If f (x) = x2 + kx + 1 for all x and f is an even function, find k whe – InterviewSolution

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1.

If f (x) = x2 + kx + 1 for all x and f is an even function, find k where k ∈ R(a) 1 (b) – 2 (c) 0 (d) – 1

Answer»

Answer: (c) = 0

f (x) = x² + kx + 1

Given, f (x) is an even function

⇒ f (x) = f (– x)

⇒ x² + kx + 1 = (– x)² + k (– x) + 1

⇒ x² + kx + 1 = x²– kx + 1

⇒ 2kx = 0

⇒ k = 0.

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