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If x2(y + z) = a2, y2(z + x) = b2, z2(x + y) = c2, xyz = a b c prove that a2 + b2 + c2 + 2abc = 1
If x2(y + z) = a2, y2(z + x) = b2, z2(x + y) = c2, xyz = a b cx2 (y + z) y2 (z + x) z2 (x + y) = a2b2c2x2y2z2 (x + y) (y + z) (z + x) = a2b2c2xyz = abc(x + y) (y + z) (z + x) = 1a2 + b2 + c2 + 2abc => x2(y + z) + y2 (z + x) + z2(x + y) + 2xyzPut y = - z0 + z2(z + x) + z2(x - z) + 2x (-z)zz3 + z2xRead more
If x2(y + z) = a2, y2(z + x) = b2, z2(x + y) = c2, xyz = a b c
x2 (y + z) y2 (z + x) z2 (x + y) = a2b2c2
x2y2z2 (x + y) (y + z) (z + x) = a2b2c2
xyz = abc
(x + y) (y + z) (z + x) = 1
a2 + b2 + c2 + 2abc
=> x2(y + z) + y2 (z + x) + z2(x + y) + 2xyz
Put y = – z
0 + z2(z + x) + z2(x – z) + 2x (-z)z
z3 + z2x + z2x – z3 – 2z2x = 0
∴ (y + z) is factors
Similarly (x + y) & (x + z) are factors
x2(y + z) + y2(z + x) + z2(x + y) + 2xyz
= k (x + y) (y + z) (x + z)
For k put x = 0 y = 1 z = 1
0 + 1 + 1 + 0 = k (1) (2) (1)
2 = 2 k
k = 1
x2(y + z) + y2(z + x) + z2(x + y) + 2xyz
= (x + y) (y + z) (z + x)
and (x + y) (y + z) (z + x) = 1
∴ x2(y + z) + y2(z + x) + z2 (x + y) + 2xyz = 1
See lessThere are 4 coins in a row and all are showing heads to start with. The coins can be flipped with the following rules :
(a) The fourth coin (from the left) can be flipped any time
(b) An intermediate coin can be changed to tail only if its immediate neighbor on the right is heads and all other coins (if any) to its right are tails.
(c) Only one coin can be flipped in one step.
The minimum number of steps required to bring all coins to show tails is _______.
(8)HHHH, HHTH, HHTT, THTT, THHT,TTHT, TTHH, TTTH, TTTT
(8)
HHHH, HHTH, HHTT, THTT, THHT,
TTHT, TTHH, TTTH, TTTT
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In an elementary school 26% of the students are girls. If there are 240 less girls than boys, then the strength of the school is _________
Girls = 26%Boys = 100 – 26 = 74%Given 74% – 26% = 24048% = 2401% = 240/48 students100% = 240/48 x 100= 500 students
Girls = 26%
Boys = 100 – 26 = 74%
Given 74% – 26% = 240
48% = 240
1% = 240/48 students
100% = 240/48 x 100
= 500 students
See lessWhen a two digit number divides 265, the remainder is 5. The number of such two digit numbers is ________
265 – 5 = 260260 = 2 × 2 × 5 × 13two digits such numbers will be=> 1 × 13 = 13=> 2 × 13 = 26=> 2 × 5 = 10=> 4 × 5 = 20=> 4 × 13 = 52=> 5 × 13 = 656
265 – 5 = 260
260 = 2 × 2 × 5 × 13
two digits such numbers will be
=> 1 × 13 = 13
=> 2 × 13 = 26
=> 2 × 5 = 10
=> 4 × 5 = 20
=> 4 × 13 = 52
=> 5 × 13 = 65
6
See lessThe four digit number 8ab9 is a perfect square. The value of a2 + b2 is
(A) 52
(B) 62
(C) 54
(D) 68
Correct option: (A) 52Explanation:932 = 8649∴ a = 6, b = 4∴ a2 + b2 = 62 + 42 = 52.
Correct option: (A) 52
Explanation:
932 = 8649
∴ a = 6, b = 4
∴ a2 + b2 = 62 + 42 = 52.
See lessVishva plays football every 4th day. He played on a Tuesday . He plays football on a Tuesday again in _________ days.
Number of days in week = 7 daysVishva plays football in = 4 daysHe will play football on a Tuesday again in 7 × 4 = 28 days
Number of days in week = 7 days
Vishva plays football in = 4 days
He will play football on a Tuesday again in 7 × 4 = 28 days
See lessSamrud had to multiply a number by 35. By mistake he multiplied by 53 and got a result 720 more. The new product is _________.
Let the number be xincorrect product = x × 53correct product = x × 35x × 53 – x × 35 = 72018x = 720x = 40New product x × 53=> 40 × 53 = 2120
Let the number be x
incorrect product = x × 53
correct product = x × 35
x × 53 – x × 35 = 720
18x = 720
x = 40
New product x × 53
=> 40 × 53 = 2120
See lessProve that x4 + 3x3 + 6x2 + 9x + 12 cannot be expressed as a product of two polynomials of degree 2 with integer coefficients.
Let x4 + 3x3 + 6x2 + 9x + 12= (x2 + Ax + B) (x2 + Cx + D)= x4 + Cx3 + Dx2 + Ax3 + ACx2 + ADx + Bx2 + BCx + BD= x4 + (A + C)x3 + (D + AC + B) x2 + (AD + BC)x + BDNow by comparing coefficientA + C = 3B + D + AC = 6AD + BC = 9BD = 12Case - I : B = 1, D = 12∴ A + C = 312A + C = 9 have no integer solutioRead more
Let x4 + 3x3 + 6x2 + 9x + 12
= (x2 + Ax + B) (x2 + Cx + D)
= x4 + Cx3 + Dx2 + Ax3 + ACx2 + ADx + Bx2 + BCx + BD
= x4 + (A + C)x3 + (D + AC + B) x2 + (AD + BC)x + BD
Now by comparing coefficient
A + C = 3
B + D + AC = 6
AD + BC = 9
BD = 12
Case – I : B = 1, D = 12
∴ A + C = 3
12A + C = 9 have no integer solution.
Case – II : B = – 1, D = – 12
C + 12 A = – 9
C + A = 3 have no integer solution.
Case – III : B = 2, D = 6
2C + 6A = 9
C + A = 3 have no integer solution.
Case – IV : B = – 2, D = – 6
2C + 6A = – 9
A + C = 3 have no integer solution.
So, x4 + 3x3 + 6x2 + 9x + 12 cannot be expressed as a product of two polynomial of degree 2 with integer coefficient.
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