$$\therefore$$ $$f(x)$$ satisfies all conditions of Rolle's theorem
therefore $$f'\left( x \right) = 0$$ has a root in $$\left( {0,1} \right)$$
i.e. $$a{x^2} + bx + c = 0$$ has at lease one root in $$(0, 1)$$
3
AIEEE 2004
MCQ (Single Correct Answer)
A function $$y=f(x)$$ has a second order derivative $$f''\left( x \right) = 6\left( {x - 1} \right).$$ 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
A
$${\left( {x + 1} \right)^2}$$
B
$${\left( {x - 1} \right)^3}$$
C
$${\left( {x + 1} \right)^3}$$
D
$${\left( {x - 1} \right)^2}$$
Explanation
$$f''\left( x \right) = 6\left( {x - 1} \right).$$ Inegrating,
we get $$f'\left( x \right) = 3{x^2} - 6x + c$$
Slope at $$\left( {2,1} \right) = f'\left( 2 \right) = c = 3$$
$$\left[ {\,\,} \right.$$ As slope of tangent at $$(2, 1)$$ is $$3$$ $$\left. {\,\,} \right]$$