M.C.Q

MCQ 1511 Mark

The value of $\frac{0.75\times0.75\times0.75+0.25\times0.25\times0.25}{0.75\times0.75-0.75\times0.25+0.25\times0.25}$ is:

A
-1
B
2
C
1
D
0

Answer

1
Solution:
$\frac{0.75\times0.75\times0.75+0.25\times0.25\times0.25}{0.75\times0.75-0.75\times0.25+0.25\times0.25}$
$=\frac{(0.75)^3+(0.25)^3}{(0.75)^2-0.75\times0.25+(0.25)^2}$
$=\frac{(0.75+0.25[(0.75)^2-0.75\times0.25+(0.25)^2]}{(0.75)^2-0.75\times0.25+(0.25)^2}$
$=0.75+0.25$
$=1$

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MCQ 1521 Mark

(x + 1) is a factor of xⁿ + 1 only if:

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

Answer

n is an odd integer
Solution:
If x + 1 is a factor of xⁿ + 1,
then, at x = -1, xⁿ + 1 = 0
(-1)ⁿ + 1 = 0
(-1)ⁿ = -1
(-1)ⁿ will be equal to -1 if and only if n is an odd integer.
If n is even, then (-1)ⁿ = 1
So, n should be an odd integer.

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MCQ 1531 Mark

If a² + b² + c² = 30 and a + b + c = 10, then the value of ab + bc + ca is:

A
30
B
35
C
40
D
25

Answer

35
Solution:
Using identity,
$\Rightarrow(\text{a}+\text{b}+\text{c})^2=\text{a}^2+\text{b}^2+\text{c}^2+2\text{bc}+2\text{bc}+2\text{ca}$
$\Rightarrow(\text{a}+\text{b}+\text{c})^2=\text{a}^2+\text{b}^2+\text{c}^2+2(\text{ab}+\text{bc}+\text{ca)}$
$\Rightarrow\text{ab}+\text{bc}+\text{ca}=\frac{(\text{a}+\text{b}+\text{c})^2-(\text{a}^2+\text{b}^2+\text{c}^2)}{2}$
$\Rightarrow\text{ab}+\text{bc}+\text{ca}=\frac{(10)^2-(30)}{2}$
$\Rightarrow\text{ab}+\text{bc}+\text{ca}=\frac{100-30}{2}$
$\Rightarrow\text{ab}+\text{bc}+\text{ca}=\frac{70}{2}$
$\Rightarrow\text{ab}+\text{bc}+\text{ca}=35$

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MCQ 1541 Mark

If x + a is a factor of x⁴ - a²x² + 3x - 6a, then a =

A
0
B
1
C
-1
D
2

Answer

0
Solution:
Put x + a = 0 ⇒ x = -a
Then, value of x⁴ - a²x² + 3x - 6a at x = -a is zero
Therefore, (-a)⁴ - a²(-a)² + 3(-a) - 6a = 0
-9a = 0 ⇒ a = 0

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MCQ 1551 Mark

If p(x) = (x - 1) (x + 1), then the value of p(2) + p(1) - p(0) is:

Answer

4
Solution:
Given: p(x) = (x - 1) (x + 1), then
p(2) + p(1) - p(0)
= (2 - 1) (2 + 1) + (1 - 1) (1 + 1) - (0 - 1) (0 + 1)
= 1 × 3 + 0 × 2 -(-1) × 1
= 3 + 0 + 1
= 4

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MCQ 1561 Mark

The value of x³ + y³ + 15xy - 125 when x + y = 5 is:

Answer

0
Solution:
Given: x + y = 5 ⇒ x = 5 - y
x³ + y³ + 15xy - 125
Putting the value of x, we get
(5 - y)³ + y³ + 15(5 - y)y - 125
= 125 - y³ - 3 × 5 × y(5 - y) + y³ + 15(5 - y)y - 125
= 125 - y³ - 75y + 15y² + y³ + 75y - 15y² - 125
= 0

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MCQ 1571 Mark

The factors of 8a³ + b³ - 6ab + 1 are:

A
(2a + b - 1)(4a² + b² + 1 - 3ab - 2a)
B
(2a - b + 1)(4a² + b² - 4ab + 1 - 2a + b)
C
(2a + b + 1)(4a² + b² + 1 -2ab - b - 2a)
D
(2a - 1 + b)(4a² + 1 - 4a - b - 2ab)

Answer

(2a + b + 1)(4a² + b² + 1 -2ab - b - 2a)
Solution:
We know the identity
a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)
So by using identity, we can write given expression as
(2a)³ + (b)³ + (1)³ - 3(2a)(b)(1)
= (2a + b + 1)[(2a)² + b² + 1² -2a × b - b × 1 - 2a × 1]
= (2a + b + 1)(4a² + b² + 1 -2ab - b - 2a)
Hence, correct option is (c).

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MCQ 1581 Mark

If $\text{x}^4+\frac{1}{\text{x}^4}=623,$ than $\text{x}+\frac{1}{\text{x}}=$

A
$27$
B
$3\sqrt{3}$
C
$25$
D
$-3\sqrt{3}$

Answer

$3\sqrt{3}$
Solution:
$\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)=623$
$\Rightarrow(\text{x}^2)^2+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}^2\times\frac{1}{\text{x}^2}=623+2\times\text{x}^2\times\frac{1}{\text{x}^2}$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=625$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=\sqrt{625}=25$
Now,
$\Rightarrow(\text{x}^2)+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}\times\frac{1}{\text{x}}=25+2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=27$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=\sqrt{27}=3\sqrt{3}$

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MCQ 1591 Mark

If (x − 1) is a factor of polynomial f(x) but not of g(x) , then it must be a factor of:

A
f(x)g(x)
B
-f(x) + g(x)
C
f(x) - g(x)
D
{f(x) + g(x)}g(x)

Answer

f(x)g(x)
Solution:
If x - 1 is a factor of f(x) then definitely f(1) = 0
And,
x - 1 is not a factor of g(x), then $\text{g(1)}\neq0.$
So, at x = 1

f(1)g(1) = 0 × g(1) = 0
-f(1) + g(1) = 0 + g(1) = g(1) ≠ 0
f(1) - g(1) = 0 - g(1) = -g(1) ≠ 0
{f(1) + g(1)}g(1) = {0 + g(1)}g(1) = {g(1)}² ≠ 0
So, at x = 1 only, f(x)g(x) = 0
Thus, (x - 1) is factor of f(x)g(x) too.

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MCQ 1601 Mark

$\sqrt3$ is a polynomial of degree:

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

Answer

0
Solution:
The degree of a constant polynomial is 0.
So, $\sqrt3$ is a polynomial of degree 0.

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MCQ 1611 Mark

(x + 1) is a factor of the polynomial.

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

Answer

x³ + x² + x + 1
Solution:
x³ + x² + x + 1 = x³ (x + 1) + 1(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 1621 Mark

The zero of the polynomial p(x) = 5x - 2 is:

A
$\frac{-2}{5}$
B
$\frac{2}{5}$
C
$\frac{5}{2}$
D
$\frac{-5}{2}$

Answer

$\frac{2}{5}$
Solution:
p(x) = 5x - 2
To find of the polynomial, we write 5x - 2 = 0
$\Rightarrow5\text{x}=2$
$\Rightarrow\text{x}=\frac{2}{5}$

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MCQ 1631 Mark

If x + y + z + = 9 and xy + yz + zx = 23, then the value of x³ + y³ + z³ - 3xyz is:

A
108
B
209
C
144
D
180

Answer

108
Solution:
Given: x + y + z = 9 and xy + yz + zx = 23
x³ + y³ + z³ - 3xyz = (x + y + z) (x² + y² + z² - xy - yz - zx)
= (x + y + y) [(x + y + z)² - 2xy - 2yz - 2zx - xy -yz - zx]
= (x + y + z) [(x + y + z)² - 3xy - 3yz - 3zx]
=(x + y + z) [(x + y + x)² - 3(xy + yz + xx)]
= (9) [(9)² - 3(23)]
= 9 × [(81 - 69)]
= 9 × 12
= 108

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MCQ 1641 Mark

(x + 1) is a factor of the polynomial:

A
x³ + x² - x + 1
B
x³ + 2x² - x - 2
C
x³ + 2x² - x + 2
D
x⁴ + x³ + x² + 1

Answer

x³ + 2x² - x - 2
Solution:
Given, x³ + 2x² - x - 2
For f(-1),
-1 + 2(-1)² - (-1) - 2
-1 + 2 -1 - 2 = 0
x = -1,
x + 1 = 0
So, (x + 1) is a factor of the polynomial x³ + 2x² - x – 2

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MCQ 1651 Mark

If (m² - 3)x² + 3mx + 3m + 1 = 0 has roots which are reciprocal of each other, then the value of m equals

A
4
B
1
C
2
D
None of these.

Answer

4
Solution:
If the root are reciprocal then the product of the roots of the equation equals to 1.
(m² - 3)x² + 3mx + 3m + 1 = 0
Product of the roots $=\frac{\text{c}}{\text{a}}$
$\frac{3\text{m}+1}{\text{m}^2-3}=1$
or m² - 3m - 4 = 0
or m² - 4m + m - 4 = 0
or (m + 1) (m - 4) = 0
m = 4 or m = -1
$\therefore$ 4

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MCQ 1661 Mark

If a² + b² + c² - ab - bc - ca = 0, than.

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

Answer

a = b = c
Solution:
Given: a² + b² + c² - ab -bc - ca = 0
⇒ 2(a² + b² + c² - ab - bc - ca) = 0
⇒ (2a² + 2b² + 2c² - 2ab - 2bc - 2ca) = 0
⇒ ({a² + b² -2ab} + {b² + c² - 2bc} + {c² + a² - 2ca}) = 0
⇒ (a - b)² + (b - c)² + (c - a)² = 0
Now, since the sum of all squares is zero.
⇒ a - b = 0 ⇒ a = b
⇒ b - c = 0 ⇒ b = c
⇒ c - a = 0 ⇒ c = a
⇒ a = b = c

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MCQ 1671 Mark

The value of (x - a)³ + (x - b)³ + (x - c)³ - 3(x - a) (x - b) (x - c) when a + b + c = 3x, is:

Answer

0
Solution:
(x - a)³ + (x - b)³ + (x - c)³ - 3(x - a) (x - b) (x - c)
= [x - a + x - b + x - c]
[(x - a²) + (x - b²) + (x - c²) - (x - a) (x - b) - (x - b) (x - c) - (x - c) (x - a)]
= [3x - ({a + b + c})]
[(x - a)² + (x - b)² + (x - c)² - (x - a) (x - b) - (x - b) (x - c) - (x - c) (x -a)]
= 3x - 3x
[(x - a)² + (x - b)² + (x - c)² - (x - a) (x - b) - (x - b) (x - c) - (x - c) (x - a)]
= 0
[0] [(x - a)² + (x - b)² + (x - c)² - (x - a) (x - b) - (x - b) (x - c) - (x - c) (x - a)

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MCQ 1681 Mark

If 3x = a + b + c, then the value of (x - a)³ + (x - b)³ + (x - c)³ - 3(x - a) (x - b) (x - c) is:

A
a + b + c
B
(a - b)(b - c)(c - a)
C
0
D
None of these.

Answer

0
Solution:
3x = a + b + c
⇒ a + b + c - 3x = 0
⇒ 3x - (a + b + c) = 0
⇒ (x - a) + (x - b) + (x - c) = 0 ...(1)
Using identity if a + b + c = 0 then, a³ + b³ + c³ - 3abc = 0
If we take x - a = A, x - b = B, x - c = C in equation (1), we get
A + B + C = 0
⇒ A³ + B³ + C³ - 3ABC= 0
⇒ (x - a)³ + (x - b)³ + (x - c)³ - 3(x - a) (x - b) (x - c) = 0
Hence, correct option is (c).

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MCQ 1691 Mark

If (x + 1) and (x - 1) are factors of px³ + x² - 2x + p then value of p and q are.

A
p = -1, q = 2
B
p = 2, q = -1
C
p = 2, q = 1
D
p = -2, q = -2

Answer

p = 2, q = -1
Solution:
Given: f(x) =px³ + x² - 2x + q
If x + 1 is a factor of f(x).
Then f(-1) = 0
p(-1)³ + (-1)² - 2(1) + q = 0
-p + 1 + 2 + q = 0
-p + q = -3
p - q = 3 ......(i)
Also, if x - 1 is a factor of f(x), then
p(1)³ + (1)² - 2(1) + q = 0
p + 1 - 2 + q = 0
p + q = 1 ......(ii)
2p = 4
p = 2
subtracting eq.(ii) from eq. (i),
we get,
-2q = 2
q = -1
q = -1
Therefore, p = 2, q = -1

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MCQ 1701 Mark

If x + y + z = 0 then x³ + y³ + z³ is :

A
3xyz
B
xyz
C
2xyz
D
0

Answer

3xyz
Solution:
x³ + y³ + z³ - 3xyz = (x + y + z) (x² + y² + z² - xy - yz - zx)
⇒ x³ + y³ + z³ - 3xyz = (0) (x² + y² + z² xy - yz - zx)
⇒ x³ + y³ + z³ - 3xyz = 0
⇒ x³ + y³ + z³ = 3xyz if x + y + z = 0, then x³ + y³ + z³ is 3xyz.

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MCQ 1711 Mark

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

Answer

8
Solution:
We have,
= (x + y)³ - (x - y)³ - 6y(x² - y²) = ky³= (x + y - x + y)³ + 3(x + y) (x - y) (x + y - x + y) - 6y(x² - y²) = ky³= 2y³ + 6y(x² - y²) - 6y(x² - y²) = ky³= 8y³ = ky³= k = 8

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MCQ 1721 Mark

A polynomial containing one nonzero term is called a ______.

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

Answer

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

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MCQ 1731 Mark

The value of $\Big(\text{x}-\frac{1}{\text{x}}\Big)\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)$ is:

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

Answer

$\text{x}^4-\frac{1}{\text{x}^4}$
Solution:
$\Big(\text{x}-\frac{1}{\text{x}}\Big)\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)$
$=\Big(\text{x}^2-\frac{1}{\text{x}^2}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)$ [Using identity (a + b) (a - b) = a² - b²]
$=\text{x}^4-\frac{1}{\text{x}^4}$ [Using identity (a + b) (a - b) = a²- b²]

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MCQ 1741 Mark

The maximum number of terms in a polynomial of degree 10 is:

A
10
B
12
C
11
D
9

Answer

11
Solution:
The maximum number of terms is one more than the power of the polynomial.
Therefore, the maximum number of terms in a polynomial of degree 10 is (10 + 1) = 11

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MCQ 1751 Mark

If $\Big(3\text{x}+\frac{1}{2}\Big)\Big(3\text{x}-\frac{1}{2}\Big)=9\text{x}^2-\text{p}$ then the value of p is:

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

Answer

$\frac{1}{4}$
Solution:
$\Big(3\text{x}+\frac{1}{2}\Big)\Big(3\text{x}-\frac{1}{2}\Big)=9\text{x}^2-\text{p}$
$9\text{x}^2-\frac{1}{4}$ $\Big(\therefore\ \big(\text{a}^2-\text{b}^2\big)=(\text{a}+\text{b})(\text{a}-\text{b})\Big)$
$=9\text{x}^2-\text{p}$
$\Rightarrow\text{p}=\frac{1}{4}$

