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(A) The area $A$ of a triangle with vertices $(0,0)$,$(x_1, y_1)$,and $(x_2, y_2)$ is given by $A = \frac{1}{2} |x_1 y_2 - x_2 y_1|$.Here,the vertices

Vedclass · Accepted Answer

$\frac{3 \sqrt{3}}{32}$

Vedclass · Answer

(A) The area $A$ of a triangle with vertices $(0,0)$,$(x_1, y_1)$,and $(x_2, y_2)$ is given by $A = \frac{1}{2} |x_1 y_2 - x_2 y_1|$.Here,the vertices are $(0,0)$,$(x, \cos x)$,and $(\sin^3 x, 0)$.Thus,$A(x) = \frac{1}{2} |x(0) - (\sin^3 x)(\cos x)| = \frac{1}{2} \sin^3 x \cos x$ for $x \in (0, \frac{\pi}{2})$.To find the maximum,we differentiate $A(x)$ with respect to $x$:$A'(x) = \frac{1}{2} [3 \sin^2 x \cos x \cdot \cos x + \sin^3 x (-\sin x)] = \frac{1}{2} [3 \sin^2 x \cos^2 x - \sin^4 x]$.Setting $A'(x) = 0$,we get $3 \sin^2 x \cos^2 x = \sin^4 x$.Since $\sin x 
eq 0$,we have $3 \cos^2 x = \sin^2 x$,which implies $	an^2 x = 3$,so $	an x = \sqrt{3}$ (since $x \in (0, \frac{\pi}{2})$).Thus,$x = \frac{\pi}{3}$.At $x = \frac{\pi}{3}$,$\sin x = \frac{\sqrt{3}}{2}$ and $\cos x = \frac{1}{2}$.Substituting these values into $A(x)$:$A = \frac{1}{2} (\frac{\sqrt{3}}{2})^3 (\frac{1}{2}) = \frac{1}{2} (\frac{3 \sqrt{3}}{8}) (\frac{1}{2}) = \frac{3 \sqrt{3}}{32}$.

If $0 < x < \frac{\pi}{2}$,then the maximum area (in sq. units) of the triangle whose vertices are $(0,0)$,$(x, \cos x)$ and $(\sin^3 x, 0)$ is

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