In order for the function #f(x)=k(x-1)+x(k+3)+2# to be a constant function, what should be the value of #k# ?

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Answer 1 · 2016-07-09T15:55:49

#k=-3/2#

A constant function of x will have same value for any real value of x.

So f(0)=f(1)

We have

#f(x)=k(x-1)+x(k+3)+2#

for x=0,

#f(0)=-k+2#

for x=1

#f(1)=k(1-1)+1(k+3)+2=k+5#

Now #f(0)=f(1)#

#=>-k+2=k+5#

#2k=2-5=-3#

#:.k=-3/2#

Alternative

Differentiating f(x) w.r.t x

#f'(x)=k+k+3=2k+3#

f(x) being constat funtion f'(x)=0

#:.2k+3=0=>k=-3/2#

Answer 2 · 2016-07-09T16:02:49

k=0

#f'=2k=#( constant )' = 0. So, k = 0.