# Algebra Practice Questions for CAT with Solutions

Question 1:
If x =2+22/3+21/3, then the value of x3-6x2+6x is:
[1] 2
[2] -2
[3] 0
[4] 4
x =2+22/3+21/3
x – 2 = 22/3+21/3
(x – 2)3 = (22/3+21/3)3
x3 – 6x2 + 12x – 8 = 22 + 3. 24/3.21/3 + 3. 22/3.22/3 + 2
x3 – 6x2 + 12x – 8 = 4 + 3.25/3 + 3.24/3 + 2
x3 – 6x2 + 12x – 8 = 6 + 6.22/3 + 6.21/3 = (12 + 6.22/3 + 6.21/3) – 6
x3 – 6x2 + 12x – 8 = 6x – 6
x3 – 6x2 + 6x = 2. Option A

Question 2:
If the roots of the equation x2-2ax+a2+a-3=0 are real and less than 3, then
[1] a < 2
[2] 2 < a < 3
[3] 3 < a < 4
[4] a > 4
For the roots to be real

4a^2 – 4(a^2 + a – 3) ≥ 0

– (a – 3) ≥ 0
a ≤ 3

Answer could be Option (a) or Option (b)

Put a = 0, we get the equation as x2 – 3 = 0. This equation has real roots and both of them are less than 3. So, a = 0 is valid solution.

a = 0, is not a part of the solution 2 < a < 3 but it is a part of a < 2. Option A

Question 3:
Find the value of $\sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt {2 + .....} } } }$
[1] -1
[2] 1
[3] 2
[4] $\frac{{\sqrt 2 + 1}}{2}$
$\sqrt { 2 + \sqrt { 2 + \sqrt { 2 + \sqrt { 2 + \ldots } } } } = x$ $\sqrt { 2 + x } = x$ $2 + x = x ^ { 2 }$ $x ^ { 2 } - x - 2 = 0$ $x = 2 , - 1$

Question 4:
If a, b and c are the roots of the equation x3 – 3x2 + x + 1 = 0 find the value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$
[1] 1
[2] -1
[3] 1/3
[4] -1/3
$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{{ab + bc + ca}}{{abc}} = \frac{{\frac{{Coefficient\;of\;x}}{{Coefficient\;of\;{x^3}}}}}{{\frac{{ - \;Const}}{{Coefficient\;of\;{x^3}}}}} = - \frac{{Coefficient\;of\;x}}{{Const}} = - 1$Option B

Question 5:
If p, q and r are the roots of the equation 2z3 + 4z2 -3z -1 =0, find the value of (1 - p) ´ (1 - q) ´ (1 - r)
[1] -2
[2] 0
[3] 2
[4] None of these
If p, q and r are the roots of the equation 2z3 + 4z2 -3z -1 =0, then

f(z) = 2z3 + 4z2 -3z -1 = (z - p) ´ (z - q) ´ (z - r)

f(1) = 2 + 4 – 3 – 1 = (1 - p) ´ (1 - q) ´ (1 - r)
(1 - p) ´ (1 - q) ´ (1 - r) = 2. Option C