Applications Of Derivatives Question 321
Question: If $ A+B=\frac{\pi }{2}, $ the maximum value of $ \cos A\cos B $ is
[AMU 1999]
Options:
A) $ \frac{1}{2} $
B) $ \frac{3}{4} $
C) 1
D) $ \frac{4}{3} $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ f(A)=\cos A\cos B=\cos A\cos ( \frac{\pi }{2}-A )=\cos A\sin A $
$ {f}’(A)={{\cos }^{2}}A-{{\sin }^{2}}A=\cos 2A $
Now, $ {f}’(A)=0\Rightarrow \cos 2A=0 $
therefore $ 2A=\frac{\pi }{2}\Rightarrow A=\frac{\pi }{4} $
Now $ {f}’’(A)=-2\sin 2A=-2\sin \frac{\pi }{2}=-2(-ve) $
Hence $ f(A) $ is maximum at $ \frac{\pi }{4} $ Maximum value = $ \cos \frac{\pi }{4}\sin \frac{\pi }{4}=\frac{1}{2} $ .
BETA
