A) \[5,-10\]
B) \[5,-5\]
C) 5, 5
D) \[10,-5\]
E) 10, 10
Correct Answer: A
Solution :
Let \[f(x)=a{{x}^{2}}+bx+4\] On differentiating w.r.t.\[x,\]we get \[f(x)=2ax+b\] For minimum, put \[f(x)=0\] \[\Rightarrow \] \[2ax+b=0\] \[\Rightarrow \] \[x=-\frac{b}{2a}\] Since, it is given that at\[x=1\]minimum value is\[-1\]. \[\therefore \] \[1=-\frac{b}{2a}\] \[\Rightarrow \] \[2a+b=0\] ...(i) and \[f(1)=a+b+4=-1\] \[\Rightarrow \] \[a+b+5=0\] ...(ii) On solving Eqs. (i) and (ii), we get \[a=5,b=-10\]
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