Entrance Exams
Govt. Exams
Differentiation is finding the rate of change of a function with respect to a variable.
We have y = 3x² and we need to use the Power Rule of differentiation.
The Power Rule states: If y = ax^n, then \(\frac{dy}{dx} = n \cdot a \cdot x^{n-1}\)
Here, a = 3 and n = 2
Therefore, the derivative of y = 3x² is 6x
Answer: A) 6x
To find the median, we must first arrange all numbers in ascending order, then locate the middle value.
We have 7 numbers: 5, 8, 3, 9, 1, 7, 6
1, 3, 5, 6, 7, 8, 9
Since n = 7 (odd number), median position = \(\frac{n+1}{2} = \frac{7+1}{2} = 4\)
Counting from left: 1st is 1, 2nd is 3, 3rd is 5, 4th is 6
Therefore, the median is 6.
The answer is (A) 6
For a regular hexagon inscribed in a circle of radius R, the side length equals R.
Here, side = 6 cm, so perimeter = 6 × 6 = 36 cm.
The circumference of the circle = 2πR = 12π cm.
Since 12π ≈ 37.7 > 36, the difference is 12π - 36 cm (circumference is greater).
The area of the ring = π(R² - r²).
Given that this equals the area of the inner circle = πr², we have R² - r² = r², which gives R² = 2r², so R = r√2.
Therefore, the ratio R:r = √2:1.
The inscribed circle in the square has diameter 8 cm, so radius = 4 cm.
When a square is inscribed in this circle, its diagonal equals the diameter (8 cm).
If the diagonal of the smaller square is 8 cm, then using diagonal = side√2, we get side = 8/√2 = 4√2 cm.
Therefore, area = (4√2)² = 32 cm².
For a quadratic equation ax² + bx + c = 0, sum of roots = -b/a and product of roots = c/a.
Here, sum = -(k-3)/2 and product = (k-5)/2.
Given that sum = (1/2) × product, we have -(k-3)/2 = (1/2) × (k-5)/2.
Solving: -(k-3)/2 = (k-5)/4, which gives -2(k-3) = k-5, leading to -2k+6 = k-5, thus k = 11.