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d link dsnt giv d cut offs for d dual courses
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rank 1873
expctd around 300 marks. guess i mustve messd up sumwhere in d answer key
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http://www.ankur-gupta.com/images/iit_jee_2007_branch_wise_cutoff_full.gif
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where can i get d last year cut-offs for the dual courses? also, what r d advantages n disadvantages of dez courses ovr d normal b-tech courses? n if i want to change my field after d first year, can i change from dual to b-tech?
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as per d last year cut-offs, i think i can get metallurgical or civil.... is any other field possible?
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as per d last year cut-offs, i think i can get metallurgical or civil.... is any other field possible?
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as per d last year cut-offs, i think i can get metallurgical or civil.... is any other field possible?
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I got air1873. wich fields can i get in iit bombay??
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I got air1873. wich fields can i get in iit bombay??
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My rank is 1873.... wat shud i expct?
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i dunno whether dis is der for jee or not (since i hav not yet done algebra) but der is dis formula 4 findin da highest power of a prime dividin n!... da formula is, if p is da prime n x is its highest power dividin n!, den x= [n/p] + [n/p^2] + [n/p^3] +.... where [] denotes da greatest integer function... as u can c, dis value bcums zero in sumtym...
now using dis formula for da sum...
da number of zeroes at da end of 100! is da highest power of 10 wich divides 100! let dis value b x. now let da highest power of 2 wich divides 100! b y n da highest power of 5 wich divides 100! b z... now x = min{y,z}. but since y is much more dan z (da power of 2 wich divides will b very high), therefore x =z.
den v get z = [100/5] + [100/25] + [100/125]
so z= 20 + 4 + 0
x=z=24
dis method may appear slightly long but wen u acutually try 2 do it, it hardly takes more dan a few seconds... for example, da number of zeroes at da end of 1000! can b easily found to b 249 usin da same logic
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hey ankur, i think your initial equation shud hav (y+-4)^2 since da circle can also be in da third or fourth quadrant wich will add two more solutions 2 ur answer
x^2 + y^2 -10x + 8y + 16 = 0
&
x^2 + y^2 +10x + 8y +16 = 0
wich is same as da answer as given by vineet
n DON007, how did u get the square root of g^2 +f^2 - c = 3... it comes out to be 5 wich is again da same answer
deep01 please step up n say y is dis answer wrong
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no... i think dats da onli catch in dis question. two balls in 1 second means da 1st ball at 0 sec n da second ball at 1sec... so tym shud b 1sec... amaron has done it perfectly right
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even i got the same solution... n i cant understand why you put this as 'a good solution' and why this answer is wrong!!
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hey shine, you have made a very silly mistake... you have written the equation 2g1g2 + 2f1f2 = c1 + c2 correctly but while substitutin da values u hav substitued g1 as -g1 n f1 as -f1... i think u must hav thot dat is u take -g1 n -f1 in the equation, u wud not hav 2 substitute -g1 n -f1 in the equation 4 da centre of the circle, but den u again changed da signs while substitutin it there... so simply take make a correction there n u wil get da answer
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hii,
i thot dis method 2 b too long and required too much of trignometry(which is my weak area) so i tried solvin it in dis way:
i took da required pt 2 b (h,k). Now da chord of contact of dis point with respect to the ellipse is given by hx/9 + ky/4 = 1.
This chord will intersect da ellipse in P n Q and dez on joinin 2 da origin will give a right angle... so all v need 2 do is to homogenize da chord with the ellipse so dat u get da eqn
x^2/9 + y^2/4 = (hx/9 + ky/4)^2 which will giv da pair of str8 lines obtained by joinin pts P n Q 2 the origin... now da angle between dez lines shud b 90 degrees... i.e. coefficient of x^2 + coefficient of y^2=0
therefore, h^2/81+ k^2/16 -1/9 -1/4 = 0
so the locus is h^2/81 + k^2/16 = 13/36...
dis does not match the answer given (the 36 term is not supposed 2 b der). however, i m unable 2 find a mistake in the same. plz tell me where hav i gone wrong!
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Well Vinu, two days back i read dis question here n tried doin it by da very method you have suggested above, but it turns out dat u get only da value of sin alpha n not da angle itslf, n also, da value is not a well known one, so it becomes really difficult to carry on da calculations further... is there any other method anyone can suggest?
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y do u say dat da points cannot lie on opposite sides of da x-axis? da circle passes thru da origin... so it is quite possible 4 da points 2 lie on opposite sides of x-axis... let (h,0) b da mid point of da chord... den write da equation of da chord wid dat as da mid-point... now since (p,q) lies on da line, it has 2 satisfy da equation... so substitutin it in da equation, we get a quadratic in h... now since der r two lines, der hav 2 b 2 real values of h... so da determinant has to be greater than zero... n u get da answer
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no need 2 even find da intercept... since da intercepts r supposed 2 b equal... simply equate da distances of da lines 4m da centre of da circle (taking da required line 2 b y=mx)... u will get a quadratic in m, solve it n get da answer
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