Algebra Practice Questions for CAT with Solutions

Question 1:
If a, b, and g are the roots of the equation x3 – 4x2 + 3x + 5 = 0, find (a + 1)(b + 1)(g + 1)
[1] -3
[2] 0
[3] 3
[4] 1
f(x) = x3 – 4x2 + 3x + 5 = (x – a)(x – b)(x – g)
f(-1) = – 1 – 4 – 3 + 5 = (– 1 – a)(– 1 – b)(– 1 – g) =  – (a + 1)(b + 1)(g + 1)
(a + 1)(b + 1)(g + 1) = 3. Option C

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Question 2:
Let A = (x – 1)4 + 3(x – 1)3 + 6(x – 1)2 + 5(x – 1) + 1. Then the value of A is:
[1] (x – 2)4
[2] x4
[3] (x + 1)4
[4] None of these
A = f(x) = (x – 1)4 + 3(x – 1)3 + 6(x – 1)2 + 5(x – 1) + 1
f(1) = 1
f(2) = 1 + 3 + 6 + 5 + 1 = 16
f(3) = 16 + 24 + 24 + 10 + 1 = 75

Now check the options which satisfy these values. Put x = 3, we get

Option A is 1. Option B is 81. Option C is 256. All of them are invalid. None of these. Option D

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Question 3:
Find the remainder when 3x5 + 2x4 – 3x3 – x2 + 2x + 2 is divided by x2 – 1.
[1] 3
[2] 2x – 2
[3] 2x + 3
[4] 2x – 1
When 3x5 + 2x4 – 3x3 – x2 + 2x + 2 is divided by (x – 1), the remainder can be obtained by putting the value of x as 1 = 3 + 2 – 3 – 1 + 2 + 2 = 5

When 3x5 + 2x4 – 3x3 – x2 + 2x + 2 is divided by (x + 1), the remainder can be obtained by putting the value of x as – 1 = – 3 + 2 + 3 – 1 – 2 + 2 = 1

Check with the options by putting in the value of x as 1 & –1

Option A is 3. Invalid

Option B is 0 and – 4. Invalid

Option C is 5 & –1. So, the valid value of the remainder is 2x + 3. Option C

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Question 4:
A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f (x) at x = 10?
[1] -105
[2] -119
[3] -159
[4] -110
If the function attains the maximum of 3 at x = 1

f(x) = p(x – 1)2 + 3

f(0) = p + 3 = 1 {It is given as 1}
p = –2
f(x) = –2(x – 1)2 + 3
f(10) = –2(10 – 1)2 + 3 = –162 + 3 = –159. Option C

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Question 5:
$x + \frac{1}{x} = 3$then, what is the value of ${x^5} + \frac{1}{{x{}^5}}.$
[1] 123
[2] 144
[3] 159
[4] 186
$\begin{array}{l}x + \frac{1}{x} = 3\\{x^2} + \frac{1}{{{x^2}}} + 2 = 9{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{\{ Squaring both sides\} }}\\{x^2} + \frac{1}{{{x^2}}} = 7{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{Equation (2)}}\\({x^2} + \frac{1}{{{x^2}}})(x + \frac{1}{x}) = 7 \times 3\\{x^3} + x + \frac{1}{x} + \frac{1}{{{x^3}}} = 21\\{x^3} + \frac{1}{{{x^3}}} = 21 - (x + \frac{1}{x}) = 21 - 3\\{x^3} + \frac{1}{{{x^3}}} = 18{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{ \ldots Equation (3)}}\\{x^4} + \frac{1}{{{x^4}}} + 2 = 49{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{\{ Squaring both sides of Equation (2)\} }}\\{x^4} + \frac{1}{{{x^4}}} = 47{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{ \ldots Equation (4)}}\\({x^4} + \frac{1}{{{x^4}}})(x + \frac{1}{x}) = 47 \times 3\\{x^5} + {x^3} + \frac{1}{{{x^3}}} + \frac{1}{{{x^5}}} = 141\\{x^5} + \frac{1}{{{x^5}}} = 141 = ({x^3} + \frac{1}{{{x^3}}}) = 141 - 18\\{x^5} + \frac{1}{{{x^5}}} = 123\end{array}$

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Algebra Practice Questions for CAT with Solutions
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