## ALGEBRA MCQs

Total Questions : 240 | Page 1 of 24 pages
Question 1. If α and β are the roots of the equation x2 + px + q = 0 and α4 and β4 are the roots of the equation  x2– rx + s = 0, then the equation x2 – 4qx + 2q2 – r = 0 has always:
1.    2 real roots
2.    2 positive roots
3.    2 negative roots
4.    1 positive and 1 negative root
Answer: Option D. -> 1 positive and 1 negative root
:
D
Given question isflexible. So, we can assume the roots and can make an other quardatic equation.
Let α = 1 and β = 2, then the expression becomes x2 – 3x + 2 = 0. So p = -3, q = 2, r = 17, s = 16. The new expression is x2 – 8x – 9 = 0. 1 root is positive and other is negative.
Question 2.  If ab+c=bc+a=ca+b=r then, r cannot take any value except :(CAT 2004)
1.    12
2.    -1
3.    12 or -1
4.    −12 or -1
Answer: Option C. -> 12 or -1
:
C
Option (c)
As ab+c=bc+a=ca+b=a+b+cb+c+c+a+a+b
=a+b+c2(a+b+c)=r=12 (Assuming a + b + c ≠ 0)
If a + b + c = 0
ab+c = aa+b+ca (by adding and subtracting a in the denominator) =a0a=aa=r=1
(similarly bc+a=ca+b= -1)
Hence r can take only 12 or -1 as values. Choice (c)
Question 3.
1.    t3+t2+t+1
2.    t3+t2-5
3.    t3+ √5t2 +√5t -2
4.    t3-1
5.    t3
:
D
Option (d)Assume t= 2The question changes to[(5-√3i)(5+√3i)]/4 [25-(√3i)2]/4, since (a-b)(a+b)=a2-b2(25+3)/4, since i2=-1=7Only answer option (d) gives 7 on substitution of t=2
Question 4. Given that p and q are the roots of the equation x2 – ax +b =0 and Dn= pn+qn. Find the value of Dn+1
3.    bDn
5.    Dn
:
A
Option (a)Best way to proceed is by assumption of values.Assume a quadratic equation. Let’s take x2+5x-6=0. Here the roots are p=1 and q=-6Thus, a = sum of roots = -5 and b= product of roots = -6D1= p1+q1 = -5D0 = 2D2 = 37Assume n=1We need to find Dn+1 = D2 = 37Look in the answer options for 37Option a is the only one which gives (-5)(-5)-(-6)(2) = 37
Question 5. What is the value of X ?
1.    3
2.    2
3.    1
4.    0
5.    none of these
:
C
WXYZ = WXY * XZ => WXY (10) + Z = WXY (10x) + WXYZ.
Comparing we get => X = 1 and Z = 0
Question 6. Given that b3+mb2 +nb + c is divisible by (b-s) if s3 + ms2 + ns + c = 0Also given that d3+d2+ed+1 is divisible by (d-1) and d3-4d2+fd-3 is divisible by (d-3)What is the value of e+f ?
1.    0
2.    -1
3.    1
4.    2
5.    none of these
:
C
Option (c)d3+d2+ed+1 is divisible by (d-1). As 1 is a root, Put d=1 in the equation and equate it to 0, to get the value of e1+1+e+1=0 e=-3Similarly put d=3 in d3-4d2+fd-3 to get the value of f 27-36+3f-3=0 3f=12 f=4e+f= -3+4=1
Question 7. The roots of the equation (x+1)(x+8) +10 = 0 are given as p and q. Find the roots of (x+p)(x+q)-4=0
1.    -1, 3
2.    7, 8
3.    2,7
4.    –2, 7
5.    –2, 3
:
C
Option c (x+1)(x+8)+10 can be expanded as x2+9x+8+10=0 x2+9x+18=0 Sum of roots p+q= -9 Product of roots pq = 18(x+p)(x+q)-4=0 can be written as x2+ (p+q)x+pq-4=0 x2-9x+18-4=0x2-9x+14=0Roots are 7 and 2
Question 8. If a and b are the roots of the equation x2+px+1=0 and ; c and d are the roots of the equation x2+qx+1=0, then (a-c)(b-c)(a+d)(b+d)=?
1.    p2-q2
2.    a2+b2
3.    q2-p2
4.    q2+p2
5.    p2
:
C
Shortcut: AssumptionOption (c)Product of roots=1 is the only constraint.Assume values that satisfy the constraint and substitute, take a= -1 and b= -1, then p= 2Assume c= - 1/2 and d= - 2; q= 5/2 Question is = (a-c)(b-c)(a+d)(b+d)=9/4Only option (c) satisfies this => q2 –p2 = 25/4 – 4 =9/4
Question 9. The largest interval for which x12 – x9 + x4 – x + 1 > 0 is:
1.    -4 < x < 0
2.    0 < x < 1
3.    -100 < x < 100
:
D