JEE Main 2026 — 24 January, Evening Shift — Question 12

at $\alpha=0 \Rightarrow \mathrm{f}(0)$

$\mathrm{x}=0, \mathrm{x}=1, \mathrm{y}^{2}=\mathrm{x}$

$\mathrm{y}=|0 \cdot \mathrm{x}-5|-|1-0 \cdot \mathrm{x}|+0 \cdot \mathrm{x}^{2}$

$\mathrm{y}=4$

$A_{1}=\int_{0}^{1}(4-\sqrt{x}) d x$

$=4 x-\left.\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right|_{0} ^{1}$

$=4-\frac{2}{3}(1)=\frac{10}{3}$ at $\alpha=1 \Rightarrow \mathrm{f}(1)$

$\mathrm{x}=0, \mathrm{x}=1, \mathrm{y}^{2}=\mathrm{x}$,

$\mathrm{y}=|\mathrm{x}-5|-|1-\mathrm{x}|+\mathrm{x}^{2}$ in $\mathrm{x} \in(0,1)$

$\mathrm{y}=5-\mathrm{x}-(1-\mathrm{x})+\mathrm{x}^{2}$ $\mathrm{y}=4+\mathrm{x}^{2}$

$A_{2}=\int_{0}^{1}\left(\left(4+x^{2}\right)-(\sqrt{x})\right) d x$ $=4 x+\frac{x^{3}}{3}-\left.\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right|_{0} ^{1}$

$=4+\frac{1}{3}-\frac{2}{3}=\frac{11}{3}$

$|f(0)+f(1)|=\left|A_{1}+A_{2}\right|=\left|\frac{10}{3}+\frac{11}{3}\right|=\left|\frac{21}{3}\right|=7$