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A function y=f(x) has a second order derivative f''(x)=6(x-1). If its graph passes through the point (2,1) and at that point the tangent to the graph is y=3x-5, then the function is
- (x+1)²
- (x-1)³
- (x+1)³
- (x-1)²
Correct answer: (x-1)³
Solution
The second derivative, when integrated, gives the first derivative, which must match the slope of the tangent line at the given point. Integrating twice and applying the conditions of the point and tangent line leads to the function being (x) = (x-1)³, which satisfies both the second derivative and the initial conditions.
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