}$$ Simplifying, we have: $$1 = \frac{4\alpha^2 \times p_2}{1 - \alpha^2}$$ The ratio of $k_{p_1}$ to $k_{p_2}$ is: $$\frac{k_{p_1}}{k_{p_2}} = \frac{p_1}{4p_2}$$ Thus, we have: $$\frac{3}{1} = \frac{p_1}{4p_2}$$ Therefore, $p_1 : p_2 = 12 : 1$ and $x = 12$.

Question 55

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Numerical

The total number of chiral compound/s from the following is ________.

Answer: 2

Solution

The molecule has a chiral center indicated by the asterisk. There is no plane of symmetry (POS) or center of symmetry (COS), making it chiral. The first structure is achiral due to the presence of a plane of symmetry. The second structure is also achiral due to the presence of a plane of symmetry.

Question 56

Chemistry · Solutions · Numerical

25 $\mathrm{mL}$ of an aqueous solution of $\mathrm{KCl}$ was found to require 20 $\mathrm{mL}$ of 1 $\mathrm{M}$ $\mathrm{AgNO_3}$ solution when titrated using $\mathrm{K_2CrO_4}$ as an indicator. What is the depression in freezing point of $\mathrm{KCl}$ solution of the given concentration? (Nearest integer). (Given: $K_f$ = 2.0 $\mathrm{K \, kg \, mol^{-1}}$) Assume 1) 100$\%$ ionization and 2) density of the aqueous solution as 1 $\mathrm{g \, mL^{-1}}$

Answer: 3

Solution

At equivalence point, mmol of KCl = mmol of AgNO_3. = 20 mmole Volume of solution = 25 ml Mass of solution = 25 gm Mass of solvent = 25 - mass of solute = 25 - [20 $\times$ 10^{-3} $\times$ 74.5] = 23.51 gm Molality of KCl = $\frac{mole of KCl}{mass of solvent in kg}$ = $\frac{20 \times 10^{-3}}{23.51 \times 10^{-3}}$ = 0.85 i of KCl = 2 $\;$ (100$\%$ ionisation) $\Delta$ $T_f$ = i $\times$ $K_f$ $\times$ m = 2 $\times$ 2 $\times$ 0.85 = 3.4 $\approx$ 3

Question 57

Chemistry · Electrochemistry · Numerical

At what pH, given half cell $\mathrm{MnO_4^-} \ (0.1 \, \mathrm{M}) \mid \mathrm{Mn^{2+}} \ (0.001 \, \mathrm{M})$ will have electrode potential of $1.282 \, \mathrm{V}$? __________ (Nearest Integer) Given $E^o_{\mathrm{MnO_4^-/Mn^{2+}}} = 1.54 \, \mathrm{V}$, $\frac{2.303RT}{F} = 0.059 \, \mathrm{V}$

Answer: 3

Solution

The balanced chemical equation is: $$\mathrm{MnO_4^- + 8H^+ + 5e^- \rightleftharpoons Mn^{2+} + 4H_2O}$$ The Nernst equation is given by: $$E = E^\circ - \frac{0.059}{5} \log \frac{[\mathrm{Mn^{2+}}]}{[\mathrm{MnO_4^-}][\mathrm{H^+}]^8}$$ Substituting the given values: $$1.282 = 1.54 - \frac{0.059}{5} \log \frac{10^{-3}}{10^{-1} \times [\mathrm{H^+}]^8}$$ Simplifying further: $$0.258 \times 5 = \frac{0.059}{[\mathrm{H^+}]^8} \log 10^{-2}$$ This implies: $$21.86 = -2 + 8 \mathrm{pH}$$ Therefore, $$\mathrm{pH} = 2.98$$ Approximately, $$\approx 3$$

Question 58

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

A and B are two substances undergoing radioactive decay in a container. The half life of A is $15 \, \mathrm{min}$ and that of B is $5 \, \mathrm{min}$. If the initial concentration of B is 4 times that of A and they both start decaying at the same time, how much time will it take for the concentration of both of them to be same?

Answer: 15

Solution

For $A$: Let $[A]_t$ be $y$ and $[A]_0$ be $x$; $k = \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{15 \, \mathrm{min}}$. $y = xe^{-kt}$ $= xe^{-\left(\frac{\ln 2}{15}\right)t}$ For $B$: $[B]_t = [B]_0 e^{-kt}$ Let $[B]_t = y$; $[B]_0 = 4x$; $k = \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{5 \, \mathrm{min}}$. $y = 4xe^{-\left(\frac{\ln 2}{5}\right)t}$ $xe^{-\left(\frac{\ln 2}{15}\right)t} 4xe^{-\left(\frac{\ln 2}{5}\right)t}$ $e^{\left(\frac{\ln 2}{5} - \frac{\ln 2}{15}\right)t} = 4$ $t \times \left[\frac{\ln 2}{5} - \frac{\ln 2}{15}\right] = \ln 4$ $t \times \ln 2 \left[\frac{1}{5} - \frac{1}{15}\right] = 2 \ln 2$ $t = 15 \, \mathrm{min}$

Question 59

Chemistry · Redox Reactions · Fill in the blank

Sum of oxidation states of bromine in bromic acid and perbromic acid is _______.

Answer: 12

Solution

$\mathrm{HBrO_3}$ (Bromic acid) Ox. State of Br = $+5$ $\mathrm{HBrO_4}$ (Perbromic acid) Ox. State of Br = $+7$ Sum of Ox. State = $12$

Question 60

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical

Number of isomeric compounds with molecular formula $\mathrm{C_9H_{10}O}$ which (i) do not dissolve in $\mathrm{NaOH}$ (ii) do not dissolve in $\mathrm{HCl}$. (iii) do not give orange precipitate with $2, 4 - \mathrm{DNP}$ (iv) on hydrogenation give identical compound with molecular formula $\mathrm{C_9H_{12}O}$ is

Answer: 2

Solution

As per the language of given question, the best possible isomeric structure is Ph - CH = CH - O - $CH_3$ (cis and trans). So, the answer is 2.

Maths

Question 61

Maths · Basics Of Mathematics · Single correct

Let $$S = \left\{ x : x \in \mathbb{R} and \left( \sqrt{3} + \sqrt{2} \right)^{x^2 - 4} + \left( \sqrt{3} - \sqrt{2} \right)^{x^2 - 4} = 10 \right\}$$ Then n (S) is equal to

Answer: (a)

Solution

Let $\left( \sqrt{3} + \sqrt{2} \right)^{x^2 - 4} = t$. $$t + \frac{1}{t} = 10$$ Therefore, $t = 5 + 2\sqrt{6}, 5 - 2\sqrt{6}$. Thus, $\left( \sqrt{3} + \sqrt{2} \right)^{x^2 - 4} = 5 + 2\sqrt{6}, 5 - 2\sqrt{6}$. This implies $x^2 - 4 = 2, -2$ or $x^2 = 6, 2$. Therefore, $x = \pm \sqrt{2}, \pm \sqrt{6}$.

Question 62

Maths · Complex Numbers and Quadratic Equations · Single correct

If the center and radius of the circle $|\frac{z-2}{z-3}|$ = 2 are respectively $\alpha, \beta$ and $\gamma$, then $3\alpha + \beta + \gamma$ is equal to

Answer: (d)

Solution

Given $\sqrt{(x-2)^2 + y^2} = 2\sqrt{(x-3)^2 + y^2}$. Expanding both sides, we have: $$x^2 + y^2 - 4x + 4 = 4x^2 + 4y^2 - 24x + 36$$ Simplifying, we get: $$3x^2 + 3y^2 - 20x + 32 = 0$$ Dividing by 3: $$x^2 + y^2 - \frac{20}{3}x + \frac{32}{3} = 0$$ This gives $(\alpha, \beta) = \left(\frac{10}{3}, 0\right)$. Now, calculate $\gamma$: $$\gamma = \sqrt{\frac{100}{9} - \frac{32}{3}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$$ Finally, compute $3(\alpha, \beta, \gamma)$: $$3\left(\frac{10}{3} + \frac{2}{3}\right) = 12$$

Question 63

Maths · Sequences and Series · Single correct

The sum to 10 terms of the series \[\frac{1}{1+1^2+1^4} + \frac{2}{1+2^2+2^4} + \frac{3}{1+3^2+3^4} + \ldots is: \]

$\frac{59}{111}$
$\frac{55}{111}$
$\frac{56}{111}$
$\frac{58}{111}$

Answer: (b)

Solution

Given $$T_r = \frac{(r^2 + r + 1) - (r^2 - r + 1)}{2(r^4 + r^2 + 1)}$$ This simplifies to $$T_r = \frac{1}{2} \left[ \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right]$$ Calculating for specific values: $$T_1 = \frac{1}{2} \left[ \frac{1}{1} - \frac{1}{3} \right]$$ $$T_2 = \frac{1}{2} \left[ \frac{1}{3} - \frac{1}{7} \right]$$ $$T_3 = \frac{1}{2} \left[ \frac{1}{7} - \frac{1}{13} \right]$$ $\vdots$ $$T_{10} = \frac{1}{2} \left[ \frac{1}{91} - \frac{1}{111} \right]$$ Summing up the series: $$\Rightarrow \sum_{r=1}^{10} T_r = \frac{1}{2} \left[ 1 - \frac{1}{111} \right] = \frac{55}{111}$$

Question 64

Maths · Binomial Theorem · Single correct

The value of $\frac{1}{1!50!}$ + $\frac{1}{3!48!}$ + $\frac{1}{5!46!}$ + $\ldots$ + $\frac{1}{49!2!}$ + $\frac{1}{51!1!}$ is

$\frac{2^{50}}{50!}$
$\frac{2^{50}}{51!}$
$\frac{2^{51}}{51!}$
$\frac{2^{51}}{50!}$

Answer: (b)

Solution

The given expression is $$\sum_{r=1}^{26} \frac{1}{(2r-1)!(51-(2r-1))!} = \sum_{r=1}^{26} {}^{51}C_{2r-1} \frac{1}{51!}$$ which simplifies to $$= \frac{1}{51!} \left\{ {}^{51}C_{1} + {}^{51}C_{3} + \ldots + {}^{51}C_{51} \right\} = \frac{1}{51!} (2^{50}).$$

Question 65

Maths · Straight Lines and Pair of Straight Lines · Single correct

The combined equation of the two lines $ax + by + c = 0$ and $a'x + b'y + c' = 0$ can be written as $(ax + by + c) (a'x + b'y + c') = 0$ The equation of the angle bisectors of the lines represented by the equation $2x^2 + xy - 3y^2 = 0$ is

$3x^2 + 5xy + 2y^2 = 0$
$x^2 - y^2 + 10xy = 0$
$3x^2 + xy - 2y^2 = 0$
$x^2 - y^2 - 10xy = 0$

