To find \( f(g(x)) \), we first evaluate \( g(x) \) based on the value of \( x \).
\[ g(x) = \begin{cases} e^x, & x \geq 0 \\ x + 1, & x \leq 0 \end{cases} \]
The function \( f \) is defined as \( f(x) = |x - 1| \). Therefore, we have: \[ f(g(x)) = |g(x) - 1|. \]
Case 1: When \( x \geq 0 \)
\[ f(g(x)) = |e^x - 1|. \]
Case 2: When \( x \leq 0 \)
\[ f(g(x)) = |x + 1 - 1| = |x| = -x \quad \text{(since \( x \leq 0 \))}. \]
Analysis of \( f(g(x)) \) - For \( x \geq 0 \), \( f(g(x)) = |e^x - 1| \) is neither one-one nor onto because it cannot cover all values in the codomain (as it is non-negative). For \( x \leq 0 \), \( f(g(x)) = -x \) is also neither one-one nor onto due to its behavior as a non-injective transformation on the interval.
Therefore, the function \( f(g(x)) \) is neither one-one nor onto.
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32