Find the Derivative \( \frac{dy}{dx} \)Given:\[ y = \cos(x^2) + \cos(2x) + \cos^2(x^2) + \cos(x^x) \]
For the curve \( \sqrt{x} + \sqrt{y} = 1 \), find the value of \( \frac{dy}{dx} \) at the point \( \left(\frac{1}{9}, \frac{1}{9}\right) \).
If \(\begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0\), then the value of \(x\) is:
Show that the following lines intersect. Also, find their point of intersection:
Line 1: \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \]
Line 2: \[ \frac{x - 4}{5} = \frac{y - 1}{2} = z \]
Let \( \vec{a} \) and \( \vec{b} \) be two co-initial vectors forming adjacent sides of a parallelogram such that: \[ |\vec{a}| = 10, \quad |\vec{b}| = 2, \quad \vec{a} \cdot \vec{b} = 12 \] Find the area of the parallelogram.
Prove that: \( \tan^{-1}(\sqrt{x}) = \frac{1}{2} \cos^{-1}\left( \frac{1 - x}{1 + x} \right), \quad x \in [0, 1] \)
Four students of class XII are given a problem to solve independently. Their respective chances of solving the problem are: \[ \frac{1}{2},\quad \frac{1}{3},\quad \frac{2}{3},\quad \frac{1}{5} \] Find the probability that at most one of them will solve the problem.
Solve the following LPP graphically: Maximize: \[ Z = 2x + 3y \] Subject to: \[ \begin{aligned} x + 4y &\leq 8 \quad \text{(1)} \\ 2x + 3y &\leq 12 \quad \text{(2)} \\ 3x + y &\leq 9 \quad \text{(3)} \\ x &\geq 0,\quad y \geq 0 \quad \text{(non-negativity constraints)} \end{aligned} \]
If \[ A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}, \] find \( A^{-1} \).
Using \( A^{-1} \), solve the following system of equations:
\[ \begin{aligned} 2x - 3y + 5z &= 11 \quad \text{(1)} \\ 3x + 2y - 4z &= -5 \quad \text{(2)} \\ x + y - 2z &= -3 \quad \text{(3)} \end{aligned} \]