The tangent at the point $( 2 , - 2 )$ to the curve, $x ^ { 2 } y ^ { 2 } - 2 x = 4 ( 1 - y )$ does not pass through the point:
(1) $( - 2 , - 7 )$
(2) $( 8,5 )$
(3) $( - 4 , - 9 )$
(4) $\left( 4 , \frac { 1 } { 3 } \right)$

If the curves $y ^ { 2 } = 6 x , 9 x ^ { 2 } + b y ^ { 2 } = 16$ intersect each other at right angles, then the value of $b$ is:
(1) $\frac { 9 } { 2 }$
(2) 6
(3) $\frac { 7 } { 2 }$
(4) 4

The tangent to the curve $y = x ^ { 2 } - 5 x + 5$, parallel to the line $2 y = 4 x + 1$, also passes through the point :
(1) $\left( \frac { 1 } { 4 } , \frac { 7 } { 2 } \right)$
(2) $\left( \frac { 7 } { 2 } , \frac { 1 } { 4 } \right)$
(3) $\left( - \frac { 1 } { 8 } , 7 \right)$
(4) $\left( \frac { 1 } { 8 } , - 7 \right)$

If the tangent to the curve, $y = x ^ { 3 } + a x - b$ at the point $( 1 , - 5 )$ is perpendicular to the line, $- x + y + 4 = 0$, then which one of the following points lies on the curve?
(1) $( 2 , - 2 )$
(2) $( 2 , - 1 )$
(3) $( - 2,1 )$
(4) $( - 2,2 )$

Let $S$ be the set of all values of $x$ for which the tangent to the curve $y = f ( x ) = x ^ { 3 } - x ^ { 2 } - 2 x$ at ( $x , y$ ) is parallel to the line segment joining the points $( 1 , f ( 1 ) )$ and $( - 1 , f ( - 1 ) )$, then $S$ is equal to
(1) $\left\{ - \frac { 1 } { 3 } , - 1 \right\}$
(2) $\left\{ - \frac { 1 } { 3 } , 1 \right\}$
(3) $\left\{ \frac { 1 } { 3 } , 1 \right\}$
(4) $\left\{ \frac { 1 } { 3 } , - 1 \right\}$

Let $x = 4$ be a directrix to an ellipse whose centre is at the origin and its eccentricity is $\frac { 1 } { 2 }$. If $P ( 1 , \beta ) , \beta > 0$ is a point on this ellipse, then the equation of the normal to it at $P$ is
(1) $4 x - 3 y = 2$
(2) $8 x - 2 y = 5$
(3) $7 x - 4 y = 1$
(4) $4 x - 2 y = 1$

If the tangent to the curve $y = x + \sin y$ at a point $(a, b)$ is parallel to the line joining $\left(0, \frac{3}{2}\right)$ and $\left(\frac{1}{2}, 2\right)$, then
(1) $b = a$
(2) $|b - a| = 1$
(3) $|a + b| = 1$
(4) $b = \frac{\pi}{2} + a$

The equation of the normal to the curve $y = ( 1 + x ) ^ { 2 y } + \cos ^ { 2 } \left( \sin ^ { - 1 } x \right)$, at $x = 0$ is
(1) $y + 4 x = 2$
(2) $y = 4 x + 2$
(3) $x + 4 y = 8$
(4) $2 y + x = 4$

Let $P(h, k)$ be a point on the curve $y = x^{2} + 7x + 2$, nearest to the line, $y = 3x - 3$. Then the equation of the normal to the curve at $P$ is
(1) $x + 3y + 26 = 0$
(2) $x + 3y - 62 = 0$
(3) $x - 3y - 11 = 0$
(4) $x - 3y + 22 = 0$

Which of the following points lies on the tangent to the curve $x^4 e^y + 2\sqrt{y+1} = 3$ at the point $(1, 0)$?
(1) $(2, 2)$
(2) $(2, 6)$
(3) $(-2, 6)$
(4) $(-2, 4)$

If the lines $x + y = a$ and $x - y = b$ touch the curve $y = x^2 - 3x + 2$ at the points where the curve intersects the $x$-axis, then $\frac{a}{b}$ is equal to ...

If the curves $x = y ^ { 4 }$ and $x y = k$ cut at right angles, then $( 4 k ) ^ { 6 }$ is equal to $\underline{\hspace{1cm}}$.