Answer: (d)

Solution

Equation of the pair of angle bisector for the homogenous equation $ax^2 + 2hxy + by^2 = 0$ is given as $$\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$$ Here $a = 2$, $h = \frac{1}{2}$ and $b = -3$. Equation will become $$\frac{x^2 - y^2}{2 - (-3)} = \frac{xy}{1/2}$$ $$x^2 - y^2 = 10xy$$ $$x^2 - y^2 - 10xy = 0$$

Question 66

Maths · Properties of Triangles · Single correct

If the orthocentre of the triangle, whose vertices are $(1, 2)$, $(2, 3)$ and $(3, 1)$ is $(\alpha, \beta)$, then the quadratic equation whose roots are $\alpha + 4\beta$ and $4\alpha + \beta$, is

$x^2 - 19x + 90 = 0$
$x^2 - 18x + 80 = 0$
$x^2 - 22x + 120 = 0$
$x^2 - 20x + 99 = 0$

Answer: (d)

Solution

Here $m_{BH} \times m_{AC} = -1$. $$\left(\frac{\beta - 3}{\alpha - 2}\right) \left(\frac{1}{-2}\right) = -1$$ $$\beta - 3 = 2\alpha - 4$$ $$\beta = 2\alpha - 1$$ $m_{AH} \times m_{BC} = -1$. $$\left(\frac{\beta - 2}{\alpha - 1}\right)(-2) = -1$$ $$2\beta - 4 = \alpha - 1$$ $$2(2\alpha - 1) = \alpha + 3$$ $$3\alpha = 5$$ $$\alpha = \frac{5}{3}, \beta = \frac{7}{3} \Rightarrow H\left(\frac{5}{3}, \frac{7}{3}\right)$$ $$\alpha + 4\beta = \frac{5}{3} + \frac{28}{3} = \frac{33}{3} = 11$$ $$\beta + 4\alpha = \frac{7}{3} + \frac{20}{3} = \frac{27}{3} = 9$$ $$x^2 - 20x + 99 = 0$$

Question 67

Maths · Mathematical Reasoning · Single correct

The negation of the expression $q \lor ((\sim q) \land p)$ is equivalent to

$(\sim p) \land (\sim q)$
$p \land (\sim q)$
$(\sim p) \lor (\sim q)$
$(\sim p) \lor q$

Answer: (a)

Solution

The expression is given as: $$\sim (q \lor ((\sim q) \land p))$$ This can be rewritten as: $$= \sim q \land \sim ((\sim q) \land p)$$ Applying distribution: $$= \sim q \land (q \lor \sim p)$$ Using distribution again: $$= (\sim q \land q) \lor (\sim q \land \sim p)$$ Simplifying further: $$= (\sim q \land \sim p)$$

Question 68

Maths · Statistics · Single correct

The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is

1072
1792
1216
1456

Answer: (a)

Solution

Given $$$ $\frac{1 + 3 + 5 + a + b}{5}$ = 5 $$$ $$$ a + b = 16 $$$ ......(1) $$$ $\sigma$^2 = $\frac{\sum x_1^2}{5}$ - $\left$( $\frac{\sum x}{5}$ $\right$)^2 $$$ $$$ 8 = $\frac{1^2 + 3^2 + 5^2 + a^2 + b^2}{5}$ - 25 $$$ $$$ a^2 + b^2 = 130 $$$ ......(2) By (1), (2) $$$ a = 7, b = 9 $$$

Question 69

Maths · Properties of Triangles · Single correct

For a triangle ABC, the value of $\cos 2A + \cos 2B + \cos 2C$ is least. If its inradius is 3 and incentre is M, then which of the following is NOT correct?

Perimeter of $\triangle ABC$ is $18\sqrt{3}$
$\sin 2A + \sin 2B + \sin 2C = \sin A + \sin B + \sin C$
$\overline{MA} \cdot \overline{MB} = -18$
area of $\triangle ABC$ is $\frac{27\sqrt{3}}{2}$

Answer: (d)

Solution

If $\cos 2A + \cos 2B + \cos 2C$ is minimum then $A = B = C = 60^\circ$. So $\triangle ABC$ is equilateral. Now in-radius $r = 3$. So in $\triangle MBD$ we have $$\tan 30^\circ = \frac{MD}{BD} = \frac{r}{a/2} = \frac{6}{a}$$ $$\frac{1}{\sqrt{3}} = \frac{1}{a} = a = 6\sqrt{3}$$ Perimeter of $\triangle ABC = 18\sqrt{3}$. Area of $\triangle ABC = \frac{\sqrt{3}}{4} a^2 = 27\sqrt{3}$.

Question 70

Maths · Relations and Functions · Single correct

Let R be a relation on $\mathbb{R}$, given by $$R = \{(a, b) : 3a - 3b + \sqrt{7} \text{ is an irrational number}\}.$$ Then R is

Reflexive but neither symmetric nor transitive
Reflexive and transitive but not symmetric
Reflexive and symmetric but not transitive
An equivalence relation

Answer: (a)

Solution

Check for reflexivity: As $3(a - a) + \sqrt{7} = \sqrt{7}$ which belongs to relation so relation is reflexive. Check for symmetric: Take $a = \frac{\sqrt{7}}{3}, b = 0$. Now $(a, b) \in R$ but $(b, a) \notin R$. As $3(b - a) + \sqrt{7} = 0$ which is rational so relation is not symmetric. Check for Transitivity: Take $(a, b)$ as $\left( \frac{\sqrt{7}}{3}, 1 \right)$ and $(b, c)$ as $\left( 1, \frac{2\sqrt{7}}{3} \right)$. So now $(a, b) \in R$ and $(b, c) \in R$ but $(a, c) \notin R$ which means relation is not transitive.

