JEE Mains Important Questions with Answers PDF - – Physics, Chemistry, Maths

JEE Mains Important Questions with Answers: Chapter-Wise Important Topics for JEE Main

In this section, we will understand the high-weighted chapters. Along with these chapters, we will see JEE Main important questions with answers. Some JEE practice questions PDF with answers are also given below so that you can get a reference for what type of questions are asked. Additionally, we have provided you with the entire JEE Main important questions with answers PDF, which you can download and practice accordingly. The following link provides the PYQs that you must solve to practice for JEE Mains 2027.

JEE Main Important Questions with Answers PDF: Physics

The following table highlights some of the most important and high-weightage Physics chapters for JEE Main 2027. Candidates can practice questions from these topics and access chapter-wise PDFs wherever available.

JEE Mains Physics Important Questions with Answers:

1. Optics

Question: A ray of light passing through an equilateral prism is having velocity $2.12 \times 10^8 \mathrm{~m} / \mathrm{s}$ in the prism material, then the minimum angle of deviation is
$\_\_\_\_$ degrees.
1) 45
2) 30
3) 28

4) 58
Correct answer: Option (2)

Solution:

$\begin{aligned} & \mu=\frac{\sin \left[\frac{\delta_{\min }+A}{2}\right]}{\sin \frac{A}{2}} \\ & \mu=\frac{C}{V}=\frac{3 \times 10^8}{2.12 \times 10^8}=\sqrt{2} \\ & \sqrt{2}=\frac{\sin \left[\frac{\delta_{\min }+60^{\circ}}{2}\right]}{\sin 30^{\circ}} \\ & \sin \left[\frac{\delta_{\min }+60^{\circ}}{2}\right]=\frac{1}{\sqrt{2}} \\ & \delta_{\min }+60^{\circ}=90^{\circ} \Rightarrow \delta_{\min }=30^{\circ}\end{aligned}$

Question: In Young's double slit experiment, the path difference, at a certain point on the screen between two interfering waves is $\frac{1}{8}$ th of the wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:

1: 0.80

2: 0.94

3: 0.85

4: 0.74

Correct answer: Option (3)

Solution:

$
\begin{aligned}
& \Delta x=\frac{\lambda}{8} \\
& \Delta \phi=\left(\frac{2 \pi}{\lambda}\right) \frac{\lambda}{8}=\frac{\pi}{4} \\
& I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \theta
\end{aligned}
$
Putting $I_1$ and $I_2=I_o$

$
\text { we get } \quad \Rightarrow I=I_0+I_0+2 \sqrt{I_0 I_0} \cos \phi=4 I_0 \cos ^2 \frac{\phi}{2}
$
At the centre $I_c=4 I_0$
and at that point $I=4 I_o \cos ^2\left(\frac{\pi}{8}\right)=I_c \cos ^2\left(\frac{\pi}{8}\right)$

$
\begin{aligned}
& \frac{I}{I_c}=\cos ^2\left(\frac{\pi}{8}\right) \\
& \approx 0.85
\end{aligned}
$

Hence, the answer is option 3.

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2. Electrostatics

Question: Within a spherical charge distribution of charge density $\rho(\mathrm{r})$, N equipotential surfaces of potential $\mathrm{V}_0, \mathrm{~V}_0+\mathrm{V}, \mathrm{V}_0+2 \mathrm{~V}, \ldots \ldots \ldots . . V_0+N \Delta V(\Delta V>0)$, are drawn and have increasing radii $\mathrm{r}_0, \mathrm{r}_1$, $r_2, \ldots \ldots \ldots . . r_N$, respectively. If the difference in the radii of the surfaces is constant for all values of $\mathrm{V}_0$ and $\Delta V$ then:

1) $\rho(r) \propto r$
2) $\rho(r)=$ constant
3) $\rho(r) \propto \frac{1}{r}$
4) $\rho(r) \propto \frac{1}{r^2}$

Correct answer: Option (3)

Solution:

