2 Marks Questions - Maths STD 9 Questions - Vidyadip

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22 questions · timed · auto-graded

Question 12 Marks

What must be subtracted from x³ - 6x² - 15x + 80 so that the result is exactly divisible by x² + x - 12?

Answer

$\therefore$ 4x - 4 should subtracted from x³ - 6x²- 15x + 80
So that the result is exactly divisible by x² + x - 12.

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Question 22 Marks

Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:
$\text{f(x)}=\text{l(x)}+\text{m},\text{x}=-\frac{\text{m}}{\text{l}}$

Answer

$\text{f(x)}=\text{l(x)}+\text{m},\text{x}=-\frac{\text{m}}{\text{l}}$
We know that,
$\text{f(x)}=\text{l(x)}+\text{m}$
Given, that $\text{x}=-\frac{\text{m}}{\text{l}}$
Substitute the value of x in f(x)
$\text{f}\Big(-\frac{\text{m}}{\text{l}}\Big)=\text{I}\Big(-\frac{\text{m}}{\text{l}}\Big)+\text{m}$
$=-\text{m}+\text{m}$
$=0$
Since, the result is 0, $\text{x}=-\frac{\text{m}}{\text{l}}$ is the root of lx + m

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Question 32 Marks

Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:
$\text{f(x)}=3\text{x}+1,\text{x}=-\frac{1}{3}$

Answer

$\text{f(x)}=3\text{x}+1,\text{x}=-\frac{1}{3}$
We know that,
$\text{f(x)}=3\text{x}+1$
Substitite $\text{x}=-\frac{1}{3}$ in f(x)
$\text{f}\Big(-\frac{1}{3}\Big)=3\Big(-\frac{1}{3}\Big)+1$
$= -1 + 1$
$=0$
Since, the result is 0 $\text{x}=-\frac{1}{3}$ is the root of 3x + 1

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Question 42 Marks

Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:
f(x) = x², x = 0

Answer

f(x) = x², x = 0
we know that , f(x) = x²
Given that value of x is '0'
Substitute the value of x in f(x)
f(0) = 0²
= 0
Since, the result is zero, x = 0 is the root of x²

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Question 52 Marks

In the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not:
f(x) = x³ - 6x² - 19x + 84, g(x) = x - 7

Answer

Let g(x) = 0
⇒ x - 7 = 0
⇒ x = 7
f(7) = 7³ - 6(7)²- 19(7) + 84
= 343 - 6(49) - 19(7) + 84
= 343 - 294 - 133 + 84
= 427 - 427 = 0
$\because$ f(7) = 0, by factor theorem x - 7 is a factor of f(x).

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Question 62 Marks

In the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not:
f(x) = x³ - 6x² + 11x - 6; g(x) = x - 3

Answer

Let g(x) = 0
⇒ x - 3 = 0
⇒ x = 3
f(3) = 3³ - 6(3)² + 11(3) - 6
= 27 - 6(9) + 33 - 6
= 60 - 60 = 0
$\because$ f(3) = 0, by factor theorem x - 3 is a factor of f(x).

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Question 72 Marks

In the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not:
f(x) = 3x⁴ + 17x³ + 9x² - 7x - 10; g(x) = x + 5

Answer

Let g(x) = 0
⇒ x + 5 = 0
⇒ x = -5
Now,
f(-5) = 3(-5)⁴ + 17(-5)³ + 9(-5)² - 7(-5) - 10
= 3(625) + 17(-125) + 9(25) + 35 - 10
= 1875 - 2125 + 225 + 35 - 10
$\because$ f(-5) = 0, by factor theorem x + 5 is a factor of f(x).

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Question 82 Marks

In the following, use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not:
f(x) = x⁵ + 3x⁴ - x³ - 3x² + 5x + 15, g(x) = x + 3

Answer

Let g(x) = 0
⇒ x + 3 = 0
⇒ x = -3
f(-3) = (-3)⁵ - 3(-3)⁴ -(-3)³ - 3(-3)² + 5(-3) + 15
= -243 + 243 + 27 - 27 - 15 + 15 = 0
$\because$ f(-3) = 0, by factor theorem x + 3 is a factor of f(x).