The tangents at the points $A(1,3)$ and $B(1,-1)$ on the parabola $y^2 - 2x - 2y = 1$ meet at the point $P$. Then the area (in unit$^2$) of the triangle $PAB$ is:
(1) 4
(2) 6
(3) 7
(4) 8

If the tangent at the point $\left( x _ { 1 } , y _ { 1 } \right)$ on the curve $y = x ^ { 3 } + 3 x ^ { 2 } + 5$ passes through the origin, then $\left( x _ { 1 } , y _ { 1 } \right)$ does NOT lie on the curve

If the line $y = 4 + kx$, $k > 0$, is the tangent to the parabola $y = x - x^2$ at the point $P$ and $V$ is the vertex of the parabola, then the slope of the line through $P$ and $V$ is
(1) $\frac{3}{2}$
(2) $\frac{26}{9}$
(3) $\frac{5}{2}$
(4) $\frac{23}{6}$

Let $S$ be the set of all the natural numbers, for which the line $\frac { x } { a } + \frac { y } { b } = 2$ is a tangent to the curve $\left( \frac { x } { a } \right) ^ { n } + \left( \frac { y } { b } \right) ^ { n } = 2$ at the point $( a , b ) , ab \neq 0$. Then
(1) $S = \phi$
(2) $n ( S ) = 1$
(3) $S = \{ 2k : k \in N \}$
(4) $S = N$

Consider a curve $y = y ( x )$ in the first quadrant as shown in the figure. Let the area $A _ { 1 }$ is twice the area $A _ { 2 }$. Then the normal to the curve perpendicular to the line $2 x - 12 y = 15$ does NOT pass through the point
(1) $( 6,21 )$
(2) $( 8,9 )$
(3) $( 10 , - 4 )$
(4) $( 12 , - 15 )$

Let $M$ and $N$ be the number of points on the curve $y ^ { 5 } - 9 x y + 2 x = 0$, where the tangents to the curve are parallel to $x$-axis and $y$-axis, respectively. Then the value of $M + N$ equals $\_\_\_\_$ .

If the equation of the normal to the curve $y = \frac { x - a } { ( x + b ) ( x - 2 ) }$ at the point $( 1 , - 3 )$ is $x - 4 y = 13$ then the value of $a + b$ is equal to $\_\_\_\_$

Q2 Suppose that the curve $y=2\cos 2x$ and the curve $y=4\cos x+k$ have a common tangent at $x=a\left(0(1) We set $f(x)=2\cos 2x$ and $g(x)=4\cos x+k$. Since we have assumed that $y=f(x)$ and $y=g(x)$ have a common tangent at $x=a$, we see that
$$f'(a)=g'(a), \quad f(a)=g(a).$$
Since $f'(a)=g'(a)$ and $0Hence the coordinates of the tangent point are $\left(\frac{\pi}{\mathbf{O}},-\mathbf{Q}\right)$, and the equation of the common tangent line is
$$y=-\mathbf{R}\sqrt{\mathbf{S}}\left(x-\frac{\pi}{\mathbf{T}}\right)-\mathbf{U}.$$
(2) We are to find the area $S$ of the region bounded by these two curves over the range $-\frac{\pi}{2}\leqq x\leqq\frac{\pi}{2}$.
Since both of these curves are symmetric with respect to the $y$-axis, by putting $b=\mathbf{V}$ and $c=\frac{\pi}{\mathbf{O}}$ we have
$$S=\mathbf{W}\int_{b}^{c}(2\cos 2x-4\cos x-k)\,dx$$
By calculating this, we obtain
$$S=\mathbf{X}\cdot\pi-\mathbf{Y}\cdot\mathbf{Z}.$$

Let $f ( x ) = x ^ { 4 } + 2 x ^ { 3 } - 12 x ^ { 2 } + 4$. We are to find the values of $p$ such that we can draw three tangents to the curve $y = f ( x )$ from the point $\mathrm { P } ( 0 , p )$ on the $y$-axis.
(i) The equation of the tangent to the curve $y = f ( x )$ at the point $( t , f ( t ) )$ is
$$y = \left( \mathbf { A } t ^ { 3 } + \mathbf { B } t ^ { 2 } - \mathbf { C D } t \right) x - \mathbf { E } t ^ { 4 } - \mathbf { F } t ^ { 3 } + \mathbf { G H } t ^ { 2 } + \mathbf { I }$$
The condition under which this straight line passes through the point $\mathrm { P } ( 0 , p )$ is that
$$p = - \mathbf { J } t ^ { 4 } - \mathbf { K } t ^ { 3 } + \mathbf { L M } t ^ { 2 } + \mathbf { N }$$
(ii) For $\mathbf { O }$ and $\mathbf { S }$ in the following statements, choose either (0) or (1) and for the other blanks, enter the correct number. (0) local minimum
(1) local maximum
When the right side of (1) is set to $g ( t )$, the function $g ( t )$ takes a $\mathbf{O}$ at $t = \mathbf { P Q }$ and $t = \mathbf { R }$. On the other hand, $g ( t )$ takes a $\mathbf { S }$ at $t = \mathbf { T }$.
Hence the values of $p$ such that we can draw three tangents to the curve $y = f ( x )$ from the point $\mathrm { P } ( 0 , p )$ are
$$p = \mathbf { U } \text { and } p = \mathbf { V } ,$$
where $\mathbf { U } < \mathbf { V }$.

