MCQOPTIONS
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| 1. |
Function \[f(x)=\frac{1-\cos 4x}{8{{x}^{2}}},\] where \[x\ne 0\]and \[f(x)=k\] where \[x=0\] is a continous function at \[x=0\]then the value of k will be [AMU 2005] |
| A. | \[k=0\] |
| B. | \[k=1\] |
| C. | \[k=-1\] |
| D. | None of these |
| Answer» C. \[k=-1\] | |