RS AGGARWAL CLASS 9 Chapter 2 POLYNOMIALS MCQ

 MULTIPLE CHOICE QUESTIONS

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Question 1:

Which of the following expressions is a polynomial in one variable?
(a) $x+\frac{2}{x}+3$
(b) $3\sqrt{x}+\frac{2}{\sqrt{x}}+5$
(c) $\sqrt{2}{x}^2 - \sqrt{3}x + 6$
(d) x¹⁰ + y⁵ + 8

Answer 1:

(c) $\sqrt{2}{x}^2 - \sqrt{3}x + 6$

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

Question 2:

Which of the following expression is a polynomial?
(a) $\sqrt{x}-1$
(b) $\frac{x-1}{x+1}$
(c) $x^2-\frac{2}{x^2}+5$
(d) $x^2+\frac{2x^{3/2}}{\sqrt{x}}+6$

Answer 2:

(d) $x^2+\frac{2x^{3/2}}{\sqrt{x}}+6$

We have:

$x^2+\frac{2x^{{\displaystyle \frac{3}{2}}}}{\sqrt{x}}+6=x^2+2x^{\frac{3}{2}}{x}^{-\frac{1}{2}}+6$
               $=x^2+2x+6$
 
It a polynomial because it has only non-negative integral powers of x.

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Question 3:

(a) $\sqrt[3]{y}+4$
(b) $\sqrt{y}-3$
(c) y
(d) $\frac{1}{\sqrt{y}}+7$

Answer 3:

(c) y

 y is a polynomial because it has a non-negative integral power 1.

Question 4:

Which of the following is a polynimial?
(a) $x-\frac{1}{x}+2$
(b) $\frac{1}{x}+5$
(c) $\sqrt{x}+3$
(d) −4

Answer 4:

(d) −4

$-$4 is a constant polynomial of degree zero.

Question 5:

Which of the following is a polynomial?
(a) x⁻² + x⁻¹ + 3
(b) x + x⁻¹ + 2
(c) x⁻¹
(d) 0

Answer 5:

(d) 0

0 is a polynomial whose degree is not defined.

Question 6:

Which of the following is quadratic polynomial?
(a) x + 4
(b) x³ + x
(c) x³ + 2x + 6
(d) x² + 5x + 4

Answer 6:

(d) x² + 5x + 4

$x^2+5x+4$ is a polynomial of degree 2. So, it is a quadratic polynomial.

Question 7:

Which of the following is a linear polynomial?
(a) x + x²
(b) x + 1
(c) 5x² − x + 3
(d) $x+\frac{1}{x}$

Answer 7:

(b) x + 1

Clearly, $x+1$ is a polynomial of degree 1. So, it is a linear polynomial.

Question 8:

Which of the following is a binomial?
(a) x² + x + 3
(b) x² + 4
(c) 2x²
(d) $x+3+\frac{1}{x}$

Answer 8:

(b) x² + 4

Clearly, $x^2+4$ is an expression having two non-zero terms. So, it is a binomial.

Question 9:

$\sqrt{3}$is a polynomial of degree
(a) $\frac{1}{2}$
(b) 2
(c) 1
(d) 0

Answer 9:

(d) 0

$\sqrt{3}$ is a constant term, so it is a polynomial of degree 0.

Question 10:

Degree of the zero polynomial is
(a) 1
(b) 0
(c) not defined
(d) none of these

Answer 10:

(c) not defined

Degree of the zero polynomial is not defined.

Question 11:

Zero of the zero polynomial is
(a) 0
(b) 1
(c) every real number
(d) not defined

Answer 11:

(d) not defined

Zero of the zero polynomial is not defined.

Question 12:

If p(x) = x = 4, then p(x) + p(−x) = ?
(a) 0
(b) 4
(c) 2x
(d) 8

Answer 12:

(d) 8

Let:
$p\left(x\right)=\left(x+4\right)$

$$\begin{gathered} \therefore p\left(-x\right)=\left(-x\right)+4 \\ =-x+4 \end{gathered}$$
Thus, we have:
$p\left(x\right)+p\left(-x\right)=\left\{\left(x+4\right)+\left(-x+4\right)\right\}$
               = 4 + 4
               =8

Question 13:

If $p\left(x\right)=x^2-2\sqrt{2}x+1$, then $p\left(2\sqrt{2}\right)=$?
(a) 0
(b) 1
(c) $4\sqrt{2}$
(d) −1

