Step 1: Understanding the Concept:

If two complex numbers are equal, then their real parts are equal and their imaginary parts are equal.

To compare them easily, the complex fraction on the right-hand side must be simplified.


Step 2: Key Formula or Approach:

Rationalize the denominator by multiplying numerator and denominator by the conjugate of \( a-i \).


Step 3: Detailed Explanation:
\[ \dfrac{a+i}{a-i} = \dfrac{(a+i)(a+i)}{(a-i)(a+i)}. \] \[ = \dfrac{a^2 + 2ai - 1}{a^2 + 1}. \]
Thus, \[ \dfrac{a+i}{a-i} = \frac{a^2-1}{a^2+1} + i\frac{2a}{a^2+1}. \]
Given: \[ \dfrac{\sqrt{3}+i}{2} = \frac{\sqrt{3}}{2} + i\frac{1}{2}. \]
Equating real and imaginary parts: \[ \frac{a^2-1}{a^2+1} = \frac{\sqrt{3}}{2}, \quad \frac{2a}{a^2+1} = \frac{1}{2}. \]
From the imaginary part: \[ 4a = a^2 + 1 \Rightarrow a^2 - 4a + 1 = 0. \]
Solving: \[ a = 2 \pm \sqrt{3}. \]
Checking these values in the real-part equation, only \[ a = \sqrt{3} \]
satisfies the given equality.


Step 4: Final Answer:

The value of \( a \) is \( \sqrt{3}. \)
Quick Tip: Always rationalize complex fractions first, then equate real and imaginary parts separately.

Step 1: Understanding the Concept:

The Mean Value Theorem states that if a function is continuous on \( [a,b] \) and differentiable on \( (a,b) \), then there exists at least one \( c \in (a,b) \) such that: \[ f'(c) = \frac{f(b)-f(a)}{b-a}. \]


Step 2: Key Formula or Approach:

Compute the average rate of change and equate it with the derivative of the function.


Step 3: Detailed Explanation:

Given \( f(x)=2x-x^2 \): \[ f(1)=2-1=1, \quad f(0)=0. \]
Average rate of change: \[ \frac{f(1)-f(0)}{1-0} = 1. \]
Derivative: \[ f'(x)=2-2x. \]
Equating: \[ 2-2c = 1 \Rightarrow 2c = 1 \Rightarrow c = \frac{1}{2}. \]
Clearly, \( \frac{1}{2} \in (0,1) \), so it satisfies the theorem.


Step 4: Final Answer:

The required value of \( c \) is \( \frac{1}{2}. \)
Quick Tip: For MVT problems, first compute the average slope, then equate it with the derivative.

Step 1: Understanding the Concept:

If two vectors are perpendicular, then their dot product is zero.

Here, the vectors are cross products.


Step 2: Key Formula or Approach:

Use the vector identity: \[ (a\times b)\cdot(c\times d) = (a\cdot c)(b\cdot d) - (a\cdot d)(b\cdot c). \]


Step 3: Detailed Explanation:

Since \( a\times b \) and \( c\times d \) are perpendicular: \[ (a\times b)\cdot(c\times d) = 0. \]
Substituting given values: \[ \lambda \cdot \lambda - (4)(9) = 0. \] \[ \lambda^2 = 36. \]
Since \( \lambda > 0 \): \[ \lambda = 6. \]


Step 4: Final Answer:

The value of \( \lambda \) is \( 6. \)
Quick Tip: Memorize the dot product identity of cross products for fast vector calculations.

Step 1: Understanding the Concept:

Certain definite integrals vanish due to symmetry when limits are \( 0 \) to \( \frac{\pi}{2} \).


Step 2: Key Formula or Approach:

Use the property: \[ \int_0^{\pi/2} f(x)\,dx = \int_0^{\pi/2} f\!\left(\frac{\pi}{2}-x\right) dx. \]


Step 3: Detailed Explanation:

Let: \[ I = 2\int_{0}^{\pi/2} \sin 2x \log(\tan x)\,dx. \]
Using substitution \( x \to \frac{\pi}{2}-x \): \[ \sin 2x \to \sin 2x, \quad \log(\tan x) \to \log(\cot x) = -\log(\tan x). \]
Hence: \[ I = -2\int_{0}^{\pi/2} \sin 2x \log(\tan x)\,dx = -I. \]
Thus: \[ I = 0. \]


Step 4: Final Answer:

The value of the integral is \( 0. \)
Quick Tip: Whenever logarithmic functions appear in symmetric limits, always test symmetry first.