Question

The positive value of the parameter 'k' for which the area of the figure bounded by the curve $y=\sin \left(kx\right),x=\frac{2\pi }{3k},x=\frac{5\pi }{3k}$ and x-axis is less than 2 can be

• A. $\frac{1}{8}<k<\frac{3}{8}$
• B. $0<k<\frac{1}{8}$
• C. 1 < k < 2
• D. $\frac{3}{8}<k<\frac{5}{8}$

Answer 1

The correct option is C 1 < k < 2

$A={\int }_{\frac{2\pi }{3k}}^{\frac{\pi }{k}}\left(\sin kx\right)dx-{\int }_{\frac{\pi }{k}}^{\frac{5\pi }{3k}}\left(\sin kx\right)dx$

$=-{\left(\frac{\cos kx}{k}\right)}_{2\pi /3k}^{\pi /k}+{\left(\frac{\cos kx}{k}\right)}_{\pi /k}^{5\pi /3k}$

$-\frac{1}{k}[-1-\cos \frac{2\pi }{3}]+\frac{1}{k}[\cos \frac{5\pi }{3}-\cos \pi ]$

$=-\frac{1}{k}[-1+\frac{1}{2}]+\frac{1}{k}[\frac{1}{2}+1]$

$=\frac{1}{2k}+\frac{3}{2k}=\frac{2}{k}<2\Rightarrow k>1$