# Algebra | Set-2

Question 1: If x + 1/x = -2, then the value of x1000 + 1/x1000 is
Solution. x= -1 satisfy the above eq.
So, (-1)1000 + 1/(-1)1000 = 1 + 1 = 2

Question 2: If a = 1 – 1/b and b = 1 – 1/c, then the value of c – 1/a is
Solution. Let a = 2 and b = -1
2 = 1 – (-1)
2 = 2 True
So, find the value of c
b = 1 – 1/c
-1 = 1 – 1/c
c =1/2
Then, c – 1/a = 1/2 – 1/2 = 0

Question 3: If a + b + c = 3, then the value of 1/(1 – a)(1 – b) + 1/(1 – b)(1 – c) + 1/(1 – c)(1 – a)
Solution 1/(1 – a)(1 – b) + 1/(1 – b)(1 – c) + 1/(1 – c)(1 – a)
=> [(1 – c) + (1 – a) + (1 – b)]/(1 – a)(1 – b)(1 – c)
=> [3 – (a + b + c)]/(1 – a)(1 – b)(1 – c)
=> 3 – 3 /(1 – a)(1 – b)(1 – c)
=> 0

Question 4: If a + 1/a = √3, then the value of a18 + a12 + a6 + 1 is
Solution. a3 + 1/a3 = (a + 1/a)3 – 3(a + 1/a)
=> 3 √3 – 3 √3
=> 0
a3 + 1/a3 = 0
a6 + 1 = 0
Then,
a18 + a12 + a6 + 1
a12(a6 + 1) + (a6 + 1)
a12 x 0 + 0 = 0

Question 5: If a = √3 + 1 / √3 -1 and b = √3 -1 / √3 + 1, then find the value of
(a2 + ab + b2)/(a2 – ab + b2) is
Solution. a = 1/b
therefore ab = 1
a + b = (√3 + 1) / ( √3 -1) + (√3 -1) / (√3 + 1)
=> (3 + 1 + 2√3 + 3 + 1 – 2√3)/ (3 – 1)
=> 8/2
=> 4
a + b = 4
a2 + b2 = 42 – 2 *(ab)
a2 + b2 = 14
Now, (a2 + ab + b2)/(a2 – ab + b2)
=>(14 + 1)/(14 -1)
=> 15/13

Question 6: If x = 8, then find value of x5 – 9x4 + 9x3 – 9x2 + 9x1 – 1
Solution. We can write it as
85 – 8*x4 – 1*x4 + 8*x3 + 1*x3 – 8*x2 – 1*x2 +8*x1+ 1*x1 – 1
Now put x = 8
85 – 8*84 – 1*84 + 8*83 + 1*83 – 8*82 – 1*82 +8*81+ 1*81 – 1
= 8 – 1
= 7

Question 7: If the sum of the square of two real numbers is 74 and their sum is 12. Then the sum of cubes of these two numbers is
Solution Let two numbers are a and b
Given, a2 + b2 = 74
a + b = 12
(a + b)2 = a2 + b2 + 2ab
122 = 74 + 2ab
144 = 74 + 2ab
ab = 35
We get a=7 and b=5
Then, a3 + b3 = 73 + 53
=343 + 125
=468

Question 8: If ab(a+b)=m, then the value of a3 + b3 + 3 m is
Solution. ab(a + b)=m
a + b = m/ab
Cubing both sides
a3 + b3 + 3ab(a + b) = m3 / a3b3
a3 + b3 + 3 m = m3 / a3b3

Question 9: If m=√7+√7+√7….. and n=√7-√7-√7…….
then among the following relation between m and n holds is
Solution. m = √(7 + m)
m2 = 7 + m
m2 – m = 7…….(1)
and n = √(7 – n)
n2 + n = 7…….(2)
from (1) and (2)
m2 – m = n2 + n
m2 – n2 – (m + n) = 0
(m + n)(m – n) – (m + n)= 0
m – n – 1 = 0

Question 10: If x2 + y2 + z2 = 2(x + y -1), then the value of x3 + y3 + z3?
Solution. x2 + y2 + z2 = 2x + 2y -2
(x2 + 1 -2x) +(y2 + 1 -2y) + (z2) = 0
(x – 1)2 + (y – 1)2 + (z)2 = 0
=> (x – 1)2 = 0
=> x = 1
(y – 1)2 = 0
=> y=1
(z)2 = 0
=> z = 0
Put value in eq
x3 + y3 + z3
13 + 13 + 03
=> 2

Question 11: If (x12 + 1 )/x6 = 6, then the value of (x36 + 1 )/x18 ?
Solution. Given
(x12 + 1 )/x6 = 6
x6 + 1 /x6 = 6
Cubing both sides
(x6 + 1 /x6)3 = 63
x18 + 1/x18 + 3 (x6 + 1 /x6) = 216
x18 + 1/x18 + 3 * 6 = 216
x18 + 1/x18 = 198
(x36 + 1)/x18 = 198

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