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UPSC NDA I 2018 Mathematics Solved Paper

Consider the following information for the next three (03) items that follow:

Three sides of a trapezium are each equal to

6 \,\mathrm{cm}

. Let

\alpha \in \left(0, \;\frac{\pi}{2}\right)

be the angle between a pair of adjacent sides.

[NDA I 2018]

Question : 76 of 120

Marks: +1, -0

\alpha

Solution: 👈: Video Solution

A=Area

=\frac{1}{2}(6+6+2x)\sqrt{36-x^{2}}

=(6+x)\sqrt{36-x^{2}}

\frac{dA}{dx}=(6+x)\left(\frac{-2x}{2\sqrt{36-x^{2}}}\right)+\sqrt{36-x^{2}}

=\sqrt{36-x^{2}}-\frac{x(6+x)}{\sqrt{36-x^{2}}}=\frac{36-6x-2x^{2}}{\sqrt{36-x^{2}}}

\frac{dA}{dx}=0\Rightarrow 36-6x-2x^{2}=0\Rightarrow x=3

\frac{d^{2}A}{dx^{2}} < 0

Now ,

\cos\alpha=\frac{x}{6}=\frac{3}{6}=\frac{1}{2}\Rightarrow \alpha=3

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