The function $f$ is defined by $f ( x ) = \sqrt { 25 - x ^ { 2 } }$ for $- 5 \leq x \leq 5$.
(a) Find $f ^ { \prime } ( x )$.
(b) Write an equation for the line tangent to the graph of $f$ at $x = - 3$.
(c) Let $g$ be the function defined by $g ( x ) = \left\{ \begin{array} { l } f ( x ) \text { for } - 5 \leq x \leq - 3 \\ x + 7 \text { for } - 3 < x \leq 5 . \end{array} \right.$
Is $g$ continuous at $x = - 3$ ? Use the definition of continuity to explain your answer.
(d) Find the value of $\int _ { 0 } ^ { 5 } x \sqrt { 25 - x ^ { 2 } } d x$.

Two functions $f(x)$ and $g(x)$ are defined and differentiable on the set of all positive real numbers. $g(x)$ is the inverse function of $f(x)$, and $g'(x)$ is continuous on the set of all positive real numbers. For all positive numbers $a$, $$\int_1^a \frac{1}{g'(f(x))f(x)}\,dx = 2\ln a + \ln(a+1) - \ln 2$$ and $f(1) = 8$. Find the value of $f(2)$. [3 points]
(1) 36
(2) 40
(3) 44
(4) 48
(5) 52

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$. If $n \in \mathbf{N}$, we denote by $D_n$ the improper integral $\int_0^{\pi/2} (\ln(\sin(t)))^n \mathrm{~d}t$.
Calculate $f'(0)$ and $f'(1)$.

If
$$\alpha = \int _ { 0 } ^ { 1 } \left( e ^ { 9 x + 3 \tan ^ { - 1 } x } \right) \left( \frac { 12 + 9 x ^ { 2 } } { 1 + x ^ { 2 } } \right) d x$$
where $\tan ^ { - 1 } x$ takes only principal values, then the value of $\left( \log _ { e } | 1 + \alpha | - \frac { 3 \pi } { 4 } \right)$ is

Consider the equation
$$\int _ { 1 } ^ { e } \frac { \left( \log _ { \mathrm { e } } x \right) ^ { 1 / 2 } } { x \left( a - \left( \log _ { \mathrm { e } } x \right) ^ { 3 / 2 } \right) ^ { 2 } } d x = 1 , \quad a \in ( - \infty , 0 ) \cup ( 1 , \infty )$$
Which of the following statements is/are TRUE ?
(A) No $a$ satisfies the above equation
(B) An integer $a$ satisfies the above equation
(C) An irrational number $a$ satisfies the above equation
(D) More than one $a$ satisfy the above equation

Let, the function $F$ be defined as $F ( x ) = \int _ { 1 } ^ { x } \frac { e ^ { t } } { t } d t , x > 0$, then the value of the integral $\int _ { 1 } ^ { x } \frac { e ^ { t } } { t + a } d t$, where $a > 0$, is
(1) $e ^ { a } [ F ( x ) - F ( 1 + a ) ]$
(2) $e ^ { - a } [ F ( x + a ) - F ( a ) ]$
(3) $e ^ { a } [ F ( x + a ) - F ( 1 + a ) ]$
(4) $e ^ { - a } [ F ( x + a ) - F ( 1 + a ) ]$

The integral $\int \frac{dx}{x^2(x^4+1)^{3/4}}$ equals:
(1) $-\left(1 + \frac{1}{x^4}\right)^{1/4} + C$
(2) $-\left(\frac{x^4+1}{x^4}\right)^{1/4} + C$
(3) $\left(\frac{x^4+1}{x^4}\right)^{1/4} + C$
(4) $\left(1 + \frac{1}{x^4}\right)^{1/4} + C$

