Algebra

(a) $a^2 - 4bc = 0$

For $bx^2 + ax + c = 0$ the coefficients are $b$ (of $x^2$), $a$ (of $x$) and $c$. The roots are equal when the discriminant is zero: $a^2 - 4bc = 0$.

(b) $\dfrac{\log 5}{2a}$

$\log_4 5 = \dfrac{\log 5}{\log 4} = a$, so $\log 4 = \dfrac{\log 5}{a}$. Since $\log 4 = 2\log 2$, we get $\log 2 = \dfrac{\log 5}{2a}$.

(a) 3

For three terms in A.P., twice the middle term equals the sum of the other two: $2(3x) = (x+2) + (4x+1) \Rightarrow 6x = 5x + 3 \Rightarrow x = 3$

(c) 550

Let the first-year production be $a$ and the yearly increase be $d$ (an A.P.). $a + 2d = 600,\qquad a + 6d = 700$ Subtracting, $4d = 100$, so $d = 25$. Then $a = 600 - 50 = 550$.

Equilibrium before tax

Demand: $P = 125 - Q$. Supply: $2P = 3Q + 30 \Rightarrow P = 1.5Q + 15$.

At equilibrium, $125 - Q = 1.5Q + 15 \Rightarrow 110 = 2.5Q \Rightarrow Q = 44$. $P = 125 - 44 = 81$ So the equilibrium quantity is 44 units and the price is Rs 81.

After a tax of Rs 5 per unit

The tax shifts the supply curve up by 5: $P = 1.5Q + 15 + 5 = 1.5Q + 20$.

Equating with demand: $125 - Q = 1.5Q + 20 \Rightarrow 105 = 2.5Q \Rightarrow Q = 42$. $P = 125 - 42 = 83$ The new equilibrium is $Q = 42$ units and $P =$ Rs 83. Price rises by Rs 2 and quantity falls by 2 units.

Who pays the tax?

Consumers now pay Rs 83 instead of Rs 81, so they bear Rs 2 of the tax per unit. Producers receive $83 - 5 =$ Rs 78 instead of Rs 81, so they bear Rs 3 per unit. Hence consumers pay 2/5 (40%) and producers pay 3/5 (60%) of the tax. The total tax revenue is $5 \times 42 =$ Rs 210.

Substituting $C$ and $I_0$ into $Y = C + I$: $Y = 60 + 0.4Y + 300 \Rightarrow 0.6Y = 360 \Rightarrow Y_e = \text{Rs. } 600$ $C_e = 60 + 0.4(600) = \text{Rs. } 300$

(a) At $t = 0$: $P(0) = \frac{50000}{1 + 49999 e^{0}} = \frac{50000}{50000} = 1$ So 1 person is initially infected.

(b) 10% of 50000 is 5000. Setting $P(t) = 5000$: $1 + 49999e^{-0.5t} = 10 \Rightarrow e^{-0.5t} = \frac{9}{49999}$ $-0.5t = \ln\frac{9}{49999} \Rightarrow t = 2\ln\frac{49999}{9} = 2\ln(5555.4) \approx 2(8.623) \approx 17.25$ The infection crosses 10% after about 17.25 days, so the country locks down on the 18th day.

(c) As $t \to \infty$, $e^{-0.5t} \to 0$, so $P(t) \to \frac{50000}{1+0} = 50000$ Eventually all 50000 people become infected.

(b) $\{2, 3\}$

$A \cap B$ is the set of elements common to both $A$ and $B$. The elements 2 and 3 appear in both sets, so $A \cap B = \{2, 3\}$.

Given $\log_2(x+3)+\log_2(x-1)=3$.

Domain: both arguments must be positive, so $x+3>0$ and $x-1>0$, i.e. $x>1$.

Using the product law, $\log_2[(x+3)(x-1)]=3$ $(x+3)(x-1)=2^3=8$ $x^2+2x-3=8 \;\Rightarrow\; x^2+2x-11=0$

By the quadratic formula, $x=\frac{-2\pm\sqrt{4+44}}{2}=\frac{-2\pm4\sqrt3}{2}=-1\pm2\sqrt3$

Since $x>1$, the root $x=-1-2\sqrt3$ is rejected ($-1+2\sqrt3\approx2.46>1$ is accepted).

$\therefore\; x=2\sqrt3-1$

Let $\frac{2x+3}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}$

Multiplying both sides by $(x-1)(x+2)$, $2x+3=A(x+2)+B(x-1)$

Put $x=1$: $5=3A \Rightarrow A=\dfrac53$.

Put $x=-2$: $-1=-3B \Rightarrow B=\dfrac13$.

Hence $\frac{2x+3}{(x-1)(x+2)}=\frac{5}{3(x-1)}+\frac{1}{3(x+2)}$

Let the two consecutive positive integers be $x$ and $x+1$.

(i) Forming the quadratic equation

The sum of their squares is 145, so $x^2+(x+1)^2=145$ Expanding: $x^2+x^2+2x+1=145$ $2x^2+2x+1=145$ $2x^2+2x-144=0$ Dividing throughout by 2, $\boxed{x^2+x-72=0}$

(ii) Solving the equation

$x^2+x-72=0$ Factorising, $x^2+9x-8x-72=0$ $x(x+9)-8(x+9)=0$ $(x+9)(x-8)=0$

So, $x+9=0 \quad \text{or} \quad x-8=0$ $x=-9 \quad \text{or} \quad x=8$

Since the integers are positive, we reject $x=-9$.

Hence, $x=8$ and the next consecutive integer is $x+1=9$

Therefore, the required integers are $\boxed{8}$ and $\boxed{9}$.

(iii) Verification

Check whether the sum of the squares of 8 and 9 is 145: $8^2+9^2=64+81=145$

Since the result is exactly 145, the answer is verified.

$\boxed{\text{The two consecutive positive integers are } 8 \text{ and } 9.}$