detailed analysis of jee main 2015

JEE Main 2015 question paper detailed analysisOVERALL PAPER:

1.Overview (general theme setting and touching upon briefly on all points). We can have aJEE 2014 vs 2015 approach

2.Class XI and XII break-up of syllabus3.Difficulty level analysis - by section and overall

o Question-wise difficulty level break-upo Section-wise difficulty level break-upo Questions that should've been attempted firsto Questions that should've been avoided in first go

4.Time management analysis - by section and overallo Question-wise time spent break-upo Section-wise time spent break-upo Questions that should've been attempted firsto Questions that should've been avoided in first go

5.Special Insights and Tricks that could've been usedo 2 questions that were wrong, why?o Speed tips that could've saved timeo Trick questions?

Seemingly difficult but easy questions Seemingly easy but difficult questions

Analysis

The question papers were available in four sets- A, B, C and D.

Version Order of Subjects

A Physics, Chemistry, Mathematics

B Chemistry, Mathematics, Physics

C Mathematics, Physics, Chemistry

D Physics, Mathematics, Chemistry

Physics: : It was the toughest amongst the three; the questions that were included in this sectionwere a little tricky. Out of the 30 questions from Physics, only five questions were theory based,the rest 25 were numerical based questions. It was a time consuming section. Some experts alsobelieve that the questions in this section were ambiguous, which increased the difficultylevel. The topic of Thermodynamics in Physics was very tough, especially the Entropy questionswere quite unexpected. About 25% of the questions are pertaining to graphs. a few areas likewave optics, properties of liquids etc. were totally absent in the paper.

Ideal Time for Physics – (Student should invest 75 min. for this section)

Chemistry Majority of the questions in the Chemistry part of the JEE Main 2015 are in the rangeof easy to moderate. The overall difficulty level of this part in this year exam is less as comparedto last year. Usually, application level questions are expected from Physical and Organicchemistry areas, but this year questions from these two areas were quite straight in nature.However, one question from Inorganic Chemistry was based on practical part and it can be rateddifficult to answer. Moreover, this year's contribution from Inorganic Chemistry was morecompared to previous years. The questions were all fact based. This section was not very timeconsuming. Most of the questions are asked directly from the NCERT texts for Chemistry.

Ideal Time for Chemistry – (Student should invest 40 min. for this section)

Mathematics: This section, however, not difficult, was found a little lengthy by the examinees. Ithad a good mix of questions. Also, two questions were not properly framed in Mathsection. One was from Probability (All options were incorrect) and other was from DifferentialEquation. (Initial condition was not given). There was one question each from Statistics, SetRelation and Mathematical Reasoning and one easy questions from height and distancetrigonometry. All these chapters are not in JEE Advanced Syllabus. Questions from InverseTrigonometry ,3-D and Vectors were easy. Few questions, which posed difficulty, weregenuinely concept wise difficult and tricky in nature.

Ideal Time for Math – (Student should invest 65 min. for this section)

Analysis of JEE (Main) 2015 Paper -1

Marks Distribution from Class XII CBSE Syllabus

Subject No. of questions Total Marks

Mathematics 14 56

Chemistry 20 80

Physics 17 68

Marks Distribution from Class XI CBSE Syllabus

Subject No. of questions Total Marks

Mathematics 16 64

Chemistry 10 40

Physics 13 52

Q.No.Topic(based on SET-A) JEE mainPaper-1 Easy Moderate Difficult

1 Kinematics

embibepredicted highweightage

2 Unit & Dimensions

embibepredicted easyto score

3 Laws of Motion

embibepredicted LowWeightage

4 Conservation of Momentum

embibepredicted easyto score (MustDo)

