Messages posted by kabishy

Let the sides be a , b ,c

r₁=S/(s-a) ...r₂=S/(s-b).........r₃=S/(s-c)

As these r in HP

2 / r₂ = 1/ r₁ + 1/r₃

putting the values we will get " b = 8 " and a +c = 16

so tan B = 4 / 3

by Area the height on "b" should be 6

by seeing the value of tan B

a = 10 ,b =8,c=6

or

a=6,b=8 ,c=10

i m in a hurry so just giving answer to 1st one .

int ( -3 to -1 ) (x+1)(x+2)(x+3) dx

Transformation x+2 =y

=int (-1 to 1 ) (y-1) y (y+1)dy

= 0

lim_x--->0 sin(x)/x is 1 .

But actually sin(x) <x

so the ratio would tend to 1 or we can say its GIF is 0

Thank u

7¹⁰⁰⁰ = (50-`1)⁵⁰⁰=⁵⁰⁰C₀ 50⁵⁰⁰-.......................................-500*50+1

=1000(k)+1

ab dekh 7 ki powers ke unit digit periodic hain 4 ke . e.g . 7 , 49 , 243 ,2401,then again 16807

So u can say without any doubt that unit digit of 7^{7^(1000)} would be 7 .

7 ^1000k+1=7 ( 7^1000k) = 7 ( 50 -1)^500k =7.(^500kC₀50^500k-................-500*50k+1)

all the first 500k numbers would be having atleast 3 zero on the last three places . 07 would be last two digit

Suggestion ::

Yaar have a knowlegde of number theory . It helps a lot .

Thank u

We just need to concentrate again on a period only let it be x belonging to 0 to 2pi

Now from x =0 to x=pi/2

everywhere sinx is less than 1 so graph would be same as sin(x) but on x=pi/2 sinx becomes 1 so every comes first point of disconcounity {sin(pi/2)}=0

again after x=pi/2 to x =pi

graph would be same as sin(x)

Then from x =pi to x=2pi ( excluding these points )

sin(x) lie b/w 0 and (-1) so its Greatest integer part would be (-1)

So just add 1 to the graph or in easier way uplift the graph to 1 unit .

Repeat the pattern further on .

Thank u

http://www.goiit.com/posts/list/integral-calculus-indefinte-integration-reloaded-998418.htm

2. using integration by parts

I = x log (Sqrt(1-x)+sqrt(1+x)) - integration ( ( 1-Sqrt(1-x²)) /( -2Sqrt(1-x²)) dx

= x log (Sq.(1-x)+Sq (1+x)) - 1/2 ( integration ( 1 - (1/Sqrt(1-x² )) dx )

= x log (Sq.(1-x)+Sq (1+x))- x/2 +sin^-1(x)/2 +constant

thank u

Check d figure ::

either CD would be of 9 length or CB .

Case I :: when BC=9

AS AC is perpendicular to BD

In triangle

AOD

p²+q²=4

In tri. AOB

q²+r²=9

Also in triangle BOC

s²+r² =81

using eq. I and III in II

gives

p²+s² =76 =CD²

CD=sqrt(76 )

Case II

CD=9

repeat the above method

BC =sqrt(86)

thank u

well

2 nd method This way is my way to solve this

look

(x sin(x)+cos(x) ) = Sqrt(1 +x² ) cos( x -tan^-1(x))

so ur question become

int ( x² sec² (x -tan^-1x ))/(1+x²) dx

put x-tan^(-1)x = t

x²/(1+x²)dx =dt

= int sec^2 t dt

= tan (x -tan^-1(x)) +C

thank u

hi

i have 2 method

1st method

I = integration ( x sec(x) x cos(x)/(xsinx +cos(x))² dx

integration by parts

=x sec(x) int ( xcosx /(xsinx +cos(x))² dx) - int [ ( xsecxtanx+secx) int (xsinx +cos(x))² ]dx

= -xsecx / ( xsinx+cosx) + int (sec^2x)dx

= -xsecx / ( xsinx+cosx) +tanx +c

thank u

2 . lim ( n --> infty) (1/n) [ ( 1/ n+1) + (2 / n+2) + .......... + (3n/4n) ]

=lim ( n --> infty) (1/n) (1/(n+1) +1/(n/2 +1)+....................+1/(n/3n+1) ]

= lim ( n --> infty) (1/n) sigma ( r from 1 to 3n ) [ 1/(n/r +1) ]

= integration ( 0 to 3 ) x / (x+1) dx ( put r/n =x )

=x - log (x+1)

after putting limits

=3-log(4)

thank u

thank to Jesus , atleast there is one who know kabi is a gal not a bhai sahab .

hello

is it greatest integer of mode x or greatest integer of x ?

If greatest integer of mode x ::

-3=<[IxI]=<2

-3=< IxI < 3

we know IxI>=0 always

so inequality becomes ::

0=< IxI <3

so -3 <x<3 is the solution

if it is greatest integer of x only

-3=<[x] <=2

-3=<x<3 will be d solution

b. f([2x+3])

-3<=[2x+3]<=2

-3<=2x+3 <3

-3<=x<0 is d solution .

thank u

Is not FTSE Part 1 question ? Don't u have the solution booklet or u r asking for the alternate methods other than the one discussed in the booklet .

hello

i m not going to solve this prob just giving the approach

first of all draw the circles so that u can actually visualise where a should lie .

we want y = 2x+a neither to touch the circles nor intersect any of them

put y = 2x +a in the eq. x²+y²-2x-2y+1 =0

u will get

5x² +4xa -6x+a² -2a +1 =0

for fulfilling the condition we want the d < 0 of this quadractic

so a²+2a -4 >0

solve this a >sqrt(5)-1 or a < - (sqrt(5)+1)

if we want such value of a so that lie b/w the circles u can easily come to conclusion

we want a < - (sqrt(5)+1) only .

then do the same thing for second circle . and find the value of a . ask me again if u have prob.

hello

we have D ( 20,25) E ( 8,16) C( 10,15 )

slope of AC = -1/2

slope of BC = 1

Thus slope of BE = 2 and that of AD = -1

thus the eq. of AD becomes y +x = 45

eq. of BE is y = 2x

the point of concurrency is x = 15 y=30 ie. coordinate of orthocenter .

thank u sorry if there is a calculation mistake as i m solving in hurry .

one more thing the approach used by the student who have given the second answer is correct .

i would say " Samajdar ko ishara kafi "

well here we say thanks by giving rates .

first of all this is a wrong forum for ur question ::

I =integration( sin @ to cosec@ ) f(x) dx.................(1)

let x =1/y

I = integration ( cosec@ to sin@ ) -f (y)dy /y² ( we know from relation f(y) = - f(1/y) / y² )

= integration ( sin@ to cosec@ ) -f(y) dy

integration ( sin@ to cosec@ ) -f(x) dx .................(2)

adding 1 and 2

I = 0

thank u

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