A) a point of inflection.
B) not a critical point.
C) a point of local minima.
D) a point of local maxima.
Correct Answer: C
Solution :
\[f\left( x \right)=\int\limits_{1}^{x}{{{t}^{2}}g\left( t \right)dt,}\] \[g\left( t \right)=\int\limits_{1}^{t}{f\left( u \right)\,du}\] \[f'(x)={{x}^{2}}g(x),\] \[g'(t)=f(t)\] and \[g(1)=0\] \[\therefore \,\,\,\,\,\,\,f'(1)=g(1)=0\] Also, \[f''(x)={{x}^{2}}g'(x)+2x\,\,g(x)\] \[f''(1)=g'(1)+2g(1)=3>0\] \[\therefore \] Local Minima at \[x=1\]You need to login to perform this action.
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