Let $C$ be the parabola $y = x ^ { 2 } + 1$ with origin O, and let P be the point $(a, 2a)$.
(1) Find the equation of the line passing through point P and tangent to parabola $C$.
The equation of the tangent line at point $(t, t ^ { 2 } + 1)$ on $C$ is
$$y = \square t x - t ^ { 2 } + 1$$
If this line passes through P, then $t$ satisfies the equation
$$t ^ { 2 } - \square a t + \text { エ } a - \text { オ } = 0$$
so $t = \square a -$ カ y. Therefore
When $a \neq$ ケ, there are 2 tangent lines to $C$ passing through P, and their equations are
$$y = ( \square a - \square ) x - \text { シ } a ^ { 2 } + \text { ス } a$$
a ⋯⋯⋯(1)
and
$$y = \text { せ } x$$
(2) Let $\ell$ be the line represented by equation (1) in (1). If the intersection of $\ell$ and the $y$-axis is $\mathrm { R } ( 0 , r )$, then $r = -$ シ $a ^ { 2 } +$ ス $a$. For $r > 0$,
ソ $< a <$ タ
タ and in this case, the area $S$ of triangle OPR is
$$S = \text { チ } \left( a ^ { \text {ツ } } - a \text { 园 } \right)$$
When ソ $< a <$ タ , examining the increase and decrease of $S$ , we find that $S$ attains its maximum value at $a = \frac { \text { ト } } { \text { ナ } }$
(3) When □ソ $< a <$ □タ, the area $T$ of the region enclosed by the parabola $C$, the line $\ell$ in (2), and the two lines $x = 0, x = a$ is
$$T = \frac { 1 } { \square } \text { ハ } a ^ { 3 } - \square a ^ { 2 } + \square$$
ト
In the range $\frac { 1 } { \text { ナ } } \leqq a <$ タ, $T$ is □ヘ.
In the range, $T$ is □ヘ. Choose the correct answer for □ヘ from the following options (0) through (5).
(0) decreasing (1) attains a local minimum but not a local maximum (2) increasing (3) attains a local maximum but not a local minimum (4) constant (5) attains both a local minimum and a local maximum

Given the functions $f ( x ) = x ^ { 3 } + 3 x ^ { 2 } - 1$ and $g ( x ) = 6 x$, it is requested:\ a) (0.5 points) Justify, using the appropriate theorem, that there exists some point in the interval [1,10] where both functions take the same value.\ b) (1 point) Calculate the equation of the tangent line to the curve $y = f ( x )$ with minimum slope.\ c) (1 point) Calculate $\int _ { 1 } ^ { 2 } \frac { f ( x ) } { g ( x ) } d x$

In the coordinate plane, a ``Bézier curve'' determined by four points $A$, $B$, $C$, $D$ refers to a polynomial function of degree at most 3 whose graph passes through points $A$ and $D$, and whose tangent line at point $A$ passes through point $B$, and whose tangent line at point $D$ passes through point $C$. Let $y = f(x)$ be the ``Bézier curve'' determined by the four points $A(0, 0)$, $B(1, 4)$, $C(3, 2)$, $D(4, 0)$. Answer the following questions.
(1) Let the equation of the tangent line to the graph of $y = f(x)$ at point $D$ be $y = ax + b$, where $a$ and $b$ are real numbers. Find the values of $a$ and $b$. (2 points)
(2) Prove that the polynomial $f(x)$ is divisible by $x^{2} - 4x$. (2 points)
(3) Find $f(x)$. (4 points)
(4) Find the value of the definite integral $\int_{2}^{6} |8f(x)|\, dx$. (4 points)

On the coordinate plane, let $\Gamma$ be the graph of the cubic function $f(x) = x^{3} - 9x^{2} + 15x - 4$. Show that $P(1, 3)$ is a point on $\Gamma$, and find the equation of the tangent line $L$ to $\Gamma$ at point $P$.