Least positive value of $c$ if $c, k, b$ are in A.P. is:
Step-by-Step Solution
Key Concept: Using the periodic property g(x+2a) = g(x) derived from g(x+a) + g(x) = 0, combined with the arithmetic progression condition c = 2k - b (where c, k, b are in A.P.), to minimize the integral ∫g(t)dt over a symmetric interval that exploits periodicity.
From the condition $g(x+a) + g(x) = 0$, we get $g(x+2a) = g(x)$, making $g(x)$ periodic with period $2a$. For any $b$ and $c$ in arithmetic progression with common difference $k$, we have $\int_b^{2k} g(t)dt = \int_a^{b+c} g(x)dx$ using periodicity. Since this integral is independent of $b$, the minimum value of $c$ must be $2a$.
Correct Answer: 4