2 Marks Questions - MATHS STD 11 Science Questions - Vidyadip

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2 Marks Questions

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24 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks

Show the following quadratic equation by factorization method:
17x² - 8x + 1 = 0

Answer

17x² - 8x + 1 = 0
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where D = b² - 4ac
= (-8)² - 4.17.1
= 64 - 68
= -4
From (A)
$\text{x}=\frac{-(-8)\pm\sqrt{-4}}{2.17}$
$=\frac{8\pm2\text{i}}{34}$
$=\frac{4\pm\text{i}}{17}$
$\therefore\text{x}=\frac{4}{17}+\frac{\text{i}}{17},\frac{4}{17}-=\frac{\text{i}}{17}$

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Question 22 Marks

Show the following quadratic equation by factorization method:
x² - x + 1 = 0

Answer

x² - x + 1 = 0
Now, completing the squares, we get
$\Big(\text{x}+\frac{1}{2}\Big)^2+\frac{3}{4}=0$
$\Rightarrow\Big(\text{x}+\frac{1}{2}\Big)^2-\Big(\frac{\sqrt{3}}{2}\text{i}\Big)^2=0$
$\Rightarrow\Big(\text{x}+\frac{1}{2}+\frac{\sqrt{3}}{2}\text{i}\Big)\Big(\text{x}+\frac{1}{2}-\frac{\sqrt{3 }}{2}\text{i}\Big)=0$
$\Rightarrow\Big(\text{x}+\frac{1}{2}+\frac{\sqrt{3}}{2}\text{i}\Big)=0\ \text{or } \Big(\text{x}+\frac{1}{2}-\frac{\sqrt{3}}{2}\text{i}\Big)=0$
$\therefore\text{x}=\frac{-1}{2}+\frac{-1}{2}+\frac{\sqrt{3}}{2}\text{ i},\frac{-1}{2}-\frac{\sqrt{3}}{2}\text{ i}$

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Question 32 Marks

Show the following quadratic equation by factorization method:
$\text{x}^2-(2\sqrt{3}+3\text{i}))\text{x}+6\sqrt{3}\text{ i}=0$

Answer

$\text{x}^2-(2\sqrt{3}+3\text{i}))\text{x}+6\sqrt{3}\text{ i}=0$
$\Rightarrow\text{x}^2-2\sqrt{3\text{x}}-3\text{ix}+6\sqrt{3}\text{i}=0$
$\Rightarrow\text{x}(\text{x}-2\sqrt{3})-3\text{i}(\text{x}-2\sqrt{3})=0$
$\Rightarrow(\text{x}-3\text{i})(\text{x}-2\sqrt{3})=0$
$\Rightarrow\text{x}=3\text,2\sqrt{3}$

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Question 42 Marks

Show the following quadratic equation:

(2 + i) x² - (5 - i) x + 2 (1 - i) = 0

Answer

(2 + i) x² - (5 - i) x + 2 (1 - i) = 0
⇒ (2 + i) x² - 2x - (3 - i) x + 2 (1 - i) = 0
⇒ x [2 + ix - 2] - (1 - i) [(2 + i) x - 2] = 0
⇒ [x - (1 - i)] [(2 + i) x - 2] = 0
Either [x - (1 - i)] = 0 or [(2 + i) x - 2] = 0
$\Rightarrow\text{x}=1- \text{i }\text{ or }\text{x}=\frac{2}{2+\text{i}}$
$\Rightarrow\text{x}=1-\text{i}\text{ or }\text{x}=\frac{2\times2-\text{i}}{(2+\text{i})(2-\text{i})}$
$\text{x}=\frac{4-2\text{i}}{4+1}=\frac{4}{5}-\frac{2}{5}\text{i}$
Thus
$\text{x}=1-\text{i},\frac{4}{5}-\frac{2}{5}\text{i}$

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Question 52 Marks

Show the following quadratic equation by factorization method:
5x² + 6x + 2 = 0

Answer

5x² + 6x + 2 = 0
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where D = b² - 4ac
= 36 - 40
= 4 - 8
= -4
From (A)
$\text{x}=-\frac{-(-6)\pm\sqrt{-4}}{2.5}$
$=\frac{6\pm2\text{i}}{10}$
$=\frac{3\pm\text{i}}{5}$
$\therefore\text{x}=\frac{3}{5}+\frac{\text{i}}{5},\frac{3}{5}-\frac{\text{i}}{5}$

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Question 62 Marks

Show the following quadratic equation by factorization method:
x² + 4x + 7 = 0

Answer

x² + 4x + 7 = 0
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where D = b² - 4ac = (-4)² - 4.1.7 = -12
From (A)
$\text{x}=-\frac{(-4)\pm\sqrt{-12}}{2}$
$=\frac{4\pm2\sqrt{3}\text{i}}{2}$
$=2\pm\sqrt{3}\text{i}$
$\therefore\text{x}=2+\sqrt{3}\text{i},2-\sqrt{3}\text{i}$

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Question 72 Marks

Show the following quadratic equation by factorization method:
x² + 2x + 2 = 0

Answer

x² + 2x + 2 = 0
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where D = b² - 4ac
= 2² - 4.1.2
= 4 - 8
= -4
From (A)
$\text{x}=-\frac{(-2)\pm\sqrt{-4}}{2}$
$=\frac{-2\pm2\text{i}}{2}$
$=-1\pm\text{i}$
$\therefore\text{x}=-1+\text{i},-1-\text{i}$

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Question 82 Marks

Show the following quadratic equation by factorization method:
4x² + 1 = 0

Answer

4x² + 1 = 0
⇒ (2x)² - i² = 0 [$\because$ i² = -1]
⇒ (2x + i) (2x - i) = 0
⇒ Either 2x + i = 0 or 2x - i = 0
$\Rightarrow\text{x}=\frac{-\text{i}}{2}\ \text{or }\text{x}=\frac{\text{i}}{2}$
$\therefore\text{x}=\frac{-\text{i}}{2},\frac{\text{i}}{2}$

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Question 92 Marks

Show the following quadratic equation by factorization method:
x² + x + 1 = 0

Answer

x² + x + 1 = 0
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where D = b² - 4ac
= (-1)² - 4.1.1
= 1 - 4
= -3
From (A)
$\text{x}=\frac{-1\pm\sqrt{-3}}{2}$
$=\frac{-1\pm\sqrt{3\text{i}}}{2}$
$\therefore\text{x}=\frac{-1}{2}+\frac{\sqrt{3}}{2}\ \text{i},\frac{-1}{2}-\frac{\sqrt{3}}{2}\ \text{i}$

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Question 102 Marks

Show the following quadratic equation by factorization method:
x²+ (1 − 2i) x − 2i = 0

Answer

x²+ (1 − 2i) x − 2i = 0

⇒ x² + x - 2i - 2i = 0

⇒ x (x + 1) - 2i (x + 1) = 0

⇒ (x + 2i) (x + 1) = 0

⇒ x = 2i, -1

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Question 112 Marks

Show the following quadratic equation:
ix² - 4x - 4i = 0

Answer

ix² - 4x - 4i = 0
⇒ ix² + 4i²x + 4i³ = 0
⇒ x² + 4ix + 4i² = 0
⇒ x² + 2ix + 2ix + 4i² = 0
⇒ x (x + 2i) + 2i (x + 2i) = 0
⇒ (x + 2i) (x + 2i)
$\therefore\text{x}=-2\text{i},-2\text{i}$

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Question 122 Marks

Show the following quadratic equation by factorization method:
13x² - 7x + 1 = 0

Answer

13x² - 7x + 1 = 0
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where D = b² - 4ac
= 7² - 4.13.1
= 49 - 52
= -3
From (A)
$\text{x}=\frac{-7\pm\sqrt{-3}}{2.13}$
$=\frac{-7\pm2\sqrt{3}\text{ i}}{26}$
$\therefore\text{x}=\frac{-7}{26}\pm\frac{\sqrt{3}}{26}\ \text{i}$

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Question 132 Marks

Show the following quadratic equation by factorization method:
x² - x + 1 = 0

Answer

x² - x + 1 = 0
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where D = b² - 4ac
= (-1)² - 4.1.1
= 1 - 4
= -3
From (A)
$\therefore\text{x}=\frac{-(-1)\pm\sqrt{-3}}{2}$
$=\frac{1\pm\sqrt{3\text{i}}}{2}$
$\therefore\text{x}=\frac{1}{2}+\frac{\sqrt{3}}{2}\ \text{i},\frac{1}{2}-\frac{\sqrt{3}}{2}\ \text{i}$

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Question 142 Marks

Show the following quadratic equation by factorization method:
6x² - 17ix - 12 = 0

Answer

6x² - 17ix - 12 = 0
⇒ 6x² - 17ix + 12i² = 0
⇒ 6x² - 9ix - 8ix + 12i² = 0
⇒ 3x (2x - 3i) -4i (2x - 3i) = 0
⇒ (3x - 4i) (2x - 3i) = 0
$\Rightarrow\text{x}=\frac{4}{3}\text{i}\ \text{or }\frac{3}{2}\text{i}$

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Question 152 Marks

Show the following quadratic equation by factorization method:
4x² - 12x + 25 = 0

Answer

4x² - 12x + 25 = 0
Now, completing the squares, we get
(2x - 3)² + 16 = 0
⇒ (2x - 3)² - 4i² = 0
⇒ (2x - 3 + 4i) = 0 or (2x - 3 - 4i) = 0
$\therefore\text{x}=\frac{3}{2}+2\text{i},\frac{3}{2}-2\text{i}$

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Question 162 Marks

Show the following quadratic equation by factorization method:
2x² + x + 1 = 0

Answer

2x² + x + 1 = 0
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where D = b² - 4ac
= 1² - 4.2.1
= 1 - 8
= -7
From (A)
$\text{x}=\frac{-1\pm\sqrt{-7}}{2.2}$
$=\frac{-1\pm\sqrt{7}\text{ i}}{4}$
$\therefore\text{x}=\frac{-1}{4}\pm\frac{\sqrt{7}}{4}\ \text{i}$

