A) Statement-1 is false, Statement-2 is true.
B) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1.
C) Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1.
D) Statement-1 is true, statement-2 is false.
Correct Answer: B
Solution :
\[f'(x)=\frac{1}{x}+2bx+a\] at \[x=-1\] \[-1-1b+a=0\] \[a-2b=1\] ... (i) at \[x=2\] \[\frac{1}{2}+4b+a=0\] \[a+4b=-\frac{1}{2}\] ... (ii) On solving (i) and (ii) \[a=\frac{1}{2},\,\,b=-\frac{1}{4}\] \[f'(x)=\frac{1}{x}-\frac{x}{2}+\frac{1}{2}=\frac{2-{{x}^{2}}+x}{2x}=\frac{-(x+1)(x-2)}{2x}\] So maxima at \[x=-1,2\]You need to login to perform this action.
You will be redirected in
3 sec