As we learned
Relation between field and potential -

$
E=\frac{-d V}{d r}
$

- wherein
$\frac{d v}{d r}$ - Potential gradient.
If P lies inside -

$
\begin{aligned}
& E_{i n}=\frac{1}{4 \pi \epsilon_0} \frac{Q r}{R^3} V_{i n}=\frac{Q}{4 \pi \epsilon_0} \frac{3 R^2-r^2}{2 R^3} \\
& E_{i n}=\frac{\rho r}{3 \epsilon_0} \quad V_{i n}=\frac{\rho\left(3 R^2-r^2\right)}{6 \epsilon_0}
\end{aligned}
$

We know $E=\frac{-d V}{d r}$

Here $\Delta v$ and $\Delta r$ are same for any pair of surfaces.
$\mathrm{E}=$ constant
Now, the electric field inside the spherical charge distribution

$
E=\frac{\rho}{3 \epsilon_0} r
$

E would be constant of $\rho r=$ constant

$
\rho(r) \propto \frac{1}{r}
$

Hence, the answer is the option (3).

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3. Properties of Solids and Liquids

Question: A man grows into a giant such that his linear dimensions increase by a factor of 9 . Assuming that his density remains the same, the stress in the leg will change by a factor of:

Solution:

As we learned in

$
\begin{aligned}
& \text { Stress }=\frac{\text { force }}{\text { Area }}=\frac{F}{A} \\
& F=\text { applied force } \\
& A=\text { area } \\
& \text { Stress(S) }=\frac{F}{A}=\frac{m g}{A} \\
& \text { Stress }=\frac{\text { Volume × density } \times g}{L^2} \\
& \text { Stress }=\frac{L^3 \rho g}{L^2}=\text { Stress } \alpha L \\
& \text { Stress }\left(\mathrm{S}^{\prime}\right)=\mathrm{ma} /\left.9\right|^* 9 \mathrm{~b}=\rho \mathrm{Va} /\left.8\right|^* 8 \mathrm{~b} \\
& \text { therefore } \mathrm{S}^{\prime} / \mathrm{S}=\left(9^3 \mathrm{~V} / 81^* \mathrm{lb}\right)^* \mathrm{lb} / \mathrm{V} \\
& \mathrm{~S}^{\prime}=9 \mathrm{~S}
\end{aligned}
$

Hence, stress in the leg will change by a factor of 9.

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4. Physics and Measurement

Question: Time (T), velocity $(C)$ and angular momentum $(H)$ are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be:

(1) $ [\mathrm{M}]=\left[\mathrm{T}^{-1} \mathrm{C}^{-2} \mathrm{~h}\right] $
(2) $ [M]=\left[T^{-1} C^2 h\right] $
(3) $ [\mathrm{M}]=\left[\mathrm{T}^{-1} \mathrm{C}^{-2} \mathrm{~h}^{-1}\right] $

(4) $ [\mathrm{M}]=\left[\mathrm{TC}^{-2} \mathrm{~h}\right] $

Correct answer: Option (1)

Solution:
Dimension of length $[\mathrm{L}]=[\mathrm{CT}]$
Dimension of mass $=[$ Angular Momentum $][$ Velocity $][$ Length $]$

$
=[\mathrm{h}][\mathrm{C}][\mathrm{CT}]=\left[\mathrm{C}^{-2} \mathrm{~T}^{-1} \mathrm{~h}\right]
$

Hence, the answer is option 1.

Question: The diameter and height of a cylinder are measured by a meter scale to be 12.6±0.1 cm and 34.2±0.1 cm, respectively. What will be the value of its volume in the appropriate significant figure?