5 Centre of Mass

embibepredictedLowWeightage

6 Rotational Mechanics embibe

predicted LowWeightage

7 Gravitation


8 Simple Harmonic Motion


9 Heat &Thermodynamics






12 Simple Harmonic Motion


13 Wave & Sound


14 Electrostatics


15 Electrostatics


16 Electrostatics


17 Electrostatics


18 Electrostaticsembibe

predicted high

weightage

19 Magnetism


20 Magnetism


21 Magnetism


22 Magnetism


23 Optics


24 Optics


25 Optics


26 Optics


27 Modern Physics


28 Modern Physics


29 Communication Systems


30 Magnetism


31 s-Block elements and Hydrogen


32 Solid State


33 Atomic Structure


34 Chemical Bonding


35 Thermochemistry and thermodynamics


36 Liquid solution


37 Chemical Equilibrium


38 Electrochemistry


39 Chemical kinetics


40 Surface chemistry


41 Periodicity


42 Metallurgy embibe

predicted highweightage

43 s-Block elements and hydrogen


44 s-Block elements and hydrogen


45 group 15 - 18


46 Transition metals


47 group 15 - 18


48 Coordination compounds


49 Coodination compounds


50 group 15 - 18


51 Stoichiometry


52General Organic Chemistry andIsomerism


53 Hydrocarbons

embibepredicted easyto score

(Must Do)

54 Alkyl and Aryl halides


55 Aldehyde and Ketone


56 Amines


57 Polymers

embibepredicted easyto score(Must Do)

58 Biomolecules


59 Chemistry in everyday life


60 Salt Analysis


61 Matrices


62 Complex Number


63 Quadratic Equation & Expression


64 Matrices embibe


65 Determinants


66 Permutation & Combination


67 Binomial Theorem


68 Progression & Series


69 Progression & Series


70 Limit,Continuity & Differentiability


71 Limit,Continuity & Differentiability


72 Straight Lines


73 Application of Derivatives


74 Indefinite Integral


75 Definite Integral


76 Area under Curves embibe


77 Differential Equation


78 Straight lines


79 Straight lines


80 Circle


81 Ellipse


82 Parabola


83 3 - D Geometry


84 3 - D Geometry


85 Vector Algebra


86 Probability


87 Statistics


88 Height & Distance


89 Inverse Trigonometric Function


90 Mathematical Reasoning


Physics – 25 Questions From High weightage and 2 from Easy to Score – (90% from the topicsmentioned important by embibe.

Chemistry – 18 Questions From High weightage and 6 from Easy to Score – (80% from thetopics mentioned important by embibe.

Math– 23 Questions From High weightage and 5 from Easy to Score – (93.3% from the topicsmentioned important by embibe.

Category Expected Cutoff Score 2015

Common Merit List (CML) 112-117

Other Backward Class (OBC - NCL) 74-80

Scheduled Caste (SC) 53-59

Scheduled Tribe (ST) 46-50

Tricks for lengthy questions –

1. A complex number z is said to be unimodular if |z| = 1. suppose z and z are complex

numbers such that is unimodular and z is not unimodular. Then the point z lies on a:

(A) Circle of radius √2(B) Straight line parallel to x-axis.

(C) Straight line parallel to y-axis.

(D) Circle of radius 2

Answer: (D)

Sol. – Put z = 0, as it is not unimodular, we can assume z = 0, so we will get as unimodular

which will give us |z | = 2. Hence Option (D) is correct.

2. The normal to the curve, x + 2xy − 3y = 0, at (1, 1):

(A) Meets the curve again in the fourth quadrant.

(B) Does not meet the curve again.

(C) Meets the curve again in the second quadrant.

(D) Meets the curve

again in the third quadrant.

Answer: (A)Its Pair of straight lines passing through origin.x + 3y = 0 or x − y = 0 so normal at (1,1) will meets the curve again in thefourth quadrant. (Draw Diagram).

3. The negation of ~s ∨ (~r ∧ s) is equivalent to:

(A) s ∧ r(B) s ∧∼ r(C) s ∧ (r ∧ ∼ s)(D) s ∨ (r ∨ ∼ s)Answer: (A)Consider it as Sets.