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MCQ 1761 Mark

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

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

Answer

(a + b + c)³ =27abc
Solution:
Let $\text{a}^{\frac{1}{3}}=\text{A},\ \text{b}^{\frac{1}{3}}=\text{B}$ and $\text{c}^{\frac{1}{3}}=\text{C}$
Now, A + B + C = 0 (given)
If A + B + C = 0, then A³ + B³ + C³ - 3ABC = 0
⇒ A³ + B³ + C³ - 3ABC = 0
⇒ A³ + B³ + C³ = 3ABC ...(1)
$\begin{Bmatrix}\text{A}=\text{a}^{\frac{1}{3}},\ \text{B}=\text{b}^{\frac{1}{3}},\ \text{C}=\text{c}^{\frac{1}{3}}\\\text{A}^3=\text{a},\ \text{B}^3=\text{b},\ \text{C}^3=\text{c}\end{Bmatrix}$
Then, equation (1) becomes
$\text{a}+\text{b}+\text{c}=3(\text{abc})^{\frac{1}{3}}$
Cubing both Sides of above equation, we get
(a + b + c)³ = 27abc
Hence, correct option is (b).

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MCQ 1771 Mark

If x² + x + 1 is a factor of the polynomial 3x³ + 8x²+ 8x + 3 + 5k, then the value of k is:

A
$0$
B
$\frac{2}{5}$
C
$\frac{5}{2}$
D
$-1$

Answer

$\frac{2}{5}$
Solution:
Let p(x) = 3x³ + 8x² + 8x + 3 + 5k and q(x) = x² + x + 1
Now,
If q(x) is a factor of p(x), then arranging p(x) in order to have q(x) in common,
p(x) = 3x³ + 3x² + 3x + 5x² + 5x + 3 + 2 - 2 + 5k [ Adding +2, -2 in p(x)]
= 3x(x² + x + 1) + 5(x² + x + 1) + 5k - 2
p(x) = (x²+ x + 1)(3x + 5) + 5k - 2 ...(1)
From equation (1), we can see if we divide p(x) by q(x), then quotient will be (3x + 5) and remainder will be (5k - 2)
But q(x) is a factor of p(x).
So, remainder = 0
⇒ 5k - 2 = 0
$\Rightarrow\text{k}=\frac{2}{5}$

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MCQ 1781 Mark

If $\text{x}^4+\frac{1}{\text{x}^4}=623,$ then $\text{x}+\frac{1}{\text{x}}=$

A
$27$
B
$25$
C
$3\sqrt3$
D
$-3\sqrt3$

Answer

$3\sqrt3$
Solution:
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2.\text{x}.\frac{1}{\text{x}}=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}$
Squaring both sides.
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}^2$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}+2.\text{x}^2.\frac{1}{\text{x}}=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}^2$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}+2=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}^2=(623)+2$
$\Rightarrow623+2=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}^2\ \Big\{\text{x}^4+\frac{1}{\text{x}^4}=623\Big\}$
$\Rightarrow625=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}^2$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2=\sqrt{625}=25$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=25+2=27$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\sqrt{27}$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=3\sqrt{3}$
Hence, correct option is (c).

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MCQ 1791 Mark

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

Answer

2
Solution:
Given: $\text{x}^3-\Big(\frac{1}{\text{x}^3}\Big)=14$
Let x = a and $\frac{1}{\text{x}}=\text{b}$
Say, $\text{x}-\frac{1}{\text{x}}=\text{A}$
Then, $\text{a}^3-\text{b}^3=14$
$\Rightarrow(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)=14$
$\Rightarrow(\text{a}-\text{b})(\{(\text{a}-\text{b})^2+2\text{ab}\}+2\text{ab})=14$
$\Rightarrow(\text{a}-\text{b})\{(\text{a}-\text{b})^2+3\text{ab}\}=14$
$\Rightarrow(\text{a}-\text{b})\{(\text{a}-\text{b})^2+3\}=14$
$\Rightarrow\text{A}(\text{A}^2+3)=14$
$\Rightarrow\text{A}(\text{A}^2+3)=14$
$\Rightarrow\text{A}^3+3\text{A}-14=0$
$\Rightarrow\text{A}^3-2\text{A}^2+2\text{A}^2-4\text{A}+7\text{A}-14=0$
$\Rightarrow\text{A}^2(\text{A}-2)+2\text{Y}(\text{Y}-2)+7(\text{Y}-2)=0$
$\Rightarrow(\text{A}-2)(\text{A}^2+2\text{A}+7)=0$
$\Rightarrow\text{A}-2=0,$
$\Rightarrow\text{A}=2$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=2$

