Question:

Step 1: Evaluate the integral $I(x) = \displaystyle\int_0^{\pi/2}|x-t|\sin t\, dt$. For $0 \leq x \leq \pi/2$, the absolute value $|x-t|$ splits the integral at $t=x$: $$I(x) = \int_0^x (x-t)\sin t\, dt + \int_x^{\pi/2}(t-x)\sin t\, dt$$ Step 2: Compute the first integral term. $$ \int_0^x (x-t)\sin t\, dt = x\int_0^x \sin t\, dt - \int_0^x t\sin t\, dt $$ Using the standard integrals $\int \sin t\, dt = -\cos t$ and $\int t\sin t\, dt = \sin t - t\cos t$ (obtained via integration by parts): $$ x[-\cos t]_0^x - [\sin t - t\cos t]_0^x $$ $$ = x(-\cos x - (-\cos 0)) - ((\sin x - x\cos x) - (\sin 0 - 0\cos 0)) $$ $$ = x(1-\cos x) - (\sin x - x\cos x) $$ $$ = x - x\cos x - \sin x + x\cos x = x - \sin x $$ Step 3: Compute the second integral term. $$ \int_x^{\pi/2}(t-x)\sin t\, dt = \int_x^{\pi/2} t\sin t\, dt - x\int_x^{\pi/2}\sin t\, dt $$ Using the same integral formulas: $$ [\sin t - t\cos t]_x^{\pi/2} - x[-\cos t]_x^{\pi/2} $$ $$ = ((\sin(\pi/2) - (\pi/2)\cos(\pi/2)) - (\sin x - x\cos x)) - x(-\cos(\pi/2) - (-\cos x)) $$ $$ = ((1 - 0) - (\sin x - x\cos x)) - x(0 + \cos x) $$ $$ = 1 - \sin x + x\cos x - x\cos x = 1 - \sin x $$ Step 4: Combine the terms for $I(x)$ and substitute into the expression for $f(x)$. For $0 \leq x \leq \pi/2$: $$ I(x) = (x - \sin x) + (1 - \sin x) = x - 2\sin x + 1 $$ Substituting this into the given form of $f(x)$: $$ f(x) = a(x - 2\sin x + 1) + bx + c $$ $$ f(x) = ax - 2a\sin x + a + bx + c $$ $$ f(x) = (a+b)x - 2a\sin x + (a+c) $$ Step 5: Compare coefficients with the definition of $f(x)$ for $0 \leq x \leq \pi/2$. The given function is $f(x) = \sin x$ for $0 \leq x \leq \pi/2$. Comparing the coefficients of $x$, $\sin x$, and the constant term: $$ -2a = 1 \implies a = -\frac{1}{2} $$ $$ a+b = 0 \implies -\frac{1}{2} + b = 0 \implies b = \frac{1}{2} $$ $$ a+c = 0 \implies -\frac{1}{2} + c = 0 \implies c = \frac{1}{2} $$ Step 6: Evaluate the derived conditions using the values of $a, b, c$. The values are $a = -1/2$, $b = 1/2$, $c = 1/2$. $$ 2a+1 = 2\left(-\frac{1}{2}\right)+1 = -1+1 = 0 $$ $$ 2b-1 = 2\left(\frac{1}{2}\right)-1 = 1-1 = 0 $$ $$ 2c-1 = 2\left(\frac{1}{2}\right)-1 = 1-1 = 0 $$ $$ 8abc-1 = 8\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)-1 = 8\left(-\frac{1}{8}\right)-1 = -1-1 = -2 $$