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Maths M.C.Q

Polynomials · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 273 questions.

Questions · Page 2 of 6

M.C.Q

MCQ 511 Mark
If p(x) = 5x - 4x2 + 3 then p(-1) = ?
  • A
    2
  • B
    -2
  • C
    6
  • D
    -6
Answer
  1. -6
    Solution:
    P(x) = 5x - 4x2 + 3
    ⇒ p(-1) = 5(-1) - 4(-1)2 + 3
    = -5 - 4 + 3
    = -6
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MCQ 521 Mark
Write the correct answer in the following:
Which one of the following is a polynomial?
  • A
    $\frac{\text{x}^2}{2}-\frac{2}{\text{x}^2}$
  • B
    $\sqrt{2\text{x}-1}$
  • C
    $\text{x}^2+\frac{3\text{x}^{\frac{3}{2}}}{\sqrt{\text{x}}}$
  • D
    $\frac{\text{x}-1}{\text{x}+1}$
Answer
  1. ​​​​​​$\text{x}^2+\frac{3\text{x}^{\frac{3}{2}}}{\sqrt{\text{x}}}$

    Solution:
  1. Now, $\frac{\text{x}^2}{2}-\frac{2}{\text{x}^2}=\frac{\text{x}^2}{2}-2\text{x}^{-2},$ it is not a polynomial, because exponent of x is -2 which is not a whole number.
  2. Now, $\sqrt{2\text{x}-1}=\sqrt{\text{2}\text{x}}^{\frac{1}{2}}-1, $ it is not a polynomial, because exponent of x is $-\frac{1}{2}$ which is not a whole number.
  3. Now, $\text{x}^2+\frac{3\text{x}^{\frac{3}{2}}}{\sqrt{\text{x}}}=\text{x}^2+3\text{x}^{\frac{3}{2}-\frac{1}{2}}=\text{x}^2+3\text{x}^{\frac{2}{2}}=\text{x}^2+3\text{x},$ it is not a polynomial, because exponent of x is which is a whole number.
  4. $\frac{\text{x}-1}{\text{x}+1},$ it is not a polynomial because it is a rational function.
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MCQ 531 Mark
The factors of a2 - 1 - 2x - x2, are.
  • A
    (a + x + 1) (a - x - 1)
  • B
    (a - x + 1) (a - x - 1)
  • C
    (a + x - 1) (a - x + 1)
  • D
    None of these.
Answer
  1. (a + x + 1) (a - x - 1)
    Solution:
    The given expression to be factorized is a2 - 1 - 2x - x2
    Take common -1 from the last three terms and then we have
    a2 - 1 - 2x - x2 = a2 - (1 + 2x + x3)
    = a2 - {(1)2 + 2.1 × x + (x)2}
    = a2 - (1 + x)2
    = (a)2 - (1 + x)2
    = {a + (1 + x)} {a - (1 + x)}
    = (a + 1 + x) (a - 1 - x)
    = (a + x + 1) (a - x - 1)
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MCQ 541 Mark
If both x - a and $\text{x}-\frac{1}{2}$ are the factors of $\text{px}^2 + 5\text{x} +\text{r},$ than:
  • A
    p = r
  • B
    2p = r
  • C
    p = 2r
  • D
    None of these.
Answer
  1. p = r
    Solution:
    If both x - a and $\text{x}-\frac{1}{2}$ are the factors of $\text{px}^2 + 5\text{x} +\text{r},$ than f(2) = 0
    $\Rightarrow\text{p}(2)^2+5(2)+\text{r}=0$
    $\Rightarrow4\text{p}+10+\text{r} = 0$
    $\Rightarrow4\text{p}+\text{r}=-10\ ...(\text{i})$
    Also, $\text{f}\Big(\frac{1}{2}\Big)=0$
    $\Rightarrow\text{p}\Big(\frac{1}{2}\Big)^2+5\Big(\frac{1}{2}\Big)+\text{r}=0$
    $\Rightarrow\frac{\text{p}}{4}+\frac{5}{2}+\text{r}=0$
    $\Rightarrow\text{p}+10+4\text{r}=0$
    $\Rightarrow\text{p}+4\text{r}=-10\ ...{\text{ii}}$
    From eq. (i) and eq. (ii), we get
    $4\text{p}+\text{r}=\text{p}+4\text{r}$
    $\Rightarrow3\text{p}=3\text{r}$
    $\Rightarrow\text{p}=\text{r}$
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MCQ 551 Mark
If $49\text{a}^2-{\text{b}}=\Big(7\text{a}+\frac{1}{2}\Big)\Big(7\text{a}-\frac{1}{2}\Big),$ then the value of b is:
  • A
    $0$
  • B
    $\frac{1}{4}$
  • C
    $\frac{1}{\sqrt2}$
  • D
    $\frac{1}{2}$
Answer
  1. $\frac{1}{4}$
    Solution:
    $\Big(7\text{a}+\frac{1}{2}\Big)\Big(7\text{a}-\frac{1}{2}\Big)=(7\text{a})^2-\Big(\frac{1}{2}\Big)^2$
    [by using identity (a + b)(a - b) = a2 - b2]
    $\Rightarrow\Big(7\text{a}+\frac{1}{2}\Big)\Big(7\text{a}-\frac{1}{2}\Big)=49\text{a}^2-\frac{1}{4}$
    $\Rightarrow49\text{a}^2-\text{b}=49\text{a}^2-\frac{1}{4}$
    $\Rightarrow\text{b}=\frac{1}{4}$
    Hence, correct option is (b).
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MCQ 561 Mark
One of the zeroes of the polynomial 2x2 + 7x - 4 is:
  • A
    $\frac{1}{2}$
  • B
    $\frac{-1}{2}$
  • C
    -2
  • D
    2
Answer
  1. $\frac{1}{2}$
    Solution:
    2x2 + 7x - 4
    = 2x2 + 8x - x - 4
    = 2x(x + 4) - 1(x + 4)
    = (2x - 1)(x + 4)
    2x - 1 = 0 and x + 4 = 0
    $\text{x}=\frac{1}{2}$ and x = -4
    Therefore, one zero of the given polynomial is $\frac{1}{2}$
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MCQ 571 Mark
Zero of the zero polynomial is:
  • A
    Every real number.
  • B
    Not defined.
  • C
    0
  • D
    1
Answer
  1. Every real number.
    Solution:
    Zero of the zero polynomial is any real number.
    e.g., Let us consider zero polynomial be 0(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 581 Mark
The zeroes of the polynomial p(x) = x(x - 2) (x + 3) are:
  • A
    0, 2, -4
  • B
    0, 2, 4
  • C
    0, 2, -3
  • D
    0
Answer
  1. 0, 2, -3
    Solution:
    p(x) = x(x - 2) (x + 3)
    ⇒ x = 0 and x - 2 = 0 and x + 3 = 0
    ⇒ x = 0, x = 2 and x = -3
    Therefore, the zeroes are 0, 2, -3
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MCQ 591 Mark
A linear polynomial.
  • A
    Has one and only one zero.
  • B
    May have more than one zero.
  • C
    May have one zero.
  • D
    May have two zero.
Answer
  1. Has one and only one zero.
    Solution:
    A polynomial which has one and only one zero, is called a linear polynomial.
