Question

if f(x)=x^2+2kx+4 and if f is an even function,find the value of k

Answer 1

Answer :

k = 0

Note :

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

★ Odd function : A function f(x) is said to be an odd function if f(-x) = – f(x) .

Solution :

Here ,

The given function is ;

f(x) = x² + 2kx + 4 .

Also ,

It is given that , the function f(x) is an even function .

Thus ,

=> f(x) = f(-x)

=> x² + 2kx + 4 = (-x)² + 2k(-x) + 4

=> x² + 2kx + 4 = x² - 2kx + 4

=> x² + 2kx + 4 - x² + 2kx - 4 = 0

=> 4kx = 0

=> k = 0

Hence , k = 0