Q85. Let $\mathrm { a } > 0$ be a root of the equation $2 x ^ { 2 } + x - 2 = 0$. If $\lim _ { x \rightarrow \frac { 1 } { \mathrm { a } } } \frac { 16 \left( 1 - \cos \left( 2 + x - 2 x ^ { 2 } \right) \right) } { ( 1 - \mathrm { a } x ) ^ { 2 } } = \alpha + \beta \sqrt { 17 }$, where $\alpha , \beta \in Z$, then $\alpha + \beta$ is equal to $\_\_\_\_$ Q86. Let the mean and the standard deviation of the probability distribution

X	$\alpha$	1	0	- 3
$\mathrm { P } ( \mathrm { X } )$	$\frac { 1 } { 3 }$	K	$\frac { 1 } { 6 }$	$\frac { 1 } { 4 }$

be $\mu$ and $\sigma$, respectively. If $\sigma - \mu = 2$, then $\sigma + \mu$ is equal to $\_\_\_\_$

Q85. The value of $\lim _ { x \rightarrow 0 } 2 \left( \frac { 1 - \cos x \sqrt { \cos 2 x } \sqrt [ 3 ] { \cos 3 x } \ldots \ldots \sqrt [ 10 ] { \cos 10 x } } { x ^ { 2 } } \right)$ is

Let $f(x) = x^3 + ax^2 + 2bx + c$, $x \in \mathbb{R}$. If $f'(1) = a$, $f''(2) = b$, and $f'''(3) = c$, then the value of $f'(5)$ is
(A) $\frac{117}{5}$
(B) $\frac{62}{5}$
(C) $\frac{675}{5}$
(D) $\frac{117}{5}$

Let $\mathrm { p } ( \mathrm { x } )$ be a differentiable function such that $\mathrm { p } ( 1 ) = 2$.
If $\operatorname { Lim } _ { t \rightarrow x } \left( \frac { t ^ { 2 } p ( x ) - x ^ { 2 } p ( t ) } { x - t } \right) = 3$, then the value of $2 p ( 2 )$.

$f ( x ) = \frac { e ^ { x } \left( e ^ { \tan x - x } - 1 \right) + \log ( \sec x + \tan x ) - x } { \tan x - x }$ If $\mathrm { f } ( \mathrm { x } )$ is continuous at $\mathrm { x } = 0$, then find $\mathrm { f } ( 0 )$

$$\lim _ { x \rightarrow 0 } \frac { \ln \left( \sec ( e x ) \sec \left( e ^ { 2 } x \right) \sec \left( e ^ { 3 } x \right) \ldots \sec \left( e ^ { 10 } x \right) \right) } { \left( e ^ { 2 } - e ^ { 2 } \cos x \right) e ^ { 2 } / ( 1 - G x ) }$$ (A) $\frac { \mathrm { e } ^ { 18 } - 1 } { \mathrm { e } ^ { 2 } - 1 }$ (B) $\frac { \mathrm { e } ^ { 20 } - 1 } { \mathrm { e } ^ { 2 } - 1 }$ (C) $\frac { e ^ { 2 } - 1 } { e ^ { 2 } - 1 }$ (D) $\frac { \mathrm { e } ^ { 22 } - 1 } { \mathrm { e } ^ { 2 } - 1 }$

We are to differentiate
$$f ( x ) = \int _ { 0 } ^ { 2 x } \left( t ^ { 2 } - x ^ { 2 } \right) \sin 3 t \, d t$$
with respect to $x$.
(1) We know that if $g ( t )$ is a continuous function and $G ( t )$ is one of its primitive functions, then
$$\int _ { 0 } ^ { 2 x } g ( t ) d t = G ( 2 x ) - G ( 0 )$$
By differentiating both sides of this equality with respect to $x$, we have
$$\frac { d } { d x } \int _ { 0 } ^ { 2 x } g ( t ) d t = \mathbf { A }$$
where $\mathbf{A}$ is the appropriate expression from among the following (0) $\sim$ (7).
(0) $g ( x )$
(1) $\frac { 1 } { 2 } g ( x )$
(2) $2 g ( x )$
(3) $g ( 2 x )$
(4) $\frac { 1 } { 2 } g ( 2 x )$
(5) $2 g ( 2 x )$ (6) $g ( x ) - g ( 0 )$ (7) $g ( 2 x ) - g ( 0 )$
(2) We know that $f ( x ) = \int _ { 0 } ^ { 2 x } t ^ { 2 } \sin 3 t \, d t - \int _ { 0 } ^ { 2 x } x ^ { 2 } \sin 3 t \, d t$.
Since
$$\frac { d } { d x } \int _ { 0 } ^ { 2 x } t ^ { 2 } \sin 3 t \, d t = \mathbf { B } x ^ { 2 } \sin \mathbf { C } x$$
and
$$\frac { d } { d x } \int _ { 0 } ^ { 2 x } x ^ { 2 } \sin 3 t \, d t = \frac { \mathbf { D } } { \mathbf { E } } x ( - \cos \mathbf { F } x + \mathbf { G } + \mathbf { H } x \sin \mathbf { I } x )$$
we obtain
$$f ^ { \prime } ( x ) = \frac { \mathbf { D } } { \mathbf { E } } x ( \cos \mathbf { J } x - \mathbf { K } + \mathbf { L } x \sin \mathbf { M } x )$$

Q1 Let $a > 0$. Consider two curves
$$C_1: y = e^{6x}$$ $$C_2: y = ax^2.$$
We are to find the condition on $a$ such that there exist two straight lines, each of which is tangent to both $C_1$ and $C_2$.
The equation of the tangent to $C_1$ at a point $(t, e^{6t})$ is
$$y = \mathbf{A}e^{6t}x - e^{6t}(\mathbf{B}t - \mathbf{C})$$
This is tangent also to $C_2$ under the condition that the quadratic equation
$$ax^2 = \mathbf{A}e^{6t}x - e^{6t}(\mathbf{B}t - \mathbf{C})$$
has just one solution. Hence, the equation
$$\mathbf{D}e^{12t} - ae^{6t}(\mathbf{E}t - \mathbf{F}) = 0$$
must hold for $a$ and $t$. From this equation we obtain
$$a = \frac{\mathbf{D}}{\mathbf{E}t - \mathbf{F}}e^{6t}$$
Let $f(t)$ denote the right side of this equation. The condition under which there exist two straight lines each of which is tangent to both $C_1$ and $C_2$, is that the straight line $s = a$ intersects the graph of $s = f(t)$ at two points.
Now, the derivative of $f(t)$ is
$$f'(t) = \frac{108e^{6t}(\mathbf{G}t - \mathbf{H})}{(\mathbf{E}t - \mathbf{F})^2}.$$
Hence the condition on $a$ that we are seeking is
$$a > \square e^{\square}.$$
Note that $\lim_{t \to \infty} \frac{e^t}{t} = \infty$.

Given the function in $x$
$$f _ { n } ( x ) = \sin ^ { n } x \quad ( n = 1,2,3 , \cdots ) ,$$
answer the following questions.
(1) Consider the cases in which the equality
$$\lim _ { x \rightarrow 0 } \frac { a - x ^ { 2 } - \left( b - x ^ { 2 } \right) ^ { 2 } } { f _ { n } ( x ) } = c$$
holds for three real numbers $a , b$ and $c$.
(i) We have $a = b$.
(ii) When $n = 2$, if $c = 6$, then $b = \frac { \mathbf { P } } { \mathbf { Q } }$.
(iii) When $n = 4$, then $b = \frac { \mathbf { R } } { \mathbf { S } }$ and $c = - \mathbf { T }$.
(2) For this $f _ { n } ( x )$, consider the definite integral
$$I _ { n } = \int _ { 0 } ^ { \frac { \pi } { 2 } } f _ { n } ( x ) \sin 2 x \, d x \quad ( n = 1,2,3 , \cdots )$$
When the integral is calculated, we have
$$I _ { n } = \frac { \mathbf { U } } { n + \mathbf { V } } .$$
Hence we obtain
$$\lim _ { n \rightarrow \infty } \left( I _ { n - 1 } + I _ { n } + I _ { n + 1 } + \cdots + I _ { 2 n - 2 } \right) = \int _ { 0 } ^ { \mathbf { W } } \frac { \mathbf { X } } { \mathbf { Y } + x } \, dx = \log \mathbf { Z }$$