Question 71

Maths · Determinants · Single correct

Let S denote the set of all real values of $\lambda$ such that the system of equations $$\lambda x + y + z = 1$$ $$x + \lambda y + z = 1$$ $$x + y + \lambda z = 1$$ is inconsistent, then $\sum_{\lambda \in S} (|\lambda|^2 + |\lambda|)$ is equal to

Answer: (d)

Solution

Given the determinant equation: $$\begin{vmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{vmatrix} = 0$$ Expanding, we have: $$(\lambda + 2) \begin{vmatrix} 1 & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{vmatrix} = 0$$ Simplifying further: $$(\lambda + 2)[1(\lambda^2 - 1) - 1(\lambda - 1) + (1 - \lambda)] = 0$$ $$(\lambda + 2)[(\lambda^2 - 2\lambda + 1)] = 0$$ $$(\lambda + 2)(\lambda - 1)^2 = 0 \Rightarrow \lambda = -2, \lambda = 1$$ At $\lambda = 1$, the system has an infinite solution. For inconsistent $\lambda = -2$. So, $$\sum (|-2|^2 + |-2|) = 6$$

Question 72

Maths · Inverse Trigonometric Functions · Single correct

Let S be the set of all solutions of the equation $$\cos^{-1}(2x) - 2\cos^{-1}\left(\sqrt{1-x^2}\right) = \pi, \ x \in \left[-\frac{1}{2}, \frac{1}{2}\right].$$ Then $$\sum_{x \in S} 2\sin^{-1}(x^2 - 1)$$ is equal to

0
$\frac{-2\pi}{3}$
$\pi - \sin^{-1}\left(\frac{\sqrt{3}}{4}\right)$
$\pi - 2\sin^{-1}\left(\frac{\sqrt{3}}{4}\right)$

Answer: (b)

Solution

Given $\cos^{-1}(2x) - 2 \cos^{-1} \sqrt{1-x^2} = \pi$. $\cos^{-1}(2x) - \cos^{-1}(2(1-x^2) - 1) = \pi$ $\cos^{-1}(2x) - \cos^{-1}(1-2x^2) = \pi$ $-\cos^{-1}(1-2x^2) = \pi - \cos^{-1}(2x)$ Taking $\cos$ both sides we get $\cos(-\cos^{-1}(1-2x^2)) = \cos(\pi - \cos^{-1}(2x))$ $1 - 2x^2 = -2x$ $2x^2 - 2x - 1 = 0$ On solving, $x = \frac{1-\sqrt{3}}{2}, \frac{1+\sqrt{3}}{2}$ As $x = [-1/2, 1/2]$, $x = \frac{1+\sqrt{3}}{2}$ = rejected So $x = \frac{1-\sqrt{3}}{2} \implies x^2 - 1 = -\sqrt{3}/2$ $= 2 \sin^{-1}(x^2 - 1) = 2 \sin^{-1}\left(\frac{-\sqrt{3}}{2}\right) = \frac{-2\pi}{3}$

Question 73

Maths · Applications of Derivatives · Single correct

Let $f(x) = 2x + \tan^{-1} x$ and $g(x) = \log_e \left( \sqrt{1 + x^2} + x \right)$, $x \in [0, 3]$. Then

There exists $\hat{x} \in [0,3]$ such that $f'(\hat{x}) < g'(\hat{x})$
max $f(x) >$ max $g(x)$
There exist $0 < x_1 < x_2 < 3$ such that $f(x) < g(x)$, $\forall x \in (x_1, x_2)$
min $f'(x) = 1 +$ max $g'(x)$

Answer: (b)

Solution

Given $f(x) = 2x + \tan^{-1}x$ and $g(x) = \ln \left( \sqrt{1+x^2} + x \right)$ and $x \in [0, 3]$. $g'(x) = \frac{1}{\sqrt{1+x^2}}$. Now, $0 \leq x \leq 3$. $0 \leq x^2 \leq 9$ $1 \leq 1+x^2 \leq 10$. So, $$2 + \frac{1}{10} \leq f'(x) \leq 3$$ $$\frac{21}{10} \leq f'(x) \leq 3 and \frac{1}{\sqrt{10}} \leq g'(x) \leq 1$$ Option (4) is incorrect. From above, $g'(x) \ln (3 + \sqrt{10})$ Option (2) is correct.

