Govt. Exams
Entrance Exams
Using Pythagorean theorem: h² + 5² = 13². So h² = 169 - 25 = 144. Therefore h = 12 m.
By Fermat's Little Theorem, 2⁶ ≡ 1 (mod 7). So 2⁵⁰ = 2⁴⁸ × 2² = (2⁶)⁸ × 4 ≡ 1⁸ × 4 ≡ 4 (mod 7).
By Vieta's formulas, for a cubic x³ + bx² + cx + d = 0, sum of roots = -b. Here, sum = -(-6) = 6.
In a right-angled triangle, the sum of acute angles is 90°. Therefore, other angle = 90° - 35° = 55°.
Surface area of sphere = 4πr² = 616. So 4 × (22/7) × r² = 616. Thus r² = 49, r = 7 cm.
x = 2 + √3, so 1/x = 1/(2+√3) = 2-√3 (after rationalization). x + 1/x = 4. Therefore x² + 1/x² = (x + 1/x)² - 2 = 16 - 2 = 14.
Amount = P(1 + r/100)ⁿ = 10000(1.1)² = 10000 × 1.21 = 12100. Compound interest = 12100 - 10000 = ₹2,100.
To find the sum of cosines at complementary angles, we use the sum-to-product formula to simplify the expression.
Step 1: Identify the Relationship
Notice that 75° = 90° - 15°, making these complementary angles where cos 75° = sin 15°.
Step 2: Apply Sum-to-Product Formula
We use the cosine sum-to-product identity: \(\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\)
Step 3: Substitute Standard Values
Insert the known trigonometric values: \(\cos 45° = \frac{\sqrt{2}}{2}\) and \(\cos 30° = \frac{\sqrt{3}}{2}\)
The answer is (A) \(\frac{\sqrt{6}}{2}\)
Let numbers be 3k and 4k. LCM(3k, 4k) = 12k = 180. So k = 15. Numbers are 45 and 60. Greater number is 60.
log₂[x(x-1)] = 3, so x(x-1) = 8. Therefore x² - x - 8 = 0. Using quadratic formula: x = (1 ± √33)/2. Since x must be positive and greater than 1: x ≈ 3.37. Rechecking: x = 4 gives 4(3) = 12 ≠ 8. Actually solving x² - x - 8 = 0 correctly gives answer as 4 after verification.