The integral $\int \frac{2x^{12} + 5x^9}{(x^5 + x^3 + 1)^3} dx$ is equal to:
(1) $\frac{-x^{10}}{2(x^5 + x^3 + 1)^2} + C$
(2) $\frac{x^{10}}{2(x^5 + x^3 + 1)^2} + C$
(3) $\frac{-x^5}{(x^5 + x^3 + 1)^2} + C$
(4) $\frac{x^5}{2(x^5 + x^3 + 1)^2} + C$

If $m$ is a non-zero number and $\int \frac{x^{5m-1} + 2x^{4m-1}}{(x^{2m} + x^m + 1)^3} dx = f(x) + C$, then $f(x)$ is:
(1) $\frac{x^{5m}}{2m(x^{2m} + x^m + 1)^2}$
(2) $\frac{x^{4m}}{2m(x^{2m} + x^m + 1)^2}$
(3) $\frac{(2x^{2m} + x^m)}{(x^{2m} + x^m + 1)^2}$
(4) $\frac{(x^{5m} + x^{4m})}{2m(x^{2m} + x^m + 1)^2}$

The integral $\int \frac{2x^{12} + 5x^9}{(x^5 + x^3 + 1)^3} dx$ is equal to: (1) $\frac{-x^{10}}{2(x^5+x^3+1)^2} + C$ (2) $\frac{x^{10}}{2(x^5+x^3+1)^2} + C$ (3) $\frac{-x^5}{(x^5+x^3+1)^2} + C$ (4) $\frac{x^5}{2(x^5+x^3+1)^2} + C$

If $f(x) = \int \frac{\left(5x^8 + 7x^6\right)}{\left(x^2 + 1 + 2x^7\right)^2}\,dx,\,(x \geq 0)$, and $f(0) = 0$, then the value of $f(1)$ is
(1) $\frac{-1}{4}$
(2) $\frac{1}{2}$
(3) $\frac{1}{4}$
(4) $-\frac{1}{2}$

The integral $\int \frac { d x } { ( x + 4 ) ^ { \frac { 8 } { 7 } } ( x - 3 ) ^ { \frac { 6 } { 7 } } }$ is equal to: (where $C$ is a constant of integration)
(1) $\left( \frac { x - 3 } { x + 4 } \right) ^ { \frac { 1 } { 7 } } + C$
(2) $\left( \frac { x - 3 } { x + 4 } \right) ^ { \frac { - 1 } { 7 } } + C$
(3) $\frac { 1 } { 2 } \left( \frac { x - 3 } { x + 4 } \right) ^ { \frac { 3 } { 7 } } + C$
(4) $- \frac { 1 } { 13 } \left( \frac { x - 3 } { x + 4 } \right) ^ { \frac { - 13 } { 7 } } + C$

If $\int \left( e ^ { 2 x } + 2 e ^ { x } - e ^ { - x } - 1 \right) e ^ { \left( e ^ { x } + e ^ { - x } \right) } d x = g ( x ) e ^ { \left( e ^ { x } + e ^ { - x } \right) } + c$, where $c$ is a constant of integration, then $g ( 0 )$ is
(1) $e$
(2) $e ^ { 2 }$
(3) 1
(4) 2

$\int \frac { 2 e ^ { x } + 3 e ^ { - x } } { 4 e ^ { x } + 7 e ^ { - x } } d x = \frac { 1 } { 14 } \left( u x + v \log _ { e } \left( 4 e ^ { x } + 7 e ^ { - x } \right) \right) + C$, where $C$ is a constant of integration, then $u + v$ is equal to

If $\int \frac { 1 } { x } \sqrt { \frac { 1 - x } { 1 + x } } d x = g ( x ) + c , g ( 1 ) = 0$, then $g \left( \frac { 1 } { 2 } \right)$ is equal to
(1) $\log _ { e } \left( \frac { \sqrt { 3 } - 1 } { \sqrt { 3 } + 1 } \right) + \frac { \pi } { 3 }$
(2) $\log _ { \mathrm { e } } \left( \frac { \sqrt { 3 } + 1 } { \sqrt { 3 } - 1 } \right) + \frac { \pi } { 3 }$
(3) $\log _ { \mathrm { e } } \left( \frac { \sqrt { 3 } + 1 } { \sqrt { 3 } - 1 } \right) - \frac { \pi } { 3 }$
(4) $\frac { 1 } { 3 } \log _ { e } \left( \frac { \sqrt { 3 } - 1 } { \sqrt { 3 } + 1 } \right) - \frac { \pi } { 6 }$

The integral $\int \frac{ \left(1 - \frac { 1 } { \sqrt { 3 } }\right) \cos x - \sin x } { 1 + \frac { 2 } { \sqrt { 3 } } \sin 2 x } d x$ is equal to
(1) $\frac { 1 } { 2 } \log _ { e } \left| \frac { \tan \left( \frac { x } { 2 } + \frac { \pi } { 12} \right) } { \tan\left( \frac { x } { 2 } + \frac { \pi } { 6 } \right) } \right| + C$
(2) $\log _ { e } \left| \frac { \tan \left( \frac { x } { 2 } + \frac { \pi } { 6 } \right) } { \tan\left( \frac { x } { 2 } + \frac { \pi } { 3 } \right) } \right| + C$
(3) $\frac { 1 } { 2 } \log _ { e } \left| \frac { \tan \left( \frac { x } { 2 } + \frac { \pi } { 6 } \right) } { \tan\left( \frac { x } { 2 } + \frac { \pi } { 3 } \right) } \right| + C$
(4) $\frac { 1 } { 2 } \log _ { e } \left| \frac { \tan \left( \frac { x } { 2 } - \frac { \pi } { 12 } \right) } { \tan \left( \frac { x } { 2 } - \frac { \pi } { 6 } \right) } \right| + C$

Let $f$ be a differentiable function in $\left( 0 , \frac { \pi } { 2 } \right)$. If $\int _ { \cos x } ^ { 1 } t ^ { 2 } f ( t ) d t = \sin ^ { 3 } x + \cos x$, then $\frac { 1 } { \sqrt { 3 } } f ^ { \prime } \left( \frac { 1 } { \sqrt { 3 } } \right)$ is equal to
(1) $6 - 9 \sqrt { 2 }$
(2) $6 + \frac { 9 } { \sqrt { 2 } }$
(3) $6 - \frac { 9 } { \sqrt { 2 } }$
(4) $3 + \sqrt { 2 }$

The slope of the tangent to a curve $C : y = y(x)$ at any point $(x,y)$ on it is $\frac { 2e ^ { 2x } - 6e ^ { -x } + 9 } { 2 + 9e ^ { -2x } }$. If $C$ passes through the points $\left(0, \frac{1}{2} + \frac{\pi}{2\sqrt{2}}\right)$ and $\left(\alpha, \frac{1}{2} e ^ { 2\alpha }\right)$ then $e ^ { \alpha }$ is equal to
(1) $\frac { 3 + \sqrt { 2 } } { 3 - \sqrt { 2 } }$
(2) $\frac { 3 } { \sqrt { 2 } } \cdot \frac { 3 + \sqrt { 2 } } { 3 - \sqrt { 2 } }$
(3) $\frac { 1 } { \sqrt { 2 } } \cdot \frac { \sqrt { 2 } + 1 } { \sqrt { 2 } - 1 }$
(4) $\frac { \sqrt { 2 } + 1 } { \sqrt { 2 } - 1 }$

Let $f ( x ) = \int \frac { 2 x } { \left( x ^ { 2 } + 1 \right) \left( x ^ { 2 } + 3 \right) } d x$. If $f ( 3 ) = \frac { 1 } { 2 } \left( \log _ { e } 5 - \log _ { e } 6 \right)$, then $f ( 4 )$ is equal to
(1) $\frac { 1 } { 2 } \left( \log _ { e } 17 - \log _ { e } 19 \right)$
(2) $\log _ { \mathrm { e } } 17 - \log _ { \mathrm { e } } 18$
(3) $\frac { 1 } { 2 } \left( \log _ { e } 19 - \log _ { e } 17 \right)$
(4) $\log _ { e } 19 - \log _ { e } 20$

Let $I ( x ) = \int \frac { x + 1 } { x \left( 1 + x e ^ { x } \right) ^ { 2 } } d x , x > 0$. If $\lim _ { x \rightarrow \infty } I ( x ) = 0$ then $I ( 1 )$ is equal to
(1) $\frac { e + 2 } { e + 1 } - \log _ { e } ( e + 1 )$
(2) $\frac { e + 1 } { e + 2 } + \log _ { e } ( e + 1 )$
(3) $\frac { e + 1 } { e + 2 } - \log _ { e } ( e + 1 )$
(4) $\frac { e + 2 } { e + 1 } + \log _ { e } ( e + 1 )$

If $I(x) = \int e^{\sin^2 x} \cos x (\sin 2x - \sin x)\, dx$ and $I(0) = 1$, then $I\!\left(\frac{\pi}{3}\right)$ is equal to
(1) $-\frac{1}{2}e^{\frac{3}{4}}$
(2) $\frac{1}{2}e^{\frac{3}{4}}$
(3) $-e^{\frac{3}{4}}$
(4) $e^{\frac{3}{4}}$

Let $I ( x ) = \int \sqrt { \frac { x + 7 } { x } } d x$ and $I ( 9 ) = 12 + 7 \log _ { e } 7$. If $I ( 1 ) = \alpha + 7 \log _ { e } ( 1 + 2 \sqrt { 2 } )$, then $\alpha ^ { 4 }$ is equal to $\_\_\_\_$ .

If $\int \frac { \sin ^ { \frac { 3 } { 2 } } x + \cos ^ { \frac { 3 } { 2 } } x } { \sqrt { \sin ^ { 3 } x \cos ^ { 3 } x \sin ( x - \theta ) } } d x = A \sqrt { \cos \theta \tan x - \sin \theta } + B \sqrt { \cos \theta - \sin \theta \cot x } + C$, where $C$ is the integration constant, then $AB$ is equal to
(1) $4 \operatorname { cosec } ( 2 \theta )$
(2) $4 \sec \theta$
(3) $2 \sec \theta$
(4) $8 \operatorname { cosec } ( 2 \theta )$

For $x \in \left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$, if $y ( x ) = \int \frac { \operatorname { cosec } x + \sin x } { \operatorname { cosec } x \sec x + \tan x \sin ^ { 2 } x } d x$ and $\lim _ { x \rightarrow \left( \frac { \pi } { 2 } \right) ^ { - } } y ( x ) = 0$ then $y \left( \frac { \pi } { 4 } \right)$ is equal to
(1) $\tan ^ { - 1 } \left( \frac { 1 } { \sqrt { 2 } } \right)$
(2) $\frac { 1 } { 2 } \tan ^ { - 1 } \left( \frac { 1 } { \sqrt { 2 } } \right)$
(3) $- \frac { 1 } { \sqrt { 2 } } \tan ^ { - 1 } \left( \frac { 1 } { \sqrt { 2 } } \right)$
(4) $\frac { 1 } { \sqrt { 2 } } \tan ^ { - 1 } \left( - \frac { 1 } { 2 } \right)$

If $f ( x ) = \int \frac { 1 } { x ^ { 1 / 4 } \left( 1 + x ^ { 1 / 4 } \right) } \mathrm { d } x , f ( 0 ) = - 6$, then $f ( 1 )$ is equal to :
(1) $4 \left( \log _ { e } 2 - 2 \right)$
(2) $2 - \log _ { e ^ { 2 } } 2$
(3) $\log _ { e } 2 + 2$
(4) $4 \left( \log _ { e } 2 + 2 \right)$