سوال ۲۸ کنکور ارشد مهندسی کامپیوتر ۱۴۰۳

The kernel $t+x$ is separable, so the integral splits into two constants and $f$ must be a linear function of $t$.

Write $f(t) = 2 + \int_0^1 (t+x)f(x)\,dx = 2 + t\underbrace{\int_0^1 f(x)\,dx}_{A} + \underbrace{\int_0^1 x f(x)\,dx}_{B}.$

So $f(t) = (2+B) + A\,t$ for unknown constants $A,B$. Substitute this linear form back to get two equations.

Equation for $A$: $A = \int_0^1\big[(2+B) + A x\big]\,dx = (2+B) + \frac{A}{2} \;\Rightarrow\; \frac{A}{2} = 2+B \;\Rightarrow\; A = 4 + 2B.$

Equation for $B$: $B = \int_0^1 x\big[(2+B) + A x\big]\,dx = \frac{2+B}{2} + \frac{A}{3}.$

Multiply by $6$ and insert $A = 4+2B$: $6B = 3(2+B) + 2(4+2B) = 14 + 7B \;\Rightarrow\; B = -14, \quad A = 4 + 2(-14) = -24.$

Hence $f(t) = (2+B) + A t = -12 - 24\,t.$

Check: $\int_0^1 f = -24 = A$ and $\int_0^1 x f = -14 = B$, and $2 + At + B = -12 - 24t = f(t)$. ✓

Set $f(\alpha) = 0$: $-12 - 24\alpha = 0 \;\Rightarrow\; \alpha = -\frac{12}{24} = -\frac{1}{2}.$

Exam trick: When the kernel splits as a sum/product of a $t$-part and an $x$-part (a separable/degenerate kernel), the unknown $f$ is forced into a finite-dimensional form (here a line $at+b$). Plug that form back and solve the small linear system for the integral constants — no calculus of variations needed.