1: $4300 \pm 80 \mathrm{~cm}^3$

2: $4264.4 \pm 81.0 \mathrm{~cm}^3$

3: $4264 \pm 81 \mathrm{~cm}^3$

4: $4260 \pm 80 \mathrm{~cm}^3$

Correct answer: Option (4)

Solution:

$\mathrm{v}=\frac{\pi \mathrm{d}^2}{4} \mathrm{~h}=4260 \mathrm{~cm}^3$

$\frac{\Delta \mathrm{v}}{\mathrm{v}}=\frac{2 \Delta \mathrm{~d}}{\mathrm{~d}}+\frac{\Delta \mathrm{h}}{\mathrm{~h}}$

$\Delta \mathrm{v}=2 \times \frac{0.1 \mathrm{v}}{12.6}+\frac{0.1 \mathrm{v}}{34.2}=\frac{0.2}{12.6} \times 4260+\frac{0.1 \times 4260}{34.2}=80$

$\therefore \text { Volume }=4260 \pm 80 \mathrm{~cm}^3$

Hence, the answer is the option is (4).

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5. Rotational Motion

Question: The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the origin. If the force acting on it is 1 N , and the distance of the particle from the origin is 5 m, the angle between the force and the position vector is (in radians) :

Correct answer: 0.52

Solution:

$T=\vec{r} \times \vec{F}$

$T=|\vec{r}| \cdot|\vec{F}| \cdot \sin \theta----(1)$

$T=2.5 \mathrm{Nm}$

$|\vec{r}|=5 \mathrm{~m}$

$|\vec{F}|=1 \mathrm{~N} \cdots-\text { put in (1) }$

$T=2.5=1 \times 5 \times \sin \theta$

$ \sin \theta=0.5=\frac{1}{2}$

$\theta=\frac{\pi}{6}$

$\theta=0.52 \text { radian }$

Hence, the answer is 0.52

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JEE Practice Questions PDF with Answers: Chemistry

The table below lists important Chemistry chapters based on previous year trends and weightage analysis. Students can use these topics for focused preparation and download chapter-wise PDFs where available.

JEE Mains Chemistry Important Questions with Answers:

1. Co-ordination Compounds

Question: The IUPAC name of $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$ is:

1) Pentaamminechloridocobalt(III) chloride

2) Pentamminecobalt(III) chloride

3) Pentaamminecobalt(II) chloride

4) Pentamminechloridocobalt(II) chloride

Correct answer: Option (1)

Solution:

- Central metal: Co

- Oxidation state of Co:x+0( from NH₃)+(−1)=+2. But total charge =+1 (complex ion is +2 , balanced by 2Cl⁻). So Co=+3.

- Correct name = Pentaamminechloridocobalt(III) chloride.

Question: The pair of compounds having metals in their highest oxidation state is :

1: $\mathrm{MnO}_2$ and $\mathrm{CrO}_2 \mathrm{Cl}_2$

2: $\left[\mathrm{NiCl}_4\right]^{2-}$ and $\left[\mathrm{CoCl}_4\right]^{2-}$

3: $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ and $\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{2-}$
4: $\left.\mathrm{FeCl}_4\right]^{-}$and $\mathrm{Co}_2 \mathrm{O}_3$

Correct answer: Option (1)

Solution:

The oxidation state of metals in respective compounds are:

(1) $\mathrm{MnO}_2$ Oxidation State $=+4$
$\mathrm{CrO}_2 \mathrm{Cl}_2$ Oxidation State $=+6$
(2) $\left[\mathrm{NiCl}_4\right]^{2-}$ Oxidation State $=+2$
$\left[\mathrm{CoCl}_4\right]^{2-}$ Oxidation State $=+2$
(3) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ Oxidation State $=+3$
$\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{2-}$ Oxidation State $=+2$
(4) $\left[\mathrm{FeCl}_4\right]^{-}$Oxidation State $=+3$
$\mathrm{Co}_2 \mathrm{O}_3$ Oxidation State $=+3$

Observing all the figures we can say that the first pair has the highest oxidation state Hence, the answer is an option (1).

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2. Chemical Thermodynamics

Question: For a reaction, $\Delta \mathrm{H}=-120 \mathrm{~kJ}$ and $\Delta \mathrm{S}=-200 \mathrm{~J} / \mathrm{K}$ at 300 K . Predict spontaneity.