Question 74

Maths · Applications of Derivatives · Single correct

Let $f(x) = \begin{vmatrix} 1 + \sin^2 x & \cos^2 x & \sin 2x \\ \sin^2 x & 1 + \cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1 + \sin 2x \end{vmatrix}$, $x \in \left[ \frac{\pi}{6}, \frac{\pi}{3} \right]$. If $\alpha$ and $\beta$ respectively are the maximum and the minimum values of $f$, then

$\beta^2 - 2\sqrt{\alpha} = \frac{19}{4}$
$\beta^2 + 2\sqrt{\alpha} = \frac{19}{4}$
$\alpha^2 - \beta^2 = 4\sqrt{3}$
$\alpha^2 + \beta^2 = \frac{9}{2}$

Answer: (a)

Solution

Perform the column operation $C_1 \rightarrow C_1 + C_2 + C_3$ on the matrix: $$f(x) = \begin{vmatrix} 2 + \sin 2x & \cos^2 x & \sin 2x \\ 2 + \sin 2x & 1 + \cos^2 x & \sin 2x \\ 2 + \sin 2x & \cos^2 x & 1 + \sin 2x \end{vmatrix}$$ This simplifies to: $$f(x) = (2 + \sin 2x) \begin{vmatrix} 1 & \cos^2 x & \sin 2x \\ 1 & 1 + \cos^2 x & \sin 2x \\ 1 & \cos^2 x & 1 + \sin 2x \end{vmatrix}$$ Perform the row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$: $$f(x) = 2 + \sin 2x \begin{vmatrix} 1 & \cos^2 x & \sin 2x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$$ This results in: $$(2 + \sin 2x)(1) = 2 + \sin 2x$$ Thus, $\sin 2x \in \left[ \frac{\sqrt{3}}{2}, 1 \right]$ Hence, $2 + \sin 2x \in \left[ 2 + \frac{\sqrt{3}}{2}, 3 \right]$

Question 75

Maths · Sequences and Series · Single correct

$\lim_{{n \to \infty}}$ ($\frac{1}{1+n}$ + $\frac{1}{2+n}$ + $\frac{1}{3+n}$ + $\ldots$ + $\frac{1}{2n}$) is equal to :-

0
$\\log_e$ 2
$\\log_e(\\frac{3}{2})$
$\\log_e(\\frac{2}{3})$

Answer: (b)

Solution

The limit is given by $$\lim_{{n \to \infty}} \left( \frac{1}{1+n} + \ldots + \frac{1}{n+n} \right) = \lim_{{n \to \infty}} \sum_{{r=1}}^{n} \frac{1}{n+r}$$ which can be rewritten as $$= \lim_{{n \to \infty}} \sum_{{r=1}}^{n} \frac{1}{n} \left( \frac{1}{1+\frac{r}{n}} \right)$$ This is equivalent to the integral $$= \int_{0}^{1} \frac{1}{1+x} \, dx = [\ln(1+x)]_{0}^{1} = \ln 2$$

Question 76

Maths · Differential Equations · Single correct

The area enclosed by the closed curve C given by the differential equation $\frac{dy}{dx} + \frac{x+a}{y-2} = 0, \ y(1) = 0$ is $4\pi$. Let P and Q be the points of intersection of the curve C and the y-axis. If normals at P and Q on the curve C intersect x-axis at points R and S respectively, then the length of the line segment RS is

$2\sqrt{3}$
$\frac{2\sqrt{3}}{3}$
2
$\frac{4\sqrt{3}}{3}$

Answer: (d)

Solution

Given $( \frac{dy}{dx} + \frac{x+a}{y-2} = 0 )$ $( \frac{dy}{dx} = \frac{x+a}{2-y} )$ $((2-y)\,dy = (x+a)\,dx)$ $( 2y - \frac{y^2}{2} = \frac{x^2}{2} + ax + c )$ $( a + c = -\frac{1}{2} )$ as $( y(1) = 0 )$ $( x^2 + y^2 + 2ax - 4y - 1 - 2a = 0 )$ $( \pi\,r^2 = 4\,\pi )$ $( r^2 = 4 )$ $( 4 = \sqrt{a^2 + 4 + 1 + 2a} )$ $((a+1)^2 = 0 )$ P, Q = $( ( 0,\; 2 \pm \sqrt{3} ) )$ Equation of normal at P, Q are $( y - 2 = \sqrt{3}\,(x-1) )$ $( y - 2 = -\sqrt{3}\,(x-1) )$ R = $( ( 1 - \frac{2}{\sqrt{3}},\; 0 ) )$ S = $( ( 1 + \frac{2}{\sqrt{3}},\; 0 ) )$ RS = $( \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3} )$

Question 77

Maths · Differential Equations · Single correct

If y = y(x) is the solution curve of the differential equation $$$ $\frac{dy}{dx}$ + y $\tan$ x = x $\sec$ x $$$, $$$ 0 $\leq$ x $\leq$ $\frac{\pi}{3}$ $$$, y(0) = 1, then $$$ y( $\frac{\pi}{6}$ ) $$$ is equal to

$\frac{\pi}{12} - \frac{\sqrt{3}}{2} \log_e ( \frac{2}{e\sqrt{3}} )$
$\frac{\pi}{12} + \frac{\sqrt{3}}{2} \log_e ( \frac{2\sqrt{3}}{e} )$
$\frac{\pi}{12} - \frac{\sqrt{3}}{2} \log_e ( \frac{2\sqrt{3}}{e} )$
$\frac{\pi}{12} + \frac{\sqrt{3}}{2} \log_e ( \frac{2}{e\sqrt{3}} )$

Answer: (a)

Solution

Given $y^3 - 12y = 32\sqrt{2}$. Here I.F. = $\sec x$. Then solution of D.E.: $$y(\sec x) = x \tan x - \ln(\sec x) + c$$ Given $y(0) = 1 \Rightarrow c = 1$. Therefore, $$y(\sec x) = x \tan x - \ln(\sec x) + 1$$ At $x = \frac{\pi}{6}$, $y = \frac{\pi}{12} + \frac{\sqrt{3}}{2} \ln \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}$.

