M.C.Q

MCQ 11 Mark

If $p(x) = 5x - 4x^2 + 3$ then $p(-1) =$ ?

A
$2$
B
$-2$
C
$6$
✓
$-6$

Answer

Correct option: D.

$-6$

$P(x) = 5x - 4x^2 + 3$
$\Rightarrow p(-1) = 5(-1) - 4(-1)^2 + 3$
$= -5 - 4 + 3$
$= -6$

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MCQ 21 Mark

Which of the following is a linear polynomial?

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

Answer

Correct option: B.

$x + 1$

A polynomial of degree $1$ is called a linear polynomial.
Options $(a),$ and $(c)$ have degree $2$,
so ther are quadratic polynomials.
option $(d)$ has a negative power, so it is not a polynomial.
The degree of $x + 1$ is $1$, so it is a linear polynomial.

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MCQ 31 Mark

The zeros of the polynomial $p(x) = x^2 - 3x$ are:

A
$0, 0$
✓
$0, 3$
C
$0, -3$
D
$3, -3$

Answer

Correct option: B.

$0, 3$

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

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MCQ 41 Mark

When $p(x) = x^3 - ax^2 + x$ is divided by $(x - a),$ the remainder is:

A
$0$
✓
$a$
C
$2a$
D
$3a$

Answer

Correct option: B.

$a$

$p(x) = x^3 - ax^2 + x$
$x - a = 0 \Rightarrow x = a$
By the remainder theorem, we know that when $p(x)$ is divided by $(x - a),$ the remainder is $p(a).$
Now, $p(a) = a^3 - ax^2 + a$
$= a^3 - a^3 + a$
$= a$

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MCQ 51 Mark

The zeros of the polynomial $p(x) = 2x^2 + 7x - 4$ are:

A
$4,\frac{-1}{2}$
B
$4,\frac{1}{2}$
✓
$-4,\frac{1}{2}$
D
$-4,\frac{-1}{2}$

Answer

Correct option: C.

$-4,\frac{1}{2}$

$p(x) = 2x^2 + 7x - 4$
Now, $p(x) = 0$
$\Rightarrow 2x^2 + 7x - 4 = 0$
$\Rightarrow 2x^2 + 8x - x - 4 = 0$
$\Rightarrow 2x(x + 4) - 1(x + 4) = 0$
$\Rightarrow (x + 4)(2x - 1) = 0$
$\Rightarrow x + 4 = 0$ and $2x - 1 = 0$
$\Rightarrow x = -4$ and $\text{x}=\frac{1}{2}$

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MCQ 61 Mark

For what value of k is the polynomial $p(x) = 2x^3 - kx^2 + 3x + 10$ exactly divisible by $(x + 2)$?

A
$-\frac{1}{3}$
B
$\frac{1}{3}$
C
$3$
✓
$-3$

Answer

Correct option: D.

$-3$

$p(x) = 2x^3 - kx^2 + 3x + 10$
$x + 2 = 0 \Rightarrow x = -2$
By the factor theorem, we know that when $p(x)$ is divided by $(x + 2),$ the remainder is $p(-2).$
Now, $p(-2) = 2(-2)^3 + k(-2)^{2 }+ 3(-2) + 10$
$\Rightarrow 0 = -16 - 4k - 6 + 10$
$\Rightarrow 0 = -12 - 4k$
$\Rightarrow 4k = -12$
$\Rightarrow k = -3$

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Question 71 Mark

Which of the following expression is a polynomial in one variable?

$\text{x}+\frac{2}{\text{x}}+3$
$3\sqrt{\text{x}}+\frac{2}{\sqrt{\text{x}}}+5$
$\sqrt2\text{x}^2-\sqrt3\text{x}+6$
$\text{x}^{10}+\text{y}^5+8$

Answer

$\sqrt2\text{x}^2-\sqrt3\text{x}+6$

Solution:
Clearly, $\sqrt2\text{x}^2-\sqrt3\text{x}+6$ is a polynomial in one variable because it has only non-negative integral powers of x.

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Question 81 Mark

$\sqrt3$ is a polynomial of degree:

$\frac{1}{2}$
2
1
0

Answer

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

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Question 91 Mark

Zero of the zero polynomial is:

0
1
Every real number.
Not defined.

Answer

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 101 Mark

If $(x + 2)$ and $(x - 1)$ are factors of the polynomial $p(x) = x3 + 10x^2 + mx + n$ then:

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

Answer

Correct option: B.

$m = 7, n = -18$

Let $f(x) = x^3 + 10x^2 + mx + n$
Now, $x + 2 = 0 \Rightarrow x = -2$
and $x - 1 = 0 \Rightarrow x = 1$
By factor theorem,
$f(-2) = 0$
$\Rightarrow (-2)^3 + 10(-2)^2 + m(-2) + n$
$\Rightarrow -8 + 40 - 2m + n = 0$
$\Rightarrow 2m - n = 32 ...(i)$
By factor theorem,
$f(1) = 0$
$\Rightarrow (1)^3 + 10(1)^2 + m(1) + n = 0$
$\Rightarrow m + n = -11 ...(ii)$
Adding $(i)$ and $(ii),$ we get
$3m = 21$
$\Rightarrow m = 7$
Substituting in $(ii),$ we get
$n = -18$

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MCQ 111 Mark

Degree of the zero polynomial is:

A
$1$
B
$0$
✓
Not defined.
D
Non of these.

Answer

Correct option: C.

Not defined.

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 $p(x) = 0 = 0x = 0x^2 = 0x^3$ 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 121 Mark

If $(x^{51} + 51)$ is divided by $(x + 1)$ then the remainder is:

A
$0$
B
$1$
C
$49$
✓
$50$

Answer

Correct option: D.

$50$

Let $f(x) = x^{51} + 51$
By the remainder theorem, when $f(x)$ is divided by $(x + 1),$ the remainder is $f(-1).$
Now, $f(-1) = [(-1)^n + 51]$
$= -1 + 51 = 50$

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Question 131 Mark

Zero of the polynomial p(x) = 2x + 5 is:

$\frac{-2}{5}$
$\frac{-5}{2}$
$\frac{2}{5}$
$\frac{5}{2}$

Answer

$\frac{-5}{2}$

Solution:
p(x) = 2x + 5
Now, p(x) = 0
⇒ 2x + 5 = 0
⇒ 2x = -5
$\Rightarrow\text{x}=-\frac{5}{2}$

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Question 141 Mark

If $\text{p}(\text{x})=\text{x}^2-2\sqrt2\text{x}+1$ then $\text{p}(2\sqrt2)=?$

0
1
$4\sqrt2$
-1

Answer

Solution:
$\text{p}(\text{x})=\text{x}^2-2\sqrt2\text{x}+1$
$\text{p}(2\sqrt2)=(2\sqrt2)^2-2\sqrt2(2\sqrt2)+1$
$=8-8+1$
$=1$

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MCQ 151 Mark

When $p(x) = x^3 - 3x^2 + 4x + 32$ is divided by $(x + 2),$ the remainder is:

A
$0$
B
$32$
C
$36$
✓
$4$

Answer

Correct option: D.

$4$

$p(x) = x^3 - 3x^2 + 4x + 32$
$x + 2 = 0 \Rightarrow x = -2$
By the renainder theorem, we know that when $p(x)$ is divided by
$(x + 2),$ the remainder is $p(-2).$
Now, $p(-2) = x^3 - 3x^2 + 4x + 32$
$= (-2)^3 - 3(-2)^2 + 4(-2) + 32$
$= -8 - 12 - 8 + 32$
$= 4$

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MCQ 161 Mark

Which of the following is a binomial?

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

Answer

Correct option: B.

$x^2 + 4$

A polynomial with two non$-$zero terms is called a binomial.
$x^2 + 4$ is the polynomial that has two non$-$zero terms.
Hence is a binomial.

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MCQ 171 Mark

Which of the following is a quadratic polynomial?

