Free Algebra 04 Practice Test - CAT 

Question 1

Find the value of      ,     given that      

A. 10
B. 6
C. 14
D. 21
E. cannot be determined

SOLUTION

Solution : C

Question 2

The value of a for which the quadratic equation 3x2+2(a2+1) x+ (a2-3a+2) =0 possesses real roots of the opposite sign lies in

A.
B.
C. (1, 2)
D. (3/2, 2)
E. [1, 2]

SOLUTION

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 roots

4(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

Question 3

Given that P ≥ 1. What is the minimum value of P2 + 625/P2 ?

A. 10
B. 25
C. 50
D. 75
E. 5

SOLUTION

Solution : C

Question 4

y = px2+12x+9 has real roots. What is the range for the value of p?

A. p<4
B. p≤ 1/4
C. p≤4
D. p≤ 3/4
E. p> 4

SOLUTION

Solution : C

Option (c)

Since the condition D≥0 is given (as the roots are real), we will find the value of p using this condition

D= b2-4ac≥0

122-(4xpx9) ≥0

144-36p≥0

144≥36p

p≤4

Question 5

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 ?

A. 0
B. -1
C. 1
D. 2
E. none of these

SOLUTION

Solution : 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 e

1+1+e+1=0 e=-3

Similarly put d=3 in d3-4d2+fd-3 to get the value of f

27-36+3f-3=0 3f=12 f=4

e+f= -3+4=1

Question 6

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 

A. -1, 3
B. 7, 8
C. 2,7
D. –2, 7
E. –2, 3

SOLUTION

Solution : 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=0

x2-9x+14=0

Roots are 7 and 2

Question 7

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 

A. aDn-bDn-1
B. aDn
C. bDn
D. aDn + bDn-1
E. Dn

SOLUTION

Solution : 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=-6

Thus, a = sum of roots = -5 and b= product of roots = -6

D1= p1+q1 = -5

D0 = 2

D2 = 37

Assume n=1

We need to find Dn+1 = D2 = 37

Look in the answer options for 37

Option a is the only one which gives (-5)(-5)-(-6)(2) = 37

Question 8

if m>0, n>0 and p>0, then both the roots of the equation mx2+nx+p=0 are

A. real and negative
B. have –ve real parts
C. rational numbers
D. cannot be determined
E. none of these

SOLUTION

Solution : B

option (b)

x=(-b±√D)/2a

D=n2-4mp<n2 (as m>0 and p>0), the roots of this equation will be –ve. If D<0, then the roots will have negative real parts

Question 9

The equation has

A. no solution
B. 1
C. 2 solutions
D. 3 solutions
E. >3 solutions

SOLUTION

Solution : A

Option (a)

Squaring both sides

 

For x=5/4 LHS≠RHS

Hence x=5/4 is not a root of the equation.

Answer=a

Question 10

  

A. t3+t2+t+1
B. t3+t2-5
C. t3+ √5t2 +√5t -2
D. t3-1
E. t3

SOLUTION

Solution : D

Option (d)

Assume t= 2

The 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

=7

Only answer option (d) gives 7 on substitution of t=2

Question 11

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)=? 

A. p2-q2
B. a2+b2
C. q2-p2
D. q2+p2
E. p2

SOLUTION

Solution : C

Shortcut: Assumption

Option (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= 2

Assume c= - 1/2 and d= - 2; q= 5/2

Question is = (a-c)(b-c)(a+d)(b+d)=9/4

Only option (c) satisfies this => q2 –p2 = 25/4 – 4 =9/4

Question 12

If the product of the roots for the equation x2 – 3kx +2e2ln(k) -1=0 is 7, then the roots of the equation are real for k= ?

A. 0
B. 1
C. 2
D. 4

SOLUTION

Solution : C

Answer = c

Product of roots = 2e2ln(k) -1 = 7; 2e2ln(k) = 8; e2ln(k) = 4; k2 = 4; k = ±2

For the roots to be real, k>0 k=2. to recheck CHECK for D>0; D= 9k2 -4 (2e2ln(k) -1) = 8>0

Question 13

There are 2 quadratic equations, a2 –a +p=0 and a2 – a + 3p=0 (p≠0). For what value of p will one root of the second equation = double the root of the first equation? 

A. 2
B. 3
C. -2
D. –3
E. none of these

SOLUTION

Solution : 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

Question 14

What is the value of X ?

A. 3
B. 2
C. 1
D. 0

SOLUTION

Solution : C

WXYZ = WXY * XZ => WXY (10) + Z = WXY (10x) + WXYZ.

Comparing we get => X = 1 and Z = 0

Question 15

Find the value of Z.

A. 2
B. 1
C. 0
D. 3

SOLUTION

Solution : C

WXYZ = WXY * XZ => WXY (10) + Z = WXY (10x) + WXYZ.

Comparing we get => X = 1 and Z = 0

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