Step 1: For increasing function, we need \( f'(x) > 0 \).
Step 2: Differentiate the function:
\[ f'(x) = \frac{d}{dx}(x^4 - 4x^3 + 4x^2 + 15) = 4x^3 - 12x^2 + 8x \] Factor the derivative:
\[ f'(x) = 4x(x^2 - 3x + 2) = 4x(x - 1)(x - 2) \]
Step 3: Solve \( f'(x) > 0 \):
\[ 4x(x - 1)(x - 2) > 0 \] Critical points: \( x = 0, 1, 2 \)
Step 4: Sign chart analysis:
Conclusion:
\( f'(x) > 0 \) for intervals \( (0, 1) \) and \( (2, \infty) \)
Therefore, f(x) is strictly increasing in (0, 1) and (2, ∞)
Similarly, f(x) is decreasing in (−∞, 0) and (1, 2)
If \( x = a(0 - \sin \theta) \), \( y = a(1 + \cos \theta) \), find \[ \frac{dy}{dx}. \]
Find the least value of ‘a’ for which the function \( f(x) = x^2 + ax + 1 \) is increasing on the interval \( [1, 2] \).
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.
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:
\[ \begin{aligned} 2x - 3y + 5z &= 11 \quad \text{(1)} \\ 3x + 2y - 4z &= -5 \quad \text{(2)} \\ x + y - 2z &= -3 \quad \text{(3)} \end{aligned} \]