M.C.Q

MCQ 11 Mark

When p(x) = 4x³ - 12x² + 11x - 5 is divided by (2x - 1), the remainder is:

A
0
B
-5
C
-2
D
2

Answer

-2
Solution:
$\text{p}(\text{x}) = 4\text{x}^3 - 12\text{x}^2 + 11\text{x} - 5$
$\text{x}-1=0\Rightarrow\text{x}=\frac{1}{2}$
By the remainder theorem, we know that when p(x) is divided by (2x - 1), the remainder is $\text{p}\Big(\frac{1}{2}\Big).$
Now, $\text{p}\Big(\frac{1}{2}\Big)= 4\text{x}^3 - 12\text{x}^2 + 11\text{x} - 5$
$=4\Big(\frac{1}{2}\Big)^3-12\Big(\frac{1}{2}\Big)^2+11\Big(\frac{1}{2}\Big)-5$
$=\frac{1}{2}-3+\frac{11}{2}-5$
$=-2$

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MCQ 21 Mark

$\sqrt{2}$ is a polynomial of degree.

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

Answer

0
Solution:
$\sqrt{2}$ is a constant term. Therefore, the degree of $\sqrt{2}$ is 0.

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MCQ 31 Mark

The zeros of the polynomial p(x) = 2x² + 5x - 3 are.

A
$1,\frac{-1}{2}$
B
$\frac{-1}{2},3$
C
$\frac{1}{2},-3$
D
$\frac{1}{2},3$

Answer

$\frac{1}{2},-3$
Solution:
The given polynomial is p(x) = 2x² + 5x - 4
Putting $\text{x}=\frac{1}{2}$ in p(x), we get
$\text{p}\Big(\frac{1}{2}\Big)=2\times\Big(\frac{1}{2}\Big)^2+5\times\frac{1}{2}-3$
$=\frac{1}{2}+\frac{5}{2}-3=3-3=0$
Putting x = -3 in p(x), we get
$\text{p}(-3)=2\times(-3)^2+5\times(-3)-3$
$= 18-15-3$
$= 0$
Therefore, x = -3 is a zero of the polynomial p(x)
Thus, $\frac{1}{2}$ and -3 are the zeros of the given polynomial p(x).

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MCQ 41 Mark

When p(x) = x⁴ + 2x³ - 3x² + x - 1 is divided by (x - 2), the remainder is:

A
21
B
0
C
-1
D
-15

Answer

21
Solution:
x⁴ + 2x³ - 3x² + x - 1
Using remainder theorem,
= (2)⁴ + 2(2)³ - 3(2)² + 2 - 1
= 16 + 16 - 12 + 2 - 1
= 34 - 13
= 21

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MCQ 51 Mark

The value of $\frac{(2.3)^3-0.027}{(2.3)^2+0.69+0.09},$ is:

A
2
B
3
C
2.327
D
2.273

Answer

2

Solution:
$\frac{(2.3)^3-0.027}{(2.3)^2+0.69+0.09}$
$=\frac{(2.3)^3-(0.3)^3}{(2.3)^2+(0.3)^3+(2.3)(0.3)}$
$=\frac{(2.3 - 0.3)\{(2.3)^2+(0.3)^2+(2.3)(0.3)\}}{((2.3)^2+(0.3)^2+(2.3)(0.3))}$
$=2.3 - 0.3$
$=2$
Hence, correct option is (a).

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MCQ 61 Mark

Write the correct answer in the following:
One of the zeroes of the polynomial 2x² + 7x - 4 is.

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

Answer

$\frac{1}{2}$
Solution:
Let p(x) = 2x² + 7x - 4
= 2x² + 8x - x- 4 [by splitting middle term]
= 2x(x + 4) -1(x + 4)
=(2x - 1)(x + 4)
For zeroes of p(x), put p(x) = 0 ⇒ (2x - 1)(x + 4) = 0
⇒ 2x - 1 = 0 and x + 4 = 0
$\Rightarrow\text{x}=\frac{1}{2}$ and $\text{x}=-4$
Hence, one of the zeroes of the polynomial p(x) is $\frac{1}{2}.$

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MCQ 71 Mark

Write the correct answer in the following:
$\sqrt{2}$ is a polynomial of degree.

