Let \( f(x) = \log x \) and \[ g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1} \] Then the domain of \( f \circ g \) is:
Step 1: Understanding domain constraints. The function \( f(x) = \log x \) requires \( x>0 \), so we must ensure \( g(x)>0 \) for \( f(g(x)) \) to be defined.
Step 2: Finding domain of \( g(x) \). Given: \[ g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1} \] The denominator is a quadratic equation: \[ 2x^2 - 2x + 1 \] Since the discriminant is negative, it is always positive.
Step 3: Solving for \( g(x)>0 \). Setting the numerator \( x^4 - 2x^3 + 3x^2 - 2x + 2>0 \), we find that \( x<0 \) satisfies this condition. Hence, \( g(x) \) is always positive. Thus, \( g(x)>0 \) for all \( x \), meaning the domain of \( f \circ g \) is \( \mathbb{R} \).
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank is _________ cm. (Take \( g = 10 \, {m/s}^2 \)).