<p>Let \(G(x, t) = \begin{cases} x(t-1), & \text{when } x \leq t \\ t(x-1), & \text{when } t < x \end{cases}\) and \(t\) is continuous function of \(x\) in \([0,1]\).</p><p>If \(g(x) = \int_0^1 f(t)G(x,t)\,dt\), then</p>
Step-by-Step Solution
Key Concept: The piecewise definition of G(x,t) requires splitting the integral at t=x. For fixed x∈[0,1], integrate over [0,x] using t(x-1) and [x,1] using x(t-1), then differentiate g(x) with respect to x.
<p><strong>Step 1:</strong> Split the integral based on the piecewise definition of G(x,t):</p><p>$$g(x) = \int_0^x f(t) \cdot t(x-1)\,dt + \int_x^1 f(t) \cdot x(t-1)\,dt$$</p><p><strong>Step 2:</strong> Rewrite by factoring:</p><p>$$g(x) = (x-1)\int_0^x t\,f(t)\,dt + x\int_x^1 (t-1)f(t)\,dt$$</p><p><strong>Step 3:</strong> Apply Leibniz rule, differentiating each term. For the first integral:</p><p>$$\frac{d}{dx}\left[(x-1)\int_0^x t\,f(t)\,dt\right] = \int_0^x t\,f(t)\,dt + (x-1) \cdot xf(x)$$</p><p><strong>Step 4:</strong> For the second integral:</p><p>$$\frac{d}{dx}\left[x\int_x^1 (t-1)f(t)\,dt\right] = \int_x^1 (t-1)f(t)\,dt - x(x-1)f(x)$$</p><p><strong>Step 5:</strong> Add both parts:</p><p>$$g'(x) = \int_0^x t\,f(t)\,dt + xf(x)(x-1) + \int_x^1 (t-1)f(t)\,dt - x(x-1)f(x)$$</p><p>$$g'(x) = \int_0^x t\,f(t)\,dt + \int_x^1 (t-1)f(t)\,dt = \int_0^1 G(x,t)f(t)\,dt$$</p><p>∴ Answer: A</p>
Correct Answer: A