JEE Novice - (34)
c a + b − x + b a + c − x + a c + b − x + a + b + c 4 x = 1
Find x .
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4 solutions
Moderator note:
Right. It's better to show why y = a b + b c + a c 4 a b c = a + b + c with a counterexample in your solution.
Bonus question: True of false? :
"For every integer Y ≥ 2 , there exist positive integers A , B , C such that Y 4 = A 1 + B 1 + C 1 ."
Hint: You may use the internet at your disposal.
So you preferred a solution than adding a comment. I see that you are improving. Nice work :3 :3 :3 :P
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This was a different approach than Chew Seong Cheong's approach, so it wasn't suited to be a comment. :) :P
Also, my comments are getting downvoted since recently by some for unknown reasons, so I thought posting a solution would be best. :D
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Posting a solution and getting upvotes is "indeed" the best. :3 :3
y = a b + b c + c a 4 a b c = ( a + b + c ) in general.
This inequation can be verified to be true easily by taking a particular case, say a = b = c = 1 .
@Calvin Lin , if you notice, I had provided a counterexample earlier in the comments section. Nonetheless, I have included the counterexample I provided in the comments in the solution now.
By the way, thanks for the bonus question. :D
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When I review I mainly just look at what's written in the solution, instead of additionally scouring all of the comments.
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Yes, I figured that out later. I hope the solution looks better now.
I'll try the bonus question tomorrow and post a solution here in the comments if I successfully prove/disprove it. :D
c a + b − x + b a + c − x + a c + b − x + a + b + c 4 x = 1 a + b + c − a − b a + b − x + a + b + c − a − c a + c − x + a + b + c − c − b c + b − x + a + b + c 4 x = 1 − a + b + c − a − b x − a − b − a + b + c − a − c x − a − c − a + b + c − c − b x − c − b + a + b + c 4 x = 1
We observe that when x = a + b + c , then the equation becomes − 1 − 1 − 1 + 4 = 1 .
More easy way, i used to solve these type of questions in
less time
is,
1) Put a,b,c any
arbitrary possible value
so that your calculation becomes easy and then solve for
x
.
2) Now,in the option,again put the values of a,b,c you have chosen.
3) Whichever value of x satisfiesin both cases,that becomes your
answer
.
NOTE
: you have to be very
careful
in this method.
Limitations
: This method works only in
multi-correct
.
akhil bansal that was a great idea !!
For me I added 3 to both sides of the equation and arrived at (1/a + 1/b + 1/c)(a+b+c-x) = 4(a+b+c-x)/(a+b+c). I arived at a+b+c-x = 0 but may I know why 1/a + 1/b + 1/c is NOT equal to 4/(a+b+c)? Still do not quite understand. Thanks!!!
We first note that we obviously have a , b , c = 0 because otherwise the given condition can't hold. So, division by a , b , c or any linear combination of them is valid.
Next, consider y = a + b + c . Then we proceed to rewrite the equation as follows:
c y − c − x + b y − b − x + a y − a − x + y 4 x = 1 ⟹ ( y − x ) ( c 1 + b 1 + a 1 ) − 3 + y 4 x = 1 ⟹ ( y − x ) ( a 1 + b 1 + c 1 ) = y 4 ( y − x ) ⟹ ( y − x ) ( a 1 + b 1 + c 1 − y 4 ) = 0
Now, let's assume that ( y − x ) = 0 . So, we would then have that,
a 1 + b 1 + c 1 − y 4 = 0 ⟹ y = a b + b c + c a 4 a b c = ( a + b + c ) in general.
An easy way to verify this inequation is by taking the example of a particular case, say a = b = c = 1 where you can see that the obtained expression gives 3 4 which is = 3 .
So, our assumption was false and we must have,
y − x = 0 ⟺ y = x = ( a + b + c )
Q . E . D