AIEEE 2004 (Offline)Question 1
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 :
Differential Equation
Solution
Concept. Integrate $f''$ twice; the tangent gives $f'(2)=3$ and the point gives $f(2)=1$. Formulas used. (1) $\int6(x-1)\,dx=3(x-1)^2+C$Step 1 - $f'$. By (1), $f'(x)=3(x-1)^2+C$. The tangent $y=3x-5$ has slope $3$ at $x=2$: $3+C=3$, so $C=0$. Step 2 - $f$. $f(x)=(x-1)^3+D$, and $f(2)=1$ gives $D=0$. Step 3 - The function. $f(x)=(x-1)^3$. Answer: (B)