A) \[a=2,b=1/2\]
B) \[a=2,b=-1/2\]
C) \[a=-2,b=-1/2\]
D) \[a=-2,b=1/2\]
Correct Answer: B
Solution :
\[y=a\log |x|+b{{x}^{2}}+x\] If\[x>0,y=a\log x+b{{x}^{2}}+x\] \[\frac{dy}{dx}=\frac{a}{x}+2bx+1\] ...(i) If \[x<0y=a\log (-x)+b{{x}^{2}}+x\] \[\frac{dy}{dx}=\frac{a}{-x}(-1)+2bx+1\] ...(ii) From Eq. (ii), \[{{\left( \frac{dy}{dx} \right)}_{x=-1}}=0\] \[\Rightarrow \] \[-a+2b(-1)+1=0\] \[\Rightarrow \] \[-a-2b=-1\] \[\Rightarrow \] \[a=1-2b\] ?.(iii) From Eq. (i), \[{{\left( \frac{dy}{dx} \right)}_{x=2}}=0\] \[\Rightarrow \] \[\frac{a}{2}+2b-2+1=0\] \[\Rightarrow \] \[\frac{1-2b}{2}+4b+1=0\] [using Eq. (iii)] \[\Rightarrow \] \[1-2b+8b+2=0\] \[\Rightarrow \] \[6b=-3\] \[\Rightarrow \] \[b=-\frac{1}{2}\] \[\therefore \] \[a=1-2\left( -\frac{1}{2} \right)=2\]
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