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Question 11:

The equation $x^2 + px + 12 = 0$ has one root as 3, and $x^2 + px + q = 0$ has equal roots. We are to find the value of $q$.

1. Using the given root:
   Substituting $x = 3$ into $x^2 + px + 12 = 0$:
   $3^2 + 3p + 12 = 0 \implies 9 + 3p + 12 = 0 \implies 3p = -21 \implies p = -7.$

2. Condition for equal roots:
   For $x^2 - 7x + q = 0$ (substituting $p = -7$), the discriminant must be zero for equal roots:
   $D = b^2 - 4ac = (-7)^2 - 4(1)(q) = 0.$ $49 - 4q = 0 \implies 4q = 49 \implies q = \frac{49}{4}.$

Question 12:

Function $f(x + y) = f(x) + f(y)$ for all $x, y \in \mathbb{N}$, and $f(1) = 5$. We need $f(1) + f(2) + \dots + f(500)$.

1. Functional form:
   Using $f(x + y) = f(x) + f(y)$, $f(n) = 5n$ since $f(1) = 5$.

2. Summing up:
   $f(1) + f(2) + \dots + f(500) = 5(1 + 2 + \dots + 500).$ Using the sum of the first $n$ natural numbers:
   $1 + 2 + \dots + 500 = \frac{500 \times 501}{2} = 125250.$ $\text{Total} = 5 \times 125250 = 626250.$

Question 13:

Find the remainder when $7^{2024} + 3^{2024}$ is divided by 5.

1. Remainder of powers modulo 5:
   $7 \equiv 2 \pmod{5}, \quad 3 \equiv 3 \pmod{5}.$ Powers of $2$ modulo 5 repeat in a cycle: $2, 4, 3, 1$.
   $2^{2024} \equiv 1 \pmod{5} \quad (\text{since } 2024 \mod 4 = 0).$ Powers of $3$ modulo 5 repeat in a cycle: $3, 4, 2, 1$.
   $3^{2024} \equiv 1 \pmod{5}.$

2. Adding remainders:
   $7^{2024} + 3^{2024} \equiv 1 + 1 \equiv 2 \pmod{5}.$

Question 14:

An infinite G.P. with sum 15 and the sum of squares of each term is 150. Find the sum of $ar^3, ar^6, ar^9, \dots$.

1. Basic G.P. formulae:
   Sum of infinite G.P.: $S = \frac{a}{1 - r} = 15.$ Sum of squares: $S_{\text{squares}} = \frac{a^2}{1 - r^2} = 150.$

2. Solving for $a$ and $r$:
   From $a = 15(1 - r)$:
   Substituting into $a^2/(1 - r^2)$:
   $\frac{(15(1 - r))^2}{1 - r^2} = 150.$ Simplify: $\frac{225(1 - r)^2}{1 - r^2} = 150 \implies 225(1 - r)^2 = 150(1 - r^2).$ Expanding and solving gives $r = \frac{1}{2}, a = 15 \cdot \frac{1}{2} = 7.5$.

3. Sum of given terms:
   The sum of $ar^3, ar^6, ar^9, \dots$ is an infinite G.P. with first term $ar^3 = 7.5 \cdot (\frac{1}{2})^3 = 0.9375$ and ratio $r^3 = (\frac{1}{2})^3 = \frac{1}{8}$:
   $S = \frac{0.9375}{1 - \frac{1}{8}} = \frac{0.9375}{\frac{7}{8}} = \frac{0.9375 \times 8}{7} = 1.07 \, (\text{approximately}).$

Would you like detailed explanations for further questions or clarifications?

5 Related Questions:

1. How does the discriminant help in determining the type of roots for quadratic equations?
2. Can you derive general solutions for functional equations?
3. How do cycles in modular arithmetic simplify large exponents?
4. How can G.P. properties be used to solve real-world problems?
5. What are the key steps to finding areas bounded by curves?

Tip:

For modular arithmetic, observe patterns in powers to simplify computations of large exponents.