S = (1, 2) …..(i)r = (2, 3)U = (1, 2, 3, 4)Draw the Venn Diagram.S = (3, 4) (negesion of s)r̅ = (1, 4) (negesion of r)(r̅ ∩ S) = 1

And S U (r̅ ∩ S) -(3, 4) U (1) = (1, 3, 4)And Compliment of (1, 3, 4) is (2) which is (s ∩ r). Hence Option (A) is correct.

4. If A = 1 2 22 1 −2a 2 b is a matrix satisfying the equation AA = 9I, where I is 3 × 3 identity

matrix, then the ordered pairs (a, b) is equal to:

(A) (−2,−1)(B) (2, −1)

(C) (−2, 1)(D) (2, 1)Answer: (A)1 2 22 1 −22 1 22 1 22 −2 =

9 0 00 9 00 0 9Multiply first row of A [1 ,2,2] by last column of . We will get + 4 + 2 = 0Only option (A) satisfies it.

5. If m is the A.M. of two distinct real numbers ( , > 1) , are threegeometric means between , ℎ + 2 + equals.

(A) 4(B) 4(C) 4(D) 4Answer: (C)From Dimension Analysis (A) is incorrect as giving dimension of (length) the power 6. FromSymmetry (C) is the answer. Links.

6. Let O be the vertex and Q be any point on the parabola, = 8 . If the point P divides theline segment OQ internally in the ratio 1 : 3, then the locus of P is:

(A) = 2(B) =(C) =(D) = 2Answer: (A)

Sol. Find any one point on the locus and substitute in answers. Like (4,2) is a point on the curveand (1,1/2) divides OQ in the ratio 1:3. Hence (1,1/2) will satisfy the options.

7. The equation of the plane containing the line 2 − 5 + = 3 ; + + 4 = 5, and parallelto the plane, + 3 + 6 = 1, is:

(A) 2 + 6 + 12 = −13(B) 2 + 6 + 12 = 13(C) + 3 + 6 = −7(D) + 3 + 6 = 7Solution (D) Let the plane be + 3 + 6 = ,

As it contains the common line of line 2 − 5 + = 3 ; + + 4 = 5, we can find any pointon the line and substitute it in the our plane and find the value of K.

Put z = 0, 2 − 5 = 3 ; + = 5We will get x =4 and y=1. Put (4, 1, 0) in + 3 + 6 = , you will get K=7.

Hence (D) is the Correct option.

8. The sum of coefficients of integral powers of x in the binomial expansion of 1 − 2√ is:

(A) (2 + 1)(B) (3 + 1)(C) (3 )(D) (3 − 1)Solution: (B)It’s an easy question but if it is not striking to you then you can do it by Simulation Technique.

Put 50 as 2 in the question and options. The question will become

The sum of coefficients of integral powers of x in the binomial expansion of 1 − 2√x is:

(A) (2 + 1)(B) (3 + 1)(C) (3 )(D) (3 − 1)1 − 2√x = 1 - 4 √x + 4x.

Sum of integral powers of x would be 5. (1 + 4 )

Only Option B is correct.

9. Locus of the image of the point (2, 3) in the line (2x − 3y + 4) + k (x − 2y + 3) = 0, k ∈R, is a :

(A) Circle of radius √3(B) Straight line parallel to x-axis.

(C) Straight line parallel to y-axis.

(D) Circle of radius √2Solution: (D)The infinite lines (2x − 3y + 4) + k (x − 2y + 3) = 0, k ∈ R, are passing through (1,2) and willhave infinite images of (2,3), Obviously the locus would not be a straight line (Draw .theDiagram ). As one of the line passes through (2,3) and image of (2,3) in that line would be (2,3)

itself, therefore (2,3) would lie on the locus. And distance of (2,3) from (1,2) is √2 hence option(D) is correct.

detailed analysis of jee main 2015

Education

easy questions

year questions

firsto questions

application level questions

numerical based questions

timeo trick questions

good mix of questions

time consuming section