A) \[f(x)\] is strictly increasing in \[x\in (\alpha ,\beta )\]
B) \[f(x)\] is strictly decreasing in \[x\in (\alpha ,\beta )\]
C) \[f(x)\] is strictly increasing in \[x\in \left( \alpha ,\frac{\alpha +\beta }{2} \right)\] and strictly decreasing in \[x\in \left( \frac{\alpha +\beta }{2},\beta \right)\]
D) \[f(x)\] is a constant function.
Correct Answer: D
Solution :
[d] \[f(x)=1-{{\cos }^{2}}(x+\alpha )+{{\sin }^{2}}(x+\beta )\] \[-2\cos (\alpha -\beta )\sin (x+\alpha )\sin (x+\beta )\] \[=1-\cos (2x+\alpha +\beta )cos(\alpha -\beta )\] \[-2\cos (\alpha -\beta )\sin (x+\alpha )\sin (x+\beta )\] \[=1-\cos (\alpha -\beta )[\cos (2x+\alpha +\beta )\] \[-\cos (2x+\alpha +\beta )+\cos (\alpha -\beta )]\] \[={{\sin }^{2}}(\alpha -\beta )\] This is a constant function.
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