| Statement-1: f has local maximum at\[x=-1\] and\[x=2\]. |
| Statement-2: \[p=\frac{1}{2}\] and \[q=\frac{-1}{4}\] |
A) Statement-1 is true, Statement-2 is false.
B) Statement-1 is true, Statement-2 is true and Statement-2 is NOT the correct explanation for Statement-1.
C) Statement-1 is true, Statement-2 is true and Statement-2 is correct explanation for Statement-1.
D) Statement-1 is false, Statement-2 is true.
Correct Answer: C
Solution :
Sign of f'(x) \[f'(x)=\frac{1}{x}+2qx+p\] At \[x=-1\Rightarrow \,p-2q=1\] At \[x=2\,\Rightarrow \,p+4q=-\frac{1}{2}\] \[\therefore \,p=\frac{1}{2},\,q=-\frac{1}{4}\] Also, \[f'(x)=\frac{1}{x}-\frac{x}{2}+\frac{1}{2}=\frac{-(x+1)(x-2)}{2x}\]
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