IMP QUESTIONS 4 10th MATHS

LAUNCHING NEW SCHOOL IN A.P

TARGET IIT/IAS SCHOOLS

Most Important 4 Mathematics in 10th

1. Statements and sets

1. Disjunction, conjunction, conditional, bi conditional truth tables definitions

2. Converse, inverse, contra positive of a conditionals

3. Page number 6 in mathematics book example 1 page number 7 total truth

Tables. And in page number 8 example 2: (~p)^q, ~(p^q)

4. Exercise -1 10^th bit and 12^th

5. Tautologies contradiction and page number 10 and 11 examples 4, 5, 6

6. If x is even then proof that x² is even.

7. Page number 16 example 2

8. Page number 19 example 1 AU (B∩C)= (AUB)∩(AUC), page number 20

Examples 2,3 page number 21 total examples up to example 10.

9. Exercise -5 3^rd bit to 8^th bit are most imp.

2. Functions mappings

1. Page number 33 examples 5, 6, 7, 8, 9

2. exercise-2 3^rd bit to 8^th bit

3. Exercise -5 2^nd bit to 12^th bit.

3. Polynomials over integers

1. Exercise -1: is important for all competitive exams (Polytechnic, IIT, and TTC)

2. Exercise -2:1st bit and 9^th, 10th, 11th, 12th bits

3. Exercise-3: 1,2,3,8 important.

4. Exercise -5 graphs .

5. Exercise-8 total important for future classes

4.Chapter -4 only for objective

5. Real numbers Chapter -5

1. Exercise -1 for objective

2. exercise-2 12^th bit to 16^th bit most important

3. exercise-7 is most important.

6. Progressions

1. exercise-1 18^th to 30 are very important

2. exercise-2 8th, 16^th, 17th, 18^th, 19^th, 20^th, 21^st.

3. Exercise-3 total imp.

4. exrcs-4 9^th to 14^th imp

5. exercise-5 7^th to 18^th are very imp.

6. Exercise -7 is imp

First paper over if you practiced above questions then you will get in public examination 50/50

Now 2^nd - paper

1. Geometry

Only theorems

2. Analytical geometry

1. Exercise -1: short answer type questions are total imp. Mainly 1, 2, 4, 5, 6, 7, 8, 9, 10,

12, 16, 20.

2. Exercise -2: 1, 2, 3, 4,5,6,9 and in long type 1, 3, 4, 5, 8, are imp

3. exercise-3 1, 2, 4, 5, 6, 10

4. Exercise -4 1, 2, 3, 4, 5, 10

In long type 1, 2, 3, 4, 6, 8, 10

Most important Trigonometry

1. exercise-1: and exercise -2 for only knowledge purpose

2. exerciser-3: 10,11, 12, 13, 14, 15, in long type 1, 3, 8, 9, 10, 12

3. exercise -4: total important for public examination and for future competitive examinations.

4. exercise-7: 2,3, 6,7, in long type 1,2,3,4,5,6,7,8,9,10 most imp.

Statistics

1. All examples and example model exercise questions imp this is very easy chapter.

Matrices

Multiplication, additions, subs tractions, various definitions, determinants, Cramer’s, inverse imp.

Computing (Very easy)

questions will give you 100% marks all the best.

Please follow the all in one for objective questions.

The above questions will give you 100% marks all the best.

any queries contact Narendra - 9848696955

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TOPICS:
ANALYTIC GEOMETRY:
Equation of Line Click here
distance of a point from a line Click here
angle between straight lines Click here
family of lines Click here
bisectors of angle between two lines Click here

Puzzle And Fun Zone

Puzzle And Fun Zone

The river and the bridge puzzle.
This is a famous puzzle which baffled a lot of mathematicians for a long time. This single puzzle is considered by some to be the one that led to the formulation of "Graph Theory".

The town of K'gsberg in Prussia (now in Russia) consists of an island and some land along the river banks. All land masses that comprise the town are connected by seven bridges. With This situation, people tired to see if it was possible for a person to take a walk through the town, starting and ending at the same pint, crossing each of the seven bridges exactly once each. Can you find that path?


