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

Q. 1 A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 60001 $ \text { Å } $ and diffraction bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright band is :
1) 3 mm
2) 9 mm (Correct)
3) 4.5 mm
4) 1.5 mm

Solution:

Fraunhofer Diffraction -

$
b \sin \theta=n \lambda
$

- wherein

Condition of nth minima.
$\mathrm{b}=$ slit width
$\theta=$ angle of deviation

$
\begin{aligned}
& a=0.1 \mathrm{~mm}=10^{-4} \mathrm{~cm} \\
& \text { lambda }=6000 A A=6 \times 10^{-7} \mathrm{~cm}
\end{aligned}
$

$
D=0.5 \mathrm{~m}
$

for 3rd dark bond, $\operatorname{asin} \theta=3 \lambda$

$
\sin \theta=\frac{3 \lambda}{a}=\frac{x}{D}
$

therefore $x=\frac{3 \lambda D}{a}$

Q. 5 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:

Option 1:
0.80

Option 2:
0.94

Option 3:
0.85

Option 4:
0.74

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.

Important Question of Optics

2. Electrostatics

Q. 1 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:

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

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).

Important Questions of Electrostatics

3. Properties of Solids and Liquids

Q. 1 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 .

Important Questions of Properties of Solids and Liquids

4. Physics and Measurement

Q. 1 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:

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

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.

Q. 2 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?

Option 1:

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

Option 2:

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

Option 3:

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

Option 4:

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

Correct Answer:

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

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).

Important Questions of Physics and Measurement

5. Rotational Motion

Q. 1 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 .

Important Questions of Rotational Motion

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

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

A) Pentaamminechloridocobalt(III) chloride

B) Pentamminecobalt(III) chloride

C) Pentaamminecobalt(II) chloride

D) Pentamminechloridocobalt(II) chloride

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.

Answer: A

Q. 2 The pair of compounds having metals in their highest oxidation state is :

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

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

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

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).

Important Questions of Co-ordination Compounds

2. Chemical Thermodynamics

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

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.

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

Option 1:

enthalpy increases but entropy decreases.

Option 2:

enthalpy remains constant but entropy increases.

Option 3:

enthalpy decreases but entropy increases.

Option 4:

Both enthalpy and entropy remain constant.

Correct Answer:

enthalpy remains constant but entropy increases.

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).

Important Questions of Chemical Thermodynamics

3. Some Basic Concepts in Chemistry

Q: 0.5 g of an impure sample of $\mathrm{CaCO}_3$ is treated with excess HCl , releasing 120 mL of CO2 at STP. Calculate \% purity of sample.

Solution:

$1 \mathrm{~mol} \mathrm{CaCO}_3 \rightarrow 1 \mathrm{~mol} \mathrm{CO}_2=22.4 \mathrm{~L}$
$120 \mathrm{~mL}=0.120 \mathrm{~L} \rightarrow$ moles CO2 $=0.120 \div 22.4=0.00536 \mathrm{~mol}$
Moles CaCO3 $=0.00536 \mathrm{~mol} \rightarrow$ mass $=0.00536 \times 100=0.536 \mathrm{~g}$
\% purity $=(0.536 \div 0.5) \times 100=107 \%$ (practically, would round $\sim 100 \%$; excess due to measurement)

Q. 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 :

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

Correct Answer:

$\mathrm{C}_3 \mathrm{H}_8$

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}
$

Important Questions of Some Basic Concepts in Chemistry

4. Hydrocarbons

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

A) Butane

B) 2-Methylpropane

C) Neopentane

D) Pentane

Solution:

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

- Only one product formed.

Answer: C

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

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

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

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).

Important Questions of Hydrocarbons

5. Organic Compounds Containing Oxygen

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

A) Tollen's Test

B) Benedict's Test

C) Both A and B

D) Fehling's Test

Solution:

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

- Ketones do not respond.

Answer: C

Q. For standardising NaOH solution, which of the following is used as a primary standard?

Option 1: Ferrous Ammonium Sulfate

Option 2: dil. HCl

Option 3: Oxalic acid

Option 4: Sodium tetraborate

Correct Answer: Oxalic acid

Solution:

As we have learned

Primary Standard -

The primary standard is a substance of known high purity that may be dissolved in a known volume of solvent to give a primary standard solution. It is a reference chemical used to measure an unknown concentration of another known chemical.

Important Questions of Organic Compounds Containing Oxygen

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

Q : The equation of a line passing through (2,3) and perpendicular to the line 3x−4y+5=0 is:

Solution:

- Slope of given line =34.

- Required line ⊥ this, so slope =−43.

- Equation: y−3=−43(x−2).

$4 x+3 y-17=0$

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

Option 1:

$28-30 \log _e 2$

Option 2:

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

Option 3:

$30-32 \log _e 2$

Option 4:

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

Correct Answer:

$30-32 \log _e 2$

Solution:

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

solve both curves,

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

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

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

2. Integral Calculus

Q: Evaluate ∫0πsin3⁡xdx.

Solution:

$\sin ^3 x=\sin x\left(1-\cos ^2 x\right)$

$\int_0^\pi \sin ^3 x d x=\int_0^\pi \sin x d x-\int_0^\pi \sin x \cos ^2 x d x$

- First integral: ∫0πsin⁡xdx=2.

- Second: let u=cos⁡x,du=−sin⁡xdx.

$\int_0^\pi \sin x \cos ^2 x d x=-\int_1^{-1} u^2 d u=\int_{-1}^1 u^2 d u=\frac{2}{3}$

So result =2−23=43.

Q.

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

Option 1:

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

Option 2:

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

Option 3:

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

Option 4:

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

Correct Answer:

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

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)

3. Limit, Continuity & Differentiability

Q: Evaluate limx→0sin⁡5x.

Solution:

$\lim _{x \rightarrow 0} \frac{\sin 5 x}{x}=\lim _{x \rightarrow 0} \frac{\sin 5 x}{5 x} \cdot 5=1 \cdot 5=5$

Q. If the maximum value of a , for which the function fa(x)=tan−1⁡2x−3ax+7 is nondecreasing in (−π6,π6), is a―, then fa―(π8) is equal to

Option 1:

$8-\frac{9 \pi}{4\left(9+\pi^2\right)}$

Option 2:

$8-\frac{4 \pi}{9\left(4+\pi^2\right)}$

Option 3:

$8\left(\frac{1+\pi^2}{9+\pi^2}\right)$

Option 4:

$8-\frac{\pi}{4}$

Correct Answer:

$8-\frac{9 \pi}{4\left(9+\pi^2\right)}$

4. Sets, Relations and Functions

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

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

Q. 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)}.

Then :

Option 1:

Both R₁ and R₂ is symmetric.

Option 2:

R₁ is not symmetric but it is transitive.

Option 3:

R2 is symmetric but it is not transitive.

Option 4:

both R1 and R₂ are transitive.

Correct Answer:

R₂ is symmetric, but it is not transitive.

Solution:

As we learned

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 the option 3.

5. Complex Numbers & Quadratic Equations

Q : If z=3+4i, find |z| and arg⁡(z).

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).

Q. 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 :

Option 1:

$-1250 \sqrt{3} i$

Option 2:

$1250 \sqrt{3} i$

Option 3:

$1365 \sqrt{3} i$

Option 4:

$-1365 \sqrt{3} i$

Correct Answer:

$-1365 \sqrt{3} i$

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.