Correct answer: Reaction is spontaneous

Solution:

$\Delta G=\Delta H-T \Delta S$

=-120-($300 \times -0.200$)

=-120+60

$ =-60 \mathrm{~kJ}$

Since $\Delta G<0 \rightarrow$ reaction is spontaneous.

Question: An ideal gas undergoes isothermal expansion at constant pressure. During the process :

1: enthalpy increases but entropy decreases.

2: enthalpy remains constant but entropy increases.

3: enthalpy decreases but entropy increases.

4: Both enthalpy and entropy remain constant.

Correct answer: (2)

Solution:

In the equation,

$\Delta H=n C_p \Delta T$

$\Delta S=n R \ln \left(\frac{V_f}{V_i}\right) \geq 0$

Enthalpy remains constant but entropy increases.

Hence, the answer is an option (2).

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3. Some Basic Concepts in Chemistry

Question: The amount of calcium oxide produced on heating 150 kg limestone ( $75 \%$ pure) is ______ kg. (Nearest integer)
Given : Molar mass (in $\mathrm{g} \mathrm{mol}^{-1}$ ) of $\mathrm{Ca}-40, \mathrm{O}-16, \mathrm{C}-12$

Solution:

Given that

Mass of limestone $=150 \mathrm{~kg}$
Purity $=75 \%\left(\right.$ pure calcium carbonate, $\left.\mathrm{CaCO}_3\right)$
Molar masses:

• $\mathrm{Ca}=40 \mathrm{~g} / \mathrm{mol}$
• $C=12 \mathrm{~g} / \mathrm{mol}$
• $O=16 \mathrm{~g} / \mathrm{mol}$

$ \mathrm{CaCO}_3 \rightarrow \mathrm{CaO}+\mathrm{CO}_2 $

${ mass ~of } ~\mathrm{CaCO}_3=\frac{150 \times 75}{100}=112.5 \mathrm{~kg}$

molar mass of $\mathrm{CaCO}_3$

$M_{\mathrm{CaCO}_3}=40+12+(3 \times 16)=40+12+48=100 \mathrm{~g} / \mathrm{mol}$

Moles of $\mathrm{CaCO}_3$
Mass in grams $=112.5 \times 1000=112500 \mathrm{~g}$

$
n=\frac{112500}{100}=1125 \mathrm{~mol}
$

Molar mass of CaO :

$M_{\mathrm{CaO}}=40+16=56 \mathrm{~g} / \mathrm{mol}$
Mass of CaO produced:

$m=n \times M_{\mathrm{CaO}}=1125 \times 56=63000 \mathrm{~g}=63 \mathrm{~kg}$

Hence, the answer is 63 kg.

Question: At 300 K and $1 \mathrm{~atm}, 15 \mathrm{~mL}$ of a gaseous hydrocarbon requires 375 mL air containing $20 \% \mathrm{O}_2$ by volume for complete combustion. After combustion, the gases occupy 330 mL . Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is :

1: $\mathrm{C}_4 \mathrm{H}_8$
2: $\mathrm{C}_4 \mathrm{H}_{10}$
3: $\mathrm{C}_3 \mathrm{H}_6$
4: $\mathrm{C}_3 \mathrm{H}_8$

Correct answer: (4)

Solution:

Volume of $\mathrm{N}_2$ in air $=375 \times 0.8=300 \mathrm{ml}$
Volume of $\mathrm{O}_2$ in air $=375 \times 0.2=75 \mathrm{ml}$

$
\begin{aligned}
& \mathrm{C}_x \mathrm{H}_y+\left(x+\frac{y}{4}\right) \mathrm{O}_2 \longrightarrow x \mathrm{CO}_2(g)+\frac{y}{2} \mathrm{H}_2 \mathrm{O}(l) \\
& 15 \mathrm{~mL} \quad 15\left(x+\frac{y}{4}\right) \\
& 0 \quad 0 \quad 15 x
\end{aligned}
$

After combustion total volume

$
\begin{aligned}
& 330=V_{N_2}+V_{\mathrm{CO}_2} \\
& 330=300+15 x \\
& x=2
\end{aligned}
$

Volume of $\mathrm{O}_2$ used

$
\begin{aligned}
& 15\left(x+\frac{y}{4}\right)=75 \\
& \left(x+\frac{y}{4}\right)=5 \\
& y=12
\end{aligned}
$

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4. Hydrocarbons

Question: Which of the following alkanes will give only one monochlorination product?