Question 78

Maths · Three Dimensional Geometry · Single correct

Let the image of the point P(2, -1, 3) in the plane $x + 2y - z = 0$ be Q. Then the distance of the plane $3x + 2y + z + 29 = 0$ from the point Q is

$\frac{22\sqrt{2}}{7}$
$\frac{24\sqrt{2}}{7}$
$2\sqrt{14}$
$3\sqrt{14}$

Answer: (d)

Solution

Equation of line PM is $$$ $\frac{x-2}{1}$ = $\frac{y+1}{2}$ = $\frac{z-3}{-1}$ = $\lambda$ $$$. Any point on the line is $$$ ($\lambda$ + 2, 2$\lambda$ - 1, -$\lambda$ + 3) $$$. For point 'm', $$$ ($\lambda$ + 2) + 2(2$\lambda$ - 1) - (3 - $\lambda$) = 0 $$$. Solving gives $$$ $\lambda$ = $\frac{1}{2}$ $$$. Point m is $$$ $\left$( $\frac{1}{2}$ + 2, 2 $\times$ $\frac{1}{2}$ - 1, -$\frac{1}{2}$ + 3 $\right$) $$$ which simplifies to $$$ $\left$( $\frac{5}{2}$, 0, $\frac{5}{2}$ $\right$) $$$. For Image Q $$$($\alpha$, $\beta$, $\gamma$)$$$, $$$ $\frac{\alpha + 2}{2}$ = $\frac{5}{2}$, $\frac{\beta - 1}{2}$ = 0, $\frac{\gamma + 3}{2}$ = $\frac{5}{2}$ $$$. Q is $$$(3, 1, 2)$$$. The distance $$$ d = $\frac{|3(3) + 2(1) + 2 + 29|}{\sqrt{3^2 + 2^2 + 1^2}}$ $$$. This simplifies to $$$ d = $\frac{42}{\sqrt{14}}$ = 3$\sqrt{14}$ $$$.

Question 79

Maths · Three Dimensional Geometry · Single correct

The shortest distance between the lines $\frac{x-5}{1} = \frac{y-2}{2} = \frac{z-4}{-3}$ and $\frac{x+3}{1} = \frac{y+5}{4} = \frac{z-1}{-5}$ is

$7\sqrt{3}$
$5\sqrt{3}$
$6\sqrt{3}$
$4\sqrt{3}$

Answer: (c)

Solution

Shortest distance between two lines $$$ $\frac{x-x_1}{a_1}$ = $\frac{y-y_1}{a_2}$ = $\frac{z-z_1}{a_3}$ $$$ and $$$ $\frac{x-x_2}{b_1}$ = $\frac{y-y_2}{b_2}$ = $\frac{z-z_2}{b_3}$ $$$ is given as $$ \frac{\left| \begin{array}{ccc} x_1-x_2 & y_1-y_2 & z_1-z_2 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array} \right|}{\sqrt{(a_1b_3-a_3b_1)^2 + (a_3b_2-a_2b_3)^2 + (a_2b_1-a_1b_2)^2}} $$ Substituting the values, we have $$ \frac{\left| \begin{array}{ccc} 5-(3) & 2-(-5) & 4-1 \\ 1 & 2 & -3 \\ 1 & 4 & -5 \end{array} \right|}{\sqrt{(-10+12)^2 + (-5+3)^2 + (4-2)^2}} $$ This simplifies to $$ \frac{\left| \begin{array}{ccc} 8 & 7 & 3 \\ 1 & 2 & -3 \\ 1 & 4 & -5 \end{array} \right|}{\sqrt{(2)^2 + (2)^2 + (2)^2}} $$ Calculating further, $$ = \frac{|8(-10+12) - 7(-5+3) + 3(4-2)|}{\sqrt{4+4+4}} $$ $$ = \frac{16+14+6}{\sqrt{12}} = \frac{36}{\sqrt{12}} = \frac{36}{2\sqrt{3}} $$ $$ = \frac{18}{\sqrt{3}} = 6\sqrt{3} $$

Question 80

Maths · Probability · Single correct

In a binomial distribution $B(n, p)$, the sum and product of the mean & variance are 5 and 6 respectively, then find $6(n + p - q)$ is equal to :-

Answer: (b)

Solution

Given $np + npq = 5$, $np \cdot npq = 6$. $np (1 + q) = 5$, $n^2 p^2 q = 6$. $n^2 p^2 (1 + q)^2 = 25$, $n^2 p^2 q = 6$. $$\frac{6}{q} (1 + q)^2 = 25$$ $6q^2 + 12q + 6 = 25q$ $6q^2 - 13q + 6 = 0$ $6q^2 - 9q - 4q + 6 = 0$ $(3q - 2)(2q - 3) = 0$ $q = \frac{2}{3}, \frac{3}{2}, q = \frac{2}{3}$ is accepted. $p = \frac{1}{3} \implies n \cdot \frac{1}{3} + n \cdot \frac{1}{3} \cdot \frac{2}{3} = 5$ $$\frac{3n + 2n}{9} = 5$$ $n = 9$ So $6(n + p - q) = 6 \left( 9 + \frac{1}{3} - \frac{2}{3} \right) = 52$

Question 81

Maths · Permutations and Combinations · Numerical

The number of words, with or without meaning, that can be formed using all the letters of the word ASSASSINATION so that the vowels occur together, is _____.

