The positive values of the parameter a for wh | vedclass

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(B) The area $A$ is given by the integral of the absolute value of the function between the given limits:$A = \int_{\frac{\pi}{6a}}^{\frac{5\pi}{6a}}

Vedclass · Accepted Answer

$(0, 1/3)$

Vedclass · Answer

(B) The area $A$ is given by the integral of the absolute value of the function between the given limits:$A = \int_{\frac{\pi}{6a}}^{\frac{5\pi}{6a}} |\cos ax| \, dx$Since the function $\cos ax$ changes sign at $ax = \frac{\pi}{2}$,i.e.,$x = \frac{\pi}{2a}$,we split the integral:$A = \int_{\frac{\pi}{6a}}^{\frac{\pi}{2a}} \cos ax \, dx + \int_{\frac{\pi}{2a}}^{\frac{5\pi}{6a}} -\cos ax \, dx$Let $t = ax$,then $dt = a \, dx$:$A = \frac{1}{a} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos t \, dt - \frac{1}{a} \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} \cos t \, dt$$A = \frac{1}{a} [\sin t]_{\frac{\pi}{6}}^{\frac{\pi}{2}} - \frac{1}{a} [\sin t]_{\frac{\pi}{2}}^{\frac{5\pi}{6}}$$A = \frac{1}{a} (1 - \frac{1}{2}) - \frac{1}{a} (\frac{1}{2} - 1) = \frac{1}{a} (\frac{1}{2}) - \frac{1}{a} (-\frac{1}{2}) = \frac{1}{2a} + \frac{1}{2a} = \frac{1}{a}$Given that the area is greater than $3$:$\frac{1}{a} > 3 \implies a Since $a$ must be positive,the range is $(0, 1/3)$.

The positive values of the parameter $a$ for which the area of the figure bounded by the curve $y = \cos ax$,$y = 0$,$x = \frac{\pi}{6a}$,and $x = \frac{5\pi}{6a}$ is greater than $3$ are:

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