1) Butane

2) 2-Methylpropane

3) Neopentane

4) Pentane

Correct answer: (3)

Solution:

- Neopentane $\left(\mathrm{C}\left(\mathrm{CH}_3\right)_4\right)$ has all equivalent hydrogens.

- Only one product formed.

Hence the correct answer is option (3)

Question: Which hydrogen in compound (E) is easily replaceable during the bromination reaction in the presence of light?

$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}=\underset{\beta}{\mathrm{CH}}=\underset{\alpha}{\mathrm{CH}}$

1: $\alpha$ - hydrogen
2: $\gamma$-hydrogen
3: $\delta$-hydrogen
4: $\beta$ - hydrogen

Correct answer: (2)

Solution:

$\gamma$-hydrogen is easily replaceable during bromination reaction in the presence of light because, in the reaction, a free radical is formed as an intermediate at the gamma position which is very stable due to allylic conjugation.

$\mathrm{CH}_3-\dot{\mathrm{C}} \mathrm{H}-\mathrm{CH}=\mathrm{CH}_2$

Hence, the answer is option (2).

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5. Organic Compounds Containing Oxygen

Question: Which test is used to distinguish between aldehydes and ketones?

1) Tollen's Test

2) Benedict's Test

3) Both A and B

4) Fehling's Test

Correct answer: (3)

Solution:

- Aldehydes are oxidised easily, giving positive Tollen's (silver mirror) and Fehling's/Benedict's (red precipitate).

- Ketones do not respond.

Hence, the answer is option (3).

Download the JEE Main Organic Compounds Containing Oxygen PDF to practise important questions: Download PDF

JEE Practice Questions with Answers: Maths

The following Mathematics chapters are among the most important for JEE Main preparation. Practising questions from these topics can help improve accuracy, speed, and overall exam performance.

JEE Mains Mathematics Important Questions with Answers

1. Co-ordinate Geometry

Question: If the eccentricity e of the hyperbola $\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$, passing through $(6,4 \sqrt{3})$, satisfies $15\left(\mathrm{e}^2+1\right)=34 \mathrm{e}$, then the length of the latus rectum of the hyperbola $\frac{\mathrm{x}^2}{\mathrm{~b}^2}-\frac{\mathrm{y}^2}{2\left(\mathrm{a}^2+1\right)}=1$ is:

1: 10

2: 20

3: 25

4: 30

Solution:

$\begin{aligned}
& \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\
& \text { It passs through }(6,4 \sqrt{3}) \\
& \Rightarrow \frac{36}{a^2}-\frac{48}{b^2}=1 \\
& \text { also } 15 e^2-34 e+15=0 \\
& \Rightarrow 15 e^2-25 e-9 e+15=0 \\
& e=\frac{5}{3} \text { or } \frac{3}{5} \Rightarrow e=\frac{5}{3} \\
& 1+\frac{b^2}{a^2}=\frac{25}{9} \Rightarrow \frac{b^2}{a^2}=\frac{16}{9} \ldots
\end{aligned}$

using (1) and (2)

$\Rightarrow \frac{36}{a^2}-\frac{48}{16 a^2} \times 9=1 \Rightarrow a=3, b=4$

Length of L.R.