Consider the two functions
$$y = x \log a x , \tag{1}$$ $$y = 2 x - 3 , \tag{2}$$
where $a > 0$, and where $\log$ is the natural logarithm.
(1) Let us find $a$ such that the graph of (1) is tangent to the graph of (2).
The equation of the tangent to the graph of (1) at the point $( t , t \log a t )$ is $\mathbf { A }$ (for A, choose the correct answer from among choices (0) $\sim$ (3) below). (0) $y = ( \log a t + 1 ) x - t$
(1) $y = ( \log a t + a ) x - t$
(2) $y = ( a \log t + 1 ) x + t$
(3) $y = ( a \log t + a ) x + t$
Hence, the graph of (1) is tangent to the graph of (2) when $a = \frac { e } { \square \mathbf{B} }$, and the coordinates of the tangent point are ( $\mathbf { C }$, $\mathbf { D }$ ).
(2) When $a = \frac { e } { \mathbf{B} }$, function (1) is minimized at $x = \square e ^ { - \mathbf { F } }$, and in this case the minimum value is $- \mathbf { G } \cdot e ^ { - \mathbf { H } }$.
(3) When $a = \frac { e } { \mathbf{B} }$, let us find the area $S$ of the region bounded by the graphs of (1) and (2) and the $x$-axis.
For the following indefinite integral, we have
$$\int x \log a x \, d x = \square + C , \quad \text { where } C \text { is an integral constant }$$
(for I, choose the correct answer from among (0) $\sim$ (3) below). (0) $\frac { 1 } { 2 } x ^ { 2 } \log a x - \frac { 1 } { 2 } x ^ { 2 }$
(1) $2 x ^ { 2 } \log a x - 2 x ^ { 2 }$
(2) $\frac { 1 } { 2 } x ^ { 2 } \log a x - \frac { 1 } { 4 } x ^ { 2 }$
(3) $2 x ^ { 2 } \log a x - 4 x ^ { 2 }$
Hence we obtain
$$S = \frac { \mathbf { J } } { \mathbf { K } }$$

Consider the function $f(x) = x^3 - 4x + 4$. Let the straight line $\ell$ be the tangent to the graph of $y = f(x)$ at the point $\mathrm{A}(-1, 7)$, and the straight line $m$ be the tangent to the graph of $y = f(x)$ that passes through the point $\mathrm{B}(0, -12)$. Also, let C be the point of intersection of $\ell$ and $m$. Let us denote the angle formed by $\ell$ and $m$ at C by $\theta$ $\left(0 < \theta < \frac{\pi}{2}\right)$. We are to find $\tan\theta$.
(1) The derivative $f'(x)$ of $f(x)$ is $$f'(x) = \mathbf{A}x^{\mathbf{B}} - \mathbf{C}.$$ Hence, the slope of $\ell$ is $\mathbf{DE}$, and the equation of $\ell$ is $$y = \mathbf{DE}x + \mathbf{F}.$$
(2) Let us denote by $a$ the $x$-coordinate of the tangent point of the graph of $y = f(x)$ and line $m$. Then the equation of $m$ can be expressed in terms of $a$ as $$y = (\mathbf{G}a^{\mathbf{H}} - \mathbf{I})x - \mathbf{J}a^{\mathbf{K}} + \mathbf{K}.$$ Since line $m$ passes through point $\mathrm{B}(0, -12)$, we see that $a = \mathbf{M}$, and the equation of $m$ is $$y = \mathbf{N}x - \mathbf{OP}.$$ Hence, the coordinates of point C, the intersection of $\ell$ and $m$, are $(\mathbf{Q}, \mathbf{R})$.
(3) Let us denote by $\alpha$ the angle between the positive direction of the $x$-axis and line $\ell$, and by $\beta$ the angle between the positive direction of the $x$-axis and line $m$. Then we see that $$\tan\alpha = \mathbf{ST}, \quad \tan\beta = \mathbf{U},$$ and hence $$\tan\theta = \frac{\mathbf{V}}{\mathbf{W}}.$$

a) (1.25 points) Let $f$ and $g$ be two differentiable functions for which the following data are known:
$$f ( 1 ) = 1 ; f ^ { \prime } ( 1 ) = 2 ; g ( 1 ) = 3 ; g ^ { \prime } ( 1 ) = 4 :$$
Given $h ( x ) = f \left( ( x + 1 ) ^ { 2 } \right)$, use the chain rule to calculate $h ^ { \prime } ( 0 )$. Given $k ( x ) = \frac { f ( x ) } { g ( x ) }$, calculate $k ^ { \prime } ( 1 )$.
b) (1.25 points) Calculate the integral $\int ( \operatorname { sen } x ) ^ { 4 } ( \cos x ) ^ { 3 } d x$. (You can use the change of variables $t = \operatorname { sen } x$.)

Consider the function
$$f ( x ) = \left\{ \begin{array} { l l l } \operatorname { sen } x & \text { if } & x < 0 \\ x e ^ { x } & \text { if } & x \geq 0 \end{array} \right.$$
a) ( 0.75 points) Study the continuity and differentiability of $f$ at $x = 0$.\ b) (1 point) Study the intervals of increase and decrease of $f$ restricted to ( $- \pi , 2$ ). Prove that there exists a point $x _ { 0 } \in [ 0,1 ]$ such that $f \left( x _ { 0 } \right) = 2$.\ c) (0.75 points) Calculate $\int _ { - \frac { \pi } { 2 } } ^ { 1 } f ( x ) d x$.

On the coordinate plane, let $\Gamma$ be the graph of the cubic function $f(x) = x^{3} - 9x^{2} + 15x - 4$. Which of the following is the derivative of $f(x)$? (Single choice, 2 points)
(1) $x^{2} - 9x + 15$
(2) $3x^{3} - 18x^{2} + 15x - 4$
(3) $3x^{3} - 18x^{2} + 15x$
(4) $3x^{2} - 18x + 15$
(5) $x^{2} - 18x + 15$

$$f ( x ) = \frac { \left( x ^ { 2 } + 5 \right) ( 2 x ) } { \sqrt [ 4 ] { x ^ { 3 } } } , \quad x > 0$$
Which one of the following is equal to $f ^ { \prime } ( x )$ ?

The sequence of functions $f _ { 1 } ( x ) , f _ { 2 } ( x ) , f _ { 3 } ( x ) , \ldots$ is defined as follows:
$$\begin{aligned} f _ { 1 } ( x ) & = x ^ { 10 } \\ f _ { n + 1 } ( x ) & = x f _ { n } ^ { \prime } ( x ) \text { for } n \geq 1 \end{aligned}$$
where $f _ { n } ^ { \prime } ( x ) = \frac { d f _ { n } ( x ) } { d x }$
Find the value of
$$\sum _ { n = 1 } ^ { 20 } f _ { n } ( x )$$

The least possible value of the gradient of the curve $y = ( 2 x + a ) ( x - 2 a ) ^ { 2 }$ at the point where $x = 1$, as $a$ varies, is
A $- \frac { 49 } { 4 }$ B - 8 C $- \frac { 25 } { 4 }$ D $\frac { 7 } { 4 }$ E $\frac { 47 } { 16 }$

The function f is given, for $x > 0$, by
$$\mathrm { f } ( x ) = \frac { x ^ { 3 } - 4 x } { 2 \sqrt { x } }$$
Find the value of $f ^ { \prime } ( 4 )$.

