Let $f , g$ and $h$ be real-valued functions defined on the interval $[ 0,1 ]$ by $f ( x ) = e ^ { x ^ { 2 } } + e ^ { - x ^ { 2 } } , g ( x ) = x e ^ { x ^ { 2 } } + e ^ { - x ^ { 2 } }$ and $h ( x ) = x ^ { 2 } e ^ { x ^ { 2 } } + e ^ { - x ^ { 2 } }$. If $a , b$ and $c$ denote, respectively, the absolute maximum of $f , g$ and $h$ on $[ 0,1 ]$, then
A) $\mathrm { a } = \mathrm { b }$ and $\mathrm { c } \neq \mathrm { b }$
B) a $=$ c and a $\neq$ b
C) $a \neq b$ and $c \neq b$
D) $a = b = c$

If $f ( x ) = \left| \begin{array} { c c c } \cos ( 2 x ) & \cos ( 2 x ) & \sin ( 2 x ) \\ - \cos x & \cos x & - \sin x \\ \sin x & \sin x & \cos x \end{array} \right|$, then
[A] $f ^ { \prime } ( x ) = 0$ at exactly three points in $( - \pi , \pi )$
[B] $f ^ { \prime } ( x ) = 0$ at more than three points in $( - \pi , \pi )$
[C] $f ( x )$ attains its maximum at $x = 0$
[D] $f ( x )$ attains its minimum at $x = 0$

If $p$ and $q$ are positive real numbers such that $p ^ { 2 } + q ^ { 2 } = 1$, then the maximum value of ( $p + q$ ) is
(1) 2
(2) $1 / 2$
(3) $\frac { 1 } { \sqrt { 2 } }$
(4) $\sqrt { 2 }$

Let $I = \int _ { a } ^ { b } \left( x ^ { 4 } - 2 x ^ { 2 } \right) d x$. If $I$ is minimum then the ordered pair $( a , b )$ is
(1) $( 0 , \sqrt { 2 } )$
(2) $( \sqrt { 2 } , - \sqrt { 2 } )$
(3) $( - \sqrt { 2 } , 0 )$
(4) $( - \sqrt { 2 } , \sqrt { 2 } )$

The minimum value of $2 ^ { \sin x } + 2 ^ { \cos x }$ is:
(1) $2 ^ { - 1 + \frac { 1 } { \sqrt { 2 } } }$
(2) $2 ^ { - 1 + \sqrt { 2 } }$
(3) $2 ^ { 1 - \sqrt { 2 } }$
(4) $2 ^ { 1 - \frac { 1 } { \sqrt { 2 } } }$

Let $f : [ - 1,1 ] \rightarrow R$ be defined as $f ( x ) = a x ^ { 2 } + b x + c$ for all $x \in [ - 1,1 ]$, where $a , b , c \in R$ such that $f ( - 1 ) = 2 , f ^ { \prime } ( - 1 ) = 1$ and for $x \in ( - 1,1 )$ the maximum value of $f ^ { \prime \prime } ( x )$ is $\frac { 1 } { 2 }$. If $f ( x ) \leq \alpha , x \in [ - 1,1 ]$, then the least value of $\alpha$ is equal to

Let $f : [ - 1,1 ] \rightarrow R$ be defined as $f ( x ) = a x ^ { 2 } + b x + c$ for all $x \in [ - 1,1 ]$, where $a , b , c \in R$ such that $f ( - 1 ) = 2 , f ^ { \prime } ( - 1 ) = 1$ and for $x \in ( - 1,1 )$ the maximum value of $f ^ { \prime \prime } ( x )$ is $\frac { 1 } { 2 }$. If $f ( x ) \leq \alpha , x \in [ - 1,1 ]$, then the least value of $\alpha$ is equal to

Let $M$ and $m$ respectively be the maximum and minimum values of the function $f ( x ) = \tan ^ { - 1 } ( \sin x + \cos x )$ in $\left[ 0 , \frac { \pi } { 2 } \right]$. Then the value of $\tan ( M - m )$ is equal to: (1) $2 - \sqrt { 3 }$ (2) $3 - 2 \sqrt { 2 }$ (3) $3 + 2 \sqrt { 2 }$ (4) $2 + \sqrt { 3 }$

If the minimum value of $f(x) = \frac{5x^2}{2} + \frac{\alpha}{x^5}$, $x > 0$, is 14, then the value of $\alpha$ is equal to
(1) 32
(2) 64
(3) 128
(4) 256

The sum of the absolute minimum and the absolute maximum values of the function $f ( x ) = \left| 3 x - x ^ { 2 } + 2 \right| - x$ in the interval $[ - 1 , 2 ]$ is
(1) $\frac { \sqrt { 17 } + 3 } { 2 }$
(2) $\frac { \sqrt { 17 } + 5 } { 2 }$
(3) 5
(4) $\frac { 9 - \sqrt { 17 } } { 2 }$

Let $f ( x ) = 2 \cos ^ { - 1 } x + 4 \cot ^ { - 1 } x - 3 x ^ { 2 } - 2 x + 10 , x \in [ - 1 , 1 ]$. If $[ a , b ]$ is the range of the function, then $4a - b$ is equal to
(1) 11
(2) $11 - \pi$
(3) $11 + \pi$
(4) $15 - \pi$

If the absolute maximum value of the function $f(x) = (x ^ { 2 } - 2x + 7) e ^ { (4x ^ { 3 } - 12x ^ { 2 } - 180x + 31)}$ in the interval $[-3,0]$ is $f(\alpha)$, then
(1) $\alpha = 0$
(2) $\alpha = - 3$
(3) $\alpha \in (-1,0)$
(4) $\alpha \in (-3,-1)$

The sum of the absolute maximum and absolute minimum values of the function $f ( x ) = \tan ^ { - 1 } ( \sin x - \cos x )$ in the interval $[ 0 , \pi ]$ is
(1) $0$
(2) $\tan ^ { - 1 } \left( \frac { 1 } { \sqrt { 2 } } \right) - \frac { \pi } { 4 }$
(3) $\cos ^ { - 1 } \left( \frac { 1 } { \sqrt { 3 } } \right) - \frac { \pi } { 4 }$
(4) $\frac { - \pi } { 12 }$

If the function $f : (-\infty, -1] \rightarrow [a, b]$ defined by $f(x) = e^{x^3 - 3x + 1}$ is one-one and onto, then the distance of the point $P(2b+4, a+2)$ from the line $x + e^{-3}y = 4$ is:

Let $f(x) = (x+3)^2(x-2)^3$, $x \in [-4, 4]$. If $M$ and $m$ are the maximum and minimum values of $f$, respectively in $[-4, 4]$, then the value of $M - m$ is:
(1) 600
(2) 392
(3) 608
(4) 108

Let $f ( x ) = 3 \sqrt { x - 2 } + \sqrt { 4 - x }$ be a real valued function. If $\alpha$ and $\beta$ are respectively the minimum and the maximum values of $f$, then $\alpha ^ { 2 } + 2 \beta ^ { 2 }$ is equal to
(1) 42
(2) 38
(3) 24
(4) 44

Let the maximum and minimum values of $\left( \sqrt { 8 x - x ^ { 2 } - 12 } - 4 \right) ^ { 2 } + ( x - 7 ) ^ { 2 } , x \in \mathbf { R }$ be M and m , respectively. Then $\mathrm { M } ^ { 2 } - \mathrm { m } ^ { 2 }$ is equal to $\_\_\_\_$

If the function $f ( x ) = \left( \frac { 1 } { x } \right) ^ { 2 x } ; x > 0$ attains the maximum value at $x = \frac { 1 } { \mathrm { e } }$ then:
(1) $\mathrm { e } ^ { \pi } < \pi ^ { \mathrm { e } }$
(2) $\mathrm { e } ^ { \pi } > \pi ^ { \mathrm { e } }$
(3) $( 2 e ) ^ { \pi } > \pi ^ { ( 2 e ) }$
(4) $\mathrm { e } ^ { 2 \pi } < ( 2 \pi ) ^ { \mathrm { e } }$

Suppose that $x$ and $y$ satisfy
$$3 x + y = 18 , \quad x \geqq 1 , \quad y \geqq 6 .$$
We are to find the maximum value and the minimum value of $x y$.
When we express $x y$ in terms of $x$, we have
$$x y = \mathbf { A B } ( x - \mathbf { C } ) ^ { 2 } + \mathbf { D E } .$$
Also, the range of values which $x$ can take is
$$\mathbf { F } \leqq x \leqq \mathbf { G } .$$
Hence, the value of $x y$ is maximized at $x = \mathbf { H }$ and its value there is $\mathbf { I J }$, and the value of $x y$ is minimized at $x = \mathbf { K }$ and its value there is $\mathbf { L M }$.

Suppose that $x$ and $y$ satisfy
$$3 x + y = 18 , \quad x \geqq 1 , \quad y \geqq 6$$
We are to find the maximum value and the minimum value of $x y$.
When we express $x y$ in terms of $x$, we have
$$x y = \mathbf { A B } ( x - \mathbf { C } ) ^ { 2 } + \mathbf { D E } .$$
Also, the range of values which $x$ can take is
$$\mathbf { F } \leq x \leqq \mathbf { G }$$
Hence, the value of $x y$ is maximized at $x = \mathbf { H }$ and its value there is $\mathbf { I J }$, and the value of $x y$ is minimized at $x = \mathbf { K }$ and its value there is $\mathbf { L M }$.

We are to find the range of the values of a real number $t$ such that the maximum value of the cubic function $$f(x) = \frac{1}{3}x^3 - \frac{t+2}{2}x^2 + 2tx + \frac{2}{3}$$ over the interval $x \leqq 4$ is greater than 6.
First of all, since the derivative of $f(x)$ is $$f'(x) = (x - \mathbf{A})(x - t),$$ we consider the problem by dividing the range of the values of $t$ as follows:
(i) When $t > \mathbf{A}$, $f(x)$ has a local maximum at $x = \mathbf{A}$ and a local minimum at $x = t$. Since $f(4) = \mathbf{B}$, we only have to find the range of the values of $t$ satisfying $f(\mathbf{A}) > 6$.
(ii) When $t = \mathbf{A}$, the maximum value of $f(x)$ over the interval $x \leqq 4$ is $f(\mathbf{C}) = \mathbf{D}$, and hence the condition is not satisfied.
(iii) When $t < \mathbf{A}$, $f(x)$ has a local maximum at $x = t$ and a local minimum at $x = \mathbf{A}$. Since $f(4) = \mathbf{B}$, we only have to find the range of the values of $t$ satisfying $f(t) > 6$.
Here, we note $$f(t) - 6 = -\frac{1}{6}(t + \mathbf{E})(t - \mathbf{EF})^2.$$
From the above, the range of the values of $t$ is to be determined.

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
$$b = \mathbf { U } .$$

Let $a$ be a real number satisfying $a \geqq 0$. We are to express the maximum value $M$ of the function $f(x) = |x^2 - 2x|$ on the range $a \leqq x \leqq a + 1$ in terms of $a$. Furthermore, we are to find the minimum value of $M$ over the range $a \geqq 0$.
(1) The function $f(x)$ can be expressed without using the absolute value symbol as follows: when $x \leqq \mathbf{M}$ or $x \geqq \mathbf{N}$, then $f(x) = x^2 - 2x$; when $\mathbf{M} < x < \mathbf{N}$, then $f(x) = -x^2 + 2x$.
Hence, the maximum value of $f(x)$ on $a \leqq x \leqq a + 1$ is the following: when $0 \leqq a \leqq \mathbf{O}$, then $M = \mathbf{P}$; when $\mathbf{O} < a \leqq \frac{\mathbf{Q} + \sqrt{\mathbf{R}}}{\mathbf{S}}$, then $M = -a^2 + \frac{\mathbf{T}}{\mathbf{S}}a$; when $a > \frac{\mathbf{Q} + \sqrt{\mathbf{R}}}{\mathbf{S}}$, then $M = a^2 - \mathbf{U}$.
(2) The minimum value of $M$ over the range $a \geqq 0$ is $\frac{\sqrt{\mathbf{V}}}{\mathbf{W}}$.

$$f(x) = x^{4} - 5x^{2} + 4$$
What is the maximum value of the function on the interval $\left[\frac{-1}{2}, \frac{1}{2}\right]$?
A) 8
B) 6
C) 4
D) 2
E) 0