M.C.Q

MCQ 1011 Mark

Let f(x) be a polynomial such that $\text{f}\Big(-\frac{1}{2}\Big)=0,$ then a factor of f(x) is:

A
2x - 1
B
2x + 1
C
x - 1
D
x + 1

Answer

2x + 1
Solution:
If f(x) is a polynomial and $\text{f}(\alpha)=0.$ Then $(\text{x}-\alpha)$ is a factor of f(x) or vice versa if $(\text{x}-\alpha)$ is a factor of f(x) then $\text{f}(\alpha)=0.$
Now,
$\text{f}\Big(\frac{-1}{2}\Big)=0$
So, at $\text{x}=\frac{-1}{2},\text{f(x)}=0$
Or at 2x = -1, f(x) = 0
Or at 2x + 1 = 0, f(x) = 0
⇒ (2x + 1) is a factor of f(x).

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MCQ 1021 Mark

The factors of x³ - x²y - xy²+ y³ are:

A
(x + y)(x² - xy + y²)
B
(x + y)(x² + xy + y²)
C
(x + y)²(x - y)
D
(x - y)²(x + y)

Answer

(x - y)²(x + y)
Solution:
x³ - x²y - xy² + y³ = x³ + y³ - xy(x + y)
Now by identity x³ + y³ = (x + y)(x² + y² - xy), we have
x³ - x²y - xy² + y³ = (x + y)(x² + y² - xy) - xy(x + y)
= (x + y)(x² + y² - xy - xy)
= (x + y)(x² + y² - 2xy)
= (x + y)(x - y)²Hence, correct option is (d).

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MCQ 1031 Mark

If $\text{x}^3+\frac{1}{\text{x}^3}=110,$ then $\text{x}+\frac{1}{\text{x}}=$

A
5
B
10
C
15
D
None of these.

Answer

5
Solution:
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3-3\Big(\text{x}+\frac{1}{\text{x}}\Big)=\text{x}^3+\frac{1}{\text{x}^3}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3-3\Big(\text{x}+\frac{1}{\text{x}}\Big)=110$
Let $\text{x}+\frac{1}{\text{x}}=\text{t}$
$\Rightarrow\text{t}^3-3\text{t}-110=0$
t =5 is one of it's solution which is real, other two solutions are imaginary
$\Rightarrow\text{x}+\frac{1}{\text{x}}=5$
Hence, correct option is (a).

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MCQ 1041 Mark

If $\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}=-1,$ then a³ - b³ =

A
$1$
B
$-1$
C
$\frac{1}{2}$
D
$0$

Answer

$0$
Solution:
$\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}=-1$
$\Rightarrow\frac{\text{a}^2+\text{b}^2}{\text{ab}}=-1$
$\Rightarrow\text{a}^2+\text{b}^2+\text{ab}=0$
Now using identity
a³ - b³= (a - b)(a² + b² + ab)
=(a - b)(0) $(\because$ a² + b² + ab = 0$)$
= 0
Hence, correct option is (d).

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MCQ 1051 Mark

The coefficient of x² in the expansion of (x + 3)⁴ is:

A
27
B
54
C
3
D
1

Answer

54
Solution:
(x + 3)⁴= (x + 3)² × (x + 3)²= [x² + 6x + 9] [x² + 6x + 9]
= x⁴ + 36x² + 81 + 12x³ + 108x + 18x²= x⁴ + 12x³ + 54x² + 108x + 81
Therefore, the coefficient of x² is 54.

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MCQ 1061 Mark

Let f(x) be a polynomial such that $\text{f}(-\frac{1}{2})=0,$ then a factor of f(x) is:

A
2x + 1
B
x - 1
C
x + 1
D
2x - 1

Answer

2x + 1
Solution:
Let f(x) be a polynomial such that $\text{f}(-\frac{1}{2})=0,$
i.e., $\text{x}+\frac{1}{2}=0$ is a factors.
On rearranging $\text{f}(-\frac{1}{2})=0 $ can be written as (2x + 1) = 0
Thus, (2x + 1) is a factor of f(x).

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MCQ 1071 Mark

The monomial of degree 50 is:

A
x + 50
B
x⁵⁰ + 1
C
50
D
2x⁵⁰

Answer

2x⁵⁰Solution:
Monomial means only one term and degree means highest power.

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MCQ 1081 Mark

If $\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}+\text{c}^{\frac{1}{3}}=0,$ than.

A
a³ + b³ + c³ = 0
B
a + b + c
C
(a + b + c)³ = 27abc
D
a + b + c = 3abc

Answer

(a + b + c)³ = 27abc
Solution:
$\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}+\text{c}^{\frac{1}{3}}=0$
$\Rightarrow\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}=-\text{c}^{\frac{1}{3}}$
$\Rightarrow\Big[(\text{a}^{\frac{1}{3}})(\text{b}^{\frac{1}{3}})\Big]^3=\Big(-\text{c}^{\frac{1}{3}}\Big)^3$
$\Rightarrow\text{a}+\text{b}+\Big[3\times\text{a}^{\frac{1}{3}}\times\text{b}^{\frac{1}{3}}\Big(\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}\Big)\Big]=-\text{c}$
$\Rightarrow\text{a}+\text{b}+3\times\text{a}^{\frac{1}{3}}\times\text{b}^{\frac{1}{3}}\Big(-\text{c}^{\frac{1}{3}}\Big)=-\text{c}$
$\Rightarrow\text{a}+\text{b}+\text{c}=3\times\text{a}^{\frac{1}{3}}\times\text{b}^{\frac{1}{3}}\times\text{c}^{\frac{1}{3}}$
$\Rightarrow(\text{a}+\text{b}+\text{c})^3=\Big(3\times\text{a}^{\frac{1}{3}}\times\text{b}^{\frac{1}{3}}\times\text{c}^{\frac{1}{3}}\Big)$
$\Rightarrow(\text{a}+\text{b}+\text{c})^3=27\text{abc}$

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MCQ 1091 Mark

If x¹⁴⁰ + 2x¹⁵¹ + k is divisible by x + 1, then the value of k is:

A
1
B
-3
C
2
D
-2

Answer

1
Solution:
Let p(x) = x¹⁴⁰ + 2x¹⁵¹ + k
Since p(x) is divisible by (x + 1),
(x + 1) is a factor of p(x).
So,
p(-1) = 0
(-1)¹⁴⁰ + 2(-1)¹⁵¹ + k = 0
1 + 2(-1) + k = 0
1 - 2 + k = 0
k - 1 = 0
k = 1

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MCQ 1101 Mark

The factors of x³ - x²y - xy² + y³, are.

A
(x + y)² (x- y)
B
(x + y) (x² - xy + y²)
C
(x - y)² (x + y)
D
(x + y) (x² + xy + y²)

Answer

(x - y)² (x + y)
Solution:
The given expression to be factorized is x³ - x²y - xy² + y³Take common x² from the first two terms and -y² from the last two terms,
That is x³ - x²y - xy² + y³ = x²(x - y) - y²(x - y)
Finally, take common (x - y) from the two terms,
That is x³ - x²y - xy² + y³ = (x - y) (x² - y²)
= (x - y) {(x² - y²)}
= (x - y) (x + y) (x - y)
= (x - y)² (x + y)

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MCQ 1111 Mark

A polynomial of degree _____ is called a cubic polynomial.

A
2
B
1
C
0
D
3.

Answer

3
Solution:
A polynomial of degree 3 is called a cubic polynomial.
Its general form is ax³ + bx² + cx + d.

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MCQ 1121 Mark

Which of the following expressions is a polynomial?

A
$\frac{\text{x}-1}{\text{x}+1}$
B
$\sqrt{\text{x}}-1$
C
$\text{x}^2+\frac{2\text{x}^{\frac{3}{2}}}{\sqrt{\text{x}}}+6$
D
$\text{x}^2-\frac{2}{\text{x}^2}+5$

Answer

$\text{x}^2+\frac{2\text{x}^{\frac{3}{2}}}{\sqrt{\text{x}}}+6$
Solution:
We have: $\text{x}^2+\frac{2\text{x}^{\frac{3}{2}}}{\sqrt{\text{x}}}+6=\text{x}^2+2\text{x}^{\frac{3}{2}}\text{x}^{\frac{-1}{2}}+6$
$=\text{x}^2+2\text{x}+6$
It is a polynomial because it has only non-negative integral powers of x

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MCQ 1131 Mark

If $\text{x}^3-\frac{1}{\text{x}^3}=14,$ then $\text{x}-\frac{1}{\text{x}}=$

Answer

2
Solution:
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\text{x}^3-\frac{1}{\text{x}^3}-3\not\text{x}\frac{1}{\not\text{x}}\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\text{x}^3-\frac{1}{\text{x}^3}=\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-\text{x}^3-\frac{1}{\text{x}^3}=0$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-14=0$
Let $\Rightarrow\text{x}-\frac{1}{\text{x}}=\text{t}$
⇒ t³ + 3t - 14 = 0
⇒ t³ - 2t² + 2t² - 4t + 7t -14 = 0
⇒ t(t - 2) + 2t(t - 2) + 7(t - 2) = 0
⇒ (t - 2)(t + 2t + 7) = 0
⇒ t² + 2t + 7 = 0 has no real roots
So, t = 2 is a solution
$\Rightarrow\text{x}-\frac{1}{\text{x}}=2$
Hence, correct option is (d).

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MCQ 1141 Mark

If x² - 1 is a factor of ax⁴+ bx³ + cx² + dx + e, then

A
a + c + e = b + d
B
a + b + e = c + d
C
a + b + c = d + e
D
b + c + d = a + e

Answer

a + c + e = b + d
Solution:
If x² - 1 is factor of p(x) = ax⁴+ bx³ + cx² + dx + e.
Then (x - 1) and (x + 1) will also be factors of p(x).
Because x² - 1 = (x - 1)(x + 1)
Then, at x = 1 and x = -1, p(x) = 0
⇒ p(1) = 0 and p(-1) = 0
⇒ a + b + c + d + e = 0 ...(1)
And
⇒ a - b + c - d + e = 0 ...(2)
Adding equations (1) and (2).
2a + 2c + 2e = 0
⇒ a + c + e = 0 ...(3)
Substracting equation (2) from (1)
2b + 2d = 0
⇒ b + d = 0 ...(4)
From equations (3) and (4), we get
a + c + e = b + d

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MCQ 1151 Mark

If 10x - 4x² - 3, then the value of p(0) + p(1) is:

A
3
B
-3
C
0
D
1

Answer

0
Solution:
10x - 4x² - 3,
p(x) -4x² + 10x - 3
⇒ p(0) + p(1) = [-4(0)² + 10(0) - 3] + [-4(1)² + 10(1) - 3]
⇒ p(0) + p(1) = [0 + 0 - 3] + [-4 + 10 - 3]
⇒ p(0) + p(1) = [-3] + [3]
⇒ p(0) + p(1) = 0

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MCQ 1161 Mark

One of the factors of (25x² - 1) + (1 + 5x)² is:

A
5 + x
B
5 - x
C
5x - 1
D
10x

Answer

10x
Solution:
(25x² - 1) + (1 + 5x)²= (5x - 1)(5x + 1) + (1 + 5x)²= (5x + 1)[(5x - 1) + (1 + 5x)]
= (5x + 1)(10x)
So, the factors of (25x² - 1) + (1 + 5x)² are (5x + 1) and 10x

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MCQ 1171 Mark

If $\text{x}^2+\frac{1}{\text{x}^2}=102,$ then $\text{x}-\frac{1}{\text{x}}=$

A
8
B
10
C
12
D
13

Answer

10
Solution:
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2(\text{x})\frac{1}{\text{x}}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$=102-2$ $\Big\{\text{x}^2+\frac{1}{\text{x}^2}=102\Big\}$
$=100$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=\sqrt{100}$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=10$
Hence, correct option is (b).

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MCQ 1181 Mark

(305 × 308) = ?

A
94940
B
93840
C
93940
D
94840

Answer

93940
Solution:
305 × 308 = (300 + 5)(300 + 8)
= (300)² + 300 × (5 + 8) + 5 × 8
= 90000 + 3900 + 40
= 93940

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MCQ 1191 Mark

f (3x - 1)⁷ = a₇x⁷ + a₆x⁶ + a₅x⁵ + ... + a₁x + a₀, then a₇ + a₆ + a₅ + ... + a₁ + a₀ =

A
0
B
1
C
128
D
64

Answer

128
Solution:
(3x - 1)⁷ = a₇x⁷ + a₆x⁶ + ... + a₁x + a_{0 ...(1)}Putting x = 1 in equation (1), we have
[3(1) - 1]⁷ = a₇ + a₆ + ..... + a₁ + a₀So,
a₇ + a₆ + a₅ + ..... + a₁ + a₀ = 2⁷= 128

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MCQ 1201 Mark

The possible expressions for the length, breadth and height of the cuboid whose volume is given by 3x³ - 12x is:

A
3x, (x + 2) and (x - 2)
B
x, (3x + 2) and (x - 2)
C
x, (x + 2) and (3x - 2)
D
None of these.

Answer

3x, (x + 2) and (x - 2)
Solution:
To find the length, breadth and height, we will factorize the given polynomial.
3x³ - 12x
= 3x[x² - 4]
= 3x[x² - (2)²]
= 3x(x + 2) (x + 2)
Therefore, the possible expressions for the length, breadth and height of the cuboid whose volume is given by 3x³ - 12x are 3x, (x + 2) and (x - 2).

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MCQ 1211 Mark

If x + 2 is a factor of x² + mx + 14, then m =

A
7
B
2
C
9
D
14

Answer

9
Solution:
If x + 2 is a factor of x² + mx + 14,
then at x = -2,
x² + mx + 14 = 0
i.e. (-2)² + m(-2) + 14 = 0
4 - 2m + 14 = 0
2m = 18
m = 9

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MCQ 1221 Mark

If x - a is a factor of x³ - 3x²a + 2a²x + b, then the value of b is:

Answer

0
Solution:
If x - a is a factor then x - a = 0
Therefore, x = a
New substitute x = a in polynomial
(a³) - 3(a)²a + 2a²(a) + b = 0
a³ - 3a³ + 2a³ + b = 0
-3a³ + 3a³ + b = 0
b = 0

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MCQ 1231 Mark

The degree of the polynomial 4x⁴ +0x³ + 0x⁵ + 5x + 7 is:

Answer

4
Solution:
4x⁴ +0x³ + 0x⁵ + 5x + 7
= 4x⁴+ 5x + 7
Here, the height power is 4.
Therefore, the degree of given polynomial is 4.

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MCQ 1241 Mark

If (x + 5) is a factor of p(x) = x³ - 20x + 5k then k = ?

A
-5
B
-3
C
3
D
5

Answer

5
Solution:
(x + 5) is a factor or p(x) = x³ - 20x +5k
$\therefore$ p(-5) = 0
⇒ (-5)³ -20 × (-5) + 5k = 0
⇒ -125 + 100 + 5k = 0
⇒ 5k = 25
⇒ k = 5

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MCQ 1251 Mark

Write the correct answer in the following:
x + 1 is a factor of the polynomial.

A
x³ + x² - x + 1
B
x³ + x² + x + 1
C
x⁴ + x³ + x² + 1
D
-x⁴ + 3x³ + 3x²+ x + 1

Answer

x³ + x² + x + 1
Solution:
Let assume (x + 1) is a factor of x³ + x² + x + 1
So, x = -1 is zero of x³ + x² + x + 1
(-1)³ + (-1)² + (-1) + 1 = 0
⇒ -1 + 1 - 1 + 1 = 0
⇒ 0 = 0
Hence, our assumption is true.

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MCQ 1261 Mark

he coefficient of x³ in 2x + x² - 5x³ + x⁴ is:

A
-1
B
2
C
1
D
-5

Answer

-5
Solution:
The coefficient of x³ in 2x + x² -5x³ + x⁴ is -5.

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MCQ 1271 Mark

If (x¹⁰⁰ + 2x⁹⁹ + k) is divisible by (x + 1) then the value of k is:

A
1
B
2
C
-2
D
-3

Answer

1
Solution:
p(x) = x¹⁰⁰ + 2x⁹⁹ + k
x + 1 = 0 ⇒ x = -1
By the factor theorem, we know that when p(x) is divided by (x + 1), the remainder is p(-1).
Now, p(-1) = (-1)¹⁰⁰ + 2(-1)⁹⁹ + k
⇒ 0 = 1 - 2 + k ...(Given that p(x) is divisible by x + 1.)
⇒ k = 1

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MCQ 1281 Mark

Write the correct answer in the following:
One of the factors of (25x² - 1) + (1 + 5x)² is.

A
5 + x
B
5 - x
C
5x - 1
D
10x

Answer

10x
Solution:
(25x² - 1) + (1 + 5x)² = (5x)² - 1² + (5x + 1)²= (5x - 1)(5x - 1) + (5x + 1)² = (5x + 1)(5x - 1 + 5x + 1)
= (5x + 1)(10x) = 10x(5x + 1)
Hence, one of the factors of (25x² - 1) + (1 + 5x)² is 10x.

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MCQ 1291 Mark

If x + 1 is a factor of the polynomial 2x² + kx + 1, then the value of 'k' is:

A
3
B
-3
C
-2
D
2

Answer

3
Solution:
If x + 1 is a factor of p(x) = 2x² + kx + 1, then p(-1) = 0
⇒ 2x² + kx + 1 = 0
⇒ 2(-1)² + k(-1) + 1 = 0
⇒ 2 - k + 1 = 0
⇒ k = 3

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MCQ 1301 Mark

If $\text{x}^4+\frac{1}{\text{x}^4}=194,$ than $\text{x}^3+\frac{1}{\text{x}}^3=$

A
64
B
52
C
76
D
None of these.

Answer

52
Solution:
$\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)=194$
$\Rightarrow(\text{x}^2)^2+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}^2\times\frac{1}{\text{x}^2}=194+2\times\text{x}^2\times\frac{1}{\text{x}^2}$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=196$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=\sqrt{196}=14$
Now,
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}\times\frac{1}{\text{x}}=14+2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=16$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=\sqrt{16}=4$
Now, $\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=(4)^3$
$\Rightarrow(\text{x})^3+\Big(\frac{1}{\text{x}}\Big)^3+3\times\text{x}\times\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)=64$
$\Rightarrow(\text{x}^3)+\Big(\frac{1}{\text{x}^3}\Big)+3(4)=64$
$\Rightarrow(\text{x}^3)+\Big(\frac{1}{\text{x}^3}\Big)=64-12 = 52$

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MCQ 1311 Mark

Which of the following is a factor of (x + y)³ - (x³ + y³)?

A
xy²
B
x² + y² + 2xy
C
3xy
D
x² + y² - xy

Answer

3xy
Solution:
Given polynomial: (x + y)³ - (x³ + y³)
= (x³ + y³ + 3x²y + 3xy²) - (x³ + y³) [$\because$ (a + b)³ = (a³ + b³ + 3a²b + 3ab²)]
=3x²y + 3xy² = 3xy(x + y)
⇒ 3xy and (x + y) are factors of given polynomial.

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MCQ 1321 Mark

If $\text{x}^2+\frac{1}{\text{x}^2}=38,$ then the value of $\text{x}-\frac{1}{\text{x}}$ is:

Answer

6
Solution:
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=38-2$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=36$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=\pm6$

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MCQ 1331 Mark

The factors of x³ - 7x - 6 are.

A
x(x - 6) (x - 1)
B
(x - 1) (x - 3) (x + 2)
C
(x + 1) (x + 2) (x - 3)
D
(x² - 6) (x - 1)

Answer

(x + 1) (x + 2) (x - 3)
Solution:
The given expression to be factorized is x³ - 7x + 6
This can be written in the form
x³ - 7x + 6 = x³ - (1 + 6)x + 6
= x³ - x - 6x + 6
Take common x from the first two terms and -6 from the last two terms.
Then we have,
x³ + 7x + 6 = x(x² - 1) - 6(x - 1)
= x{(x)² - (1)²} - 6{x - 1}
= x(x + 1)(x - 1) - 6(x - 1)
Finally, take common (x - 1) from the above expression,
x³ - 7x + 6 = (x - 1) {(x + 1) - 6}
= (x - 1) (x² + x - 6)
= (x - 1) (x² + 3x - 2x - 6)
= (x - 1) {x(x + 3) - 2(x + 3)}
= (x - 1) (x + 3) (x - 2)

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MCQ 1341 Mark

If $\text{p}(\text{x})=\text{x}^2-2\sqrt{2}\text{x}+1,$ than $\text{p}(2\sqrt{2})$ is equal to:

A
$0$
B
$4\sqrt{2}$
C
$8\sqrt{2}+1$
D
$1$

Answer

$1$
Solution:
$\text{p}(\text{x})=\text{x}^2-2\sqrt{2}\text{x}+1,$
$\Rightarrow\text{p}(2\sqrt{2})=(2\sqrt{2})^2-2\sqrt{2}(2\sqrt{2})+1$
$\Rightarrow\text{p}(2\sqrt{2})=8-8+1$
$\Rightarrow\text{p}(2\sqrt{2})=1$

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MCQ 1351 Mark

The value of $\frac{(2.3)^3-0.027}{(2.3)^20.69+0.09}.$

A
2.327
B
2.273
C
2
D
3

Answer

2
Solution:
The given expresstion is
$\frac{(2.3)^3-0.027}{(2.3)^20.69+0.09}$
This can be written in the form
$\frac{(2.3^3)-(0.3)^3}{(2.3)^2+2.3\times0.3+(0.3)^2}$
Assume a = 2.3 and b = 0.3. then the given expression can be rewritten as $\frac{\text{a}^3-\text{b}^3}{\text{a}^2+\text{ab}+\text{b}^2}$
Recall the formula for difference of two cubes
a³ - b³ = (a - b) (a² + ab + b²)
Using the above formula, the expression becomes $\frac{\text{(a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)}{\text{a}^2+\text{ab}+\text{b}^2}$
Note that both a and b are positive, unequal. so, neither a³ - b³ nor any factor of it can be zero.
Therefore we can cancel the term (a² + ab + b²) from both numerator and denominator. then the expression becomes
$\frac{(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)}{\text{a}^2+\text{ab}+\text{b}^2}=\text{a}-\text{b}$
$=2.3-0.3$
$=2$

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MCQ 1361 Mark

Write the correct answer in the following :
Which of the following is a factor of (x + y)³ - (x³+ y³)?

A
x² + y² + 2xy
B
x² + y² - xy
C
xy²
D
3xy

Answer

3xy
Solution:
(x + y)³ - (x³+ y³) = x³+ y³ + 3xy(x + y) - x³ - y³[(a + b)³ = a³ + b³ + 3ab(a + b)]
= 3xy(x + y)
So, 3xy is a factor of (x + y)³ - (x³+ y³).

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MCQ 1371 Mark

Write the correct answer in the following :
If $49\text{x}^2 -\text{b}=\Big(7\text{x}+\frac{1}{2}\Big)\Big(7\text{x}-\frac{1}{2}\Big),$ the value of b is.

A
$0$
B
$\frac{1}{\sqrt2}$
C
$\frac{1}{4}$
D
$\frac{1}{2}$

Answer

$\frac{1}{4}$
Solution:
$49\text{x}^2 -\text{b}=\Big(7\text{x}+\frac{1}{2}\Big)\Big(7\text{x}-\frac{1}{2}\Big)$
$\Rightarrow49\text{x}^2 -\text{b}=\Big(7\text{x}\Big)^2-\Big(\frac{1}{2}\Big)^2$
$49^2-\frac{1}{4} [\therefore(\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2]$
So, we get $\text{b}=\frac{1}{4}.$

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MCQ 1381 Mark

The factors of 12x² - x - 6 are :

A
(3x + 2) (4x - 3)
B
(12x - 1) (x + 6)
C
(3x - 2) (4x + 3)
D
(12x + 1) (x - 6)

Answer

(3x + 2) (4x - 3)
Solution:
12x² - x - 6
= 12x² - 9x + 8x - 6
= 3x(4x - 3) + 2(4x - 3)
= (3x + 2) (4x - 3)

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MCQ 1391 Mark

The factors of x² - 9 is:

A
(x - 3) (x - 3)
B
(x + 3) ( x + 3)
C
(x + 3) (x - 3)
D
(x - 3) (x + 9)

Answer

(x + 3) (x - 3)
Solution:
x² - 9
= x² - 3²Using identity a² - b² =(a + b) (a - b)
= (x + 3) (x - 3)

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MCQ 1401 Mark

A polynomial containing three non-zero terms is called a ________.