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

Answer

0
Solution:
$\sqrt{2}$ is a constant polynomial. The only term here is $\sqrt{2}$ which can be written as $\sqrt{2}\text{x}^\circ.$ So, the exponent of x is zero. Therefore, the degree of the polynomial is 0.

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MCQ 81 Mark

If x - 2 is a factor of x² + 3ax - 2a, then a =

A
1
B
-2
C
2
D
-1

Answer

-1
Solution:
As (x - 2) is a factor of f(x) = x² + 3ax - 2a
i.e., f(2) = 0
(2)² + 3a(2) - 2a = 0
4 + 6a - 2a = 0
= -1

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MCQ 91 Mark

If both (x + 2) and (2x + 1) are factors of ax² + 2x + b, than the value of a - b is:

A
1
B
2
C
0
D
-1

Answer

0
Solution:
$\text{p}(\text{x})=\text{ax}^2+2\text{x}+\text{b}$
$\Rightarrow\text{a}(-2)^2 + 2(-2) + b = 0$
$\Rightarrow4\text{a} - 4+\text{b}=0\ .... (\text{i})$
Also,
$\text{p}\Big(\frac{-1}{2}\Big)=0$
$\Rightarrow\text{a}\Big(\frac{-1}{2}\Big)^2+2\Big(\frac{-1}{2}\Big)+\text{b}=0$
$\Rightarrow\frac{\text{a}}{4}-1+\text{b}=0$
$\Rightarrow\text{a}-4+4\text{b}=0\ ...(\text{ii})$
Subtracting eq. (ii) from eq. (i), we get
$3\text{a}+0-3\text{b}=0$
$\Rightarrow3(\text{a}-\text{b})=0$
$\Rightarrow\text{a}-\text{b}=0$

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MCQ 101 Mark

If x + y = 8 and xy = 15, than x² + y²

A
32
B
1
C
34
D
36

Answer

34
Solution:
x² + y² = (x + y)² - 2xy
⇒ x² + y² = (8)² - 2 × 15
⇒ x² + y² = 64 - 30
⇒ x² + y² = 34

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MCQ 111 Mark

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

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

Answer

0
Solution:
Here, $\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}=1$
$\Rightarrow\frac{\text{a}^2+\text{b}^2}{\text{ab}}=1$
$\Rightarrow\text{a}^2+\text{b}^2=\text{ab}$
$\Rightarrow\text{a}^2+\text{b}^2-\text{ab}=0$
Using, $\text{a}^2+\text{b}^2=(\text{a}+\text{b})(\text{a}^2+\text{b}^2-\text{ab})$
$=(\text{a}+\text{b})(0)$
$=0$

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MCQ 121 Mark

If a + b + c = 0 then $\Big(\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}\Big)=?$

A
1
B
0
C
-1
D
3

Answer

3
Solution:
a + b + c = 0 ⇒ a³ + b³ + c³ = 3abc
Thus, we have:
$\Big(\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}\Big)=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}$
$=\frac{3\text{abc}}{\text{abc}}$
$=3$

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MCQ 131 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:
The given polynomial is p(x) = x² + x - 6
Putting x = 2 in p(x), we get
p(2) = 2² + 2 - 6 = 4 + 2 - 6 = 0
Therefore, x = 2 is a zero of the polynomial p(x).
Putting x = -3 in p(x), we get
p(-3) = (-3)² - 3 - 6 = 9 - 9 = 0
Therefore, x = -3 is a zero of the polynomial p(x)
Thus, 2 and -3 are the zeros of the given polynomial p(x).

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MCQ 141 Mark

If x + 2 and x - 1 are the factors of x³ + 10x² + mx + n, then the values of m and n are respectively

A
5 and -3
B
17 and -8
C
7 and -18
D
23 and -19

Answer

7 and -18
Solution:
If (x + 2) and (x - 1) are factors of polynomial x³ + 10x² + mx + n,
then x = -2, x = +1 will satisfy the polynomial.
Let p(x) = x³ + 10x² + mx + n
Then, p(-2) = 0
(-2)³ + 10(-2)² + m(-2) + n = 0
-8 + 40 - 2m + n = 0
32 - 2m + n = 0 ...(1)
And, p(1) = 0
(1)³ + 10(1)² + m(1) + n = 0
1 + 10 + m + n = 0
11 + m + n = 0 ...(2)
Substracting equation (1) from equation (2), we get
-21 + 3m = 0
3m = 21
m = 7
Substituting m = 7 in equation (2),
11 + 7 + n = 0
18 + n = 0
n = -18

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MCQ 151 Mark

(x + 1) is a factor of xⁿ +1 only if.

A
N is a negative integer.
B
N is an odd integer.
C
N is an even integer.
D
N is a positive integer.

Answer

N is an odd integer.
Solution:
The linear polynomial (x - 1) is a factor of xn + 1, only if,
f(-1) = (-1)ⁿ + 1 = 0
If n is odd integer, then f(-1) = -1 + 1 = 0

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MCQ 161 Mark

One factor of x⁴ + x² - 20 is x² + 5. The other factor is:

A
x² - 4
B
x - 4
C
x²- 5
D
x + 4

Answer

x² - 4
Solution:
x⁴ + x² - 20
= x⁴ + 5x² - 4x² - 20
=x²(x² + 5) - 4(x² + 5)
= (x² + 5)(x² - 4)
So, other factor is x² - 4.

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MCQ 171 Mark

If (x + 1) is a factor of the polynomial (2x² + kx) then k = ?

A
4
B
-3
C
2
D
-2

Answer

2
Solution:
Let p(x) = 2x² + kx
Since (x + 1) is a factor of p(x),
= P(-1) = 0
⇒ 2(-1)² + k(-1) = 0
⇒ 2 - k = 0
⇒ k = 2

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MCQ 181 Mark

The degree of the polynomial (x³ - 2) (x² - 11) is:

Answer

5
Solution:
(x³ - 2) (x² - 11)
= x³(x² - 11) - 2(x² - 11)
= x⁵ - 11x³ - 2x² + 22
Here, the highest power is 5.
Therefore, the degree is 5.

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MCQ 191 Mark

If x⁵¹ + 51 is divided by x + 1, then the remainder is:

A
50
B
51
C
0
D
1

Answer

50
Solution:
x⁵¹ + 51
If x⁵¹ + 51 is divided by x + 1, then using remainder theorem
(-1)⁵¹ + 51
= -1 + 51
= 50

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MCQ 201 Mark

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

A
-5
B
-1
C
1
D
2

Answer

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

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MCQ 211 Mark

Which of the following is a binomial?

A
x² + x + 3
B
x² + 4
C
2x²
D
$\text{x}+3+\frac{1}{\text{x}}$

Answer

x² + 4
Solution:
A polynomial with two non-zero terms is called a binomial.
x² + 4 is the polynomial that has two non-zero terms.
Hence is a binomial.

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MCQ 221 Mark

A polynomial of degree ____ is called a linear polynomial.

Answer

1
Solution:
A polynomial of degree 1 is called a linear polynomial.
Its general form is ax + b

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MCQ 231 Mark

Which of the following is a polynomial in one variable?

A
$\text{x}^{10}+\text{y}^5+\text{8}$
B
$\sqrt{2}\text{x}^2-\sqrt{3}\text{x}+6$
C
$3\sqrt{\text{x}}+\frac{2}{\sqrt{\text{x}}}+5$
D
$\text{x}+\frac{2}{\text{x}}+3$

Answer

$\sqrt{2}\text{x}^2-\sqrt{3}\text{x}+6$
Solution:
Clearly, $\sqrt{2}\text{x}^2-\sqrt{3}\text{x}+6$ is a polynomial in one variable because it has only non-negative integral powers of x.

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MCQ 241 Mark

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

A
2
B
-2
C
1
D
-3

Answer

1
Solution:
Zero of (x + 1) = -1.
By Remainder theorem,
P(x) = x¹⁴⁰ + 2x¹⁵¹ + k
p(1) = (-1)¹⁴⁰ + 2(-1)¹⁵¹ + k = 0
1 + 2(-1) + k = 0
1 - 2 + k = 0
-1 + k = 0
k = 1.

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MCQ 251 Mark

The value of the polynomial 5x - 4x² + 3, when x = -1 is:

A
6
B
-6
C
1
D
-1

Answer

-6
Solution:
5x - 4x² + 3
= -4x² + 5x + 3
Putting x = -1 in the given polynomial, we get
-4(-1)² + 5 (-1) + 3
= -4 - 5 + 3
= -9 + 3
= -6

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MCQ 261 Mark

Which of the following polynomials has (-3) as a zero?

A
x² + 3
B
x² - 3x
C
x² - 9
D
(x - 3)

Answer

x² - 9
Solution:
x² - 9
x² - 3²= (x + 3) (x - 3) [Using identity a² - b² = (a + b) (a - b)]
Then the zeroes are x + 3 = 0 and x - 3 = 0
⇒ x = -3 and x = 3

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MCQ 271 Mark

Which of the following expression is a monomial.

A
4x³
B
x⁶ + 2x² + 2
C
3 + x
D
None of these.

Answer

4x³Solution:
4x³ because monomial means only one term in an expression.

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MCQ 281 Mark

When p(x) = 4x³ - 12x² + 11x -5 is divided by (2x - 1), the remainder is:

A
2
B
0
C
-5
D
-2

Answer

-2
Solution:
$2\text{x}-1=0\Rightarrow\text{x}=\frac{1}{2}$
By the remainder theorem, we know that when p(x) is divided by (2x - 1), the remainder is $\text{p}\Big(\frac{1}{2}\Big)$
Now, we have:
$\text{p}\Big(\frac{1}{2}\Big)=4\times\Big(\frac{1}{2}\Big)^3-12\times\Big(\frac{1}{2}\Big)^2+11\times\frac{1}{2}-5$
$=\frac{1}{2}-3+\frac{11}{2}-5$
$=-2$

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MCQ 291 Mark

Which of the following is a binomial ?

A
x² + 4
B
x² + x + 3
C
2x²
D
$\text{x}+3+\frac{1}{\text{x}}$

Answer

x² + 4
Solution:
Clearly, x² + 4 is an expresstion having two non-zero terms.
So, it is a binomial.

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MCQ 301 Mark

Which of the following is a polynomial?

A
$\text{x}-\frac{1}{\text{x}}+2$
B
$\frac{1}{\text{x}}+5$
C
$\sqrt{\text{x}}+3$
D
$-4$

Answer

$-4$
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, -4 being a real number is a polynomial.

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MCQ 311 Mark

Zero of the zero polynomial is:

A
1
B
0
C
Not defined.
D
Any real number.

Answer

Any real number.
Solution:
Zero of the polynomial is any real number.
e.g., Let us consider zero polynomial x - k, where k is a real number.
For determining the zero, put x - k = 0
⇒ x = k Hence, zero of the zero polynomial be any real number.

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MCQ 321 Mark

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

A
4
B
3
C
2
D
-2

Answer

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

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MCQ 331 Mark

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

A
2
B
12
C
10
D
13

Answer

10
Solution:
$\text{x}^2+\frac{1}{\text{x}^2}=102,$
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)-2\times\text{x}\times\frac{1}{\text{x}}=102-2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=100$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=\sqrt{100}=10$

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MCQ 341 Mark

8 is a polynomial of degree.

A
8
B
0
C
1
D
None of these.

Answer

0
Solution:
Since 8 is a constant term.
Therefore its degree is 0.

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MCQ 351 Mark

Write the correct answer in the following:
If x⁵¹ + 51 is divided by x + 1, the remainder is.

A
0
B
1
C
49
D
50

Answer

50
Solution:
If p(x) is divided by x + a, then the remainder is p(-a).
Here p(x) = x⁵¹ + 51 is divided by x + 1, then
x = -1
Remainder = p(-1) = (-1)⁵¹ + 51 = 50 = -1 + 51 = 50

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MCQ 361 Mark

Write the correct answer in the following:
If x + 1 is a factor of the polynomial 2x² + kx, then the value of k is.

A
-3
B
4
C
2
D
-2

Answer

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

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MCQ 371 Mark

The Zero of the polynomial (x - 2)² - (x + 2)² is:

A
1
B
-2
C
0
D
2

Answer

0
Solution:
(x - 2)² - (x + 2)²= (x - 2 + x + 2) (x - 2 - x - 2) [Using identity a² - b² = (a + b) (a - b)]
= (2x) (-4)
= -8x
Then the zero is,
-8 = 0
⇒ x = 0

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MCQ 381 Mark

The value of (249)² - (248)² is:

A
1²
B
477
C
487
D
497

Answer

497
Solution:
(249)² - (248)²We know
a² - b² = (a + b)(a - b)
So,
(249)² - (248)²(249 - 248)(249 + 248)
= 497

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MCQ 391 Mark

Write the correct answer in the following:
If p(x) = x + 3, then p(x) + p(-x) is equal to.

A
3
B
2x
C
0
D
6

Answer

6
Solution:
We have p(x) = x + 3, then
p(-x) = -x + 3
Therefore, p(x) + p(-x) = x + 3 + (-x + 3) = x + 3 - x + 3 = 6

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MCQ 401 Mark

Write the correct answer in the following:
The factorisation of 4x² + 8x + 3 is.

A
(x + 1)(x + 3)
B
(2x + 1)(2x + 3)
C
(2x + 2)(2x + 5)
D
(2x –1)(2x –3)

Answer

(2x + 1)(2x + 3)
Solution:
Now, 4x² + 8x + 3= 4x² + 6x + 2x + 3 [by splitting middle term]
= 2x(2x + 3) + 1 (2x + 3)
= (2x + 3)(2x + 1)

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MCQ 411 Mark

The factors of x² + 4y² + 4y - 4xy -2x - 8, are.

A
(x + 2y - 4) (x + 2y + 2)
B
(x - 2y - 4) (x - 2y + 2)
C
(x - y + 2) (x - 4y - 4)
D
None of these.

Answer

(x - 2y - 4) (x - 2y + 2)
Solution:
The value of x² + 4y² + 4y - 4xy - 2x - 8 is.
=x² + 4y² - 4xy + 4y - 2x - 8
= (x - 2y)² + 2(2y - x) - 8
= (x - 2y)² - 2(x - 2y) - 8
Let A = x - 2y
Thus, A² - 2A - 8
=(A - 4) (A + 2)
Re-substitute the value of A.
Thus x² + 4y² + 4y - 4xy - 2x - 8
= (x - 2y - 4) (x - 2y + 2)

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MCQ 421 Mark

A polynomial of degree 3 in x has at most.

A
3 terms
B
1 terms
C
5 terms
D
4 terms

Answer

3 terms
Solution:
3 terms of not more than the power of 3

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MCQ 431 Mark

The expression (a - b)³ + (b - c)³ + (c - a)³ can be factorized as.

A
3(a - b) (b - c) (c - a)
B
3a³b³bc³
C
[a - (b + c)]³
D
3abc

Answer

3(a - b) (b - c) (c - a)
Solution:
Here, a - b + b - c + c - a = 0
Therefore, (a - b)³ + (b - c)³ + (c - a)³= 3(a - b) (b - c) (c - a)

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MCQ 441 Mark

If P(x) = x³ - 1, then the value of p(1) + p(-1) is:

A
2
B
1
C
-2
D
0

Answer

-2
Solution:
P(x) = x³ - 1,
Then the value of P(1) + P(-1)
= (1)³ - 1 + (-1)³ - 1
= 1 - 1 - 1 - 1 = 1 - 3 = -2

