Find the equation of the tangent to the curve x2 − xy − y2 + 4 = 0 at K(2,4). The sum Sn of the first n terms of a linear sequence (A.P.) is given by Sn = n2 − 8n. If the n^th term of the sequence is

here are the solutions to the math problems:

Q1) Find the equation of the tangent to the curve $x^2 - xy - y^2 + 4 = 0$ at $K(2,4)$.

Step 1: Differentiate the equation implicitly with respect to $x$.

$$\frac{d}{dx}(x^2) - \frac{d}{dx}(xy) - \frac{d}{dx}(y^2) + \frac{d}{dx}(4) = 0$$ $$2x - \left(1 \cdot y + x \cdot \frac{dy}{dx}\right) - 2y \frac{dy}{dx} + 0 = 0$$ $$2x - y - x \frac{dy}{dx} - 2y \frac{dy}{dx} = 0$$

Step 2: Rearrange the equation to solve for $\frac{dy}{dx}$.

$$2x - y = x \frac{dy}{dx} + 2y \frac{dy}{dx}$$ $$2x - y = (x + 2y) \frac{dy}{dx}$$ $$\frac{dy}{dx} = \frac{2x - y}{x + 2y}$$

Step 3: Substitute the coordinates of point $K(2,4)$ into $\frac{dy}{dx}$ to find the gradient ($m$) of the tangent.

$$m = \frac{2(2) - 4}{2 + 2(4)} = \frac{4 - 4}{2 + 8} = \frac{0}{10} = 0$$

Step 4: Use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, with $m=0$ and $(x_1, y_1) = (2,4)$.

$$y - 4 = 0(x - 2)$$ $$y - 4 = 0$$ $$y = 4$$

The equation of the tangent is $\boxed{y = 4}$.

Q2) The sum $S_n$ of the first $n$ terms of a linear sequence (A.P.) is given by $S_n = n^2 - 8n$. If the $n^{th}$ term of the sequence is greater than 30, find the least possible value of $n$.

Step 1: Find the formula for the $n^{th}$ term, $T_n$, using the relationship $T_n = S_n - S_{n-1}$.
First, find $S_{n-1}$:

$$S_{n-1} = (n-1)^2 - 8(n-1)$$ $$S_{n-1} = (n^2 - 2n + 1) - (8n - 8)$$ $$S_{n-1} = n^2 - 2n + 1 - 8n + 8$$ $$S_{n-1} = n^2 - 10n + 9$$

Now, find $T_n$:

$$T_n = (n^2 - 8n) - (n^2 - 10n + 9)$$ $$T_n = n^2 - 8n - n^2 + 10n - 9$$ $$T_n = 2n - 9$$

Step 2: Set up the inequality given that the $n^{th}$ term is greater than 30.

$$T_n > 30$$ $$2n - 9 > 30$$

Step 3: Solve the inequality for $n$.

$$2n > 30 + 9$$ $$2n > 39$$ $$n > \frac{39}{2}$$ $$n > 19.5$$

Step 4: Determine the least possible integer value of $n$.
Since $n$ must be an integer and $n$ must be greater than 19.5, the least possible integer value for $n$ is 20.
The least possible value of $n$ is $\boxed{20}$.

Q3) Resolve $\frac{4-x}{x^2 - 5x - 24}$ into partial fractions.

Step 1: Factorize the denominator.
We need two numbers that multiply to -24 and add to -5. These numbers are -8 and 3.

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