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MCQ 601 Mark
When p(x) = x3 - ax2 + x is divided by (x - a), the remainder is:
  • A
    0
  • B
    a
  • C
    2a
  • D
    3a
Answer
  1. a
    Solution:
    p(x) = x3 - ax2 + x
    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) = a3 - ax2 + a
    = a3 - a3 + a
    = a
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MCQ 611 Mark
The value of $\frac{(0.87)^3+(0.13)^3}{(0.87)^2-(0.87\times0.13)+(0.13)^2}$ is:
  • A
    0
  • B
    0.13
  • C
    0.87
  • D
    1
Answer
  1. 1
    Solution:
    $\frac{(0.87)^3+(0.13)^3}{(0.87)^2-(0.87\times0.13)+(0.13)^2}$
    $=\frac{(0.87+0.13)[(0.87)^2-(0.87\times0.13)+(0.13)^2]}{(0.87)^2-(0.87\times0.13)+(0.13)^2}$
    $=0.87+0.13$
    $=1$
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MCQ 621 Mark
If a + b + c = 0, then $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}=$
  • A
    0
  • B
    1
  • C
    -1
  • D
    3
Answer
  1. 3
    Solution:
    a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
    If a + b + c = 0, then
    a3 + b3 + c3 - 3abc = 0
    ⇒ a3 + b3 + c3 = 3abc ...(1)
    Now, consider $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}$
    Multiplying dividing by a. b. and c in $\frac{\text{a}^2}{\text{bc}}.\frac{\text{b}^2}{\text{ca}}$ and $\frac{\text{c}^2}{\text{ab}}$ respectively.
    we get
    $\frac{\text{a}^3}{\text{abc}}+\frac{\text{b}^3}{\text{bca}}+\frac{\text{c}^3}{\text{cab}}$
    $=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}$
    $=\frac{3\text{abc}}{\text{abc}}$ ....[From (1)]
    $=3$
    Hence, correct option is (d).
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MCQ 631 Mark
Where p(x) = x4 + 2x3 - 3x2 - 1 is divided by (x - 2), the remainder is:
  • A
    0
  • B
    -1
  • C
    -15
  • D
    21
Answer
  1. 21
    Solution:
    p(x) = x4 + 2x3 - 3x2 + x - 1
    x - 2 = 0 ⇒ x = 2
    By the remainder theorem, we know that when p(x) is divided by
    (x - 2), the remainder is p(2).
    Now, p(2) = x4 + 2x3 - 3x2 + x - 1
    = (2)4 + 2(2)3 - 3(2)2 + 2 - 1
    = 16 +16 - 12 + 2 - 1
    = 21
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MCQ 641 Mark
If both x - 2 and $\text{x} -\frac{1}{2}$ are the factors of px2 + 5x + r, then.
  • A
    2p = r
  • B
    p = 2r
  • C
    p = r
  • D
    None of these.
Answer
  1. p = r
    Solution:
    If both x - 2 and $\text{x}-\frac{1}{2}$ are the factors of f(x) = px2 + 5x + r, then f(2) = 0
    ⇒ p(2)2 + 5(2) + r = 0
    ⇒ 4p + 10 + r = 0
    ⇒ 4p + r = -10 .......(i)
    Also, $\text{f}\Big(\frac{1}{2}\Big)=0$
    $\Rightarrow\text{p}\Big(\frac{1}{2}\Big)^2+5\Big(\frac{1}{2}\Big)+\text{r}=0$
    $\Rightarrow\text{p}+10+4\text{r}=0$
    $\Rightarrow\text{p}+4\text{r}=-10\ .....(\text{ii})$
    From eq. (i) and eq. (ii), we get
    $4\text{p}+\text{r}=\text{p}+4\text{r}$
    $\Rightarrow3\text{p}=3\text{r}$
    $\Rightarrow\text{p}=\text{r}$
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MCQ 651 Mark
If x3 + 6x2 + 4x +k is exactly divisible by x + 2 then k =
  • A
    -8
  • B
    -7
  • C
    -6
  • D
    -10
Answer
  1. -8
    Solution:
    -8
    f(x) = x3 + 6x2 + 4x + k
    f(-2) = 0
    $\therefore$ (-2)3 + 6(-2)2 + 4(-2) + k = 0
    $\therefore$ -8 + 6(4) + (-8) + k = 0
    24 - 16 + k = 0
    k + 8 = 0
    k = -8
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MCQ 661 Mark
If $\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}=-1$ then (a3 - b3) = ?
  • A
    -3
  • B
    -2
  • C
    -1
  • D
    0
Answer
  1. 0
    Solution:
    $\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}=-1$
    $\Rightarrow\frac{\text{a}^2+\text{b}^2}{\text{ab}}=-1$
    ⇒ a2 + b2 = -ab
    ⇒ a2 + b2 + ab = 0
    Thus, we have:
    (a3 - b3) = (a - b)(a2 + b2 + ab)
    = (a - b) × 0
    = 0
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MCQ 671 Mark
If both x - 2 and $\text{x}-\frac{1}{2}$ are factor of px2 + 5x + r, then
  • A
    p = r
  • B
    p + r = 0
  • C
    2p + r = 0
  • D
    p + 2r = 0
Answer
  1. p = r
    Solution:
    Let f(x) = px2 + 5x + r
    Now,
    If x - 2 and $\text{x}-\frac{1}{2}$ are factors of f(x).
    Then at x = 2 and $\text{x}-\frac{1}{2},$ f(x) = 0.
    So, f(2) = 0, $\text{f}\Big(\frac{1}{2}\Big)=0$
    ⇒ p(2)2 + 5(2) + r = 0
    And, $\text{p}\Big(\frac{1}{2}\Big)^2+5\Big(\frac{1}{2}\Big)+\text{r}=0$
    ⇒ 4p + r + 10 = 0 ...(1)
    And 4r + p + 10 = 0 ...(2)
    Subtracting equation (2) from (1), we have
    3p - 3r = 0
    ⇒ p = r
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MCQ 681 Mark
If p(x) = x + 3, then p(x) + p(-x) is equal to:
  • A
    2x
  • B
    3
  • C
    0
  • D
    6
Answer
  1. 6
    Solution:
    p(x) = x + 3 and p(-x) = -x + 3
    Then, p(x) + p(-x)
    = x + 3 - x + 3
    = 6
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MCQ 691 Mark
If a - b = -8 and ab = -12, then a3 - b3 =
  • A
    -244
  • B
    -240
  • C
    -224
  • D
    -260
Answer
  1. -224
    Solution:
    a - b = -8
    (a - b)2 = 64
    a2 + b2 - 2ab = 64
    a2 + b2 - 2ab + 3ab = 64 + 3ab
    a2 + b2 + ab = 64 + 3(-12)
    a2 + b2 + ab = 64 - 36
    a2 + b2 + ab = 28
    Now
    a3 - b3 = (a - b)(a2 + b2 + ab)
    = (-8)(28)
    = -224
    Hence, correct option is (c).
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MCQ 701 Mark
If p(x) = x + 4 then p(x) + p(-x) = ?
  • A
    4
  • B
    2x
  • C
    0
  • D
    8
Answer
  1. 8
    Solution:
    Let: p(x) = (x + 4)
    $\therefore$ p(-x) = (x + 4)
    = -x + 4
    Thus, we have: p(x) + p(-x) ={(x + 4) + (-x + 4)}
    = 4 + 4
    = 8
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MCQ 711 Mark
When polynomial x2 + 3x2 + 3x + 1 is divided by x + 1, the remainder is:
  • A
    0
  • B
    -6
  • C
    1
  • D
    8
Answer
  1. 0
    Solution:
    p(-1) = (-1)3 + 3(-1)2 + 3(-1) + 1
    = -1 + 3 × 1 - 3 + 1
    = -1 + 3 - 3 + 1
    = 0
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MCQ 721 Mark
Which of the following is a linear polynomial?
  • A
    x + 1
  • B
    x + x2
  • C
    5x2 - x + 3
  • D
    $\text{x}+\frac{1}{\text{x}}$
Answer
  1. x + 1
    Solution:
    x + 1 is a linear polynomial because its degree is 1.
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MCQ 731 Mark
If (3x - 1)7 = a7x7 + a6x6 + a5x5 + ___ + a1x + a0, then a7 + a6 + a5 + ____ + a1 + a0 =
  • A
    0
  • B
    128
  • C
    1
  • D
    64
Answer
  1. 128
    Solution:
    Given that,
    (3x - 1)7 = a7x2 + a5x5 + ___ + a1x + a0
    Putting x = 1
    We get
    (3 × 1 - 1)7 = a6(1)5 + a5(1)5 ____ + a1(1) + a0
    ⇒ a6 + a5 + ____ + a1 + a0 = 27 = 128
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MCQ 741 Mark
If the volume of a cuboid is 3x2 - 27, then its possible dimensions are:
  • A
    3, x2, - 27x
  • B
    3, x - 3, x + 3
  • C
    3, x2, 27x
  • D
    3, 3, 3
Answer
  1. 3, x - 3, x + 3
    Solution:
    Volume of a cuboid of side a, b and c = abc
    Now, Volume = 3x2 - 27 (given)
    abc = 3(x2 - 9)
    abc = 3(x - 3)(x + 3)
    So, possible dimensions are 3, x - 3 and x + 3
    Hence, correct option is (b).
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MCQ 751 Mark
The coefficient of x2 in the polynomial 2x3 + 4x2 + 3x + 1.
  • A
    1
  • B
    2
  • C
    4
  • D
    3
Answer
  1. 4
    Solution:
    A coefficient is a multiplicative factor in a term of a polynomial or any expression.
    Therefore, the coefficient of x2 in the polynomial 2x3 + 4x2 + 3x + 1 is 4.
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MCQ 761 Mark
If x - 3 is a factor of x2 - ax - 15, then a =
  • A
    -2
  • B
    5
  • C
    -5
  • D
    3
Answer
  1. -2
    Solution:
    x - 3 is a factor of x2 - ax - 15,
    then at x = 3,
    x2 - ax - 15 = 0
    i.e. (3)2 - a(3) - 15 = 0
    9 - 3a - 15 = 0
    a = -2
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MCQ 771 Mark
(x - y) (x + y) (x2 + y2) (x4 + y4) is equal to:
  • A
    x16 - y16
  • B
    x8 - y8
  • C
    x8 + y8
  • D
    x16 + y16
Answer
  1. x8 - y8
    Solution:
    (x - y)(x + y) = x2 - y2 [by identity (a + b)(a - b) = a2 - b2]
    (x2 - y2)(x2 + y2) = x4 - y4
    (x4 - y4)(x4 + y4) = x8 - y8
    Now,
    (x - y)(x + y)(x2 + y2)(x4 + y4)\
    = (x2 - y2)(x2 + y2)(x4 + y4)
    = (x4 - y4)(x4 + y4)
    = x8 - y8
    Hence, correct option is (b).
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MCQ 781 Mark
3x3 + 2x2 + 3x + 2 = ?
  • A
    (3x - 2)(x2 - 1)
  • B
    (3x - 2)(x2 + 1)
  • C
    (3x + 2)(x2 - 1)
  • D
    (3x + 2)(x2 + 1)
Answer
  1. (3x + 2)(x2 + 1)
    Solution:
    3x3 + 2x2 + 3x + 2
    = x2(3x + 2) + 1(3x + 2)
    = (3x + 2)(x2 + 1)
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MCQ 791 Mark
The value of $\frac{(0.013)^3+(0.007)^3}{(0.013)^2-0.013\times0.007+(0.007)^2},$ is:
  • A
    0.006
  • B
    0.02
  • C
    0.0091
  • D
    0.00185
Answer
  1. 0.02
    Solution:
    By using identity a3 + b3 = (a + b)(a2 + b2 - ab), we have
    $\frac{(0.013)^3+(0.007)^3}{(0.013)^2-0.013\times0.007+(0.007)^2}$
    $=\frac{\{(0.013)+(0.007)\}(0.013)^2-(0.013)(0.007)+(0.007)^2}{(0.013)^2-(0.013)(0.007)+(0.007)^2}$
    $=0.013+0.007$
    $=0.020$
    $=0.02$
    Hence, correct option is (b).
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MCQ 801 Mark
The value of (102)3 is:
  • A
    1061280
  • B
    1061208
  • C
    1001208
  • D
    1820058
Answer
  1. 1061208
    Solution:
    (102)3 = (100 + 2)3
    = (100)3 + (2)3 + 3 × 100 × 2(100 + 2)
    = 1000000 + 8 + 60000 + 1200
    = 1061208
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MCQ 811 Mark
$\sqrt{3}$ is a polynomial of degree.
  • A
    1
  • B
    2
  • C
    0
  • D
    $\frac{1}{2}$
Answer
  1. 0
    Solution:
    $\sqrt{3}$ is a constant term, so it is a polynomial of degree 0.
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MCQ 821 Mark
If x + 3 is a factor of x2 - ax - 15, then a =
  • A
    -5
  • B
    -2
  • C
    5
  • D
    3
Answer
  1. -2
    Solution:
    Put x - 3 = 0, then x = 3
    Therefore, value of x2 - ax - 15 at x = 3 is zero
    ⇒ 32 - 3a - 15 = 0
    ⇒ -6 - 3a = 0
    ⇒ a = -2
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MCQ 831 Mark
The factors of a2 - 1 - 2x - x2 are:
  • A
    (a - x + 1)(a - x - 1)
  • B
    (a + x - 1)(a - x + 1)
  • C
    (a + x +1)(a - x + 1)
  • D
    None of these.
Answer
  1. (a + x +1)(a - x + 1)
    Solution:
    a2 - 1 - 2x - x2
    = a2 - (1 + 2x + x2)
    = a2 - (1 + x)2
    = [a - (1 + x)][a + (1 + x)]
    = (a - x - 1)(a + x + 1)
    Hence, correct option is (c).
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MCQ 841 Mark
The value of k for which x - 1 is a factor of 4x3 + 3x2 - 4x + k, is:
  • A
    3
  • B
    1
  • C
    -2
  • D
    -3
Answer
  1. -3
    Solution:
    Let p(x) = 4x3 + 3x2 - 4x + k
    Now,
    if (x - 1) is a factor of p(x), then at x = 1, p(x) = 0
    So, p(1) = 0
    ⇒ 4(1)3 + 3(1)2 - 4(1) + k = 0
    ⇒ 4 + 3 - 4 + k = 0
    ⇒ k = -3
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MCQ 851 Mark
If $3\text{x}+\frac{2}{\text{x}}=7,$ then $\Big(9\text{x}^2-\frac{4}{\text{x}^2}\Big)=$
  • A
    25
  • B
    35
  • C
    49
  • D
    30
Answer
  1. 35
    Solution:
    $\Big(3\text{x}+\frac{2}{\text{x}}\Big)^2=9\text{x}^2+\frac{4}{\text{x}^2}+12\ ...(1)$
    $\Big(3\text{x}-\frac{2}{\text{x}}\Big)^2=9\text{x}^2+\frac{4}{\text{x}^2}-12\ ...(2)$
    Subtracting eq. (1) from eq. (2). we get
    $\Big(3\text{x}-\frac{2}{\text{x}}\Big)^2-\Big(3\text{x}+\frac{2}{\text{x}}\Big)^2=-24$
    $\Rightarrow\Big(3\text{x}-\frac{2}{\text{x}}\Big)^2=(7)^2-24=25$
    $\Rightarrow3\text{x}-\frac{2}{\text{x}}=5$
    Now $\Big(3\text{x}+\frac{2}{\text{x}}\Big)-\Big(3\text{x}-\frac{2}{\text{x}}\Big)=7\times5$
    $\Big(9\text{x}^2-\frac{4}{\text{x}^2}\Big)=35$
    Hence, correct option is (b).
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MCQ 861 Mark
(x + 1) is a factor of the polynomial:
  • A
    x3 + x2 - x + 1
  • B
    x4 + x3 + x2 + 1
  • C
    x3 + x2 + x + 1
  • D
    x4 + 3x3 + 3x2 + x + 1
Answer
  1. x3 + x2 + x + 1
    Solution:
    x3 + x2 + x + 1
    = x3 (x + 1) + 1 (x + 1)
    = (x3 + 1) (x + 1)
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MCQ 871 Mark
If a + b + c = 9 and ab + bc + ca =23, then a3 + b3 + c3 - 3abc =
  • A
    108
  • B
    207
  • C
    669
  • D
    729
Answer
  1. 108
    Solution:
    Given, a + b + c = 9
    Hence, (a + b + c)2 = 81
    So, a2 + b2 + c2 + 2ab + 2bc + 2ca = 81
    i.e. a2 + b2 + c2 + 2(ab + bc + ca) = 81
    i.e. a2 + b2 + c2 + 2(23) = 81
    i.e. a2 + b2 + c2 = 81 - 46 = 35
    Now, a3 + b3 + c3 - 3abc
    = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
    = (a + b + c)[(a2 + b2 + c2) - (ab + bc + ca)]
    = (9)[35 - 23]
    = 9 × 12
    = 108
    Hence, correct option is (a).
