Q171 / 270 · NDA 2016 (Phase 1) · Mathematics
Consider the function $\mathrm{f}(\mathrm{x})=|\mathrm{x}-1|+\mathrm{x}^2$, where $\mathrm{x} \in \mathrm{R}$.
What is the area of the region bounded by x -axis, the curve $y=f(x)$ and the two ordinates $x=\frac{1}{2}$ and $x=1$ ?
- Correct answerA$\frac{5}{12}$ square unit
- B$\frac{5}{6}$ square unit
- C$\frac{7}{6}$ square units
- D2 square units
Solution
$f(x)= \begin{cases}x^2+x-1 & x \geq 1 \\ x^2-x+1 & x \lt 1\end{cases} \quad \forall x \in R$
Hence area required for given region is
$\begin{aligned}
A_1 & =\int_{1 / 2}^1 f(x) d x \\
& =\int_{1 / 2}^1\left(x^2-x+1\right) d x \\
& =\left[\frac{1}{3} x^3-\frac{x^2}{2}+x\right]_{1 / 2}^1 \\
& =\left(\frac{1}{3} \times 1-\frac{1}{2}+1\right)-\left(\frac{1}{3} \times \frac{1}{8}-\frac{1}{2} \times \frac{1}{4}+\frac{1}{2}\right) \\
A_1 & =\frac{5}{12} \text { square unit. }
\end{aligned}$
Hence, the correct answer is option: 'A'.