Answer: 50400

Solution

Vowels: A, A, A, I, I, O Consonants: S, S, S, S, N, N, T Therefore, the total number of ways in which vowels come together $$= \frac{8!}{4!2!} \times \frac{6!}{3!2!} = 50400$$

Question 82

Maths · Sequences and Series · Fill in the blank

Let $a_1 = 8$, $a_2$, $a_3$, $\ldots$, $a_n$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170, then the product of its middle two terms is ______.

Answer: 754

Solution

Given $a_1 + a_2 + a_3 + a_4 = 50$. $$\Rightarrow 32 + 6d = 50$$ $$\Rightarrow d = 3$$ And, $a_{n-3} + a_{n-2} + a_{n-1} + a_n = 170$. $$\Rightarrow 32 + (4n - 10) \cdot 3 = 170$$ $$\Rightarrow n = 14$$ $a_7 = 26$, $a_8 = 29$. $$\Rightarrow a_7 \cdot a_8 = 754$$

Question 83

Maths · Sequences and Series · Fill in the blank

The number of 3-digit numbers, that are divisible by either 2 or 3 but not divisible by 7 is ____.

Answer: 514

Solution

Divisible by 2 gives 450. Divisible by 3 gives 300. Divisible by 7 gives 128. Divisible by 2 and 7 gives 64. Divisible by 3 and 7 gives 43. Divisible by 2 and 3 gives 150. Divisible by 2, 3, and 7 gives 21. Therefore, the total numbers are calculated as follows: $$450 + 300 - 150 - 64 - 43 + 21 = 514.$$

Question 84

Maths · Binomial Theorem · Fill in the blank

The remainder when $19^{200} + 23^{200}$ is divided by 49, is ____.

Answer: 29

Solution

$(21+2)^{200}+(21-2)^{200}$ $\Rightarrow 2\left[ {}^{200}C_{0}\,21^{200} +{}^{200}C_{2}\,21^{198}\cdot2^{2} +\ldots +{}^{200}C_{198}\,21^{2}\cdot2^{198} +2^{200} \right]$ $\Rightarrow 2\left[49I_{1}+2^{200}\right]$ $=49I_{1}+2^{201}$ Now, $2^{201}=(8)^{67}$ $=(1+7)^{67}$ $=49I_{2}+{}^{67}C_{0}+{}^{67}C_{1}\cdot7$ $=49I_{2}+470$ $=49I_{2}+49\times9+29$ $\therefore$ Remainder is $29$

Question 85

Maths · Continuity and Differentiability · Fill in the blank

If $f(x) = x^2 + g'(1)x + g''(2)$ and $g(x) = f(1)x^2 + xf'(x) + f''(x)$, then the value of $f(4) - g(4)$ is equal to _______

Answer: 14

Solution

Given $f(x) = x^4 + g'(1)x + g''(2)$. $f'(x) = 2x + g'(1)$. $f''(x) = 2$. $g(x) = f(1) x^2 + x [2x + g'(1)] + 2$. $g'(x) = 2f(1) x + 4x + g'(1)$. $g''(x) = 2f(1) + 4$. $g''(x) = 0$. $2f(1) + 4 = 0$. $f(1) = -2$. $-2 = 1 + g'(1) = g'(1) = -3$. So, $f'(x) = 2x - 3$. $f(x) = x^2 - 3x + c$. $c = 0$. $f(x) = x^2 - 3x$. $g(x) = -3x + 2$. $f(4) - g(4) = 14$

Question 86

Maths · Integrals · Numerical

If \[ \int_0^1 (x^{21} + x^{14} + x^7)(2x^{14} + 3x^7 + 6)^{1/7} \, dx = \frac{1}{l}(11)^{m/n} \] where $l$, $m$, $n \in \mathbb{N}$, $m$ and $n$ are coprime then $l + m + n$ is equal to ____.

Answer: 63

Solution

Given the integral $$\int \left( x^{20} + x^{13} + x^6 \right) \left( 2x^{21} + 3x^{14} + 6x^7 \right)^{1/7} \, dx$$ Let $$2x^{21} + 3x^{14} + 6x^7 = t$$ Then, $$42(x^{20} + x^{13} + x^6) \, dx = dt$$ Thus, $$\frac{1}{42} \int_0^{11} t^{1/7} \, dt = \left( \frac{t^{8/7}}{8/7} \times \frac{1}{42} \right) \bigg|_0^{11}$$ This simplifies to $$= \frac{1}{48} \left( t^{8/7} \right) \bigg|_0^{11} = \frac{1}{48} (11)^{8/7}$$ Let $$l = 48, \ m = 8, \ n = 7$$ Therefore, $$l + m + n = 63$$

Question 87

Maths · Applications of Integrals · Numerical

Let A be the area bounded by the curve $y = x |x - 3|$, the x-axis and the ordinates $x = -1$ and $x = 2$. Then $12A$ is equal to _____.

Answer: 62

Solution

Given $$A = \int_{-1}^{0} (x^2 - 3x) \, dx + \int_{0}^{2} (3x - x^2) \, dx$$ This implies $$A = \left[ \frac{x^3}{3} - \frac{3x^2}{2} \right]_{-1}^{0} + \left[ \frac{3x^2}{2} - \frac{x^3}{3} \right]_{0}^{2}$$ Therefore, $$A = \frac{11}{6} + \frac{10}{3} = \frac{31}{6}$$ Thus, $$12A = 62$$

Question 88

Maths · Differential Equations · Numerical

Let f : ℝ → ℝ be a differentiable function such that f'(x) + f(x) = ∫₀² f(t) dt. If f(0) = e⁻², then 2f(0) − f(2) is equal to _____.

