A) \[0\]
B) \[-2\]
C) \[-4\]
D) \[4\]
Correct Answer: A
Solution :
[a] Since \[f\left( x \right)=g\left( x \right)|(x-1)(x-2)...(x-10)|-2\]is differentiable \[\forall \,\,x\in R,\] we have \[g\left( x \right)=0\]for \[x=1,2,3,.....,10\] So, \[g\left( x \right)=0\]for all real x as degree of \[g(x)\] is 9. \[\therefore \,\,\,\,f(x)=-2\] \[\therefore \,\,\,\frac{d}{dx}\left( f\left( {{x}^{2}}+g(x) \right) \right)\] at \[x=0\] is zero.You need to login to perform this action.
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