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MCQ 451 Mark

If $\text{x}-\frac{1}{\text{x}}=\frac{15}{4},$ than $\Big(\text{x}+\frac{1}{\text{x}}\Big)=$

A
$\frac{1}{4}$
B
$\frac{13}{4}$
C
$4$
D
$\frac{17}{4}$

Answer

$\frac{17}{4}$
Solution:
$\Rightarrow\text{x}+\frac{1}{\text{x}}=\frac{15}{4}$
Now, $\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\Big(\frac{15}{4}\Big)^2$
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)-2\times\text{x}\times\frac{1}{\text{x}}=\frac{225}{16}$
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)=\frac{225}{16}+2$
$\Rightarrow(\text{x})^2+\Big(\frac{1}{\text{x}^2}\Big)=\frac{257}{16}$
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}\times\frac{257}{16}+2\times\text{x}\times\frac{1}{\text{x}}\\=\frac{257}{16}+2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\frac{257+32}{16}=\frac{289}{16}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)=\sqrt{\frac{289}{16}}=\frac{17}{4}$

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MCQ 461 Mark

The expanded form of (x + y - z)² is:

A
x² + y² + z² + 2xy + 2yx + 2xz
B
x² + y² + z²- 2xy - 2yz - 2zx
C
x² + y² + z² + 2xy + 2xy + 2zx
D
x² + y² + z² + 2xy + 2yz - 2xz

Answer

x² + y² + z²- 2xy - 2yz - 2zx
Solution:
(x + y - z)² = (x)² + (y)² + (-z)² + 2 × x × y + 2 × y × (-z) + 2 × (-z) ×x
= x² + y² + z² + 2xy - 2yz - 2zx

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MCQ 471 Mark

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

A
1
B
27
C
18
D
9

Answer

27
Solution:
(x + 3)³= x³ + (3)³ + 3 × x × 3(x + 3)
= x³ +27 + 9x² +27x
= x³ + 9x² +27x + 27
Therefore, the coefficient of x, in the expansion of (x + 3)³ is 27.

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MCQ 481 Mark

A polynomial containing one nonzero term is called a ________.

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

Answer

Monomial.
Solution:
A polynomial containing one nonzero term is called a monomial.
Example: 3x, 5x², y³

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MCQ 491 Mark

If x + 2 and x - 1 are the factor of x³ + 10x² + mx + n, then the values of m and n are respectively.

A
5 and -3
B
7 and -18
C
23 and -19
D
17 and -8

Answer

7 and -18
Solution:
It is given (x + 2) and (x - 1) are the factors of the polynomial f(x)= x³ + 10x² + mx + n
i.e., f(-2) = 0 and f(1) = 0
New,
f(-2) = (-2)³ + 10(-2)² + m(-2) + n = 0
-8 + 40 - 2m + n =0
⇒ -2m + n = -32
⇒ 2m - n = 32 ....(i)
f(1) = (1)³ + 10(1)² + m(1) + n = 0
1 + 10 + m + n = 0
m + n = -11 .....(ii)
Solving equation (i) and (ii) we get,
m = 7 and n = -18

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MCQ 501 Mark

If (x + y)³ - (x - y)³ - 6y(x² - y²) = ky², then k =

Answer

8
Solution:
Let x + y = A and x - y = B
Now, (A - B)³ = A³ - B³ - 3AB(A - B)
⇒ [(x + y) - (x - y)]³ = (x + y)³ - (x - y)³ - 3(x + y)(x - y)[(x + y) - (x - y)]
= (x + y)³ - (x - y)³ - 3(x² - y²)(2y)
= (x + y)³ - (x - y)³ - 6y(x² - y²)
But, (x + y)³ - (x - y)³ - 6y(x² - y²) = ky³⇒ [(x + y) - (x - y)]³ = (2y)³ = k8y³⇒ (2y)³ = ky³⇒ 8y³ = ky³⇒ k = 8
Hence, correct option is (d).

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