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MCQ 881 Mark
Zero of the zero polynomial is:
  • A
    0
  • B
    1
  • C
    Every real number.
  • D
    Not defined.
Answer
  1. Not defined.
    Solution:
    Let p(x) be a polynomial. If $\text{p}(\alpha)=0,$ then we say that $\alpha$ is a zero of a polynomial.
    A polynomial consisting of one term, namely zero only, is called a zero polynomoial.
    So, the zero of a zero polynomial is not defined.
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MCQ 891 Mark
The value of (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 is:
  • A
    3(a + b) (b + c) (c + a) (a - b) (b - c) (c - a)
  • B
    3(a - b) (b - c) (c - a)
  • C
    3(a + b) (b + c) (c + a)
  • D
    None of these.
Answer
  1. 3(a + b) (b + c) (c + a) (a - b) (b - c) (c - a)
    Solution:
    (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3
    Here,
    a2 - b2 + b2 - c2 + c2 - a2 = 0
    Therefore,
    (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 = 3(a2 - b2) (b2 - c2) (c2 - a2)
    [Since x3 + y3 + z3 = 3xyz if x + y + z =0]
    (a2 - b2)3 + (b2 - c2)3 + (c2 - a2)3 =
    3(a + b) (b + c) (c + a) (a - b) (b - c) (c - a)
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MCQ 901 Mark
The factors of x4 + x2 + 25 are:
  • A
    (x2 + 3x + 5)(x2 - 3x + 5)
  • B
    (x2 + 3x + 5)(x2 + 3x − 5)
  • C
    (x2 + x +5)(x2 - x + 5)
  • D
    None of these.
Answer
  1. (x2 + 3x + 5)(x2 - 3x + 5)
    Solution:
    For making perfect square to x4 + x2 + 25
    We add +10x2 and -10x2 to it.
    = x4 + x2 + 25
    = x4 + x2 + 25 + 10x2 - 10x2
    = [x4 + 10x2 + 25] - 9x2
    = (x2 + 5)2 + (3x)2
    = [(x2 + 5) + 3x][(x2 + 5) - 3x]
    = (x2 + 3x + 5)(x2 - 3x + 5)
    Hence, correct option is (a).
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MCQ 911 Mark
One of the factors of (16y2 - 1) + (1 - 4y)2 is:
  • A
    (4 - y)
  • B
    (4 + y)
  • C
    8y(4y - 1)
  • D
    (4y + 1)
Answer
  1. 8y(4y - 1)
    Solution:
    (16y2 - 1) + (1 - 4y)2
    = [(4y)2 - (1)2 + (-4y - 1)]2
    = [(4y - 1) (4y + 1) + (4y - 1)2]
    = (4y - 1)[4y + 1 + 4y - 1]
    = 8y(4y - 1)
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MCQ 921 Mark
The factors of x4 + x2 + 25, are.
  • A
    (x2 + x + 5) (x2 - x + 5)
  • B
    (x2 + 3x + 5) (x2 - 3x + 5)
  • C
    (x2 + 3x + 5) (x2 + 3x - 5)
  • D
    None of these.