Answer 13:

(b) 1

$$\begin{gathered} p\left(x\right)=x^2-2\sqrt{2} x+1 \\ \therefore p\left(2\sqrt{2}\right)={\left(2\sqrt{2}\right)}^2-2\sqrt{2}\times \left(2\sqrt{2}\right)+1 \end{gathered}$$
             = 8 $-$ 8 + 1
             = 1

Question 14:

If p(x) = 5x – 4x² + 3 then p(–1) =?
(a) 2
(b) –2
(c) 6
(d) –6

Answer 14:

p(x) = 5x – 4x² + 3

Putting x = –1 in p(x), we get

p(–1) = 5 × (–1) – 4 × (–1)² + 3 = –5 – 4 + 3 = –6

Hence, the correct answer is option (d).

 

Question 15:

If (x⁵¹ + 51) is divided by (x + 1) then the remainder is
(a) 0
(b) 1
(c) 49
(d) 50

Answer 15:

Let f(x) = x⁵¹ + 51

By remainder theorem, when f(x) is divided by (x + 1), then the remainder = f(−1).

Putting x = −1 in f(x), we get

 f(−1) = (−1)⁵¹ + 51 = −1 + 51 = 50

∴ Remainder = 50

Thus, the remainder when (x⁵¹ + 51) is divided by (x + 1) is 50.

Hence, the correct answer is option (d).

Question 16:

If (x + 1) is a factor of (2x² + kx), then k = ?
(a) 4
(b) −3
(c) 2
(d) –2

Answer 16:

(c) 2

$$\begin{gathered} \left(x+1\right) \text{is a factor of} 2x^2+kx. \\ \mathrm{So}, -1 \mathrm{is} \mathrm{a} \mathrm{zero} \mathrm{of} 2x^2+kx. \\ \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ 2\times {\left(-1\right)}^2+k\times \left(-1\right)=0 \\ \Rightarrow 2-k=0 \\ \Rightarrow k=2 \end{gathered}$$

Question 17:

When p(x) = x⁴ + 2x³ − 3x² + x − 1 is divided by (x − 2), the remainder is
(a) 0
(b) −1
(c) −15
(d) 21

Answer 17:

(d) 21

$x-2=0\Rightarrow x=2$
By the remainder theorem, we know that when p(x) is divided by (x $-$ 2), the remainder is p(2).
Thus, we have:
$$\begin{gathered} p\left(2\right)=2^4+2\times 2^3-3\times 2^2+2-1 \\ =16+16-12+1 \\ =21 \end{gathered}$$

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Question 18:

When p(x) = x³ − 3x² + 4x + 32 is divided by (x + 2), the remainder is
(a) 0
(b) 32
(c) 36
(d) 4

Answer 18:

(d) 4

$x+2=0\Rightarrow x=-2$
By the remainder theorem, we know that when p(x) is divided by (x + 2), the remainder is p($-$2).
Now, we have:
$$\begin{gathered} p\left(-2\right)={\left(-2\right)}^3-3\times {\left(-2\right)}^2+4\times \left(-2\right)+32 \\ =-8-12-8+32 \\ =4 \end{gathered}$$

Question 19:

When p(x) = 4x³ − 12x² + 11x − 5 is divided by (2x − 1), the remainder is
(a) 0
(b) −5
(c) −2
(d) 2

Answer 19:

(c) −2

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

Question 20:

When p(x) = x³ – ax² + x is divided by (x – a), the remainder is
(a) 0
(b) a
(c) 2a
(d) 3a

Answer 20:

By remainder theorem, when p(x) = x³ – ax² + x is divided by (x – a), then the remainder = p(a).

Putting x = a in p(x), we get

p(a) = a³ – a × a² + a = a³ – a³ + a = a

∴ Remainder = a

Hence, the correct answer is option (b).

Question 21:

When p(x) = x³ + ax² + 2x + a is divided by (x + a), the remainder is
(a) 0
(b) −1
(c) −15
(d) 21

Answer 21:

(c) −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).
Thus, we have:
$$\begin{gathered} p\left(-a\right)={\left(-a\right)}^3+a\times {\left(-a\right)}^2+2\times \left(-a\right)+a \\ =-a^3+a^3-2a+a \\ =-a \end{gathered}$$

Question 22:

(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 22:

(b) x³ − 2x² − x − 2

Let:
f(x) = x³ − 2x² − x − 2
By the factor theorem, (x + 1) will be a factor of f(x) if f ($-$1) = 0.
We have:
$$\begin{gathered} f\left(-1\right)={\left(-1\right)}^3+2\times {\left(-1\right)}^2-\left(-1\right)-2 \\ =-1+2+1-2 \\ =0 \end{gathered}$$
Hence, (x + 1) is a factor of $f\left(x\right)=x^3+2x^2-x-2$.

Question 23:

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

(a) $\frac{-2}{5}$

(b) $\frac{-5}{2}$

(c) $\frac{2}{5}$

(d) $\frac{5}{2}$

Answer 23:

The zero of the polynomial p(x) can be obtained by putting p(x) = 0.

$$\begin{gathered} p\left(x\right)=0 \\ \Rightarrow 2x+5=0 \\ \Rightarrow 2x=-5 \\ \Rightarrow x=-\frac{5}{2} \end{gathered}$$
Hence, the correct answer is option (b).

Question 24:

The zeros of the polynomial p(x) = x² + x – 6 are
(a) 2, 3
(b) –2, 3
(c) 2, –3
(d) –2, –3

Answer 24:

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 zeroes of the given polynomial p(x).

Hence, the correct answer is option (c).

Question 25:

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

(a) $\frac{1}{2}, 3$

(b) $\frac{1}{2}, -3$

(c) $\frac{-1}{2}, 3$

(d) $1,\frac{-1}{2}$

Answer 25:

The given polynomial is p(x) = 2x² + 5x – 3.

Putting $x=\frac{1}{2}$ in p(x), we get

$p\left(\frac{1}{2}\right)=2\times {\left(\frac{1}{2}\right)}^2+5\times \frac{1}{2}-3=\frac{1}{2}+\frac{5}{2}-3=3-3=0$

Therefore, $x=\frac{1}{2}$ is a zero of the polynomial p(x).

Putting x = –3 in p(x), we get

$p\left(-3\right)=2\times {\left(-3\right)}^2+5\times \left(-3\right)-3=18-15-3=0$

Therefore, x = –3 is a zero of the polynomial p(x).

Thus, $\frac{1}{2}$ and –3 are the zeroes of the given polynomial p(x).

Hence, the correct answer is option (b).

Question 26:

The zeros of the polynomial p(x) = 2x² + 7x – 4 are

(a) $4, \frac{-1}{2}$

(b) $4, \frac{1}{2}$

(c) $-4, \frac{1}{2}$

(d) $-4, \frac{-1}{2}$

Answer 26:

The given polynomial is p(x) = 2x² + 7x – 4.

Putting $x=\frac{1}{2}$ in p(x), we get

$p\left(\frac{1}{2}\right)=2\times {\left(\frac{1}{2}\right)}^2+7\times \frac{1}{2}-4=\frac{1}{2}+\frac{7}{2}-4=4-4=0$

Therefore, $x=\frac{1}{2}$ is a zero of the polynomial p(x).

Putting x = –4 in p(x), we get

$p\left(-4\right)=2\times {\left(-4\right)}^2+7\times \left(-4\right)-4=32-28-4=32-32=0$

Therefore, x = –4 is a zero of the polynomial p(x).

Thus, $\frac{1}{2}$ and –4 are the zeroes of the given polynomial p(x).

Hence, the correct answer is option (c).

Question 27:

If (x + 5) is a factor of p(x) = x³ − 20x + 5k, then k = ?
(a) −5
(b) 5
(c) 3
(d) −3

Answer 27:

(b) 5

$$\begin{gathered} \left(x+5\right) \text{is a factor of }p\left(x\right)=x^3-20x+5k. \\ \therefore p\left(-5\right)=0 \\ \Rightarrow {\left(-5\right)}^3-20\times \left(-5\right)+5k=0 \\ \Rightarrow -125+100+5k=0 \\ \Rightarrow 5k=25 \\ \Rightarrow k=5 \end{gathered}$$

Question 28:

If (x + 2) and (x − 1) are factors 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 28:

(b) m = 7, n = −18

Let:
$p\left(x\right)=x^3+10x^2+mx+n$
Now,
$x+2=0\Rightarrow x=-2$
(x + 2) is a factor of p(x).
So, we have p($-$2)=0
$$\begin{gathered} \Rightarrow {\left(-2\right)}^3+10\times {\left(-2\right)}^2+m\times \left(-2\right)+n=0 \\ \Rightarrow -8+40-2m+n=0 \\ \Rightarrow 32-2m+n=0 \\ \Rightarrow 2m-n=32 .....\left(\mathrm{i}\right) \end{gathered}$$
Now,
$x-1=0\Rightarrow x=1$
Also, 
(x $-$ 1) is a factor of p(x).
We have:
p(1) = 0
$$\begin{gathered} \Rightarrow 1^3+10\times 1^2+m\times 1+n=0 \\ \Rightarrow 1+10+m+n=0 \\ \Rightarrow 11+m+n=0 \\ \Rightarrow m+n=-11 .....\left(\mathrm{ii}\right) \\ \mathrm{From} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{get}: \\ 3m=21\Rightarrow m=7 \end{gathered}$$
By substituting the value of m in (i), we get n = −18.
∴ m = 7 and n = −18

Question 29:

If (x¹⁰⁰ + 2x⁹⁹ + k) is divisible by (x + 1), then the value of k is
(a) 1
(b) 2
(c) −2
(d) −3

Answer 29:

(a) 1

Let:
$p\left(x\right)=x^{100}+2x^{99}+k$
Now,
$x+1=0\Rightarrow x=-1$
$$\begin{gathered} \left(x+1\right) \text{is divisible by }p\left(x\right). \\ \therefore p\left(-1\right)=0 \\ \Rightarrow {\left(-1\right)}^{100}+2\times {\left(-1\right)}^{99}+k=0 \\ \Rightarrow 1-2+k=0 \\ \Rightarrow -1+k=0 \\ \Rightarrow k=1 \end{gathered}$$

Question 30:

For what value of k is the polynomial p(x) = 2x³ − kx + 3x + 10 exactly divisible by (x + 2)?
(a) $-\frac{1}{3}$
(b) $\frac{1}{3}$
(c) 3
(d) −3

Answer 30:

(d) −3

Let:
$p\left(x\right)=2x^3-k{x}^2+3x+10$
Now,
$x+2=0\Rightarrow x=-2$
$$\begin{gathered} p\left(x\right) \text{is completely divisible by }\left(x+2\right)\text{.} \\ \therefore p\left(-2\right)=0 \\ \Rightarrow 2\times {\left(-2\right)}^3-k\times {\left(-2\right)}^2+3\times \left(-2\right)+10=0 \\ \Rightarrow -16-4k-6+10=0 \\ \Rightarrow -12-4k=0 \\ \Rightarrow 4k=-12 \\ \Rightarrow k=\frac{-12}{4} \\ \Rightarrow k=-3 \end{gathered}$$

Question 31:

The zeroes of the polynomial p(x) = x² − 3x are
(a) 0, 0
(b) 0, 3
(c) 0, −3
(d) 3, −3

Answer 31:

(b) 0, 3

Let:
$p\left(x\right)=x^2-3x$
Now, we have:
$p\left(x\right)=0\Rightarrow x^2-3x=0$
         $$\begin{gathered} \Rightarrow x\left(x-3\right)=0 \\ \Rightarrow x=0 \text{and} x-3=0 \\ \Rightarrow x=0 \text{and} x=3 \end{gathered}$$

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Question 32:

The zeroes of the polynomial p(x) = 3x² − 1 are
(a) $\frac{1}{3}$ and 3
(b) $\frac{1}{\sqrt{3}} \mathrm{and} \sqrt{3}$
(c) $\frac{-1}{\sqrt{3}} \mathrm{and} \sqrt{3}$
(d) $\frac{1}{\sqrt{3}}\mathrm{and}\frac{-1}{\sqrt{3}}$

Answer 32:

(d) $\frac{1}{\sqrt{3}} \mathrm{and} \frac{-1}{\sqrt{3}}$

Let:
$p\left(x\right)=3x^2-1$

$$\begin{gathered} \mathrm{To} \mathrm{find} \mathrm{the} \mathrm{zeroes} \mathrm{of} p\left(x\right), \mathrm{we} \mathrm{have}: \\ p\left(x\right)=0\Rightarrow 3x^2-1=0 \end{gathered}$$
         $$\begin{gathered} \Rightarrow 3x^2=1 \\ \Rightarrow x^2=\frac{1}{3} \\ \Rightarrow x=\pm \frac{1}{\sqrt{3}} \\ \Rightarrow x=\frac{1}{\sqrt{3}} \text{and} x=-\frac{1}{\sqrt{3}} \end{gathered}$$
Hence, the correct answer is option D.