If $x^{100} + 2 x^{99} + k$ is divisible by $(x + 1)$ , then the value of $k$ is:

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Solution

The correct option is A 1
Here, $x + 1$ is a factor of $x^{100} + 2 x^{99} + k$ .
$⟹ f (- 1) = 0$
As, $x + 1 = 0$
$x = - 1$
$f (- 1) = (- 1)^{100} + 2 (- 1)^{99} + k = 0$
$⟹ 1 - 2 + k = 0$
$⟹ - 1 + k = 0$
$∴ k = 1$ .
Hence, Option A is correct.

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