1. **Problem:** If the zeroes of the quadratic polynomial $ax^2 + bx + \frac{2a}{b}$ are reciprocal of each other, find the value of $b$. Step 1: Let the roots be $\alpha$ and $\beta$. Since they are reciprocals, $\alpha \beta = 1$. Step 2: For quadratic $ax^2 + bx + \frac{2a}{b}$, product of roots is $\frac{c}{a} = \frac{\frac{2a}{b}}{a} = \frac{2}{b}$. Step 3: Since product $=1$, we have $$\frac{2}{b} = 1 \implies b = 2.$$ 2. **Problem:** Given $\cot A = \frac{7}{8}$, find $$\frac{(1 + \sin A)(1 - \sin A)}{(1 + \cos A)(1 - \cos A)}.$$ Step 1: Recall that $(1 + \sin A)(1 - \sin A) = 1 - \sin^2 A = \cos^2 A$. Step 2: Similarly, $(1 + \cos A)(1 - \cos A) = 1 - \cos^2 A = \sin^2 A$. Step 3: Thus, the expression simplifies to $$\frac{\cos^2 A}{\sin^2 A} = \cot^2 A = \left(\frac{7}{8}\right)^2 = \frac{49}{64}.$$ 3. **Problem:** Determine whether the system of equations $$2x + 3y = 5$$ and $$4x + 6y = 10$$ is inconsistent, consistent or dependent consistent. Step 1: Notice that the second equation is exactly twice the first. Step 2: This means they represent the same line, so infinitely many solutions exist. Step 3: Hence, the system is dependent consistent. 4. **Problem:** Find the distance of the point $(3,4)$ from the origin. Step 1: Distance formula: $$d = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$$ 5. **Problem:** If $x = 5$ is a solution of $$2x^2 + (k-1)x + 10 = 0,$$ find $k$. Step 1: Substitute $x=5$: $$2(5)^2 + (k - 1)(5) + 10 = 0$$ $$50 + 5k - 5 + 10 = 0$$ $$5k + 55 = 0 \implies 5k = -55 \implies k = -11.$$ 6. **Problem:** In an AP, if $a_{21} - a_7 = 84$, find the common difference $d$. Step 1: General term of AP: $$a_n = a_1 + (n-1)d.$$ Step 2: Compute difference: $$a_{21} - a_7 = [a_1 + 20d] - [a_1 + 6d] = 14d = 84.$$ Step 3: Solve for $d$: $$d = \frac{84}{14} = 6.$$ 7. **Problem:** Find the 4th term from the end of the AP: $$-11, -8, -5, ..., 49.$$ Step 1: Find $n$ by using last term formula: $$a_n = a_1 + (n-1)d.$$ Here, $a_1 = -11$, $d = -8 - (-11) = 3$, and $a_n = 49$. Step 2: Solve for $n$: $$49 = -11 + (n-1)3 \implies 60 = 3(n-1) \implies n -1 = 20 \implies n = 21.$$ Step 3: The 4th term from end is term number $21 - 4 + 1 = 18$. Step 4: Calculate $a_{18}$: $$a_{18} = -11 + (18 -1)3 = -11 + 51 = 40.$$ 8. **Problem:** Find coordinates of point $A$ where $AB$ is diameter of a circle with center $(2, -3)$ and $B = (1, 4)$. Step 1: Midpoint $M$ of diameter is center: $$M = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right) = (2, -3).$$ Step 2: Write equations: $$\frac{x_A + 1}{2} = 2 \implies x_A +1 =4 \implies x_A =3,$$ $$\frac{y_A +4}{2} = -3 \implies y_A +4 = -6 \implies y_A = -10.$$ **Final answers:** 1. $b = 2$ 2. $\frac{49}{64}$ 3. Dependent consistent 4. 5 units 5. $k = -11$ 6. $d=6$ 7. 40 8. $(3, -10)$