A
Binomial.
B
None of these.
C
Monomial.
D
Trinomial.

Answer

Trinomial.
Solution:
A polynomial containing three non-zero terms is called a trinomial.
Example: 5x² + 2x + 3, ax² = bx + c, 3x + 2y - 3

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MCQ 1411 Mark

The factors of x³ - 1 + y³ + 3xy are:

A
(x - 1 + y)(x² + 1 + y² + x + y - xy)
B
(x + y + 1)(x² + y² + 1 - xy - x - y)
C
(x - 1 + y)(x² - 1 - y² + x + y + xy)
D
3(x + y - 1)(x² + y² - 1)

Answer

(x - 1 + y)(x² + 1 + y² + x + y - xy)
Solution:
By using identity
a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)
We can write,
x³ - 1 + y³ + 3xy
= (x³) + (-1)³ + (y³) - 3(-1)(x)(y)
= [x + (-1) + y][x² + (-1)² + y² - x(-1) - y(-1) - xy]
= (x - 1 + y)(x² + 1 + y²+ x + y - xy)
Hence, correct option is (a).

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MCQ 1421 Mark

Which of the following is a polynomial?

A
x^-2 + x^-1 + 3
B
x + x^-1 + 2
C
x^-1
D
0

Answer

0
Solution:
A polynomial is an algebraic expression in which the variables involved have only non-negative integrals powers.
Option (a), (b) and (c) have negative and non-integral powers,
So they are not polynomials.
We know that, exery real number is a constant polynomial.
So, 0 being a real number is a polynomial.

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MCQ 1431 Mark

The value of $\frac{(\text{a}^2-\text{b}^2)^3(\text{b}^2-\text{c}^2)+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}$ is:

A
3(a - b) (b - c) (c - a)
B
3(a + b) (b + c) (c + a)
C
3(a + b) (b + c) (c + a) (a - b) (b - c) (c - a)
D
None of these.

Answer

3(a + b) (b + c) (c + a)
Solution:
$\frac{(\text{a}^2-\text{b}^2)^3(\text{b}^2-\text{c}^2)+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}$
$=\frac{3(\text{a}^2-\text{b}^2)(\text{b}^2-\text{c}^2)(\text{c}^2-\text{a}^2)}{3(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$ $[\text{Since}\text{ x}^3+\text{y}^3+\text{z}^3=3\text{xyz},\text{ if}\text{ x}+\text{y}+\text{z}=0]$
$=\frac{3(\text{a}-\text{b})(\text{a}+\text{b})(\text{b}-\text{c})(\text{b}+\text{c})(\text{c}-\text{a})(\text{c}+\text{a})}{3(\text{a}-\text{b})(\text{b}-\text{c}(\text{c})-\text{a})}$
$=3(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$

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MCQ 1441 Mark

If x³ - 3x² 3x - 7 = (x + 1) (ax² + bx + c), then a + b + c =

A
4
B
-10
C
12
D
3

Answer

4
Solution:
First multiply
(x + 1) (ax² + bx + c)
= ax³ + bx² + cx + ax² + bx + c
= ax³ + bx² + ax² + cx + c
= ax³ + (b + a)x² + (c + b)x + c
Comparing it with
x³ - 3x² + 3x - 7
a = 1
b + a = -3 ⇒ b + 1 + -3 ⇒ b = -4
c + b = 3 ⇒ c - 4 = 3 ⇒ c = 7
c = -7 should be 7
as if we put x = -1 in
x³ - 3x² + 3x - 7
-1 - 3 - 3 - 7 =14
so x + 1 can not be factor so x + 1 will be factor if x³ - 3x² + 3x - 7 is actually
x³ - 3x² + 3x + 7
then -1 -3 -3 + 7 = 0
Hence, we can say that
a = 1
b = -1
c = 7
so, a + b + c = 4

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MCQ 1451 Mark

The zeros of the polynomial p(x) = x² + x - 6 are:

A
2, 3
B
-2, 3
C
2, -3
D
-2, -3

Answer

2, -3
Solution:
Let p(x) be a polynomial. If $\text{p}(\alpha)=0,$ then we say that $\alpha$ is a zero of a polynomial.
p(x) = x² + x - 6
Now, p(x) = 0
⇒ x² + x - 6
⇒ x² + 3x - 2x - 6 = 0
⇒ x(x + 3) - 2(x + 3) = 0
⇒ (x - 2)(x + 3) = 0
⇒ (x - 2) = 0 or (x + 3) = 0
⇒ x = 2 or x = -3
$\therefore$ 2 and -3 are the zeroes of the polynomial p(x).