Solution will be given in the next Edition! So Keep Waiting...

IIT JEE Part Test 2

IIT JEE Part Test 2

Instructions:

For straight objective type with only one correct answer. +3 for correct answer, -1 for wrong answer, 0 for not attempting

Reasoning type
If STATEMENT-I is true and STATEMENT-II is correct explanation of STATEMENT-I mark (a)
If STATEMENT-I is true and STATEMENT-II is true but not the correct explanation of STATEMENT-I mark (b)
If STATEMENT-I is true and STATEMENT-II is false mark (c)
If STATEMENT-I is false and STATEMENT-II is true mark (d)
+3 for correct, -1 for wrong, 0 for not attempting

Linked comprehension type. Answer the questions according to the information given. Only one option is correct. +4 for correct answer and -1 for wrong answer, 0 for not attempting

Match the columns +1 for correct and -1 for wrong

Q1) Straight Objective Type

find the summation of the series to infinite terms
1/3 + 1/3 + 7/27 + 5/27 + ....

a) 1
b) 3/2
c) 2
d) none of these

Q2) The greatest value of x²y³z⁴ (x+2y+3z=1, x,y,z>0) is

a) 2¹⁰/3¹⁵
b) 2⁷/3¹⁹
c) 2⁶/3¹⁸
d) 2¹³/3¹²

Q3) For the expansion (1+x)¹⁰
C₀+C₃+C₆+C₉=

a) 340
b) 341
c) 342
d) none of these

Q4) for x,y in R (4xy?1)



find f'(1)

a) 1
b) 1/2
c) 1/5
d) 2/5

Q5)

a) 1
b) 1/e
c) e
d) none of these

Q6) find the area bounded by the curves
x²+y²=4
x²+y²-2|x|-2|y|+4=0

a) 4p
b) 8p-8
c) 4p+2
d) 8p-16

Q7) Let x,y be the real numbers satisfying the equation x²+2x+y²+2y-47=0
If the maximum and minimum values of x²+y²-4x-6y+13 are M and m respectively, then the numerical value of M-m is

a) 14
b) 10
c) 5
d) none of these

Q8) Let a,b,c be the unit vectors such that a+b+c=0. Then the area of the triangle formed by a,b and c is

a) 0
b) 1 units
c) can't be determined
d) none of these

Solution to IIT JEE Part Test 2

1 (b) it can be written as1/3 + 3/9 + 7/27 + 15/81....(2ⁿ-1)/3ⁿ=(2/3)ⁿ-1/3ⁿ(2/3)*1/(1-2/3) - (1/3)*1/(1-1/3)2-1/2=3/2

2 (b) AM??GM(x/2+x/2+2y/3+2y/3+2y/3+3z/4+3z/4+3z/4+3z/4)/9 ??[(x/2)²(2y/3)³(3z/4)⁴]^1/9

3 (b) putting x=1: C₀+C₁+C₂+.......+C₁₀=2¹⁰putting x=?? C₀+C₁??C₂??sup>2+.......+C₁₀??sup>10 =(1+??¹⁰=??sup>20=??sup>2putting x=??sup>2: C₀+C₁??sup>2+C₂??sup>4+.......+C₁₀??sup>20 =(1+??sup>²)¹⁰=??sup>10=??adding all three we get3(C₀+C₃+C₆+C₉)=2ⁿ+????sup>2= 2¹⁰-1Hence C₀+C₃+C₆+C₉=(2¹⁰-1)/3=1023/3=341

4 (d) putting x=y=0 we get f(0)=0putting y=-x we get f(x)+f(-x)=0 f'(x)=lim(h??) (f(x+h)-f(x))/h=(f(x+h)+f(-x))/h given f(x)=-f(-x)=f[(x+h-x)/(1+4(x+h)x)]=f[(h)/(1+4x²+4hx)]/h=As h tends to zerof'(x)=2/(1+4x²)f'(1)=2/5

5 (b) it is e^[tan(??/2)lntan(??/4)][tan(??/2)lntan(??/4)]=lntan(??/4)/cot(??/2)Using L'Hospital rule[sec²(??/4).??4]/[-tan(??/4).cosec²(??/2)]=-1

6 x²+y²-2|x|-2|y|+4??0is not a real circle this is being cancelled. For this circle solution isx²+y²-2|x|-2|y|-4??0brown area is to be found area of yellow + area of brown+area of yellow=2*2=4 units area of yellow + area of brown=??²/4=??rea of brown 2??4area will be in all 4 quadrants A=4(2??4)=8(??2)

7 (d) x²+y²-4x-6y+13=(x-2)²+(y-3)²P(2,3) lies inside the circlelet the distance between centre of circle and P be xlet M'=r+x and m'=r-xM=M'²=r²+x²+2rxm=m'²=r²+x²-2rxM-m=4rx=4*7*5=140

8 (d) It will be an equilateral triangle with the side length 1so area=??/4 units

9 (c) seating can be as 1,3,5; 1,3,6; 1,3,7; 1,3,8; 1,4,6; 1,4,7 ; 1,4,8; 1,5,7; 1,5,8; 1,6,8total no. of ways=103 seats can be chosen is ⁸C₃=56 ways probability is 10/56=5/28

10 (c) The line itself is a tangent to the circle

11 (a) [aX(bXc)]=(a.c)b-(a.b)c(bXc).[aX(bXc)]=(bXc).[(a.c)b-(a.b)c]=0

12 (c) |a+b+c|²??03+2(a.b+b.c+c.a)??0(a.b+b.c+c.a)??(-3/2)|a-b|²+|b-c|²+|c-a|²=2[(a.a)+(b.b)+(c.c)] -2(a.b+b.c+c.a)=6-2(a.b+b.c+c.a)

13 (a) the tangents are x??a=0

14 (b) A²-4A-5I=0

15 (a) A²-4A-5I=0multiplying by A^-1A-4I=5A^-1

16 (c) X=A^-1B

17 (d) sin^-1sin^-1x+cos^-1cos^-1xfirst one exists for[-0.84,0.84] and second one for [0.54,1]it exists for [0.54,0.84][??5,??4] lies in this interval

18 (d) it is true for x=1/??=0.707

19 (d) sin^-1sinx+cos^-1sinxsin^-1sinx+cos^-1cos(??2-x)x+??2-x=??2

20 (a) shifting the centre of circle to origin circle becomes x²+y²=25 and chord is x+y=5so coordinates are O(0,0), P(5,0), Q(0,5) and R(5,5)area OPQ=PQR=25/2centroid of OPQ (5/3,5/3) and that of PQR is (10/3,10/3)shifting origin back centroid of OPQ(14/3,-7/3) and of PQR (19/3,-2/3)

21 (a)

22 (d)

23 (c)

24 (b) first oneI=??sinxdx even function I=(0 to pi) 2??sinxdx=2??pi-x)sinxdx2I=2??i.sinxdxI=(0 to pi) pi??inxdxI=(0 to pi/2) 2pi??inxdx=2pisecond odd function so it is zerothird(-pi to 0){sinx}dx+(0 to pi){sinx}dx(-pi to 0)(1+sinxdx+(0 to pi)sinxdx(-pi to 0)1dx+(-pi to 0)sinxdx+(0 to pi)sinxdx(-pi to 0)1dx+(-pi to pi)sinxdxsecond is odd function so zero(-pi to 0)1 dx=pifourth one(-1 to 0) (-1)dx+(0 to 1)0.dx+(1 to 2)1dx=0

25 (d)

26 (a)

27 (d)

28 (b) A=30??, B=60?? and C=90??tan(B/2)tan(C/2)+tan(C/2)tan(A/2)+tan(A/2)tan(B/2)=1 for any angles4sin(A/2)sin(B/2)sin(C/2)=cosA+cosB+cosC-14cos(A/2)cos(B/2)cos(C/2)=sinA+sinB+sinCcot15??=2+??

29 (a)

30 (c)

31 (d)

32 (b) let x length of rod is inside watertaking moment about corner you can find the answerVdg.x/2=mg.l/2xAdg.x/2=lA??.l/2x²=l².(??d)=l²/2=1/2x=1/??height of water=1/2??cos^-1(1/??)=45??