A
$x + 4$
B
$x^3 + x$
C
$x^3 + 2x + 6$
✓
$x^2 + 5x + 4$

Answer

Correct option: D.

$x^2 + 5x + 4$

A polynomial of degree $2$ is called a quadratic polynomial.
Options $(a), (b)$ and $(c)$ have degrees $1, 3$ and $3$ respectively,
so they are not quadratic polynomials.
The degree of $x^2 + 5x + 4$ is $2,$ so it is a quadratic polynomial.

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MCQ 181 Mark

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

A
$4$
B
$-3$
✓
$2$
D
$-2$

Answer

Correct option: C.

$2$

Let $p(x) = 2x^2 + kx$
Since $(x + 1)$ is a factor of $p(x),$
$= P(-1) = 0$
$\Rightarrow 2(-1)^2 + k(-1) = 0$
$\Rightarrow 2 - k = 0$
$\Rightarrow k = 2$

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MCQ 191 Mark

When $p(x) = 4x^3 - 12x^2 + 11x - 5$ is divided by $(2x - 1),$ the remainder is:

A
$0$
B
$-5$
✓
$-2$
D
$2$

Answer

Correct option: C.

$-2$

$\text{p}(\text{x}) = 4\text{x}^3 - 12\text{x}^2 + 11\text{x} - 5$
$\text{x}-1=0\Rightarrow\text{x}=\frac{1}{2}$
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 201 Mark

The zeros of the polynomial $p(x) = 2x^2 + 5x -3$ are:

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

Answer

Correct option: B.

$\frac{1}{2},-3$

Let $p(x)$ be a polynomial. If $\text{p}(\alpha)=0,$ then we say that $\alpha$ is a zero of a polynomial.
$p(x) = 2x^2 + 5x -3$
Now, $p(x) = 0$
$\Rightarrow 2x^2 + 5x -3 = 0$
$\Rightarrow 2x^2 + 6x - x - 3 = 0$
$\Rightarrow 2x(x + 3) - 1(x + 3) = 0$
$\Rightarrow (2x - 1)(x + 3) = 0$
$\Rightarrow (2x - 1) = 0$ or $(x + 3) = 0$
$\Rightarrow\text{x}=\frac{1}{2}$ or $x = -3$
$\therefore\frac{1}{2}$ and $-3$ are the zeroes of the polynomial $p(x).$

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Question 211 Mark

Which of the following is a polynomial?

$\text{x}-\frac{1}{\text{x}}+2$
$\frac{1}{\text{x}}+5$
$\sqrt{\text{x}}+3$
$-4$

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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Question 221 Mark

If p(x) = x + 4 then p(x) + p(-x) = ?

Answer

Solution:
p(x) = x + 4
p(-x) = -x + 4
p(x) + p(-x) = (x + 4) + (-x + 4)
= x + 4 - x + 4
= 8

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MCQ 231 Mark

If $(x^{100} + 2x^{99} + k)$ is divisible by $(x + 1)$ then the value of $k$ is:

✓
$1$
B
$2$
C
$-2$
D
$-3$

Answer

Correct option: A.

$1$

$p(x) = x^{100} + 2x^{99} + k$
$x + 1 = 0 \Rightarrow x = -1$
By the factor theorem, we know that when $p(x)$ is divided by $(x + 1),$ the remainder is $p(-1).$
Now, $p(-1) = (-1)^{100} + 2(-1)^{99} + k$
$\Rightarrow 0 = 1 - 2 + k ...($Given that $p(x)$ is divisible by $x + 1.)$
$\Rightarrow k = 1$

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MCQ 241 Mark

The zeros of the polynomial $p(x) = 3x^2 - 1$ are:

A
$\frac{1}{3}$ and $3$
✓
$\frac{1}{\sqrt3}$ and $\frac{1}{\sqrt3}$
C
$\frac{-1}{\sqrt3}$ and $\sqrt3$
D
$\frac{1}{\sqrt3}$ and $ \frac{-1}{\sqrt3}$

Answer

Correct option: B.

