We are given the points \( \left( \frac{11}{2}, \alpha \right) \) that lie inside or on the boundary of the triangle formed by the lines \( x + y = 11 \), \( x + 2y = 16 \), and \( 2x + 3y = 29 \).
Step 1: Find the equation of the triangle
We first solve the system of equations for the lines forming the triangle.
Step 2: Solve for the points of intersection
We solve these systems of linear equations to find the boundaries of the triangle and determine the limits for \( \alpha \), the y-coordinate of the point \( \left( \frac{11}{2}, \alpha \right) \).
The values of \( \alpha \) that satisfy the condition for the points to lie inside or on the triangle will give the smallest and largest values of \( \alpha \).
Step 3: Find the product of the smallest and largest values of \( \alpha \)
Once the smallest and largest values of \( \alpha \) are identified, we compute their product.
After solving, we find that the product of the smallest and largest values of \( \alpha \) is \( 33 \).
Final Answer: \( 33 \).
Let $ a_1, a_2, a_3, \ldots $ be in an A.P. such that $$ \sum_{k=1}^{12} 2a_{2k - 1} = \frac{72}{5}, \quad \text{and} \quad \sum_{k=1}^{n} a_k = 0, $$ then $ n $ is:
The sum $ 1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + ... $ upto $ \infty $ terms, is equal to
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is: