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Find all positive integers x such that there exists a positive integer y satisfying: 1/x + 1/y = 1/7
Answer.
1/x + 1/y = 1/7
(x+y)/xy = 1/7
7x + 7y = xy
xy - 7x - 7y = 0
x(y-7) - 7(y-7) = 49
(x-7)(y-7) = 49
Thus satisfying values are:
49 * 1
1 * 49
7 * 7
So values of x and y can be, (42, 8), (8,
42), (14, 14).
Find 3x2y2 if x and y are integers such that y2 + 3x2y2 = 30x2 + 517.
Answer.
y2 + 3x2y2 - 30x2 – 517 = 0
3x2(y2-10) + 1(y2-10) + 10 – 517 = 0
(3x2+1)(y2-10) = 507
(3x2+1)(y2-10) = 3*13*13
To x and y be integers,
3x2+1 = 13 and y2-10 = 39
3x2 = 12 and y2 = 49
x2 = 4 and y2 = 49
To find 3x2y2 we will substitute the value of x2 and y2. 3*4*49 = 588.
Fortuna audentes iuvat, sed timidos repellit. Non in casu, sed in virtute
felicitas sita est. Obstacles cedere fortis animus debet, neque segnescere
periculis. Nam ex magnis difficultatibus maximae glories nascuntur. Quamvis
tenuis ignis, si alitur, magnus potest evadere; quamvis robusta arbor, si
negligitur, facile exarescit. Ita et ingenium nostrum, nisi colitur,
languescit; contra vero, si exercetur, ad summa quaeque contendere potest.
Ergo, animi fortitudo et labor assiduus ad res pulchras et magnas sunt
necessariae. Surge igitur, amice, et amplectere vitam cum ardore! Fiducia in te
ipso habe et nihil impossibile tibi videatur?
Let p(x) be a cubic
polynomial such that
(p(x))2 + (x4 + 2x2 + 2x)2 = f(x) = (x2+1)(x2+4)(x2-
2x+2)(x2+2x+2). Find the value of |p(-3)|.
Answer.
Put f(-3)
(p(-3))2 + ((-3)4 + 2(-3)2 + 2(-3))2 = f(-3) = ((-3)2+1)
((-3)2+4)((-3)2-2(-3)+2)((-3)2+2(-3)+2)
(p(-3))2 + (81 + 18 - 6)2 = (9+1)(9+4)(9 + 6 +2)(9 – 6
+ 2)
(p(-3))2 + 8649 =
(10)(13)(17)(5)
(p(-3))2 = 11050 –
8649
(p(-3))2 = 2401
p(-3) = ± 49
We have to find |p(-3)| . So, |± 49| = 49 = |p(-3)|
Answer.
Let a4
+ a2b2 + b4 = 900 ------- eq(1)
& a2
+ ab + b2 = 45 ------- eq(2)
Simplify
eq(1),
Add and
Subtract a2b2 for square completion,
(a2)2
+ 2(a2)(b2) + (b2)2 - (ab)2
= 900
(a2
+ b2)2 - (ab)2 = 900
(a2
+ b2 + ab)(a2 + b2 - ab ) = 900 ------- eq(3)
We know
that, a2 + b2 = 45 – ab ------- eq(4)
Put eq(4) in
eq(3)
(45 - ab +
ab)(45 - ab - ab ) = 900
(45 - 2ab)(45)
= 900
45 - 2ab =
20
-2ab = -25
2ab = 25
Let a1 + a2, ….., be a sequence defined by a1 = 1 for n>=1, an+1 = √(an2 – 2an + 3) + 1. Find a513.
First simplify:
√(an2 – 2an + 3) + 1
√(an2
– 2an + 1 + 2) + 1
√((an
– 1)2 + 2) + 1
Lets first find
out a2:
Put n = 1
a1+1 =
√(a12 – 2a1 + 3) + 1
a2 = √((1-1)2 + 2) + 1
a2
= √2 + 1
Likewise put
n = 2
a2+1
= √((a2 – 1)2 + 2) + 1
a3 =
√((√2 + 1 – 1)2 + 2) + 1
a3 =
√(4) + 1
a3 =
3
a3+1
= √((a3 – 1)2 + 2) + 1
a4
= √((3 – 1)2 + 2) + 1
a4
= √6 + 1
a4 + 1 =
√((a4 – 1)2 + 2) + 1
a5
= √((√6 + 1 – 1)2 + 2) + 1
a5
= √8 + 1
Now, we can notice that as we put the values of n
we get the result in the surd in the form of:
√(2*n) + 1
To get a513,
we have to put n = 512
So, the result directly come out to be
= √(2*512) + 1
= √1024 + 1
= 32 + 1
= 33
So, a513
= 33
Chapterwise weightage of social science. 1. History Chapter 1 - 5 marks Chapter 2 - 6 marks Chapter 3 - 7 marks Map - 2 marks 2. Democra...