For the curve $y = x ^ { 3 } - a x + b$, the slope of the line perpendicular to the tangent line at the point $( 1,1 )$ is $- \frac { 1 } { 2 }$. For the two constants $a$ and $b$, find the value of $a + b$. [4 points]

The tangent line to the curve $y = x ^ { 3 } - 3 x ^ { 2 } + 2 x + 2$ at point $\mathrm { A } ( 0,2 )$ is perpendicular to a line passing through point A. What is the $x$-intercept of this line? [3 points]
(1) 4
(2) 6
(3) 8
(4) 10
(5) 12

15. The tangent line to the curve $y = e ^ { x }$ at the point $( 0,1 )$ is perpendicular to the tangent line to the curve $y = \frac { 1 } { x } ( x > 0 )$ at point P. The coordinates of P are $\_\_\_\_$

If $st = 1$, then the tangent at $P$ and the normal at $S$ to the parabola meet at a point whose ordinate is
(A) $\frac{(t^2+1)^2}{2t^3}$
(B) $\frac{a(t^2+1)^2}{2t^3}$
(C) $\frac{a(t^2+1)^2}{t^3}$
(D) $\frac{a(t^2+2)^2}{t^3}$

Consider $f(x) = \tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$, $x \in \left(0, \frac{\pi}{2}\right)$. A normal to $y = f(x)$ at $x = \frac{\pi}{6}$ also passes through the point:
(1) $(0, 0)$
(2) $\left(0, \frac{2\pi}{3}\right)$
(3) $\left(\frac{\pi}{6}, 0\right)$
(4) $\left(\frac{\pi}{4}, 0\right)$

Consider $f(x) = \tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$, $x \in \left(0, \frac{\pi}{2}\right)$. A normal to $y = f(x)$ at $x = \frac{\pi}{6}$ also passes through the point: (1) $(0, 0)$ (2) $\left(0, \frac{2\pi}{3}\right)$ (3) $\left(\frac{\pi}{6}, 0\right)$ (4) $\left(\frac{\pi}{4}, 0\right)$

Let C be a curve given by $y ( x ) = 1 + \sqrt { 4 x - 3 } , x > \frac { 3 } { 4 }$. If $P$ is a point on C , such that the tangent at $P$ has slope $\frac { 2 } { 3 }$, then a point through which the normal at $P$ passes, is :
(1) $( 1,7 )$
(2) $( 3 , - 4 )$
(3) $( 4 , - 3 )$
(4) $( 2,3 )$

The normal to the curve $y ( x - 2 ) ( x - 3 ) = x + 6$ at the point where the curve intersects the $y$-axis passes through the point:
(1) $\left( \frac { 1 } { 2 } , - \frac { 1 } { 3 } \right)$
(2) $\left( \frac { 1 } { 2 } , \frac { 1 } { 3 } \right)$
(3) $\left( - \frac { 1 } { 2 } , - \frac { 1 } { 2 } \right)$
(4) $\left( \frac { 1 } { 2 } , \frac { 1 } { 2 } \right)$

The normal to the curve $y(x - 2)(x - 3) = x + 6$ at the point where the curve intersects the $y$-axis passes through the point:
(1) $\left(-\dfrac{1}{2}, -\dfrac{1}{2}\right)$
(2) $\left(\dfrac{1}{2}, \dfrac{1}{2}\right)$
(3) $\left(\dfrac{1}{2}, -\dfrac{1}{3}\right)$
(4) $\left(\dfrac{1}{2}, \dfrac{1}{3}\right)$

If $\beta$ is one of the angles between the normals to the ellipse $x ^ { 2 } + 3 y ^ { 2 } = 9$ at the points $( 3 \cos \theta , \sqrt { 3 } \sin \theta )$ and $( - 3 \sin \theta , \sqrt { 3 } \cos \theta ) ; \theta \in \left( 0 , \frac { \pi } { 2 } \right) ;$ then $\frac { 2 \cot \beta } { \sin 2 \theta }$ is equal to :
(1) $\frac { 1 } { \sqrt { 3 } }$
(2) $\frac { \sqrt { 3 } } { 4 }$
(3) $\frac { 2 } { \sqrt { 3 } }$
(4) $\sqrt { 2 }$

If $\beta$ is one of the angles between the normals to the ellipse, $x ^ { 2 } + 3 y ^ { 2 } = 9$ at the points ( $3 \cos \theta , \sqrt { 3 } \sin \theta$ ) and $( - 3 \sin \theta , \sqrt { 3 } \cos \theta ) ; \in \left( 0 , \frac { \pi } { 2 } \right) ;$ then $\frac { 2 \cot \beta } { \sin 2 \theta }$ is equal to
(1) $\sqrt { 2 }$
(2) $\frac { 2 } { \sqrt { 3 } }$
(3) $\frac { 1 } { \sqrt { 3 } }$
(4) $\frac { \sqrt { 3 } } { 4 }$

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) $x - 3y + 22 = 0$

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) $2 y + x = 4$

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

Let the function $f ( x ) = 2 x ^ { 2 } - \log _ { e } x , x > 0$, be decreasing in $( 0 , a )$ and increasing in $( a , 4 )$. A tangent to the parabola $y ^ { 2 } = 4 a x$ at a point $P$ on it passes through the point $( 8 a , 8 a - 1 )$ but does not pass through the point $\left( - \frac { 1 } { a } , 0 \right)$. If the equation of the normal at $P$ is $\frac { x } { \alpha } + \frac { y } { \beta } = 1$, then $\alpha + \beta$ is equal to $\_\_\_\_$.

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) $( 12 , - 15 )$

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 $\_\_\_\_$