To solve the problem, we need to find the value of G41+G42+G43+G21G23 where G1, G2, G3 are the geometric means of two distinct positive numbers A and B.
1. Define the Numbers:
Let the two distinct positive numbers be A and B.
2. Find the Geometric Means:
The geometric means G1, G2, G3 can be defined as follows:
3. Calculate G41, G42, and G43:
4. Sum the Fourth Powers:
Now we sum these results: G41+G42+G43=AB3+A2B2+A3B
5. Calculate G21G23:
Now, we need to calculate G21G23:
6. Combine All Terms:
Now we combine all the terms: G41+G42+G43+G21G23=(AB3+A2B2+A3B)+A2B2
This simplifies to: AB3+2A2B2+A3B
7. Factor the Expression:
We can factor this expression: =AB(A2+2AB+B2)=AB(A+B)2
Final Answer:
Thus, the expression G41+G42+G43+G21G23 is equal to: AB(A+B)2
If the function f(x) = \(\sqrt{x^2 - 4}\) satisfies the Lagrange’s Mean Value Theorem on \([2, 4]\), then the value of \( C \) is}
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to:
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is: