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Algebra | Set-2

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  • Difficulty Level : Medium
  • Last Updated : 24 Feb, 2022
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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.

a = 1 – 1/b

=>ab=b-1

=>1/a=b/(b-1) ——–(1)

And

 b =1-1/c =>

b + 1/c = 1

=> bc + 1 = c

=> bc – c  =  -1

=> c(b – 1)  = -1

=> c  = 1/(1 – b) ———–(2)

putting the values of 1/a and c from above 1 and 2 in c – 1/a,

1/(1 – b)- b/(b-1) =    (b+1)/(1-b)                                                            

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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