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MCQ 1801 Mark

The factors of x³ - 1 + y³ + 3xy are.

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

Answer

(x - 1 + y) (x² + 1 + y² + x + y - xy)
Solution:
The given expression to be factorized is x³ - 1 + y³ + 3xy
This can be written in the form
x³ - 1 + y³ + 3xy = (x)² + (-1)³ + (y)³ - 3(x) × (-1).(y)
Recall the formula a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)
Using the above formula, we have,
x³ - 1 + y³ + 3xy
= (x + (-1) + y} {(x)² + (-1)² + (y)² - (x) × (-1) - (-1) × (y) - (y) × (x)}
= (x - 1 + y) (x² + 1 + y² + x + y - xy)

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MCQ 1811 Mark

(a - b)³ + (b - c)³ + (c - a)³ =

A
(a + b + c)(a² + b² + c² - ab - bc - ca)
B
(a - b)(b - c)(c - a)
C
3(a - b)(b - c)(c - a)
D
None of these.

Answer

3(a - b)(b - c)(c - a)
Solution:
Let
a - b = A
b - c = B
c - a = C
Now (A + B + C)³ = A³ + B³ + C³ + 3(A + B)(B + C)(C + A)
⇒ A³ + B³ + C³ = (A + B + C)³- 3(A + B)(B + C)(C + A)
Now putting values of A, B and C. we get
$(\text{a} - \text{b})^3 + (\text{b} - \text{c})^3 + (\text{c} - \text{a})^3\\=(\not\text{a}-\not\text{b}+\not\text{b}-\not\text{c}+\not\text{c}-\not\text{a})^3\\-3(\text{a}-\not\text{b}+\not\text{b}-\text{c})(\text{b}-\not\text{c}+\not\text{c}-\text{a})(\text{c}-\not\text{a}+\not\text{a}-\text{b})$
⇒ (a - b)³+ (b - c)³ + (c - a)³ = 0 - 3 (a - c)(b - a)(c - b)
⇒ (a - b)³+ (b - c)³ + (c - a)³ = 3(a - b)(b - c)(c - a)
Hence, correct option is (c).

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MCQ 1821 Mark

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

A
(a - b)(b - c)(c - a)
B
3(a - b)(b - c)(c - a)
C
-3(a - b)(b - c)(c - a)
D
(a + b + c)(a² + b² + c² - ab - bc - ca)

Answer

3(a - b)(b - c)(c - a)
Solution:
By we know that a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)
If a + b + c = 0, then
a³ + b³ + c³ = 3abc
In given expression,
Let a - b = A, b - c = B, c - a = C
Now, a - b + b - c + c - a = 0
i.e. A + B + C = 0
⇒ A³ + B³ + C³ = 3ABC
⇒ (a - b)³ + (b - c)³ + (c - a)³ = 3(a - b)(b - c)(c - a)
Hence, correct option is (b).

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MCQ 1831 Mark

If f(x) = x² - 5x + 1, that the value of f(2) + f(-1) is:

A
2
B
1
C
-2
D
-1

Answer

2
Solution:
f(x) = x² - 5x + 1
f(2) + f(-1)
= (2)² - 5 × 2 + 2 + (-1)² - 5 × (-1) + 1
= 4 - 10 + 1 + 1 + 5 + 1
= 12 - 10
= 2

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MCQ 1841 Mark

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

A
1
B
0
C
a
D
-a

Answer

-a
Solution:
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).
Thus, we have: P(-a) = (-a)³ + a × (-a)² + 2 × (-a) + a
= -a³ + a³ - 2a + a
= -a

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MCQ 1851 Mark

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

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

Answer

a⁶ - b⁶Solution:
(a + b) (a - b) (a² - ab + b²) (a² + ab +b²)
⇒ {(a + b) (a² + b² - ab)} {(a - b) (a² + b² + ab)}
⇒ (a³ + b³) (a³ - b³)
⇒ (a⁶ - b⁶)

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MCQ 1861 Mark

If x² + kx -3 = (x - 3) (x + 1), than the value of 'k' is:

A
-2
B
2
C
-3
D
3

Answer

-2
Solution:
x² + kx -3 = (x - 3) (x + 1),
⇒ x² + kx - 3 = x² + (-3 + 1) x + (-3) × 1
⇒ x² + kx - 3 = x² = 2x - 3
On comparing the term, we get = -2

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MCQ 1871 Mark

Degree of the zero polynomial is:

A
1
B
0
C
Not defined.
D
Non of these.

Answer

Not defined.
Solution:
A polynomial consisting of one term, namely zero only, is called a zero polynomial.
So, a zero polynomial can be defined as p(x) = 0.
This can also be written as p(x) = 0 = 0x = 0x² = 0x³ and so on.
So, it is not possible to determine the degree.
Hence, the degree of a zero polynomial is not defined.

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MCQ 1881 Mark

If a + b + c = 9 and ab + bc + ca = 23, then a² + b² + c² =

A
127
B
35
C
58
D
None of these.

Answer

35
Solution:
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Hence, 9² = a² + b² + c² + 2 × 23
⇒ a² + b² + c² = 35

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MCQ 1891 Mark

If x - 1 is the factor of p(x) = x³ - 23x² + kx - 120, then the value of 'k' is:

A
120
B
142
C
124
D
140

Answer

142
Solution:
If x - 1 is the factor of p(x), then
p(1) = 0
(1)³ - 23(1)² + k(1) - 120 = 0
1 - 23 + k - 120 = 0
1 - 143 + k = 0
-142 + k = 0
k = 142

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MCQ 1901 Mark

If x + a is a factor of x⁴ - a²x² + 3x - 6a, then a is:

A
0
B
-1
C
1
D
2

Answer

0
Solution:
x + a is a factor of polynomial f(x) = x⁴ - a²x² + 3x - 6a,
Then at x = -a, p(x) = 0
⇒ (-a)⁴ - a²(-a)² + 3(-a) - 6a = 0
⇒ a⁴ - a⁴ - 3a - 6a = 0
⇒ -9a = 0
⇒ a = 0

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MCQ 1911 Mark

The degree of the zero polynomial is.

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

Answer

Not defined.
Solution:
The general form of a polynomial is a_nxⁿ, where n is a natural number.
For zero polynomial a_n = 0.
Since the largest value of n for which an is non-zero is negative infinity (all the integers are bigger than negative infinity).
Therefore, the degree of zero polynomials is not defined.

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MCQ 1921 Mark

The coefficient of x² in (2 - 3x²) (x² - 5) is:

A
17
B
-10
C
1
D
-17

Answer

17
Solution:
(2 - 3x³) (x² - 5)
= 2x² - 10 - 3x⁴ + 15x²= -3x⁴ + 17x² - 10
Therefore, the coefficient of x² is 17.

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MCQ 1931 Mark

If $\text{x}+\frac{1}{\text{x}}=5,$ then $\text{x}^2+\frac{1}{\text{x}^2}=$

A
25
B
10
C
23
D
27

Answer

23
Solution:
By using identity (a + b)² = a² + b² + 2ab.
we have,
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\Big(\frac{1}{\text{x}}\Big)^2+2\times\not\text{x}\times\frac{1}{\not\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow(5)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$ $\Big\{\text{x}+\frac{1}{\text{x}}=5\text{ given}\Big\}$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=25-2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=23$
Hence, correct option is (c).

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MCQ 1941 Mark

Which of the following statement is false?

The degree of a zero polynomial is defined.
The degree of a zero polynomial is zero.
The zero of a zero polynomial is not defined.
The degree of a constant polynomial is not defined.

A
iv
B
i
C
ii
D
iii

Answer

i
Solution:
The degree of a zero polynomial is not defined.

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MCQ 1951 Mark

The factorisation of 4x² + 8x + 3 is:

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

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 1961 Mark

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

A
0
B
1
C
49
D
50

Answer

50
Solution:
When the polynomial p(x) is divided by q(x) i. e. $(\text{x}\pm\alpha)$ then $\text{p}(\mp\alpha)$ is the remainder.
If $\text{x}\pm\alpha$ is the factor of polynomial, then remainder is '0'.
So,
If x⁵¹ + 51 is divided x + 1.
Remainder = (-1)⁵¹ + 51 = -1 + 51 = 50.

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MCQ 1971 Mark

If (x + 2) and (x - 1) are factor of (x³ + 10x² + mx + n) then:

A
m = 5, n = -3
B
m = 7, n = -18
C
m = 17, n = -8
D
m = 23, n = -19

Answer

m = 7, n = -18
Solution:
Let:
p(x) = x³ + 10x² + mx + n
Now,
x + 2 = 0 ⇒ x = -2
(x + 2) is a factor of p(x).
So, we have p(-2)² + m × (-2) + n = 0
⇒ (-2)³ + 10 × (-2)² + m × (-2) + n = 0
⇒ -8 + 40 - 2m + n = 0
⇒ 32 - 2m + n = 0
⇒ 2m - n = 32 ...(i)
Now,
x - 1 = 0 ⇒ x = 1
Also,
(x - 1) is a factor of p(x)
We have:
p(1) = 0
⇒ 1³ + 10 × 1² + m × 1 + n = 0
⇒ 1 + 10 + m + n = 0
⇒ 11 + m + n = 0
⇒ m + n = -11 ...(ii)
From (i) and (ii),
We get:
3m = 21 ⇒ m = 7
By substituting the value of m in (i), we get n = -18
$\therefore\ $m = 7 and n = -18

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MCQ 1981 Mark

The value of x³ - 8y³ - 36xy - 216, when x = 2y + 6 is:

Answer

0
Solution:
x³ - 8y³ - 36xy - 216
Putting x = 2y + 6
(2y + 6)³ - 8y³ - 36 (2y + 6) y - 216
= 8y³ + 216 + 3 × 2y × 6(2y + 6) - 8y³ - 36(2y + 6) y - 216
= 8y³ + 216 + 72y² + 216y - 8y³ - 72y² - 216y - 216
= 0

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MCQ 1991 Mark

The value of 'a' for which (x + a) is a factor of the polynomial x³ + ax² + a + 6 is:

A
1
B
0
C
2
D
-2

Answer

-2
Solution:
If (x + a) is a factor of the polynomial x³ + ax² + a + 6, than p(-a) = 0
⇒ (-a)³ + a(-a)² - 2(-a) + a + 6 = 0
⇒ -a³ + a³ + 2a + a + 6 = 0
⇒ 3a = -6
⇒ a = -2

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MCQ 2001 Mark

4a² + b² + 4ab + 8a + 4b + 4 = ?

A
(2a - b + 2)²
B
(2a + b + a)²
C
(a + 2b + 2)²
D
None of these.

Answer

(2a + b + a)²Solution:
We know that,
(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
2a² + b² + 4 + 4ab + 8a + 4b + 4
= 4a² + b² + 4 + 4ab + 8a + 4b
= (2a²) + b² + 2² + 2(2a)b + 2(2a) (2) + 2(2b)
= (2a + b + 2)²

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Maths M.C.Q