If f(x) = f(0) + xf'(0) + (x² / 2!) f''(θx) and f(x) = (1 - x)^(5/2), then the value of θ as x → 1 is -

For $f(x) = (1 - x)^{5/2}$, we have $f(0) = 1$, $f'(0) = -\frac{5}{2}$, and $f''(\theta x) = \frac{15}{4}(1 - \theta x)^{1/2}$. Substituting into $f(x) = f(0) + xf'(0) + \frac{x^2}{2!}f''(\theta x)$ and taking $x \to 1$ gives $0 = 1 - \frac{5}{2} + \frac{15}{8}\sqrt{1 - \theta} \implies \sqrt{1 - \theta} = \frac{4}{5} \implies 1 - \theta = \frac{16}{25} \implies \theta = \frac{9}{25}$.