Answer
  1. (x2 + 3x + 5) (x2 - 3x + 5)
    Solution:
    The given expression to be factorized is x4 + x2 + 25
    This can be written in the form
    x4 + x2 + 25 = (x2) + 2 × x2 × 5 + (5)2 - 9x2
    = {(x2)2 + 2x2 × 5 + (5)2} - (3x)2
    = (x2 + 52) - (3x)2
    = (x2 + 5)2 - (3x)2
    = (x2 + 5 + 3x) (x2 + 5 - 3x)
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MCQ 931 Mark
One of the zeroes of the polynomial 2x2 + 7x - 4 is:
  • A
    $-2$
  • B
    $\frac{1}{2}$
  • C
    $\frac{-1}{2}$
  • D
    $2$
Answer
  1. $\frac{1}{2}$
    Solution:
    2x2 + 7x - 4
    = 2x2 + 8x - x - 4
    = 2x(x + 4) -1(x + 4)
    = (2x - 1) (x + 4)
    2x - 1 = 0 and x + 4 = 0
    $\text{x}=\frac{1}{2}$ and $\text{x}-4$
    Therefore, one zero of the given polynomial is $\frac{1}{2}$
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MCQ 941 Mark
The degree of constant function is:
  • A
    0
  • B
    3
  • C
    1
  • D
    2
Answer
  1. 0
    Solution:
    The degree of any constant term 5 (say)
    We can write 5 as 5 × 1 = 5xº [Since aº = 1]
    Therefore, the degree of any constant term is 0
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MCQ 951 Mark
If x + 2 is a factor of x2 + mx + 14 then m =
  • A
    7
  • B
    14
  • C
    9
  • D
    2
Answer
  1. 9
    Solution:
    Firstly, we will divide x2 + mx + 14 by x + 2
    When we divide them remainder comes to be 18 - 2m ....(i)
    Since x + 2 is a factor of x2 + mx + 14
    Therefore remainder should be zero ....(ii)
    From 1 and 2
    18 - 2m = 0
    2m = 18
    m = 9
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MCQ 961 Mark
If x2 + kx + 6 = (x + 2) (x + 3), for all x,then the value of k is:
  • A
    3
  • B
    -1
  • C
    1
  • D
    5
Answer
  1. 5
    Solution:
    x2 + kx + 6 = (x + 2) (x + 3),
    ⇒ x2 + kx + 6 = x2 + (2 + 3)x + 2 × 3
    ⇒ x2 + kx + 6 = x2 + 5x + 6
    On comparing the terms,
    We get k = 5
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MCQ 971 Mark
The zeros of the polynomial p(x) = x2 - 3x are:
  • A
    0, 3
  • B
    0, 0
  • C
    3, -3
  • D
    0, -3
Answer
  1. 0, 3
    Solution:
    p(x) = x2 - 3x
    Now, we have
    p(x) = 0 ⇒ x2 - 3x = 0
    ⇒ x(x - 3) = 0
    ⇒ x = 0 and x - 3 = 0
    ⇒ x = 0 and x = 3
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MCQ 981 Mark
The remainder when x31 - 31 is divided by x + 1 is:
  • A
    -32
  • B
    31
  • C
    30
  • D
    0
Answer
  1. -32
    Solution:
    x31 - 31
    Using remainder theorem.
    = (-1)31 - 31
    = -1 - 31
    = -32
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MCQ 991 Mark
75 × 75 + 2 × 75 × 25 + 25 × 25 is equal to:
  • A
    10000
  • B
    6250
  • C
    7500
  • D
    3750
Answer
  1. 10000
    Solution:
    Given expression is 75 × 75 + 2 × 75 × 25 + 25 × 25
    Let 75 = a and 25 = b
    Then, we have
    a × a + 2 × a × b + b × b
    = a2 + 2ab + b2
    = (a + b)2
    = (75 + 25)2
    = (100)2
    = 10000
    Hence, correct option is (a).
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MCQ 1001 Mark
When p(x) = x3 - 3x2 + 4x + 32 is divided by (x + a), the remainder is:
  • A
    4
  • B
    0
  • C
    36
  • D
    32
Answer
  1. 4
    Solution:
    x + 2 = 0 ⇒ x = -2
    By the remainder theorem, we know that when p(x) is divided by (x + 2), the remainder is p(-2).
    Now, we have:
    p(-2) = (-2)3 - 3 × (-2)2 + 4 × (-2) + 32
    = -8 - 12 - 8 + 32
    = 4
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M.C.Q - Page 2 - Maths STD 9 Questions - Vidyadip