33 (b) current at time t=0 is through inductor=6A-Ldi/dt=(R₁+R₂)i=-4idi/dt=-10idi/i=-10dtln(i₂/i₁)=-10tI=6e^-10t

34 (b) 1/u + 1/v =1/f-1/30 +1/v=1/20v=12 cm from polev'=42 V=42*2/3=28 cm

35 (a) 1.5/v1 -1.75/?? =-(1.5-1.75)/R1.5/v1=0.25/Rv1=6R1.75/v - 1.5/v1=(1.75-1.5)/R1.75/f -1.5/6R=0.25/R1.75/f=1/2Rf=3.5R1.75/f=(1/R)(1/4)

36 (c) PE=-Ke²/rKE=(1/2)mv²=Ke²/2rTE=PE+KE=-Ke²/2rso |KE|=|TE|TE in the first excited state=-13.6/4=-3.4 eVKE=3.4 eV

IIT JEE Part Test 1

IIT JEE Part Test 1

Instructions:

For straight objective type with only one correct answer. +3 for correct answer, -1 for wrong answer, 0 for not attempting

Objective type with one or more than one correct answer. +4 if all the correct answer are marked, there is no negative mark in this section

Reasoning type
If STATEMENT-I is true and STATEMENT-II is correct explanation of STATEMENT-I mark (a)
If STATEMENT-I is true and STATEMENT-II is true but not the correct explanation of STATEMENT-I mark (b)
If STATEMENT-I is true and STATEMENT-II is false mark (c)
If STATEMENT-I is false and STATEMENT-II is true mark (d)
+3 for correct, -1 for wrong, 0 for not attempting

Linked comprehension type. Answer the questions according to the information given. Only one option is correct. +4 for correct answer and -1 for wrong answer, 0 for not attempting

Q1) Straight Objective Type

=

a) π/2√2
b) π/√2
c) 1
d) π/4√2

Q2) Find lim(n→∞) 1/√(n²+n) + 1/√(n²+2n)+.....+ 1/√(n²+2n²)

a) 1
b) 2(√2 - 1)
c) 2(√3 - 1)
d) 2√2 - 1)

Q3) The number of different matrices that can be with the numbers 1,2,3,4 each having four elements is

a) 4⁴
b) 2.4⁴
c) 3.4⁴
d) none of these

Q4) The value of cot^-1(-3)+cot^-1(-2)=

a) π/4
b) 3π/4
c) 5π/4
d) none of these

Q5) a²+b²+c²=ca+ab√3
then the triangle is

a) equilateral
b) right angled
c) isosceles
d) none of these

Solution to IIT JEE Part Test 1

1 (d) putting x=tanθit simplifies tofirst terms simplifies to cos2θ 1+tan⁴θ=(sec⁴θ)(sin⁴θ+cos⁴θ)=(sec⁴θ)[(sin²θ+cos²θ)²-2sin²θcos²θ]=(sec⁴θ)[1-sin²(2θ)/2]so it becomesnow put sin2θ=√2sint and solve

2 (c) it can be written as

3 (c) four places can be filled up in 4*4*4*4=4⁴ places matrices possible are 2X2, 4X1 and 1X4so total no. of ways=3.4⁴

4 (d) cot^-1(-3)+cot^-1(-2)=π-cot^-13+π-cot^-12=2π-(cot^-13+cot^-12)=2π-π/4=7π/4

5 (b) a²+b²+c²=ca+ab√3(√3a/2-b)²+(a/2-c)²=0so √3a=2b=2√3c=k(assume)b²+c²=a²

6 (c) let A is (3,-4) and B be (5,-2)slope of AB=1, so the diagonals are parallel to the axes let O be the centre then O will be at distance 2 from A i.e. O is (3,-2)

7 (d) , (c) from second|Z-4|=|Z-8|implies Re(Z)=6from first|Z-16i|/|Z-8i|=5/3put Z=6+iy

