@sampras : yar mera to written hua den GD tha usi mei lag gyi .... :(
@sampras said:given : x=(rt3+1)/2 ...... 2x-1 = rt3 ..... squaring... 2x^2-2x-1 = 0now given eqn ... 4x^3+2x^2-8x+7 = (2x^2-2x-1)(2x+3) + 10 = 10 ..... Ans
thanx........yr 1 query solve to karo..........
like 2x^2-2x-1 = 0, if find the D and roots then x will be (rt3+1)/2 and (rt3-1)/2
then if we take thz equation as
2x(x-1) = 1 and then x = 1/2 and 3/2
so what i have done wrong in latter case nd i knw that it z wrong smwhr..
plz explain..
@Brooklyn said: @sampras : yar mera to written hua den GD tha usi mei lag gyi ....
hmmm matlab ... GTO nahi hua...okay try again will guide if need... i have very long experience... in this area..
@sampras said: @pathakrahul Dr sahab kaha ho chhup chhup ke like pe click kar rahe ho...good morning
nahi bhai abhi aaya hun aapka diya hua soln dekh raha tha.thanks aapne qn ka soln de diya....i was trying but could not be able to evaluate........ thanks again
@pathakrahul said:nahi bhai abhi aaya hun aapka diya hua soln dekh raha tha.thanks aapne qn ka soln de diya....i was trying but could not be able to evaluate........ thanks again
cmmon yaar abhi bhi thanks ki jaroorat hai kya... always welcome bro... kaisi chal rahi hai taiyari...
@Brooklyn said:@sampras : CGLE nikal lu kafi hailife mei aaram bhi bura nhi hota
wo toh niklega hi for sure ... but cracking SSB is something much much more than it... you have pride, talent, respect of the uniform...the feeling of being securityman of the country...everything one looks for...
If X^a.x^b.x^c = 1 then the value of a^3 + b^3 + c^3
is
(A) 9 (B) abc
(C) a + b + c (D) 3abc
If a^2 + b^2 + c^2 = 2(a – b – c) – 3, then the
value of (a – b + c) is
(A) -1 (B) 3
(C) 1 (D) -2
bit confused abt method....please post something abt method...
@sampras said: @abhi_1234Q. when (67^67 +67) is divided by 68 the remainder is ...please explain the logic and method involved .... as confused in Binomial expansion...
(67^67 + 67)= [(68 - 1)^67 + 67]
on dividing
we can write as [(68-1)^67]/68 + 67/68
first part we get remainder -1 second part 67(still process is not completed)
finally (-1 + 67)/68 remainder is 66
on dividing
we can write as [(68-1)^67]/68 + 67/68
first part we get remainder -1 second part 67(still process is not completed)
finally (-1 + 67)/68 remainder is 66
@JS-M said:thanx........yr 1 query solve to karo..........like 2x^2-2x-1 = 0, if find the D and roots then x will be (rt3+1)/2 and (rt3-1)/2then if we take thz equation as2x(x-1) = 1 and then x = 1/2 and 3/2so what i have done wrong in latter case nd i knw that it z wrong smwhr..plz explain..
bhai going first hand seems you are also correct ... needs to review minutely... will come back if come thru any error
@abhi_1234 said:(67^67 + 67)= [(68 - 1)^67 + 67] on dividing we can write as [(68-1)^67]/68 + 67/68 first part we get remainder -1 second part 67(still process is not completed) finally (-1 + 67)/68 remainder is 66
haa bhai yehi mai kar raha tha... but as you said logic lagaao rem negative kaise ho sakta hai... got entangled and then confused... any way do enlighten me... after you are done..ab time bhi nahi hai self try karne ka ... readymade paka pakaya mil jaaye toh achha ... nahi toh bubbling ki practice kar raha hu....
@NailedNCrushed said:If X^a.x^b.x^c = 1 then the value of a^3 + b^3 + c^3
is
(A) 9 (B) abc
(C) a + b + c (D) 3abc
If a^2 + b^2 + c^2 = 2(a – b – c) – 3, then the
value of (a – b + c) is
(A) -1 (B) 3
(C) 1 (D) -2
bit confused abt method....please post something abt method...
illegible qn .can u plz put the qn again
@ lalit5800 and sumit1234567890 u both are correct ...but can you tell me what is the trick to solve it????
@NailedNCrushed said:If X^a.x^b.x^c = 1 then the value of a^3 + b^3 + c^3
is
(A) 9 (B) abc
(C) a + b + c (D) 3abc
If a^2 + b^2 + c^2 = 2(a – b – c) – 3, then the
value of (a – b + c) is
(A) -1 (B) 3
(C) 1 (D) -2
bit confused abt method....please post something abt method...
1. as we cn see.......a+b+c = 0
then a^3+b^3+c^3 = 3abc..........direct result
2. (a-1)^2+(b+1)^2+(c+1)^2 = 0 .......put each square value equal to zero
u will get a=1 b=-1and c=-1.........
Guys i have not received my CGL admit card so can i take the printout in exam?????
@sampras aasan aur chota tarika
x=(rt3 + 1)/2 and x-1= (rt3 - 1)/2 ,therefore x(x-1)= 1/2 ....eq1
(2x-1)^2=3 => 2x^2 -2x-1=0 => 2x^2=2x+1....eq.2
Now 4x^3+2x^2 - 8x + 7= 2x(2x^2+x-4) + 7
put 2x^2 value from 2 here
2x(3x-3) + 7 = 6x(x-1) + 7 substitute x(x-1) from 1 and get answr...
x=(rt3 + 1)/2 and x-1= (rt3 - 1)/2 ,therefore x(x-1)= 1/2 ....eq1
(2x-1)^2=3 => 2x^2 -2x-1=0 => 2x^2=2x+1....eq.2
Now 4x^3+2x^2 - 8x + 7= 2x(2x^2+x-4) + 7
put 2x^2 value from 2 here
2x(3x-3) + 7 = 6x(x-1) + 7 substitute x(x-1) from 1 and get answr...
@sampras said:Q. if x = (root3 + 1)/2 then 4x^3+2x^2 - 8x + 7 is....
1.x^a*x^b*x^c=1 => x^(a+b+c)=1
So there is only 1 possibilty when x^(a+b+c)=1 i.e a+b+c=0
Now if a+b+c=0
a^3+b^3+c^3=3abc.........its a standard result u can memorise it
2.
a^2 + b^2 + c^2 = 2(a – b – c) – 3
=> a^2 + b^2 + c^2-2a+2b+2c+3=0
=> (a^2-2a+1)+(b^2+2b+1)+(c^2+2c+1)=0
Now the term in bracket can be converted to whole squares as follows
(a-1)^2+(b+1)^2+(c+1)^2=0
Now since we have square term in the expression so for the expression
to be 0 all the term should be seperately equal to 0
i.e. a-1=0,b+1=0,c+1=0
=> a=1,b= -1,c=-1
so a-b+c=1 and hence the ans
@NailedNCrushed said:If X^a.x^b.x^c = 1 then the value of a^3 + b^3 + c^3
is
(A) 9 (B) abc
(C) a + b + c (D) 3abc
If a^2 + b^2 + c^2 = 2(a – b – c) – 3, then the
value of (a – b + c) is
(A) -1 (B) 3
(C) 1 (D) -2
bit confused abt method....please post something abt method...
for 1) a+b+c=0
hence 3abc ans as : a^3 + b^3 + c^3 = 3abc