1.If the functionf(x)={(1+cosx)csc2x,0<x<π2μ,x=π2(1+cosx)cos2xsin2x,π2<x<πf(x) = \begin{cases} (1+\cos x)\csc^2 x, & 0 < x < \frac{\pi}{2} \\ \mu, & x = \frac{\pi}{2} \\ \frac{(1+\cos x)\cos^2 x}{\sin^2 x}, & \frac{\pi}{2} < x < \pi \end{cases}f(x)=⎩⎨⎧(1+cosx)csc2x,μ,sin2x(1+cosx)cos2x,0<x<2πx=2π2π<x<πis continuous at x=π2x = \frac{\pi}{2}x=2π, then 9+6logeμ+μ6−e6/49 + 6\log_e \mu + \mu^6 - e^{6/4}9+6logeμ+μ6−e6/4 is equal toa.11b.8c.2e4+82e^4 + 82e4+8d.10Login to continueOnly logged in users canattempt or see the solution.
