Messages posted by KAB

Yes harshiit.The answer cannot be -root2.

Even 2root2 cannot be the answer.

Use this link http://www.padowan.dk/graph/ to download free graphing software!!!!!!!!

We CANNOT get an area of pi/2 with lower limit root2

If the upper limit is root2 then lower limit shud be 1.00000001472 (Approximately) in order to get an area of pi/2

So the problem itself is wrong!

Check the graph.Click on it and save it to get a clear picture.

You can use the same graphing software to find out area,draw tangents etc.

Eg:

Consider (x-1)(x-2)(x-3)

Now expanding it we get x³-(1+2+3)x²+(1*2+2*3+3*1)x-6

Similarly when we expand (x-1)(x-2).......(x-100)

We get x¹⁰⁰-(1+2+3....100)x⁹⁹+(Sum of product of all first 100 integers taken two at a time)x⁹⁸+................

Given g(x)=f ^-1(x)

or f(g(x))=x........(1)

Now f(x)=_{[ 0]}^{[x ]} 1/root(1+t³)dt

So f '(x)=1/root(1+x³) [Newton leibnitz rule]

Now differentiate (1) wrt x

f '(g(x))g '(x)=1

ie g '(x)/root(1+(g(x))³)=1

g '(x)=root(1+(g(x))³)

or g ''(x)=3(g(x))² g '(x)/2(root(1+(g(x))³))

or g ''(x)=3(g(x))²/2

So g ''(x)/(g(x))²=3/2

So the answer is a)

The given series is,

(x+1)(x+4)(x+9)....(x+400)

=(x+1)(x+2²)(x+3²)......(x+20²)

=x²⁰+(1+2²+3²+....+20²)x¹⁹+........

So co-efficient of x¹⁹ is 1+2²+3²+......+20²

=20*21*41/6 (Since 1+2²+3²+......+n²=n(n+1)(2n+1)/6)

=2870

No. of onto functions from set A(containing n elements) onto a set B(containing r elements) is equal to the no. of ways of dividing n things into r groups so that no group remains empty if n>=r and zero if n<r.

So the no. of onto functions= rⁿ-^rC₁(r-1)ⁿ+^rC₂(r-2)ⁿ+.....+(-1)^r-1 ^rC_r-1

Here r=n

Total no. of functions from S to S is nⁿ

So required probability=(nⁿ-ⁿC₁(n-1)ⁿ+ⁿC₂(n-2)ⁿ+.....+(-1)^n-1 ⁿC_n-1)/nⁿ

The given series T=2^{1/2+2/4+3/8+4/16+..........}

Now S=1/2+2/4+3/8+.........is an AGP upto infinity with d=1 and r=1/2

So T=2^{(1/2)(1+2/2+3/4+4/8+.........)}

=2^(1/2)(2+2)

=4

If we expand (x-1)(x-2).......(x-100) we will get

(x-1)(x-2).......(x-100)=x¹⁰⁰-(1+2+3....100)x⁹⁹+(Sum of product of all first 100 integers taken two at a time)x⁹⁸+................

Now we have to find "Sum of product of all first 100 integers taken two at a time" which is the required co-efficient.

(1+2+3+....100)²=1²+2²+3²+.....+100²+2(Sum of product of all first 100 integers taken two at a time)

So Sum of product of all first 100 integers taken two at a time=((1+2+3+....100)²-(1²+2²+3²+.....+100²))/2

Now use 1+2+3+....+n=n(n+1)/2 and 1²+2²+3²+.....+n²=n(n+1)(2n+1)/6

Here n=100

So we get the required co-efficient=12582075

When Q₃ is at R net charge at R is Q₁+Q₃

So potential energy E₁=K(Q₁+Q₃)Q₂/d where K=1/4*pi*E₀

When Q₃ is at S net charge at S is Q₂+Q₃

So potential energy E₂=KQ₁(Q₂+Q₃)/d

So change in P.E.=E₂-E₁

=KQ₃(Q₁-Q₂)/d

Q2)

The first term of an AP a_1,a_2,a₃,.......... is equal to unity .

At what value of common difference is a₁ a₃+ a₃ a₂is minimum ?

Given a₁=1

Let d be the common difference.

Then a₂=1+d and a₃=1+2d

Let S=a₁ a₃+ a₃ a₂

So a₁ a₃+ a₃ a₂=1+2d+1+2d+d+2d²

or S=2d²+5d+2

So S to be minimum dS/dd=0

So 4d+5=0

or d= -5/4

Q1)

Show that for an AP a₁, a₂ , a₃ .....................

S = a²₁- a²₂+ a²₃- a²₄+ .............+ a²_2k-1 - a²_2k

= [k/(2k-1)] (a²₁- a²_2k)

Solution:

S=(a₁-a₂)(a₁+a₂)+(a₃-a₄)(a₃+a₄)+................+(a_2k-1-a_2k)(a_2k-1+a_2k)

Let common difference of the series be d.

