Read the statement carefully and state 'T' for true and 'F' for false.
(i) The value of `2+ (1)/(2+(1)/(2+ ...oo))` is `sqrt2`
(ii) A line segment AB of length 2 m is divided at C into two parts such that `AC^(2)= AB.CB`. The length of the part CB is `3+ sqrt5`.
(iii) Every quadratic equation can have at most two real roots.
(iv) A real number a is said to be a root of the quadratic equation `ax^(2) +bx+c= 0`, if `a alpha^(2) + b alpha +c = 0`

Options

A. `{:(i,ii,iii,iv),(F,T,T,T):}`
B. `{:(i,ii,iii,iv),(F,T,T,F):}`
C. `{:(i,ii,iii,iv),(T,F,F,T):}`
D. `{:(i,ii,iii,iv),(F,F,T,T):}`

Answer

To solve the given statements, we will analyze each one step by step and determine whether they are true (T) or false (F).

### Statement Analysis

**(i)** The value of \( 2 + \frac{1}{2 + \frac{1}{2 + \ldots}} \) is \( \sqrt{2} \).

**Step 1:** Let \( x = 2 + \frac{1}{2 + \frac{1}{2 + \ldots}} \). This implies \( x = 2 + \frac{1}{x} \).

**Step 2:** Multiply both sides by \( x \) to eliminate the fraction:
\[ x^2 = 2x + 1 \]

**Step 3:** Rearrange the equation:
\[ x^2 - 2x - 1 = 0 \]

**Step 4:** Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -2, c = -1 \):
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ x = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \]

**Step 5:** Since \( x \) must be positive, we take \( x = 1 + \sqrt{2} \).

**Conclusion:** The statement is **False (F)** because \( 1 + \sqrt{2} \neq \sqrt{2} \).

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**(ii)** A line segment \( AB \) of length 2 m is divided at \( C \) into two parts such that \( AC^2 = AB \cdot CB \). The length of the part \( CB \) is \( 3 + \sqrt{5} \).

**Step 1:** Let \( AC = x \) and \( CB = 2 - x \).

**Step 2:** According to the given condition:
\[ x^2 = 2 \cdot (2 - x) \]

**Step 3:** Expand and rearrange:
\[ x^2 = 4 - 2x \]
\[ x^2 + 2x - 4 = 0 \]

**Step 4:** Use the quadratic formula:
\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
\[ x = \frac{-2 \pm \sqrt{4 + 16}}{2} = \frac{-2 \pm \sqrt{20}}{2} = \frac{-2 \pm 2\sqrt{5}}{2} = -1 \pm \sqrt{5} \]

**Step 5:** Since \( x \) must be positive, we take \( x = -1 + \sqrt{5} \).

**Step 6:** Thus, \( CB = 2 - x = 2 - (-1 + \sqrt{5}) = 3 - \sqrt{5} \).

**Conclusion:** The statement is **False (F)** because \( CB \neq 3 + \sqrt{5} \).

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**(iii)** Every quadratic equation can have at most two real roots.

**Conclusion:** This statement is **True (T)**. A quadratic equation can have either two real roots, one real root (repeated), or no real roots (complex).

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**(iv)** A real number \( a \) is said to be a root of the quadratic equation \( ax^2 + bx + c = 0 \) if \( a \alpha^2 + b \alpha + c = 0 \).

**Conclusion:** This statement is **True (T)**. If \( a \) is a root of the quadratic equation, substituting \( a \) into the equation will yield zero.

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### Final Answers:
1. (i) F
2. (ii) F
3. (iii) T
4. (iv) T