To solve the problem, we need to determine the value of \( |a + b + c| \). We are given a function \( f(x) = ae^{2x} + be^x + cx \) and specific conditions that the function must satisfy.
Therefore, the answer is 8.
The function is given by:
\(f(x) = ae^{2x} + be^x + c\).
Given \(f(0) = -1\):
\(f(0) = a + b + c = -1\).
Differentiate \(f(x)\):
\(f'(x) = 2ae^{2x} + be^x\).
Given \(f'( \log_e 2) = 21\):
\(8a + 2b = 21 \Rightarrow 4a + b = 10.5\).
Evaluate the integral:
\(\int_{0}^{\log_e 4} (f(x) - cx) dx = \int_{0}^{\log_e 4} (ae^{2x} + be^x + c - cx) dx = \frac{39}{2}\).
Break into parts and evaluate each:
\(\frac{15a}{2} + 3b + c \log_e 4 - c \times \frac{(\log_e 4)^2}{2} = \frac{39}{2}\).
Solve for \(a\), \(b\), and \(c\). The value of \(|a + b + c|\) is:
\(15a + 6b = 39\)
\(15a - 6a - 6 = 39\)
\(9a = 45 \Rightarrow a = 5\)
\(b = -6\)
\(c = 21 - 40 + 12 = -7\)
\(a + b + c = -8\)
\(|a + b + c| = 8\)
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)