1another one...
2) if (b2-4ac)2(1+ 4a2) < 64a2 , a<0 ,
then maximum value of quadratic expression ax2+bx+c is
a) 0 b) 2 c) -1 d) -2
g(x) = x5 + x4 + x3 + x2 +x +1
wats the remainder wen the polynomial g(x12) is divided by the polynomial g(x) ?
a) 6
b) 5 - x
c) 4- x - x2
d) 3 - x + x2 - x3
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16 Answers
1
62ax2+bx+c
maxima at x=-b/2a
maximum value = c-b2/4a
does this help?
1par aaspaas mein joh kachraa hai.... usko kya kare??
am having prob wid them only :P
62to get rid of that kachra..
use
(b2-4ac)2(1+ 4a2) < 64a2 , a<0 ,
(b2-4ac)2< 64a2/(1+ 4a2)
(b2-4ac)< 8a/√(1+ 4a2)
divide by 4a..
(b2/4a-c)< 2/√(1+ 4a2)
so does this clear the kachra?
1arey i got something...
(b2-4ac)2(1+ 4a2) < 64a2
so, (b2-4ac)2(1+ 4a2)/64a2< 1
=> (b2-4ac)2/(2a)2 [(1+ 4a2)/4] < 1
=> m2 [a2 + 1/4] <1
after this ?
341g(x12) = (x12)5+ (x12)4+ (x12)3+ (x12)2+ (x12)+1
= [(x12)5-1]+ [(x12)4-1]+[ (x12)3-1]+ [(x12)2+ (x12)-1]+6
Each of the bracketed expressions is divisible by (x6-1) and hence by (x5+x4+x3+x2+x+1)
So the remainder is 6
341Actually you can do this very easily by a powerful method called polynomial congruences. It works very similar to congruence in number theory.
Suppose we are talking of divisibility by a certain polynomial P(x).
Suppose f(x) ≡ g(x) mod P(x)
Then just as in number theory for any polynomial F(x), we have
F(f(x)) ≡ F(g(x)) mod P(x)
Here P(x) = x5+x4+x3+x2+x+1
Notice that x6 ≡ 1 mod P(x)
So that x12 ≡ 1 mod P(x)
So g(x12) ≡ g(1) = 6 mod P(x)
If you can pick up this technique many such problems become very very easy
1m2 [ a2 + 1/4 ] < 1
=> m2 < 1/[ a2 + 1/4 ]
=> -2/√1+4a2 < m < 2/√1+4a2
if a=0, m < 2
but a<0 => a2<1 but >0 => √1+4a2 >1
so,,,, a <2 always.
plz correct me if wrong...
1great prophet !!
thanks a ton :)
and lots of thanx to nishant bhaiya for the second sum :)
62yup.. and then i also reminded of the Chinese remainder theorem...
f(x)/(x-a) gives a remainder as . f(a)
1sorry to say...
prophet will u explain that mod thing explicitly ??
i din get you [2]