Question

if f(x) = cos [e square]x + cos [-e square]x where [x] stands for greatest integer function then​

Answer 1

SOLUTION

GIVEN

$\sf{f(x) = \cos \big[ \: {\pi}^{2} \big] x+ \cos \big[ - \: {\pi}^{2} \big]x }$

Where [ x ] stands for greatest integer function

TO DETERMINE

$\displaystyle \sf{1. \: \: \: f \bigg( \frac{\pi}{2} \bigg)}$

$\displaystyle \sf{2. \: \: \: f \bigg( \frac{\pi}{4} \bigg)}$

$\displaystyle \sf{3. \: \: \: f (0)}$

EVALUATION

Here it is given that

$\sf{f(x) = \cos \big[ \: {\pi}^{2} \big] x+ \cos \big[ - \: {\pi}^{2} \big]x }$

$\sf{Now \: \: \pi = 3.14}$

$\sf{ \therefore \: \: 9 < {\pi}^{2} < 10 } \: \: and \: \: - 10 < {\pi}^{2} < - 9$

$\sf{ \big[ \: {\pi}^{2} \big] x = 9x \: \: \: and \: \: \big[ - {\pi}^{2} \big]x = - 10x }$

Now

$\sf{f(x) = \cos \big[ \: {\pi}^{2} \big] x+ \cos \big[ - \: {\pi}^{2} \big]x }$

$\sf{ \implies \: f(x) = \cos 9x+ \cos ( - 10)x }$

$\sf{ \implies \: f(x) = \cos 9x+ \cos 10x }$

1.

$\displaystyle \sf{ f \bigg( \frac{\pi}{2} \bigg)}$

$\displaystyle \sf{ = \cos \bigg( \frac{9\pi}{2} \bigg) + \cos \bigg( \frac{10\pi}{2} \bigg) }$

$= 0 - 1$

$= - 1$

2.

$\displaystyle \sf{ f \bigg( \frac{\pi}{4} \bigg)}$

$\displaystyle \sf{ = \cos \bigg( \frac{9\pi}{4} \bigg) + \cos \bigg( \frac{10\pi}{4} \bigg) }$

$\displaystyle \sf{ = \frac{1}{ \sqrt{2} } }$

3.

$\displaystyle \sf{ f (0)}$

$\displaystyle \sf{ = \cos 0+ \cos 0 }$

$= 1 + 1$

$= 2$

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