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Question 172 Marks

Show the following quadratic equation:
x² - x + (1 + i) = 0

Answer

x² - x + (1 + i) = 0
x² - ix - (1 - i) x + i (1 - i) = 0
(x - i) (x - (1 - i)) = 0
x = i, 1 - i

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Question 182 Marks

Show the following quadratic equation:
x² + 4ix - 4 = 0

Answer

x² + 4ix - 4 = 0
⇒ x² + 4ix + 4i² = 0
⇒ x² + 2ix + 2ix + 4i² = 0
⇒ x (x + 2i) + 2i (x + 2i) = 0
⇒ (x + 2i) (x + 2i) = 0
⇒ x = -2i, -2i

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Question 192 Marks

Show the following quadratic equation by factorization method:
27x² - 10x + 1 = 0

Answer

27x² - 10x + 1 = 0
We will apply discriminant rule,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}\ ...(\text{A})$
Where D = b² - 4ac
= (-10)² - 4.27.1
= 100 - 108
= -4
From (A)
$\text{x}=\frac{-(-10)\pm\sqrt{-8}}{54}$
$=\frac{10\pm2\sqrt{2}\text{i}}{54}$
$\frac{5\pm\sqrt{2}\text{i}}{27}$
$\therefore\text{x}=\frac{5}{27}+\frac{\sqrt{2\text{i}}}{27},\frac{5}{27}-\frac{\sqrt{2}}{27}\text{i}$

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Question 202 Marks

Show the following quadratic equation by factorization method:
x² + 10ix - 21 = 0

Answer

x² + 10ix - 21 = 0
⇒ x² + 10ix + 21i² = 0
⇒ x² + 7ix + 3ix + 21i² = 0
⇒ (x + 3i) (x + 7i) = 0
$\therefore\text{x}=-3\text{i},-7\text{i}$

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Question 212 Marks

Show the following quadratic equation by factorization method:
21x² + 9x + 1 = 0

Answer

21x² + 9x + 1 = 0
Comparing the given equation with the general form
ax² + bx + c = 0, we get a = 21, b = 9, c = 1
Substituting a and b in,
$\alpha=\frac{-\text{b}+\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$ and $\beta=\frac{-\text{b}-\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$
$\alpha=\frac{-9+\sqrt{81+84}}{42}$ and $\beta=\frac{-9-\sqrt{81-84}}{42}$
$\Rightarrow\alpha=\frac{-9+\sqrt{-3}}{42}$ and $\beta=\frac{-9-\sqrt{-3}}{42}$
$\Rightarrow\alpha=\frac{-9+\text{i}\sqrt{3}}{42}$ and $\beta=\frac{-9-\text{i}\sqrt{3}}{42}$
The roots are $\text{x}=\frac{-9}{42}\pm\frac{\text{i}\sqrt{3}}{42}$

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Question 222 Marks

Show the following quadratic equation:
x² - (2 + i) x - (1 - 7i) = 0

Answer

x² - (2 + i) x - (1 - 7i) = 0
⇒ x² - (2 + i) x - (1 - 7i) = 0
⇒ x² - (3 - i) x + (1 - 2i) x - (1 - 7i) = 0
⇒ x (x - (3 - i)) + (1 - 2i) (x - (3 - i)) = 0
⇒ [ x + (1 - 2i)] [x - (3 - i)] = 0
⇒ x = -1 + 2i, 3 - i

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Question 232 Marks

Show the following quadratic equation:
$\text{x}^2-(3\sqrt{2}+2\text{i})\text{x}+6\sqrt{2}\text{ i}=0$

Answer

$\text{x}^2-(3\sqrt{2}+2\text{i})\text{x}+6\sqrt{2}\text{ i}=0$
$\Rightarrow\text{x}^2-3\sqrt{2}\text{x}-2\text{ix}+\sqrt{2}\text{i}=0$
$\Rightarrow\text{x}(\text{x}-3\sqrt{2})-2\text{i}(\text{x}-3\sqrt{2})=0$
$\Rightarrow(\text{x}-2\text{i})(\text{x}-3\sqrt{2})=0$
$\Rightarrow\text{x}=2\text{i}\ \text{or }3\sqrt{2}$

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Question 242 Marks

Show the following quadratic equation:
x²− (5 − i) x +(18 + i) = 0

Answer

x²− (5 − i) x +(18 + i) = 0
⇒ x² - 5x - ix + 18 + i = 0
⇒ x²- (3 - 4i) x - (2 + 3i) x + (18 + i) = 0
⇒ x (x - (3 - 4i)) - (2 + 3i) (x - (3 - 4i)) = 0
⇒ (x - (2 + 3i)) (x - (3 - 4i)) = 0
⇒ x = 2 + 3i or 3 - 4i

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