Question Number 1

Point P and Q lies on the same circle with center at (0, 0).
Thus, (s² + t²) = (-√3)² + 1² = 3 + 1 = 4

Again line segments OP and OQ are perpendicular.
Thus (slope of OP)*(slope of OQ) = -1

Slope of OP = 1/(-√3) = -(1/√3)
=> Slope of OQ = (t - 0)/(s - 0) = t/s = (-1)/(-1/√3) = √3
=> t = √3s

Thus, (s² + (√3s)²) = 4
=> (s² + 3s²) = 4
=> s² = 1
=> s = ±1

As point Q lies in the first quadrant s = 1.

The correct answer is B.

Question Number 2:

Remember that √(x²) is always a positive quantity. Thus for positive x, its value is x but for negative x, its value is -x i.e. in other words it is equal to |x|.

Thus, √(x²)/x = |x|/x

The correct answer is E.

Question Number 3:

Let's analyze each of the options individually:

• 1. (a + b)² ≠ (a² + b²) => f(a + b) ≠ f(a) + f(b)
  2. (a + b + 1) ≠ (a + 1) + (b + 1) = (a + b + 2) => f(a + b) ≠ f(a) + f(b)
  3. √(a + b) ≠ (√a + √b) => f(a + b) ≠ f(a) + f(b)
  4. 2/(a + b) ≠ (2/a) + (2/b) = 2(a + b)/ab => f(a + b) ≠ f(a) + f(b)
  5. (-3(a + b)) = (-3a) + (-3b) => f(a + b) = f(a) + f(b)

The correct answer is E.

Question Number 4:

Maximum number of codes that can be generated = (Maximum number of 4-letter code) + (Maximum number of 5-letter code)

As there is no restrictions regarding repetition of letters,

• Maximum number of 4-letter code = (26)^4
  Maximum number of 5-letter code = (26)^5

Maximum number of codes that can be generated = (26^4) + (26^5) = (26^4)*(1 + 26) = 27*(26^5)

The correct answer is C.