If *f *(*x*)=*x^99*+ 99*x *+ *k *is divisible by *x *+ 1 , then *k *=

A. −100

B. −98

C. 98

D. 100

E. 198

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`f(x) = x^99 +99x+k`

Using remainder theorem if fully `f(x)` is fully divided by ` (x-a)` then` f(a) = 0` ;

`f(-1) = 0`

`f(-1) = -(1)^99+99(-1)+k = 0`

`-1-99+k = 0`

`K = 100`

*So the answer is k = 100. Correct answer is at option D).*

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