Which of the following is an expression for the first derivative with respect to $x$ of
$$\frac { x ^ { 3 } - 5 x ^ { 2 } } { 2 x \sqrt { x } }$$
A $- \frac { \sqrt { x } } { 2 }$
B $\frac { \sqrt { x } } { 4 }$
C $\frac { 3 x - 5 } { 4 \sqrt { x } }$
D $\frac { 3 \sqrt { x } - 5 } { 4 \sqrt { x } }$
E $\frac { 3 \sqrt { x } - 10 } { 3 \sqrt { x } }$
F $\frac { 3 x ^ { 2 } - 10 x } { 3 \sqrt { x } }$

The Fundamental Theorem of Calculus (FTC) tells us that for any polynomial f :
$$\frac { \mathrm { d } } { \mathrm {~d} x } \left( \int _ { 0 } ^ { x } \mathrm { f } ( t ) \mathrm { d } t \right) = \mathrm { f } ( x )$$
A student calculates $\frac { \mathrm { d } } { \mathrm { d } x } \left( \int _ { x } ^ { 2 x } t ^ { 2 } \mathrm {~d} t \right)$ as follows: (I) $\quad \int _ { x } ^ { 2 x } t ^ { 2 } \mathrm {~d} t = \int _ { 0 } ^ { 2 x } t ^ { 2 } \mathrm {~d} t - \int _ { 0 } ^ { x } t ^ { 2 } \mathrm {~d} t$ (II) By FTC, $\frac { \mathrm { d } } { \mathrm { d } x } \left( \int _ { 0 } ^ { x } t ^ { 2 } \mathrm {~d} t \right) = x ^ { 2 }$ (III) By FTC, $\frac { \mathrm { d } } { \mathrm { d } x } \left( \int _ { 0 } ^ { 2 x } t ^ { 2 } \mathrm {~d} t \right) = ( 2 x ) ^ { 2 } = 4 x ^ { 2 }$ (IV) So $\frac { \mathrm { d } } { \mathrm { d } x } \left( \int _ { x } ^ { 2 x } t ^ { 2 } \mathrm {~d} t \right) = 4 x ^ { 2 } - x ^ { 2 }$ (V) giving $\frac { \mathrm { d } } { \mathrm { d } x } \left( \int _ { x } ^ { 2 x } t ^ { 2 } \mathrm {~d} t \right) = 3 x ^ { 2 }$
Which of the following best describes the student's calculation?
A The calculation is completely correct.
B The calculation is incorrect, and the first error occurs on line (I).
C The calculation is incorrect, and the first error occurs on line (II).
D The calculation is incorrect, and the first error occurs on line (III).
E The calculation is incorrect, and the first error occurs on line (IV). F The calculation is incorrect, and the first error occurs on line (V).

$$\lim_{x \rightarrow 1} \frac{1-\sqrt{x}}{\ln x}$$
What is the value of this limit?
A) $\frac{-1}{2}$
B) $0$
C) $\frac{1}{2}$
D) $1$
E) $2$

$$\lim _ { x \rightarrow 0 } \frac { x + \arcsin x } { \sin 2 x }$$
What is the value of this limit?
A) 0
B) 1
C) $\frac { 2 } { 3 }$
D) $\frac { 4 } { 3 }$
E) $\frac { 1 } { 6 }$

$$\lim _ { x \rightarrow \infty } \left( \sqrt { x ^ { 2 } + 2 x + 1 } - \sqrt { x ^ { 2 } + 1 } \right)$$
What is the value of this limit?
A) $\frac { 1 } { 2 }$
B) $\frac { 3 } { 2 }$
C) $\frac { 5 } { 2 }$
D) 1
E) 2

$$f ( x ) = \sin ^ { 2 } \left( 3 x ^ { 2 } + 2 x + 1 \right)$$
Given this, what is the value of $f ^ { \prime } ( 0 )$?
A) $2 \cos 2$
B) $2 \cos 3$
C) $6 \sin 1$
D) $4 \sin 2$
E) $2 \sin 2$

$$f ( x ) = 2 x ( x - 1 ) ^ { 3 } + ( x - 1 ) ^ { 4 }$$
What is the value of the third derivative of the function at the point $x = 1$?
A) 10
B) 12
C) 14
D) 16
E) 18