$\frac{4\left(a^2+1\right)}{b}=\frac{4(10)}{4}=10$

Question: The area enclosed by the curves xy+4y=16 and x+y=6 is equal to :

1: $28-30 \log _e 2$

2: $30-28 \log _{\mathrm{e}} 2$

3: $30-32 \log _e 2$

4: $32-30 \log _{\mathrm{e}} 2$

Correct answer: (3)

Solution:

$y(x+4)=16 \& y=6-x$

solve both curves,

$(6-x)(x+4)=16$

$6 x-x^2+24-4 x=16$

$x^2-2 x-8=0 \Rightarrow x=-2,4$

Download the JEE Main Co-ordinate Geometry PDF to practise important questions: Download PDF

2. Integral Calculus

Question: The value of the integral $\int_{-1}^1\left(\frac{x^3+|x|+1}{x^2+2|x|+1}\right) d x$ is equal to

1: $3 \log _e 2$

2: $2 \log _e 2$

3: $5 \log _e 3$

4: $3 \log _e 3$

Correct answer: (2)

Solution:

$\begin{aligned} & I=\int_{-1}^1 \frac{x^3+|x|+1}{x^2+2|x|+1} \cdot d x \\ & I=\int_{-1}^1 \frac{x^3}{x^2+2|x|+1} \cdot d x+\int_{-1}^1 \frac{|x|+1}{x^2+2|x|+1} \cdot d x \\ & I=0+\int_{-1}^1 \frac{|x|+1}{|x|^2+2|x|+1} \cdot d x \\ & 2 \int_0^1 \frac{1}{x+1} \cdot d x \\ & I=2[\ln |x+1|]_0^1 \\ & I=2 \ln (2)\end{aligned}$

Question: The integral ∫1(x−1)3(x+2)54 dx is equal to : (where C is a constant of integration)

1: $\frac{3}{4}\left(\frac{x+2}{x-1}\right)^{\frac{5}{4}}+C$

2: $\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C$

3: $\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{5}{4}}+C$

4: $\frac{3}{4}\left(\frac{x+2}{x-1}\right)^{\frac{1}{4}}+C$

Correct answer: (2)

Solution:

$\begin{aligned} & I=\int \frac{1}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}} d x \\ & I=\int \frac{1}{(x-1)^2\left(\frac{x+2}{x-1}\right)^{\frac{5}{4}}} d x \\ & \text { Let } \frac{x+2}{x-1}=t \Rightarrow \frac{(x-1)-(x+2)}{(x-1)^2} d x=d t\end{aligned}$

$\begin{aligned} & \Rightarrow I=-\frac{1}{3} \int \frac{d t}{t^{\frac{5}{4}}}=-\frac{1}{3} \times \frac{t^{-\frac{5}{4}+1}}{-\frac{5}{4}+1}+c=\frac{4}{3} t^{-\frac{1}{4}}+c \\ & =\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+c\end{aligned}$

option (2)

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3. Limit, Continuity & Differentiability

Question: Absolute Maximum Value of $f(x)=\frac{1}{|x-4|+1}+\frac{1}{|x+8|+1}$ is :

1: $\frac{14}{13}$

2: $\frac{15}{13}$

3: $\frac{8}{5}$

4: $\frac{13}{14}$

Correct answer: (1)

Solution:

$\text { Say } g(x)=\frac{1}{|x-4|+1} \Rightarrow \frac{1}{|x+8|+1}=g(x+12)$

Now for $\mathrm{x} \in(-\infty,-8)$ both $\mathrm{g}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x}+12)$ are increasing hence maximum value can't occur in this interval.
Similarly for $\mathrm{x} \in(4, \infty)$ both $\mathrm{g}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x}+12)$ are decreasing hence maximum value can't occur in this interval.
So, now for all values of $\mathrm{x} \in(-8,4)$

$f(x)=\frac{1}{x+9}+\frac{1}{5-x}=\frac{14}{(5-x)(x+9)}$

which will have maximum either at $\mathrm{x}=4$ or $\mathrm{x}=- 8$ or at minima of $(5-\mathrm{x})(\mathrm{x}+9)$. Also minima of $(5-x)(x+9)$ does not exist.

$\Rightarrow f(4)=1+\frac{1}{13}=\frac{14}{13}=f(-8) .$

So, absolute maximum is $\frac{14}{13}$.

