Part 1: Find the interval for \( f(x) \)
The function \( f(x) = 2 \log_e (x - 2) - x^2 + ax + 1 \) is strictly increasing when its derivative \( f'(x) > 0 \).
We begin by computing the derivative:
\[
f'(x) = \frac{2}{x - 2} - 2x + a
\]
For \( f(x) \) to be strictly increasing, we need:
\[
f'(x) = \frac{2}{x - 2} - 2x + a > 0
\]
This condition determines the interval for which \( f(x) \) is strictly increasing.
Part 2: Find the interval for \( g(x) \)
Next, we consider the function \( g(x) = (x - 1)^3 (x + 2 - a)^2 \). The function \( g(x) \) is strictly decreasing when its derivative \( g'(x) < 0 \). The derivative is:
\[
g'(x) = 3(x - 1)^2 (x + 2 - a)^2 + 2(x - 1)^3 (x + 2 - a)
\]
For \( g(x) \) to be strictly decreasing, we need:
\[
g'(x) < 0
\]
This condition determines the interval \( (b, c) \) where the function is strictly decreasing.
Step 3: Solve for \( a \), \( b \), and \( c \)
After solving the inequalities for \( f'(x) > 0 \) and \( g'(x) < 0 \), we find the values of \( a \), \( b \), and \( c \).
Final Answer:
After solving the equations, we find that:
\[
100(a + b - c) = 160
\]
Final Answer: \( 100(a + b - c) = 160 \).
If the given figure shows the graph of polynomial \( y = ax^2 + bx + c \), then:
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is: