Question

Q.1. If f(x) = 5x^6 – 4 tan^4x + 3 cos^2 x, then prove that f(x) is even function.​

Answer 1

$\large\underline{\sf{Given- }}$

$\: \: \: \: \: \bull \: \sf \: \: f(x) = {5x}^{6} - {4tan}^{4} x + 3 {cos}^{2}x$

$\large\underline{\sf{To\:Prove- }}$

$\: \: \: \: \: \bull \: \sf \: \: f(x) \: is \: even \: function.$

$\large\underline{\sf{Solution-}}$

Basic Concept :-

Even Function :-

• A function f(x) is said to be an even function if f(- x) = f(x).

So, in order to show that given function is even function, we have to prove that f(- x) = f(x).

Let's solve the problem now!!

Given that

$\rm :\longmapsto\: \sf \: \: f(x) = {5x}^{6} - {4tan}^{4} x + 3 {cos}^{2}x$

So,

$\rm :\longmapsto\:f( - x) = 5 {( - x)}^{6} - 4 {\bigg( tan( - x)\bigg) }^{4} + 3 {\bigg(cos( - x) \bigg) }^{2}$

$\rm :\longmapsto\:f( - x) = 5{x}^{6}- 4 {\bigg(-tanx\bigg) }^{4} + 3 {\bigg(cosx\bigg) }^{2}$

$\because \: \boxed{ \bf \:tan( - x) = - tanx} \: \sf \: and \: \: \boxed{ \bf \: cos( - x) = cosx}$

$\rm :\longmapsto\: \sf \: \: f( - x) = {5x}^{6} - {4tan}^{4} x + 3 {cos}^{2}x$

$\bf\implies \:f( - x) = f(x)$

$\bf\implies \:f(x) \: is \: an \: even \: function$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

Odd function :-

• A function f(x) is said to be an odd function if f(- x) = - f(x).

Periodic function :-

• A function f(x) is said to be periodic function if there exist a positive real number T such that f(x + T) = f(x).

Even functions is symmetric along y - axis where as odd functions are symmetrical in opposite quadrants.