Step 1: Understanding the Question:

The objective is to simplify a Boolean algebraic expression involving the OR, AND, and NOT operations to identify its equivalent logic gate representation.


Step 2: Key Formula or Approach:

We use the Distributive Law: \( (X + Y)(W + Z) = XW + XZ + YW + YZ \).

Complement Laws: \( A \cdot \overline{A} = 0 \) and \( A + \overline{A} = 1 \).


Step 3: Detailed Explanation:

Given expression: \( (\overline{A} + \overline{B}) \cdot (A + B) \)

Expanding using the distributive property:
\[ = \overline{A} \cdot A + \overline{A} \cdot B + \overline{B} \cdot A + \overline{B} \cdot B \]

Applying the property \( X \cdot \overline{X} = 0 \):
\[ = 0 + \overline{A}B + A\overline{B} + 0 \]
\[ = \overline{A}B + A\overline{B} \]

This simplified expression is the standard Boolean form of the EXCLUSIVE-OR (XOR) gate.


Step 4: Final Answer.

The identity \( (\overline{A} + \overline{B}) \cdot (A + B) \) is equal to \( A \oplus B \).
Quick Tip: Remember that \( (A+B)(\overline{A}+\overline{B}) \) is the complement of \( (AB + \overline{A}\overline{B}) \).
XOR is \( A\overline{B} + \overline{A}B \) and XNOR is \( AB + \overline{A}\overline{B} \).
In multiple-choice questions, you can also substitute values (e.g., \( A=0, B=1 \)) to quickly check the output.

Step 1: Understanding the Question:

For a function to be differentiable at a point, it must be continuous at that point, and the left-hand derivative must equal the right-hand derivative.


Step 2: Key Formula or Approach:

Continuity at \( x = c \): \( \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) \).

Differentiability at \( x = c \): \( f'(c^-) = f'(c^+) \).


Step 3: Detailed Explanation:

Condition 1: Continuity at \( x = -1 \)
\[ \lim_{x \to -1^-} (ax^2 + b) = \lim_{x \to -1^+} (bx^2 + ax + 4) \]
\[ a(-1)^2 + b = b(-1)^2 + a(-1) + 4 \]
\[ a + b = b - a + 4 \]
\[ 2a = 4 \implies a = 2 \]


Condition 2: Differentiability at \( x = -1 \)

Differentiating the function parts:
\[ f'(x) = \begin{cases} 2ax, & x < -1
2bx + a, & x > -1 \end{cases} \]

Equating LHD and RHD at \( x = -1 \):
\[ 2a(-1) = 2b(-1) + a \]
\[ -2a = -2b + a \]
\[ 3a = 2b \]

Substitute \( a = 2 \) into the equation:
\[ 3(2) = 2b \implies 6 = 2b \implies b = 3 \]


Step 4: Final Answer.

The values are \( a = 2 \) and \( b = 3 \).
Quick Tip: Always check for continuity first.
If a function is differentiable, it MUST be continuous.
Using this sequence often allows you to find one constant immediately, which simplifies finding the second one.

Step 1: Understanding the Question:

This is a Linear Programming Problem (LPP).

We need to find the maximum value of the objective function \( Z = x + 3y \) within the region defined by the given inequalities.


Step 2: Key Formula or Approach:

We use the Corner Point Method. The feasible region is a polygon, and the optimal value of a linear objective function occurs at one of the vertices (corner points) of this region.


Step 3: Detailed Explanation:

Identify the feasible region bounded by:

1) \( x + y \leq 4 \)

2) \( 0 \leq x \leq 3 \)

3) \( 0 \leq y \leq 2 \)


Find the intersection points (vertices):

- Point \( P_1 \): Origin \( (0, 0) \).

- Point \( P_2 \): Intersection of \( x = 3 \) and \( y = 0 \): \( (3, 0) \).

- Point \( P_3 \): Intersection of \( x = 3 \) and \( x + y = 4 \): \( 3 + y = 4 \implies y = 1 \). So, \( (3, 1) \).

- Point \( P_4 \): Intersection of \( y = 2 \) and \( x + y = 4 \): \( x + 2 = 4 \implies x = 2 \). So, \( (2, 2) \).

- Point \( P_5 \): Intersection of \( x = 0 \) and \( y = 2 \): \( (0, 2) \).


Evaluate \( Z = x + 3y \) at each vertex:

- At \( (0, 0) \): \( Z = 0 + 3(0) = 0 \)

- At \( (3, 0) \): \( Z = 3 + 3(0) = 3 \)

- At \( (3, 1) \): \( Z = 3 + 3(1) = 6 \)

- At \( (2, 2) \): \( Z = 2 + 3(2) = 2 + 6 = 8 \)

- At \( (0, 2) \): \( Z = 0 + 3(2) = 6 \)


The maximum value obtained is 8.


Step 4: Final Answer.

The maximum value of the function is 8.
Quick Tip: To maximize \( x + ky \), focus on the vertex with the largest \( y \)-coordinate while satisfying the \( x + y \leq C \) constraint.
Since the coefficient of \( y \) is 3 (higher than \( x \)), we try to push \( y \) to its maximum limit (\( y = 2 \)) and then find the corresponding max \( x \).