To find $\frac{1}{\beta - \alpha}$, we first need to determine the range of the function $fog(x) = f(g(x))$.
1. Calculate $fog(x)$: \[ fog(x) = f\left(g(x)\right) = f\left(\frac{2 - 3x}{1 - x}\right) \] Substitute $g(x)$ into $f(x)$: \[ fog(x) = \frac{2\left(\frac{2 - 3x}{1 - x}\right) + 3}{5\left(\frac{2 - 3x}{1 - x}\right) + 2} \] Simplify the expression: \[ fog(x) = \frac{\frac{4 - 6x + 3 - 3x}{1 - x}}{\frac{10 - 15x + 2 - 2x}{1 - x}} = \frac{7 - 9x}{12 - 17x} \]
2. Determine the range of $fog(x)$ for $x \in [2, 4]$: - Calculate $fog(2)$: \[ fog(2) = \frac{7 - 9(2)}{12 - 17(2)} = \frac{7 - 18}{12 - 34} = \frac{-11}{-22} = \frac{1}{2} \] - Calculate $fog(4)$: \[ fog(4) = \frac{7 - 9(4)}{12 - 17(4)} = \frac{7 - 36}{12 - 68} = \frac{-29}{-56} = \frac{29}{56} \]
3. Identify $\alpha$ and $\beta$:
- The range of $fog(x)$ is $\left[\frac{1}{2}, \frac{29}{56}\right]$.
- Therefore, $\alpha = \frac{1}{2}$ and $\beta = \frac{29}{56}$.
4. Calculate $\frac{1}{\beta - \alpha}$: \[ \beta - \alpha = \frac{29}{56} - \frac{1}{2} = \frac{29}{56} - \frac{28}{56} = \frac{1}{56} \] \[ \frac{1}{\beta - \alpha} = \frac{1}{\frac{1}{56}} = 56 \] Therefore, the correct answer is (4) 56.
Let A be the set of 30 students of class XII in a school. Let f : A -> N, N is a set of natural numbers such that function f(x) = Roll Number of student x.
Give reasons to support your answer to (i).
Find the domain of the function \( f(x) = \cos^{-1}(x^2 - 4) \).
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).