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MCQ 1461 Mark

The product (a + b)(a - b)(a² - ab + b²)(a² + ab + b²) is equal to:

A
a⁶ + b⁶
B
a⁶ - b⁶
C
a³ - b³
D
a³+ b³

Answer

a⁶ - b⁶Solution:
(a + b)(a - b)(a² - ab + b²)(a² + ab + b²)
= (a² - b²)(a² + b² - ab)(a² + b² - ab)
= (a² - b²) $\Big\{$(a² + b²)² - (ab)²$\Big\}$
= (a² - b²) {a⁴ + b⁴ + 2a²b² - a²b²}
= (a² - b²) {a⁴ + b⁴ + a²b²}
= {a⁶ + a²b⁴ + a⁴b² - b²a⁴ - b⁶ - b⁴a²}
= a⁶ - b⁶Hence, correct option is (b).

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MCQ 1471 Mark

If $3\text{x}+\frac{2}{\text{x}}=7,$ than $\Big(9\text{x}^2-\frac{4}{\text{x}^2}\Big)=$

A
25
B
35
C
30
D
49

Answer

35
Solution:
$\Rightarrow3\text{x}+\frac{2}{\text{x}}=7$
Now, $3\text{x}^2+2-7\text{x}=0$
$\Rightarrow3\text{x}^2-6\text{x}-\text{x}+2=0$
$\Rightarrow(\text{x}-2)(3\text{x}-1)=0$
$\Rightarrow\text{x}=2$ or $\text{x}=\frac{1}{4}$ (neglected)
$\Rightarrow3\text{x}-\frac{2}{\text{x}}=6-1=5$
$\Rightarrow3\text{x}+\frac{2}{\text{x}}=6+1=7$
Now,
$(3\text{x})^2-\Big(\frac{2}{\text{x}}\Big)^2=\Big(3\text{x}-\frac{2}{\text{x}}\Big)\Big(3\text{x}+\frac{2}{\text{x}}\Big)$
$\Rightarrow(3\text{x})^2-\Big(\frac{2}{\text{x}}\Big)^2=5(7)=35$

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MCQ 1481 Mark

When p(x) = x³ + ax² + 2x +a is divided by (x + a), the remainder is:

A
0
B
a
C
-a
D
2a

Answer

-a
Solution:
p(x) = x³ + ax² + 2x +a
x + a = 0 ⇒ x = -a
By the remainder theorem, we know that when p(x) is divided by (x + a), the remainder is p(-a).
Now, p(-a) = x³ + ax² + 2x +a
= (-a)³ + a(-a)² + 2(-a) +a
= -a³ + a³ - 2a + a
= -a

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MCQ 1491 Mark

If (x + 5) is a factor of = x³ - 20x + 5k then k = ?

A
-5
B
5
C
3
D
-3

Answer

5
Solution:
p(x) = x³ - 20x + 5k
Now, x + 5 = 0 ⇒ x = (-5)
By factor theorem,
p(-5) = 0
⇒ (-5)³ - 20(-5) + 5k = 0
⇒ -125 + 100 + 5k = 0
⇒ -25 + 5k = 0
⇒ 5k = 25
⇒ k = 5

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MCQ 1501 Mark

Which of the following is a polynomial ?

A
$\sqrt[3]{\text{y}}+4$
B
$\sqrt{\text{y}}-3$
C
$\text{y}$
D
$\frac{1}{\sqrt{\text{y}}}+7$

Answer

$\text{y}$
Solution:
A polynomial is an algebraic expression in which the variables involved have only non-negative integrals powers.
Option (a), (b) and (d) have negative and non-integral powers,
So they are not polynomials.

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Maths M.C.Q