So d=a₂-a₁=a₃-a₂=.......=a_2k-a_2k-1..................(1)

Also a_2k=a₁+(2k-1)d

So -d=(a₁-a_2k)/(2k-1).................(2)

So S= -d(a₁+a₂+....+a_2k) (From (1))

= -d((2k/2)(a₁+a_2k))

=(k/(2k-1))(a₁²-a_2k²)

(From(2))

Let Primary waves travel distance d in time t.

Then d=8t

or t=d/8..............(1)

Secondary waves travel same distance in time t+60

So d=5(t+60)

d=5d/8+300 (From (1))

or d=800km

q1)

Since f(x) is the least degree polynomial such that nf(n)-1=0 when n=1,2,3,4,5

we can say nf(n)-1=t(n-1)(n-2)(n-3)(n-4)(n-5) where t is a constant.

When n=0 we get t=1/120

So f(x)=(1+((x-1)(x-2)(x-3)(x-4)(x-5)/120))/x

When x=0 f(0) is 0/0 form.So we must find the limiting value.

So lim(x -->0)f(x)=(1/120)(120+60+40+30+24) (By L'Hospitals rule)

=274/120

=137/60

So f(0)=lim(x --->0)f(x)=137/60

Yaar relax!You will only lose 1 mark for it.It is given in the brochure.There is absolutely no possibility of answer sheet being cancelled!

In what form do you want the final answer?????

Do you want to express (99 - 70

2)^1/3as x^1/3- y^1/3

Let (99 - 70

2)^1/3=x^1/3- y^1/3

Then 99 - 70

2=x-y-3(xy)^1/3(x^1/3- y^1/3)

=x-y-3(xy)^1/3(99 - 70

2)^1/3

So x-y-99+70

2=3(xy(99-70

2))^1/3

So (x-y-99+70

2)³=27xy(99-70

2)

Simplify LHS and equate the co-efficients of rational and irrational terms.

You will get two equations in x and y.Solve them and you will get x^1/3- y^1/3

You can also use binomial expansion to find (99 - 70

2)^1/3 if you want an approximate answer.

If the question was (99 - 70 x 2^1/2)^1/2then the solution can be easily obtained by this method.

The reaction is as follows:

Phenol is dissolved in sodium hydroxide solution to give a solution of sodium phenoxide.

The solution is cooled in ice, and cold benzenediazonium chloride solution is added. There is a reaction between the diazonium ion and the phenoxide ion and a yellow-orange solution or precipitate is formed.

The product is one of the simplest of what are known as azo compounds, in which two benzene rings are linked by a nitrogen bridge.

Since phenol should lose H⁺ ion ,basic medium is required.

Here is the actual method.

ab=a+b

So a²b²=a²+b²+2ab

So a²+b²-a²b²= -2ab..........(1)

Now a/b+b/a-ab=(a²+b²-a²b²)/ab

= -2ab/ab(From 1)

= -2

By inspection a=2 and b=2

So a/b+b/a-ab=1+1-4= -2

Use the equation

=(D-d)/((n-1)d)

Where ,

is the dissociation constant

D is the vapour density of phosphine before dissociation

d is the vapour density of phosphine after dissociation.

n is the no. of moles of gaseous products obtained by the dissociation of 1 mole of reactant.

Here PH₃

(1/4)P₄+(3/2)H₂

Here D=(31+3)/2=34/2=17 , d=? ,

=0.4 , n=1/4+3/2=1.75

So using

=(D-d)/((n-1)d) we get,

17-d=0.3d

or d=17/1.3

or d=13.07

So vapour density of phosphine at 300K and 2.5atm is 13.07

Here is the solution for the problem.

Let initial no. of moles of CO₂ and H₂ be x

CO₂ + H₂

H₂O + CO

x x - -

At equi. 0.48 0.48 0.96 0.96

So x-0.96=0.48

So x=1.44moles

Now K=(0.96/2)*(0.96/2)/(0.48/2)*(0.48/2)

or K= 4

Let y moles of H₂ be added to the reacting mixture.Let z moles of CO₂ react.At equi concentration of CO is 0.6M.So the no. of moles of CO is 1.2(Since volume is 2 litre)

Then CO₂ + H₂

H₂O + CO

Initially 1.44 1.44+y - -

At equi 1.44-z 1.44+y-z z z But z=1.2 mole

So 0.24 0.24+y 1.2 1.2

Now K=[ H₂O][CO]/[CO₂][H₂]

ie 4=1.2*1.2/0.24*(0.24+y)

or 0.24+y=1.5

or y=1.26mole

So 1.26 mole of H₂ should be added.

Since the angle subtended at the centre is 60⁰ the triangle formed by the chord and two radii is equilateral.

Centre of the circle is (-2,3) and radius is 5

Let (h,k) be the midpoint of the chord.Now the distance of midpoint of the chord from the centre is L=

((h+2)²+(k-3)²)

Also L=5

3/2 because L is the altitude of the triangle that I had mentioned above and each side of the triangle is 5.

So the locus is (x+2)²+(y-3)²=75/4

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