A) \[\frac{1}{6}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{4}\]
D) \[\frac{1}{12}\]
Correct Answer: D
Solution :
| \[{{f}_{k}}=\frac{1}{4}\left( {{\sin }^{k}}x+{{\cos }^{k}}x \right)\] | \[{{f}_{6}}(x)=\frac{1}{6}\left( {{\sin }^{6}}x+{{\cos }^{6}}x \right)\] |
| \[{{f}_{4}}(x)=\frac{1}{4}\left( {{\sin }^{6}}x+{{\cos }^{4}}x \right)\] | \[{{f}_{6}}K=\frac{1}{6}\left[ 1-\frac{3}{4}{{\sin }^{2}}2x \right]\] |
| \[{{f}_{4}}(x)=\frac{1}{4}\left[ 1-\frac{{{\sin }^{2}}2x}{2} \right]\] |
You need to login to perform this action.
You will be redirected in
3 sec
