$$\therefore$$ $$f(x)$$ 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 $$(0, 1)$$
2
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 $$3$$ $$\left. {\,\,} \right]$$
Given $${{dy} \over {dx}} = 2 \Rightarrow {9 \over 2} = 2 \Rightarrow y = {9 \over 2}$$
Puting in $${y^2} = 18x \Rightarrow x = {9 \over 8}$$
$$\therefore$$ Required point is $$\left( {{9 \over 8},{9 \over 2}} \right)$$
4
AIEEE 2003
MCQ (Single Correct Answer)
If the function $$f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1,$$ where $$a>0,$$ attains its maximum and minimum at $$p$$ and $$q$$ respectively such that $${p^2} = q$$ , then $$a$$ equals