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Question 92 Marks

In the following two polynomials, find the value of a, if x - a is factor:
x⁶ - ax⁵ + x⁴ - ax³ + 3x - a + 2

Answer

x⁶ - ax⁵ + x⁴ - ax³ + 3x - a + 2
Let,
x - a = 0
⇒ x = a
$\because$ x - a is a factor of the polynomial f(x)
$\therefore$ f(a) = 0
f(a) = 0
⇒ a⁶ - a × a⁵ + a⁴ - a × a³ + 3a - a + 2 = 0
⇒ a⁶ - a⁶ + a⁴ - a⁴ + 3a - a + 2 = 0
⇒ 2a + 2 = 0
⇒ 2a = -2
⇒ a = -1

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Question 102 Marks

In the following two polynomials, find the value of a, if x + a is a factor.
x⁴ - a²x² + 3x - a

Answer

x⁴ - a²x² + 3x - a
Let,
x + a = 0
⇒ x = -a
$\because$ (x + a) is a factor of f(x) = x⁴ - a²x² + 3x - a
$\therefore$ f(-a) = 0
f(-a) ⇒ (-a)⁴ - a²(-a)² + 3(-a) - a = 0
⇒ a⁴ + a⁴ - 3a - a = 0
⇒ -4a = 0
⇒ a = 0

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Question 112 Marks

In the following two polynomials, find the value of a, if x + a is a factor.
x³ + ax² - 2x + a + 4

Answer

x³ + ax² - 2x + a + 4
Let,
x + a = 0
⇒ x = -a
$\because$ (x + a) is a factor of f(x) = x³ + ax² - 2x + a + 4
$\therefore$ p(-a) = 0
p(-a) ⇒ (-a)³ + (-a)² - 2(-a) + a + 4 = 0
⇒ -a³ + a³ + 2a + a + 4 = 0
⇒ 3a + 4 = 0
⇒ 3a = -4
$\Rightarrow\ \text{a}=\frac{-4}{3}$

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Question 122 Marks

In the following two polynomials, find the value of a, if x - a is factor:
x⁵ - a²x³ + 2x + a + 1

Answer

x⁵ - a²x³ + 2x + a + 1
Let,
x - a = 0
⇒ x = a
$\because$ (x - a) is a factor of f(x) = x⁵ - a²x³ + 2x + a + 1
$\therefore$ f(a) = 0
f(a) ⇒ a⁵ - a² × a³ + 2a + a + 1 = 0
⇒ a⁵ - a⁵ + 3a + 1
⇒ 3a = -1
$\Rightarrow\ \text{a}=-\frac{1}{3}$

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Question 132 Marks

If x + 1 is a factor of x³ + a, then write the value of a.

Answer

As (x + 1) is a factor of polynomial f(x) = x³ + a.
i. e. f(-1) = 0
(-1)³+ a = 0
⇒ a = 1
Thus, the value of a = 1.

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Question 142 Marks

If $\text{x}=\frac{1}{2}$ is a zero of the polynomial f(x) = 8x³ + ax² - 4x + 2, find the value of a.

Answer

Since $\text{x}=\frac{1}{2}$ is a zero of polynomial f(x).
Therefore $\text{f}\Big(\frac{1}{2}\Big)=0$
$\Rightarrow\ 8\Big(\frac{1}{2}\Big)^3+\text{a}\Big(\frac{1}{2}\Big)^2-4\Big(\frac{1}{2}\Big)+2=0$
$\Rightarrow\ 1+\frac{\text{a}}{4}-2+2=0$
$\Rightarrow\ \text{a}=-4$
The value of a is -4.