Answer: 1

Solution

dy/dx + y = k y·$e^x$ = k·$e^x$ + c f(0) = $e^{-2}$ ⇒ c = $e^{-2}$ - k ∴ y = k + $(e^{-2} - k)e^{-x}$ Now, k = ∫₀² [k + $(e^{-2} - k)e^{-x}]$ dx ⇒ k = $e^{-2}$ - 1 ∴ y = $(e^{-2} - 1) + e^{-x}$ f(2) = 2$e^{-2}$ - 1, f(0) = $e^{-2}$ 2f(0) - f(2) = 1

Question 89

Maths · Vector Algebra · Numerical

Let $\vec{v} = \alpha \hat{i} + 2 \hat{j} - 3 \hat{k}$, $\vec{w} = 2 \alpha \hat{i} + \hat{j} - \hat{k}$, and $\vec{u}$ be a vector such that $|\vec{u}| = \alpha > 0$. If the minimum value of the scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is $-\alpha \sqrt{3401}$, and $|\vec{u} \cdot \hat{i}|^2 = \frac{m}{n}$ where m and n are coprime natural numbers, then m + n is equal to ______.

Answer: 3501

Solution

Given $$$ $\frac{\overrightarrow{a} \times \overrightarrow{c}}{\overrightarrow{a} \cdot \overrightarrow{b}}$ = 3 $$$. $$$ [$\overrightarrow{u}$ $\overrightarrow{v}$ $\overrightarrow{w}$] = $\overrightarrow{u}$ $\cdot$ ($\overrightarrow{v}$ $\times$ $\overrightarrow{w}$) $$$ The minimum value of $$$ $\left$| $\overrightarrow{u}$ $\right$| $\left$| $\overrightarrow{v}$ $\times$ $\overrightarrow{w}$ $\right$| $\cos$ $\theta$ $$$ is $$$-$\alpha$ $\sqrt{3401}$ $$$. Therefore, $$$ $\cos$ $\theta$ = -1 $$$. $$$ $\left$| $\overrightarrow{u}$ $\right$| = $\alpha$ $$$ (Given) $$$ $\left$| $\overrightarrow{v}$ $\times$ $\overrightarrow{w}$ $\right$| = $\sqrt{3401}$ $$$ $$$ $\overrightarrow{v}$ $\times$ $\overrightarrow{w}$ = $\begin{vmatrix}$ $\hat{i}$ & $\hat{j}$ & $\hat{k}$ $\\$ $\alpha$ & 2 & -3 $\\$ 2$\alpha$ & 1 & -1 $\end{vmatrix}$ $$$ $$$ $\overrightarrow{v}$ $\times$ $\overrightarrow{w}$ = $\hat{i}$ - 5$\alpha$ $\hat{j}$ - 3$\alpha$ $\hat{k}$ $$$ $$$ $\left$| $\overrightarrow{v}$ $\times$ $\overrightarrow{w}$ $\right$| = $\sqrt{1 + 25\alpha^2 + 9\alpha^2}$ = $\sqrt{3401}$ $$$ $$$ 34$\alpha$^2 = 3400 $$$ $$$ $\alpha$^2 = 100 $$$ $$$ $\alpha$ = 10 $$$ (as $$$ $\alpha$ > 0 $$$) So $$$ $\overrightarrow{u}$ = $\lambda$ $\left$( $\hat{i}$ - 5$\alpha$ $\hat{j}$ - 3$\alpha$ $\hat{k}$ $\right$) $$$ $$$ $\left$| $\overrightarrow{u}$ $\right$| = $\sqrt{\lambda^2 + 25\alpha^2 \lambda^2 + 9\alpha^2 \lambda}$ $$$ $$$ $\alpha$^2 = $\lambda$^2 $\left$( 1 + 25$\alpha$^2 + 9$\alpha$^2 $\right$) $$$

Question 90

Maths · Three Dimensional Geometry · Fill in the blank

$A(2,6,2)$, $B(-4,0,\lambda)$, $C(2,3,-1)$ and $D(4,5,0)$, $|\lambda|\leq5$, are the vertices of a quadrilateral $ABCD$. If its area is $18$ square units, then $5-6\lambda$ is equal to $\underline{\hspace{2cm}}$.

Answer: 11

Solution

Given points A(2, 6, 2), B(-4, 0, $\lambda$), C(2, 3, -1), D(4, 5, 0). The area is given by $$Area = \frac{1}{2} \left| \overrightarrow{BD} \times \overrightarrow{AC} \right| = 18$$ The cross product is calculated as $$\overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -3 & -3 \\ 8 & 5 & -\lambda \end{vmatrix}$$ This results in $$(3\lambda + 15) \hat{i} - \hat{j} (-24) + \hat{k} (-24)$$ Thus, $$\overrightarrow{AC} \times \overrightarrow{BD} = (3\lambda + 15) \hat{i} + 24 \hat{j} - 24 \hat{k}$$ The magnitude is $$\sqrt{(3\lambda + 15)^2 + (24)^2 + (24)^2} = 36$$ Solving gives $$\lambda^2 + 10\lambda + 9 = 0$$ The solutions are $$\lambda = -1, -9$$ Given $|\lambda| \leq 5$, we have $\lambda = -1$. Finally, $$5 - 6\lambda = 5 - 6(-1) = 11$$