JEE Main 12 Jan 2019 Evening Question 4
Question: If $ {{\sin }^{4}}\alpha +4{{\cos }^{4}}\beta +2=4\sqrt{2}\sin \alpha \cos \beta ; $ $ \alpha ,\beta \in [0,\pi ], $ then $ \cos (\alpha +\beta )-\cos (\alpha -\beta ) $ is equal to [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ -1 $
B) 0
C) $ \sqrt{2} $
D) $ -\sqrt{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
Since, $ A.M.\ge G.M. $
$ \therefore $ $ \frac{{{\sin }^{4}}\alpha +4{{\cos }^{4}}\beta +1+1}{4} $ $ \ge {{({{\sin }^{4}}\alpha .4{{\cos }^{4}}\beta .1.1)}^{1/4}} $
$ \Rightarrow $ $ {{\sin }^{4}}\alpha +4{{\cos }^{4}}\beta +2\ge 4\sqrt{2}\sin \alpha \cos \beta $
But according to question, we have $ {{\sin }^{4}}\alpha +4{{\cos }^{4}}\beta +2=4\sqrt{2}\sin \alpha \cos \beta $
$ \Rightarrow $ $ A.M.=G.M.\Rightarrow sin^{4}\alpha =1=4{{\cos }^{4}}\beta $
$ \Rightarrow $ $ sin\alpha =1,\cos \beta =\pm \frac{1}{\sqrt{2}} $
$ \therefore $ $ sin\alpha =1,and,\sin \beta =\frac{1}{\sqrt{2}} $ as $ \alpha ,\beta \in [0,\pi ] $
Now, $ \cos (\alpha +\beta )-\cos (\alpha -\beta )=-2sin\alpha \beta $ $ =-2.1.\frac{1}{\sqrt{2}}=-\sqrt{2} $