8 (c) , (b) , (a) y(2x³-y²)dx+x(y²-x³)dy=0x³(2ydx-xdy)+y²(xdy-ydx)=0((2x/y)dx-(x²/y²)dy+((1/x)dy-(y/x)dx)=0d(x²/y)+d(y/x)=0d(x²/y)+(y/x))=0x²/y+y/x=C(1,1) lies of the curve so C=2(a), (b), (c) lies of this curve

9 (b) , (a) shifting origin to (2,-1) we getX=x-2 and Y=y+1circle as X²+Y²=5 and hyperbola as X²-5Y²=125the equation of common tangent be Y=mX?√(5(1+m²) and Y=m'X?√(125m²-25)now m'=m and 5(1+m²)=25(5m²-1)1+m²=25m²-524m²=6m=?1/2so the tangents are2Y=?X?5so the equation are2(y+1)=?(x-2)?5

10 (b) , (a) π/2-cos^-1x+π/2-cos^-1(1-x)=cos^-1x2cos^-1x=π-cos^-1(1-x)cos^-1(2x²-1)=cos^-1(x-1)2x²-1=x-1

11 (d) the solution will be 2π which is not in the domain RHS≥2LHS≤2so the equation will be 2sin²(x/4)cos²(x/2)=2sin²(x/4)=1 and cos²(x/2)=1

12 (d) it is not always isosceles it can be right angled toosin2A-sin2B=02cos(A+B)sin(A-B)=0so either A=B or A+B=π/2

13 (b) f'(x)=2xsin(1/x)-cos(1/x)f'(0)=lim(h→0)[f(0+h)-f(0)]/h=[h²sin(1/h)]/hlim(h→0)hsin(1/h)=0

14 (d) This point is not the closest as the line intersects the curve so minimum distance=0and point will be the intersection of line and curve

15 (c) The given equation is an ellipse with focus (?(2/3)√10,0) and a=√5ae=(2/3)√10e=(2/3)√2AlternatelyThe given equation is an ellipse so the eccentricity<1hence>

16 (c) it is square with coordinates z=?1/√2?i/√2area=2

17 (c) z=?1/√2?i/√2these are the roots of x⁴+1=0

18 (c) let t₁ and t₂ be P and Q then slope of OP=2/t₁ and that of OQ=2/t₂OQ is perpendicular to OPso t₁t₂=-4equation of PQ is (y-4t₁)/(x-2t₁²)=2/(t₁+t₂)putting y=0 it gives x=8so the point is (8,0)

19 (c) 2x=2(t₁²+t₂²) and 2y=4(t₁+t₂)x=t₁²+t₂²=(t₁+t₂)²-2t₁t₂x=(y/2)²+8

20 (c) A²=|(1/4)*(4t₁⁴+16t₁²)*(4t₂⁴+16t₂²)|A²=4(t₁t₂)²[16+4(t₁²+t₂²)+(t₁t₂)²]A²=64*4[t₁²+t₂²+8]A²=64*4[(t₁+t₂)²-2t₁t₂+8]A²=64*4[(t₁+t₂)²+16]for minimum area t₁+t₂=0A=8*2*4

21 (c) The plane passes through (-1,-2,-1) and normal of plane is perpendicular to the lines so the equation of plane isa(x+1)+b(y+2)+c(z+1)=0where3a+b+2c=0a+2b+3c=0

22 (d) The minimum distance is the distance between the line L₂ i.e. point(2,-2,3) and plane P₁|2-14-15+10|/√75

23 (c) The plane passes through (2,-2,3)and contain normal of the plane P₁ let the equation be a(x-2)+b(y+2)+c(z-3)=0DR of normal of lines is (1,7,-5)a+2b+3c=0a+7b-5c=0

24 (c) let x be the compression(1/2)kx²=mg(x+h)x=0.1 m

25 (b) Let the pressure at layer between 2ρ and 3ρ is P₁ and that between ρ and 2ρ is P₂P₂=P₀+ρg(h/2)=P₀+ρgh/2P₁=P₂+(2ρ)g(h/3)=P₂+2ρgh/3=P₀+ρgh/2+2ρgh/3P₀+(1/2)(3ρ)v²=P₁+(3ρ)g(h/4-h/6)=P₁+ρgh/4=P₀+ρgh/2+2ρgh/3+ρgh/4=P₀+(17/12)ρghv²=(17/18)gh

