MCQ
If $f(x) = \cos [{\pi ^2}]x + \cos [ - {\pi ^2}]x$, then
- A$f\left( {\frac{\pi }{4}} \right) = 2$
- B$f( - \pi ) = 2$
- C$f(\pi ) = 1$
- ✓$f\left( {\frac{\pi }{2}} \right) = - 1$
✓
Answer
Correct option: D.
$f\left( {\frac{\pi }{2}} \right) = - 1$
d
(d) $f(x) = \cos \,[{\pi ^2}]x + \cos \,[ - {\pi ^2}]\,x$
(d) $f(x) = \cos \,[{\pi ^2}]x + \cos \,[ - {\pi ^2}]\,x$
$f(x) = \cos (9x) + \cos ( - 10x)$$ = \cos (9x) + \cos (10x)$
$ = 2\cos \left( {\frac{{19x}}{2}} \right)\cos \left( {\frac{x}{2}} \right)$
$f\left( {\frac{\pi }{2}} \right) = 2\cos \left( {\frac{{19\pi }}{4}} \right)\cos \left( {\frac{\pi }{4}} \right)$;
$f\left( {\frac{\pi }{2}} \right) = 2 \times \frac{{ - 1}}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} = - 1$.
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