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JEE Advanced 2018 · Paper 1

A full paper in the 2018 Paper 1 pattern: Physics, Chemistry and Mathematics, 18 questions each, one 180-minute clock. Take it as a timed mock or open it in learn mode and check every answer.

54questions

180 minone clock

180maximum marks

3/35/16easy / medium / hard

Start timed mock Study with solutions

Sections and marking

Each subject has the same 3 sections, 60 marks per subject.

Section	Type	Questions	Marking
Section 1	One or more correct	6	+4 all correct, partial +3/+2/+1, -2 otherwise
Section 2	Numerical value	8	+3, no negative
Section 3	Paragraph based (2 sets)	4	+3 / -1

Topics in this paper

Physics · 18 questions

Rotation3
Thermodynamics1
EMI1
Wave optics1
Modern physics1
Waves1
Projectile motion + momentum conservation1
Fluid dynamics1
SHM1
Electrostatics1
Current electricity1
Magnetism1
AC circuits1
Thermal1

Chemistry · 18 questions

Coordination compounds1
Aromatic reactions and ethers/phenols1
Chemical equilibrium1
p-Block oxoacids1
Stereochemistry of alkene addition1
Electrochemistry1
Ionic equilibrium (Ksp and buffers)1
Colligative properties1
Chemical bonding and molecular shapes1
Stereoisomerism1
Magnetic properties of complexes1
Thermodynamics (entropy)1
Solid state1
Quantitative organic analysis1

Mathematics · 18 questions

Matrices1
Integral equations1
Conditional probability1
Ellipse1
3D geometry1
Hyperbola1
Definite integrals1
Differential equations1
Series1
Trigonometric equations1
Expectation1
Circle and tangent to parabola1
Combinatorics1
Limits1

Sample question from this paper

Mathematics · Matrices · one or more options correct

Let

A=\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}

and let

I

be the

3\times3

identity matrix. Which of the following statements is (are) correct?

A. The $(1,3)$ entry of $A^{10}$ equals $45$
B. The $(1,3)$ entry of $A^{-1}$ equals $-1$
C. The sum of all nine entries of $A^{4}$ equals $17$
D. $(A-I)^2$ is the zero matrix

Show answer and solution

Answer: A, C
Write

A=I+N

with

N=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}

, so

N^2

has a single entry

1

at position

(1,3)

and

N^3=0

. Since

I

and

N

commute,

A^n=I+nN+\binom n2N^2

. (A) The

(1,3)

entry of

A^{10}

\binom{10}{2}=45

, so it is true. (B)

A^{-1}=I-N+N^2

because

(I+N)(I-N+N^2)=I+N^3=I

, hence the

(1,3)

entry is

+1

, not

-1

, so it is false. (C)

A^4=I+4N+6N^2

has entry sum

3+4\cdot2+6\cdot1=17

, so it is true. (D)

(A-I)^2=N^2\neq0

, so the minimal polynomial is

(x-1)^3

and the statement is false.

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