26 (c) heat required for ice to reach 0?C=1*0.5*20=10 cal heat required to melt ice=1*80=80 cal90 cal of heat is extracted from water, let the temperature of water be T5*1*(30-T)=90T=12?Ctwo liquids at temperature of 12?C and 0?C mix so the final temperature is5*1*(12-T)=1*1*TT=10?C

27 (c) current in the circuit is v/3Rapplying kirchoffs law in above loop iR+v-v-v'=0v'=iR=v/3

28 (c) v/u=v/f - 12=-x/f - 13=-(x+25)/f -1f=-25 cm and x=75 cm1/u₁=1/f - 1/v=-1/25 + 1/75=-2/751/u₂=-1/25 + 1/100=-3/100u₂-u₁=75/2 - 100/3 =(225-200)/6=25/6

29 (d) μsin30=sinθ=0.72θ=[ 1 ]+30[ 1 ]=θ-302[ 1 ]=2(θ-30)

30 (d) , (a) Let f1 be the friction between the plank and cylinder and f2 between cylinder and ground let the acceleration of CM of cylinder be af1+f2=m₁a=20a(f1-f2)R=Ia/R=m₁aR/2=10aRf1=15a and f2=5aF-f1=m₂*2a=10a50-15a=10aa=2 m/s²

31 (d) , (c) after collision velocity of A is vv(CM)=v/2for maximum compression, velocity of A=velocity of B=v/2KE of system=m(v/2)²=mv²/4mv²/4+(1/2)kx²=(1/2)mv²x=v√m/2k

32 (c) , (b) when entering, the area of the portion of loop in the field increases so current will be anticlockwise to oppose the increase in magnetic flux it is zero when loop is inside the field and clockwise when leaving the field

Short Cuts and Tricks for IIT JEE!

Short Cuts and Tricks for IIT JEE!

Before reading further, short cuts and tricks can only help you to get those extra marks in IIT JEE. It will never help you get a selection. You can at best only expect to marginally save time and expect a few more marks. If you are reading this article for the heavenly potion that will get you through IIT JEE, read no more.

The whole analysis here is based on examples on how to solve a few problems! Your mind should be very alert to see things that are not ordinarily visible to a lot of other students! This specially helps in Mathematics.

Example 1:
One of the most common tricks in all summation questions is to put n=1.

IIT JEE 2000,equals:

A)

Now In this question, simply put n=r=2. We will get 1+2.2+1=6

The four options are respectively 3, 6, 12, and 6. So A and C are out of option! Now put n=3 and r=2 you will have solved this one. Isnt it?

Example 2:

Now lets do some differentiability questions! When you are given to find the differentiability of a function involving |x| or some term like |x-2| etc then simply check the derivative at the point of contention. If it is not zero, then the function is not differentiable! Lets see an example!

y=||x|-1| (IIT JEE 2005)

Here, the points of contentions are clearly -1,0,1. Remove the || sign. Just take y=x-1. Is the derivative at any of these points zero? No because derivative is 1! dy/dx=d(x-1)/dx=1. No need of checking anything else! This works in 99% of cases!

Another example: IIT JEE 1999: The function is not differentiable at..

Now see there are 3 points of contention, 1,2 and 0. Take 0 first .. it is to come from cos(|x|). Simply see the derivative of cos x at zero. It is sin 0=0 . Hence at zero it is differentiable!

Now look at its derivative at x=1 is zero! so the original function is differentiable at x=1!! Also derivative is not zero at x=2. Hence the original function is not differentiable at x=2! Verify it yourself!

Why this happens if very logical. dont think that there is any rocket science applicable here!

