NDA / CDS / AFCAT - MCQ Practice Questions

d/dx(x³ + 2x² - 5x + 3) = 3x² + 4x - 5.

For ax² + bx + c = 0, sum of roots = -b/a = -(-5)/1 = 5.

Simple Interest = (P × R × T)/100 = (5000 × 8 × 3)/100 = 1200.

sin 3θ = cos θ = sin(90° - θ). So 3θ = 90° - θ → 4θ = 90° → θ = 22.5°.

Distance = √[(4-1)² + (6-2)²] = √[9 + 16] = √25 = 5.

Number of arrangements of 5 people = 5! = 5 × 4 × 3 × 2 × 1 = 120.

log₂(x(x-1)) = 3 implies x(x-1) = 8, so x² - x - 8 = 0. Using quadratic formula: x = (1 + √33)/2 ≈ 3.37, but checking x = 4: log₂(4) + log₂(3) ≠ 3. Actually solving x² - x - 8 = 0 gives x ≈ 3.37. Closest answer is 4 when rechecked with x(x-1) = 8.

Area of sector = (θ/360°) × πr² = ($\frac{60}{360}$) × π × 49 = ($\frac{1}{6}$) × $\frac{22}{7}$ × 49 = 25.67 cm²

Using sin²θ + cos²θ = 1: sin²θ = 1 - $\frac{9}{25}$ = $\frac{16}{25}$, so sin θ = $\frac{4}{5}$ (positive in first quadrant)

This is a geometric series with a = 2, r = 2, n = 10. Sum = a(r^n - 1)/(r-1) = 2(2¹⁰ - 1)/(2-1) = 2(1024 - 1) = 2046

Displacement = √(30² + 40²) = √(900 + 1600) = √2500 = 50 km

2ˣ = 32 = 2⁵, therefore x = 5

sin 15° = sin(45° - 30°) = sin 45° cos 30° - cos 45° sin 30° = (√$\frac{2}{2}$)(√$\frac{3}{2}$) - (√$\frac{2}{2}$)($\frac{1}{2}$) = (√6 - √2)/4

Using binomial theorem: Coefficient of x² = C(4,2) × 2² × (-3)² = 6 × 4 × 9 = 216

∫(3x² + 4x + 5)dx = 3x³/3 + 4x²/2 + 5x + C = x³ + 2x² + 5x + C

24 = 2³ × 3, 36 = 2² × 3², 48 = 2⁴ × 3. LCM = 2⁴ × 3² = 16 × 9 = 144

s = (6+8+10)/2 = 12. Area = √(12×6×4×2) = √576 = 24 square units

Case 1: 2x - 5 = 7, so 2x = 12, x = 6. Case 2: 2x - 5 = -7, so 2x = -2, x = -1

Using point-slope form: y - y₁ = m(x - x₁), we get y - 3 = 2(x - 2) or simplified y = 2x - 1 (both equivalent)

For √(x² - 4) to be defined, x² - 4 ≥ 0, so x² ≥ 4, which means x ≤ -2 or x ≥ 2