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Question 152 Marks

If f(x) = x⁴ - 2x³ + 3x² - ax - b when divided by x - 1, the remainder is 6, then find the value of a + b.

Answer

When polynomial f(x) = x⁴ - 2x³ + 3x² - ax - b divided by (x - 1)
The remainder is 6.
i. e. f(1) = 6
(1)⁴ - 2(1)³ + 3(1)² - a(1) - b = 6
1 - 2 + 3 - a - b = 6
2 - (a + b) = 6
(a + b) = -4
Thus, the value of a + b = -4.

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Question 162 Marks

If f(x) = 2x³ - 13x² + 17x + 12, Find:
f(-3)

Answer

The given polynomial is f(x) = 2x³ - 13x² + 17x + 12
f(-3)
we need to substitute the '(-3)' in f(x)
f(-3) = 2(-3)³ - 13(-3)² + 17(-3) + 12
= (2 × (- 27)) - (13 × 9) - ( 17 × 3) + 12
= -54 - 117 - 51 + 12
= -210
therefore f(-3) = -210

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Question 172 Marks

If f(x) = 2x³ - 13x² + 17x + 12, Find:
f(2)

Answer

The given polynomial is f(x) = 2x³ - 13x² + 17x + 12
f(2)
we need to substitute the '2' in f(x)
f(2) = 2(2)³ - 13(2)² + 17(2) + 12
= (2 × 8) - (13 × 4) + (17 × 2) + 12
= 16 - 52 + 34 + 12
= 10
therefore f(2) = 10

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Question 182 Marks

If f(x) = 2x³ - 13x² + 17x + 12, Find:
f(0)

Answer

The given polynomial is f(x) = 2x³ - 13x² + 17x + 12
f(0)
we need to substitute the '(0)' in f(x)
f(0) = 2(0)³ - 13(0)² + 17(0) + 12
= (2 × 0) - ( 13 × 0) + (17 × 0) + 12
= 0 - 0 + 0 + 12
= 12
therefore f(0) = 12

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Question 192 Marks

Give one example each of a binomial of degree 35, and of a monomial of degree 100

Answer

Given, to write the examples for binomial and monomial with the given degrees.
Example of a binomial with degree 35 - 7x³⁵ - 5
Example of a monomial with degree 100 - 2t¹⁰⁰

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Question 202 Marks

Find the remainder when x³ + 3x³ + 3x + 1 is divided by:
x

Answer

Here,
f(x) = x³ + 3x² + 3x + 1
By remainder theorem
x = 0
substitute the value of x in f(x)
f(0) = 0³+ 3(0)² + 3(0) + 1
= 0 + 0 + 0 + 1
= 1

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Question 212 Marks

Find the remainder when x³ + 3x³ + 3x + 1 is divided by:
x + 1

Answer

Here, f(x) = x³ + 3x² + 3x + 1
By remainder theorem
x + 1 = 0
⇒ x = -1
substitute the value of x in f(x)
f(-1) = (−1)³ + 3(−1)² + 3(−1) + 1
= -1 + 3 - 3 + 1
= 0

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Question 222 Marks

Find the remainder when x³ + 3x³ + 3x + 1 is divided by:
$\text{x}-\frac{1}{2}$

Answer

Here, f(x) = x³ + 3x² + 3x + 1
By remainder theorem
$\text{x}-\frac{1}{2}$
$\Rightarrow\ \text{x}=\frac{1}{2}$
substitute the value of x in f(x)
$\text{f}\Big(\frac{1}{2}\Big)=\Big(\frac{1}{2}\Big)^3+3\Big(\frac{1}{2}\Big)^2+3\Big(\frac{1}{2}\Big)+1$
$=\Big(\frac{1}{2}\Big)^3+3\Big(\frac{1}{2}\Big)^2+3\Big(\frac{1}{2}\Big)+1$
$=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1$
$=\frac{1+6+12+8}{8}$
$=\frac{27}{8}$

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2 Marks Questions - Maths STD 9 Questions - Vidyadip