Example 3

There are questions where the function is split into many parts and you have to analyze the differentiability of the function at the point where the form is different . Example

f(x) = sinx; x>0
x; x<0

You have to check if the function is differentiable. What do you do? First of all check the continuity. Then no need to do any dirty limits! Just find the derivative of both sides. Derivative of x+ will be cos 0 = 1. Derivative of x- will be 1. Both are equal . Hence if the function is continuous, then it is differentiable!

IITJEE 2001 Question:
f:R->R is a function defined by f(x) = max {x, x^3}. The set of points where f is not differentiable is ?
Now the points of contention are x=x^3 the points given by x=1,-1 and 0. Check the derivative of x and x^3. 1 and 3x^2. Which are not equal for 1 or -1 or 0. Hence it is not differentiable at any of these three points!

Example 4 Find ¹²C₆
A. 924 B. 994
C. 1150 D.842
The Rank 5000 method: 12x11x10x9x8x7/(6x5x4x3x2x1) = 11x12x7=924
The Rank 500 method: It is a multiple of 11! which is the odd terms minus even terms should be a multiple of 11

Example 5. Find the value of
1/2!+1/4!+1/6!+....... (AIEEE 2004)

A. (e²-1) / 2 B. (e²-2) / e
C. (e²-1) / 2e D. (e-1)² / 2e

If you thought this was unsolvable if you did not know the expansion of e^x..... Think again! It is .. The first two terms are the only non negligible terms!

= 0.5 + 0.05 = 0.55

A .(e^2-1) / 2 > 1 easily
B .(e^2-2) / e > 1 again
C.(e^2-1) / 2e > 1 again!
D.(e-1)^2 / 2e was the only option with value less than 1!


Example 6The coefficient of xⁿ in the expansion of (1+x)(1-x)ⁿ is (AIEEE 2004)
A. (n-1) B. n (-1)^n-1
C. (n-1)²(-1)^n-1 D. (1-n)(-1)ⁿ

Easiest method: n=0
New Question: Co-efficient of x⁰ in the 1+x
the co-efficient of 1 in this expression is 1.

A. (n-1) = -1
B. n (-1)^n-1= 0
C. (n-1)²(-1)^n-1= -1
D. (1-n)(-1)ⁿ = 1

http://reflections.targetiit.com/2009_04/2.html

Last Week before IIT JEE

You have only one aim now. Maximize your Rank at IIT JEE. All that you have to do in the next 15 days have to focus on how to score more! You don't have to study for the sake of loving Mathematics or Physics or Chemistry. This is not a time for bravery. It is a time for you to focus singly on scoring more! Stop studying the subject that you love. It is time to start studying subject that you can make maximum difference in marks and the simple answer is the subject that you have studied least!

• Read as many different things as possible. don't get stuck with a few things.
• The best problems to solve are Past IIT JEE questions. If you have not solved them yet, make sure you do that
• Make a list of topics that you are half confident and by the help of someone or the other, make sure that you complete the full syllabus of those topic. It could be a big differentiator if you are able to solve even 2-3 of those questions.
• This is not the time for cramming new material, but to make sure that what you know is perfect. There is infinitely too much to study so it does not matter if you don't know some of the syllabus. Just make sure that you know perfectly what you do.
• Review concepts. Look over your summary notes if you have made them. Just recall all the tricks that you must have used.
• Try to judge for yourself which topics you have prepared well. There will be topics that you simply don't and have never solved even a single problem on. Make sure not to even touch it! Unless you have a very good friend who volunteers to help you out with that topic in two/ three hours!
• Body Clock! Make sure your body clock is inline with the actual examination time. Sleep early on the last 2-3 days at least, so that you don't have issues sleeping in time on the previous night of IIT JEE. It will be disastrous if you could not sleep till 3 in the morning and would have to go to give the exam without at least 4 hours sleep. Ideally a 6-8 hour sleep is recommended!
• Take some time each day to relax. Have a good meal. Take a walk in the fresh air. Find time for exercise. The change of pace will refresh you and the physical activity will help you relax and sleep at night.
• Solve as many problems as you can. don't get study solving one problem for 2 hours. There is not that much time. It is better you solved 100 problems in an hour by just going over them one by one in your mind than solving 1 very tough problem in an hour. The time for doing all that is behind you.