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UPSC NDA I 2016 Mathematics Solved Paper

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For the next two (2) items that follow :

Consider the curves

f(x)=x\lvert x\rvert-1

and

g(x)=\begin{cases} \frac{3x}{2}, & x>0 \\ 2x, & x\le 0 \end{cases}

[NDA I 2016]

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Question : 13 of 120

Marks: +1, -0

What is the area bounded by the curves ?

$\frac{17}{6}$ square units
$\frac{8}{3}$ square units
$2$ square units
$\frac{1}{3}$ square units

Solution:

Area

=\int\limits_{-1}^{2}[g(x)-f(x)]dx

=\int\limits_{-1}^{0}[g(x)-f(x)]dx+\int\limits_{0}^{2}[g(x)-f(x)]dx

=\int\limits_{-1}^{0}(2x+x^{2}+1)dx+\int\limits_{0}^{2}\left[\frac{3x}{2}-x^{2}+1\right]dx

=I_{1}+I_{2}

I_{1}=\left(\frac{2x^{2}}{2}+\frac{1}{3}x^{3}+x\right)_{-1}^{0}

=(0+0+0)-\left((-1)^{2}+\frac{1}{3}(-1)^{3}+(-1)\right)

=-\left(1-\frac{1}{3}-1\right)=\frac{1}{3}

I_{2}=\int\limits_{0}^{2}\left[\frac{3x}{2}-x^{2}+1\right]dx

=\left[\frac{3}{2}\cdot\frac{x^{2}}{2}-\frac{1}{3}x^{3}+x\right]_{0}^{2}

=\left(\frac{3}{4}\times4-\frac{1}{3}\times8+2\right)-0

=5-\frac{8}{3}=\frac{7}{3}

Area

=I_{1}+I_{2}=\frac{1}{3}+\frac{7}{3}

Area

=\frac{8}{3}

square units

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