If $$u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } $$
then the difference between the maximum and minimum values of $${u^2}$$ is given by
$$\sqrt x $$ is non periodic function and $$\cos \left( {something} \right)$$ is a periodic function so here in $$\cos \sqrt x $$ $$ \to $$ inside periodic function there is non periodic function which always produce non periodic function.
$${{{\cos }^2}x}$$ is a periodic function with period $$\pi $$
Note : (1) When $$n$$ is odd then the period of $${\sin ^n}\theta $$, $${\cos ^n}\theta $$, $${\csc ^n}\theta $$, $${\sec ^n}\theta $$ = $$2\pi $$
(2) When $$n$$ is even then the period of $${\sin ^n}\theta $$, $${\cos ^n}\theta $$, $${\csc ^n}\theta $$, $${\sec ^n}\theta $$ = $$\pi $$
(3) When $$n$$ is even/odd then the period of $${\tan ^n}\theta $$, $${\cot ^n}\theta $$ = $$\pi $$
(3) When $$n$$ is even/odd then the period of $$\left| {{{\sin }^n}\theta } \right|$$, $$\left| {{{\cos }^n}\theta } \right|$$, $$\left| {{{\csc }^n}\theta } \right|$$, $$\left| {{{\sec }^n}\theta } \right|$$, $$\left| {{{\tan }^n}\theta } \right|$$, $$\left| {{{\cot }^n}\theta } \right|$$ = $$\pi $$
$$\cos \sqrt x + {\cos ^2}x$$ = non periodic function + periodic function = non periodic function
3
AIEEE 2002
MCQ (Single Correct Answer)
The number of solution of $$\tan \,x + \sec \,x = 2\cos \,x$$ in $$\left[ {0,\,2\,\pi } \right]$$ is
A
2
B
3
C
0
D
1
Explanation
Given equation is $$\tan \,x + \sec \,x = 2\cos \,x$$