Question

(ii) If f(x) = x²+Kx+1, for all x and f is an even function, then find k.
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Answer 1

$\bf \color{brown}{ \underline{ \underline{Solution :}}} \\ \\ \\ \\ \tt \:being\: \: a \: \: even \: \: function \: \: we \: \: can \: \: say \\ \\ \tt \: f ( - x) =f(x) \\ \\ \implies \tt {( - x)}^{2} + k( - x) + 1 = {x}^{2} + kx + 1 \\ \\ \implies \tt \: \cancel {x}^{2} - kx + \cancel{1} = \cancel{x}^{2} + kx + \cancel{ 1} \\ \\ \implies \tt \: 2kx = 0 \\ \\ \therefore { \green{\underline{ \boxed{ \tt{k = 0}}}}}$