Let's solve the problem to find the maximum and minimum values of the function \( f(x) = (x + 3)^2 (x - 2)^3 \) in the interval \([-4, 4]\). We need to find \( M - m \), where \( M \) is the maximum value, and \( m \) is the minimum value.
The first step is to find the critical points of the function by differentiating it with respect to \( x \) and setting the derivative to zero.
Using the product rule of differentiation, if \( u(x) = (x + 3)^2 \) and \( v(x) = (x - 2)^3 \), then:
Applying the product rule:
\(f'(x) = u'(x) v(x) + u(x) v'(x) = 2(x + 3)(x - 2)^3 + 3(x + 3)^2(x - 2)^2\)
Setting \( f'(x) = 0 \), we solve:
\(2(x + 3)(x - 2)^3 + 3(x + 3)^2(x - 2)^2 = 0\)
Factor out \((x + 3)(x - 2)^2\):
\((x + 3)(x - 2)^2 [2(x - 2) + 3(x + 3)] = 0\)
This simplifies to:
\((x + 3)(x - 2)^2 (5x + 9) = 0\)
Solve each factor:
Evaluate \( f(x) \) at the critical points and endpoints of the interval \( [-4, 4] \):
From the above calculations, we find that \( M = 392 \) and \( m = -216 \). Therefore, the value of \( M - m = 392 - (-216) = 608 \).
Thus, the correct answer is \(608\).
To find the maximum and minimum values of \( f(x) \):
Take the derivative \( f'(x) \) and find the critical points.
Evaluate \( f(x) \) at critical points and endpoints \( x = -4, -3, -2, -1, 1, 2, 3, 4 \).
The maximum value \( M = 392 \) and the minimum value \( m = -216 \).
The value of \( M - m \) is:
\[ M - m = 392 - (-216) = 608. \]
Let the function, \(f(x)\) = \(\begin{cases} -3ax^2 - 2, & x < 1 \\a^2 + bx, & x \geq 1 \end{cases}\) Be differentiable for all \( x \in \mathbb{R} \), where \( a > 1 \), \( b \in \mathbb{R} \). If the area of the region enclosed by \( y = f(x) \) and the line \( y = -20 \) is \( \alpha + \beta\sqrt{3} \), where \( \alpha, \beta \in \mathbb{Z} \), then the value of \( \alpha + \beta \) is:
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged.
Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor \( \frac{R}{\sqrt{R^2 + \omega^2 L^2}} \), where \( \omega \) is the frequency of the supply across resistor \( R \) and inductor \( L \). If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage.
In light of the above statements, choose the most appropriate answer from the options given below:
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Electromagnetic waves carry energy but not momentum.
Reason (R): Mass of a photon is zero.
In the light of the above statements, choose the most appropriate answer from the options given below: