If the set of all values of \( a \), for which the equation \( 5x^3 - 15x - a = 0 \) has three distinct real roots, is the interval \( (\alpha, \beta) \), then \( \beta - 2\alpha \) is equal to
The given equation is: \[ 5x^3 - 15x - a = 0 \] Let \( f(x) = 5x^3 - 15x \).
Now, differentiate \( f(x) \): \[ f'(x) = 15x^2 - 15 = 15(x - 1)(x + 1) \] Thus, the critical points of the function are \( x = 1 \) and \( x = -1 \).
Next, to find the condition for three distinct real roots, we need to find the values of \( a \) such that the graph of \( f(x) \) intersects the x-axis at three points.
Plotting the function shows that the values of \( a \) must lie in the interval \( (-10, 10) \).
Therefore, \( \alpha = -10 \) and \( \beta = 10 \).
Finally, calculate \( \beta - 2\alpha \): \[ \beta - 2\alpha = 10 - 2(-10) = 10 + 20 = 30 \] Thus, the value of \( \beta - 2\alpha \) is \( 30 \).
\(\text{The number of solutions of the equation}\)\(\left(\frac{9}{x}-\frac{9}{\sqrt{x}}+2\right)\left(\frac{2}{x}-\frac{7}{\sqrt{x}}+3\right)=0\mathrm \; {is:}\)
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: