Exams > Cat > Quantitaitve Aptitude
ALGEBRA MCQs
Total Questions : 240
| Page 4 of 24 pages
Answer: Option A. -> no solution
:
C
Answer = option (c)To have 2 roots of the opposite sign, the product of the roots need to be negative and the equation should have real roots4(a2+1)2 -12(a2-3a+2) ≥0 and (a2-3a+2)/3< 0 i.e. (a-1)(a-2)<0 or 1<a<2
:
C
Answer = option (c)To have 2 roots of the opposite sign, the product of the roots need to be negative and the equation should have real roots4(a2+1)2 -12(a2-3a+2) ≥0 and (a2-3a+2)/3< 0 i.e. (a-1)(a-2)<0 or 1<a<2
Answer: Option C. -> -2
:
C
Option (c)
Go from answer options
1) Take p=2
a2 –a +2=0 and a2 – a + 6=0
Roots are = 1+√(1+24)2 and √(1−24)2
Hence, this option is ruled out
Take p = -2
a2 –a -2=0 and a2 – a -6=0
Roots of 1st equation = 2,-1 and roots of second equation = 3, -2
:
C
Option (c)
Go from answer options
1) Take p=2
a2 –a +2=0 and a2 – a + 6=0
Roots are = 1+√(1+24)2 and √(1−24)2
Hence, this option is ruled out
Take p = -2
a2 –a -2=0 and a2 – a -6=0
Roots of 1st equation = 2,-1 and roots of second equation = 3, -2
Answer: Option C. -> 0
:
C
WXYZ = WXY * XZ => WXY (10) + Z = WXY (10x) + WXYZ.
Comparing we get => X = 1 and Z = 0
:
C
WXYZ = WXY * XZ => WXY (10) + Z = WXY (10x) + WXYZ.
Comparing we get => X = 1 and Z = 0
Answer: Option C. -> p≤4
:
C
Option (c)Since the condition D≥0 is given (as the roots are real), we will find the value of p using this conditionD= b2-4ac≥0122-(4xpx9) ≥0144-36p≥0144≥36pp≤4
:
C
Option (c)Since the condition D≥0 is given (as the roots are real), we will find the value of p using this conditionD= b2-4ac≥0122-(4xpx9) ≥0144-36p≥0144≥36pp≤4
Answer: Option C. -> Both (a) and (b)
:
C
Let the roots be P and P.
Sum of the roots = 2P = - 2k/9
P = - k/9
and product of roots = P2 = 4/9
(- k/9)2 = 4/9
k2/81 = 4/9
k2 = 36
k =± 6
:
C
Let the roots be P and P.
Sum of the roots = 2P = - 2k/9
P = - k/9
and product of roots = P2 = 4/9
(- k/9)2 = 4/9
k2/81 = 4/9
k2 = 36
k =± 6
Answer: Option A. -> No root
:
A
Option (a)
x(x - 1) - 2 = (x - 1) - 2
x2−x−2=x−1−2
x2−2x+1=0
(x−1)2=0
x = 1.
x = 1 is the only solution but it will make the denominator 0
Hence, there are no roots.
:
A
Option (a)
x(x - 1) - 2 = (x - 1) - 2
x2−x−2=x−1−2
x2−2x+1=0
(x−1)2=0
x = 1.
x = 1 is the only solution but it will make the denominator 0
Hence, there are no roots.
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