Q2 Consider the quadratic function
$$f ( x ) = \frac { 3 } { 4 } x ^ { 2 } - 3 x + 4$$
Let $a$ and $b$ be real numbers satisfying $0 < a < b$ and $2 < b$. We are to find the values of $a$ and $b$ such that the range of the values of the function $y = f ( x )$ on $a \leqq x \leqq b$ is $a \leqq y \leqq b$.
Since the equation of the axis of symmetry of the graph of $y = f ( x )$ is $x = \mathbf { M }$, we divide the problem into two cases as follows:
(i) $\mathbf{M} \leqq a$;
(ii) $0 < a < \mathbf{M}$.
In the case of (i), since the values of $f ( x )$ increase with $x$ on $a \leqq x \leqq b$, the equations $f ( a ) = a$ and $f ( b ) = b$ have to be satisfied. By solving these, we obtain $a = \frac { \mathbf { N } } { \mathbf { O } }$ and $b = \mathbf { P }$. However, this $a$ does not satisfy (i).
In the case of (ii), since the minimum value of $f ( x )$ on $a \leqq x \leqq b$ is $\mathbf { Q }$, we have
$$a = \mathbf { R } .$$
This satisfies (ii). Then since $f ( a ) = \frac { \mathbf { S } } { \mathbf { T } } < b$, we have $f ( b ) = b$. Hence, we obtain
$$b = \mathbf { U } .$$