$\frac{1}{\sqrt3}$ and $\frac{1}{\sqrt3}$

$p(x) = 3x^2 - 1$
Now$, p(x) = 0$
$\Rightarrow 3x^{2 }- 1 = 0$
$\Rightarrow 3x^2 = 1$
$\Rightarrow\text{x}^2=\frac{1}{3}$
$\Rightarrow\text{x}=\frac{1}{\sqrt3}$ and $ \frac{1}{\sqrt3}$

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Question 251 Mark

Which of the following is a polynomial?

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

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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MCQ 261 Mark

Which of the following expression is a polynomial?

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

Answer

Correct option: D.

$\text{x}^2+\frac{2\text{x}^\frac{3}{2}}{\sqrt{\text{x}}}+6$

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.
In option $(d),$
$\text{x}^2+\frac{2\text{x}^\frac{3}{2}}{\sqrt{\text{x}}}+6=\text{x}^2+2\text{x}^{\frac{3}{2}-\frac{1}{2}}+6$
$x^2 + 2x^1 + 6$ Which is a polynomial.

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MCQ 271 Mark

Which of the following is a polynomial?

A
$x^{-2} + x^{-1} + 3$
B
$x + x^{-1} + 2$
C
$x^{-1}$
✓
$0$

Answer

Correct option: D.

$0$

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 281 Mark

Where $p(x) = x^4 + 2x^3 - 3x^2 - 1$ is divided by $(x - 2),$ the remainder is:

A
$0$
B
$-1$
C
$-15$
✓
$21$

Answer

Correct option: D.

$21$

$p(x) = x^4 + 2x^3 - 3x^2 + x - 1$
$x - 2 = 0 \Rightarrow x = 2$
By the remainder theorem, we know that when $p(x)$ is divided by
$(x - 2),$ the remainder is $p(2).$
Now, $p(2) = x^4 + 2x^3 - 3x^2 + x - 1$
$= (2)^4 + 2(2)^3 - 3(2)^2 + 2 - 1$
$= 16 +16 - 12 + 2 - 1$
$= 21$

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MCQ 291 Mark

If $(x + 5)$ is a factor of $= x^3 - 20x + 5k$ then $k =$ ?

A
$-5$
✓
$5$
C
$3$
D
$-3$

Answer

Correct option: B.

$5$

$p(x) = x^3 - 20x + 5k$
Now, $x + 5 = 0 \Rightarrow x = (-5)$
By factor theorem,
$p(-5) = 0$
$\Rightarrow (-5)^3 - 20(-5) + 5k = 0$
$\Rightarrow -125 + 100 + 5k = 0$
$\Rightarrow -25 + 5k = 0$
$\Rightarrow 5k = 25$
$\Rightarrow k = 5$

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MCQ 301 Mark

The zeros of the polynomial $p(x) = x^2 + x - 6$ are:

A
$2, 3$
B
$-2, 3$
✓
$2, -3$
D
$-2, -3$

Answer

Correct option: C.

$2, -3$

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

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MCQ 311 Mark

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

A
$x^3 + x^2 - x + 1$
✓
$x^3 + 2x^2 - x - 2$
C
$x^3 + 2x^2 - x + 2$
D
$x^4 + x^3 + x^2 + 1$

Answer

Correct option: B.

$x^3 + 2x^2 - x - 2$

Given$, x^3 + 2x^2 - x - 2$
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 $x^3 + 2x^2 - x – 2$

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MCQ 321 Mark

When $p(x) = x^3 + ax^2 + 2x +a$ is divided by $(x + a),$ the remainder is:

A
$0$
B
$a$
✓
$-a$
D
$2a$

Answer

Correct option: C.

$-a$

$p(x) = x^3 + ax^2 + 2x +a$
$x + a = 0 \Rightarrow x = -a$
By the remainder theorem, we know that when $p(x)$ is divided by $(x + a),$ the remainder is $p(-a).$
Now, $p(-a) = x^3 + ax^2 + 2x +a$
$= (-a)^3 + a(-a)^2 + 2(-a) +a$
$= -a^3 + a^3 - 2a + a$
$= -a$

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