# Free Algebra 04 Practice Test - CAT

### Question 1

Find the value of , given that

#### SOLUTION

Solution :C

### Question 2

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

#### 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(a

^{2}+1)^{2}-12(a^{2}-3a+2) ≥0 and (a^{2}-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 P^{2} + 625/P^{2 ?}

#### SOLUTION

Solution :C

### Question 4

y = px^{2}+12x+9 has real roots. What is the range for the value of p?

#### 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= b

^{2}-4ac≥012

^{2}-(4xpx9) ≥0144-36p≥0

144≥36p

p≤4

### Question 5

Given that b^{3}+mb^{2} +nb + c is divisible by (b-s) if s^{3} + ms^{2} + ns + c = 0Also given that d^{3}+d^{2}+ed+1 is divisible by (d-1) and d^{3}-4d^{2}+fd-3 is divisible by (d-3)What is the value of e+f ?

#### SOLUTION

Solution :C

Option (c)

d

^{3}+d^{2}+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=-3

Similarly put d=3 in d

^{3}-4d^{2}+fd-3 to get the value of f27-36+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

#### SOLUTION

Solution :C

Option c

(x+1)(x+8)+10 can be expanded as x

^{2}+9x+8+10=0 x^{2}+9x+18=0Sum of roots p+q= -9

Product of roots pq = 18

(x+p)(x+q)-4=0 can be written as x

^{2}+ (p+q)x+pq-4=0 x^{2}-9x+18-4=0x

^{2}-9x+14=0Roots are 7 and 2

### Question 7

Given that p and q are the roots of the equation x^{2} – ax +b =0 and D_{n}= p^{n}+q^{n}. Find the value of D_{n+1}

_{n}-bD

_{n-1}

_{n}

_{n}

_{+}bD

_{n-1}

_{n}

#### SOLUTION

Solution :A

Option (a)

Best way to proceed is by assumption of values.

Assume a quadratic equation. Let’s take x

^{2}+5x-6=0. Here the roots are p=1 and q=-6Thus, a = sum of roots = -5 and b= product of roots = -6

D

_{1}= p^{1}+q^{1}= -5D

_{0}= 2D

_{2}= 37Assume n=1

We need to find D

_{n+1}= D_{2}= 37Look 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 mx^{2}+nx+p=0 are

#### SOLUTION

Solution :B

option (b)

x=(-b±√D)/2a

D=n

^{2}-4mp<n^{2}(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

#### 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

^{3}+t

^{2}+t+1

^{3}+t

^{2}-5

^{3}+ √5t

^{2}+√5t -2

^{3}-1

^{3}

#### 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)=a^{2}-b^{2}(25+3)/4, since i

^{2}=-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 x^{2}+px+1=0 and ; c and d are the roots of the equation x^{2}+qx+1=0, then (a-c)(b-c)(a+d)(b+d)=?

^{2}-q

^{2}

^{2}+b

^{2}

^{2}-p

^{2}

^{2}+p

^{2}

^{2}

#### 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 => q

^{2}–p^{2}= 25/4 – 4 =9/4

### Question 12

If the product of the roots for the equation x^{2} – 3kx +2e^{2ln(k)} -1=0 is 7, then the roots of the equation are real for k= ?

#### SOLUTION

Solution :C

Answer = c

Product of roots = 2e

^{2ln(k)}-1 = 7; 2e^{2ln(k)}= 8; e^{2ln(k) }= 4; k^{2 }= 4; k = ±2For the roots to be real, k>0 k=2. to recheck CHECK for D>0; D= 9k

^{2}-4 (2e^{2ln(k)}-1) = 8>0

### Question 13

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

#### SOLUTION

Solution :C

Option (c)

Go from answer options

1) Take p=2

a

^{2}–a +2=0 and a^{2}– a + 6=0Roots are = 1+√(1+24)2 and √(1−24)2

Hence, this option is ruled out

Take p = -2

a

^{2}–a -2=0 and a^{2}– a -6=0Roots of 1

^{st}equation = 2,-1 and roots of second equation = 3, -2

### Question 14

**What is the value of X ?**

#### 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.**

#### SOLUTION

Solution :C

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

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