A) \[a=-2,b=1,c=1\]
B) \[a=1,b=1,c=-2\]
C) \[a=1,b=-2,c=1\]
D) \[a=-1,b=2,c=1\]
Correct Answer: C
Solution :
| \[y'=3a{{x}^{2}}+2bx+c\] \[\Rightarrow \,\,\,a+b+c=0\] | ||
| \[y'(0,0)=c=1\] \[\Rightarrow \,\,\,a+b=-1\] | ||
| \[y=a{{x}^{3}}+b{{x}^{2}}+cx\] | ||
| \[\frac{dy}{dx}=3a{{x}^{3}}+2bx+c\] \[\therefore \] \[{{\left( \frac{dy}{dx} \right)}_{(0,0)}}=c=1\] | ||
| and \[{{\left( \frac{dy}{dx} \right)}_{(1,0)}}=3a+2b+c=0\] | ||
| \[\therefore \,\,\,\,3a+2b=-1\] | ...(1) | |
| Point (1, 0) lies on curve \[\therefore \,\,\,a+b+c=0\] | ||
| \[\therefore \,\,\,a+b=-1\] | ...(2) | |
| By (1) & (2) \[a=1,\,\,b=-2\] | ||
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