Hence, the correct answer is option (1).

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4. Sets, Relations and Functions

Question: If A={1,2,3},B={a,b}, find the number of relations from A to B.

Correct answer: 64

Solution:

- Total ordered pairs possible =|A|×|B|=3×2=6.

- Each pair can be either in relation or not → total =26=64.

Answer: 64 relations

Question: Consider the following two relations on the set A={a,b,c},

R₁={(c,a),(b,b),(a,c),(c,c),(b,c),(a,a)}

and R₂={(a,b),(b,a),(c,c),(c,a),(a,a),(b,b),(a,c)}.

1: Both R₁ and R₂ is symmetric.

2: R₁ is not symmetric but it is transitive.

3: R2 is symmetric, but it is not transitive.

4: Both R1 and R₂ are transitive.

Correct answer: (3)

Solution:

REFLEXIVE RELATION: A relation R in A is said to be reflexive if aRa,∀a∈A

SYMMETRIC RELATION:
A relation $R$ on a set $A$ is said to be symmetric if

$a R b \Rightarrow b R a, \quad \forall a, b \in A .$
TRANSITIVE RELATION:
A relation $R$ on a set $A$ is said to be transitive if

$a R b \text { and } b R c \Rightarrow a R c, \quad \forall a, b, c \in A \text {. }$

Now,

R1:{(c,a)(b,b)(a,c)(c,c),(b,c)(a,a)}

Reflexive: R1 is reflexive since ( a,a), ( b,b ) and ( c,c ) are present.

Symmetric: (b,c) is present, but (c,b) is not, so not symmetric

Transitive: (b,c) and (c, a) belong to the relation, but (b, a) does not. So, it is not transitive.

R₂={(a,b),(b,a),(c,c),(c,a),(a,a),(b,b),(a,c)

Reflexive: since ( a,a ), ( b,b ) and ( c,c ) are present, it is reflexive

Symmetric: (a,b) and (b, a) both are present. Also (c, a) and (a,c) are present. So, it is symmetric

Transitive: (b, a), (a,c) are present but (b,c) is not present, so not transitive

Hence, the answer is option 3.

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5. Complex Numbers & Quadratic Equations

Question: If z=3+4i, find |z| and arg⁡(z).

Correct answer: (3)

Solution:

$\begin{gathered}|z|=\sqrt{3^2+4^2}=\sqrt{9+16}=5 \\ \arg (z)=\tan ^{-1}\left(\frac{4}{3}\right)\end{gathered}$

So, |z|=5,arg⁡(z)=tan−1⁡(43).

Question: Let z=1+ai be a complex number, a>0, such that z3 is a real number. Then the sum $1+z+z^2+\cdots+z^{11}$ is equal to :

1: $-1250 \sqrt{3} i$

2: $1250 \sqrt{3} i$

3: $1365 \sqrt{3} i$

4: $-1365 \sqrt{3} i$

Correct answer: (4)

Solution:

$z=1+a i, a>0$

given that z₃ is a real number.

$\begin{aligned} z^3 & =(1+a i)^3 \\ & =1-3 a^2+3 a i-a^3 i \\ & =1-3 a^2+\left(3 a-a^3\right) i\end{aligned}$

since, z is real number, so imaginary part is 0

$\begin{aligned} & \Rightarrow 3 a-a^3=0 \\ & \Rightarrow a=\sqrt{3}, \quad a>0 \\ & \Rightarrow z=1+i \sqrt{3} \\ & \Rightarrow z=2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)\end{aligned}$

Now,

$\begin{aligned} & 1+z+z^2+\cdots+z^{11}=\frac{1-z^{12}}{1-z} \\ & \frac{1-2^{12}(\cos 4 \pi+i \sin 4 \pi)}{1-(1+i \sqrt{3})}=\frac{1-2^{12}}{-i \sqrt{3}} \\ & =\frac{4095}{i \sqrt{3}}=-1365 i \sqrt{3}\end{aligned}$

Hence, the answer is option 4.

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