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NIMCET 2016 — Mathematics PYQ

NIMCET | Mathematics | 2016

The area of the region bounded by the lines $y = ∣ x - 1∣$ and $y = 3 - ∣ x ∣$ is

Choose the correct answer:

Explanation

Concept:
The area between the curves $y 1 = f (x)$ and $y 2 = g (x)$ is given by:
Area enclosed = $∫x2x1(y1 - y2)dx$
Where, $x 1$ and $x 2$ are the intersections of curves $y 1$ and $y 2$
Calculation:

Shaded area has to be calculated
Curve $1 : y = ∣ x - 1∣$
⇒ $y = 1 - x for x < 1$
⇒ $y = x - 1 for x ≥ 1$
Curve $2 : y = 3 - ∣ x ∣$
⇒ $y = 3 + x for x < 0$
⇒ $y = 3 - x for x ≥ 0$

\begin{aligned}
& \mathrm{Area~enclosed~(A)=\int_{x_{1}}^{x_{2}}(y_{1}-y_{2})dx} \\
& \Rightarrow\mathrm{A}=\int_{-1}^{2}3-|\mathrm{x}|-|\mathrm{x}-1|\mathrm{dx} \\
& \Rightarrow\mathrm{A=\left|\int_{-1}^{0}3+x-(1-x)dx\right|+\left|\int_{0}^{1}3-x-(1-x)dx\right|+\left|\int_{1}^{2}3-x-(x-1)dx\right|} \\
& \Rightarrow\mathrm{A=\left|\int_{-1}^{0}(2+2x)dx\right|+\left|\int_{0}^{1}2dx\right|+\left|\int_{1}^{2}(4-2x)dx\right|} \\
& \Rightarrow\mathrm{A=\left|[2x+x^{2}]_{-1}^{0}\right|+\left|[2x]_{0}^{1}\right|+\left|[4x-x^{2}]_{1}^{2}\right|} \\
& \Rightarrow\mathrm{A=1+2+1=4~sq.~units}
\end{aligned}