1.Consider f(x)={sin(2πsecx)ex−1−x;x≠0, x≠(2n+1)π2, n∈Zk;x=0f(x) = \begin{cases} \frac{\sin(2\pi \sec x)}{e^x - 1 - x} & ; x \neq 0, \, x \neq (2n+1)\frac{\pi}{2},\, n \in \mathbb{Z} \\ k & ; x = 0 \end{cases}f(x)={ex−1−xsin(2πsecx)k;x=0,x=(2n+1)2π,n∈Z;x=0Value of kkk for which f(x)f(x)f(x) is continuous at x=0x = 0x=0, is:a.000b.π\piπc.2π2\pi2πd.4π4\pi4πLogin to continueOnly logged in users canattempt or see the solution.
