Let $f : \left[ - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right] \rightarrow R$ be a differentiable function such that $f ( 0 ) = \frac { 1 } { 2 }$, If $\lim _ { x \rightarrow 0 } \frac { x \int _ { 0 } ^ { x } f ( t ) d t } { e ^ { x ^ { 2 } } - 1 } = \alpha$, then $8 \alpha ^ { 2 }$ is equal to :
(1) 16
(2) 2
(3) 1
(4) 4

Let for some function $\mathrm { y } = f ( x ) , \int _ { 0 } ^ { x } t f ( t ) d t = x ^ { 2 } f ( x ) , x > 0$ and $f ( 2 ) = 3$. Then $f ( 6 )$ is equal to
(1) 1
(2) 3
(3) 6
(4) 2

Let $f$ be a real valued continuous function defined on the positive real axis such that $g ( x ) = \int _ { 0 } ^ { x } \mathrm { t } f ( \mathrm { t } ) \mathrm { dt }$. If $\mathrm { g } \left( x ^ { 3 } \right) = x ^ { 6 } + x ^ { 7 }$, then value of $\sum _ { r = 1 } ^ { 15 } f \left( \mathrm { r } ^ { 3 } \right)$ is :
(1) 270
(2) 340
(3) 320
(4) 310

Let $f:(0,\infty) \rightarrow \mathbf{R}$ be a twice differentiable function. If for some $\mathrm{a} \neq 0$, $\int_0^1 f(\lambda x)\,\mathrm{d}\lambda = \mathrm{a}f(x)$, $f(1) = 1$ and $f(16) = \frac{1}{8}$, then $16 - f'\left(\frac{1}{16}\right)$ is equal to \_\_\_\_ .

If $6 \left( \int _ { 1 } ^ { \mathbf { x } } \mathbf { f } ( \mathbf { t } ) \mathbf { d t } \right) = 3 \left( \mathbf { x } \mathbf { f } ( \mathbf { x } ) + \mathbf { x } ^ { 3 } - 4 \right)$, then find the value of $\mathbf { f } ( 2 ) - \mathbf { f } ( 3 )$

$\int _ { 0 } ^ { 36 } \mathbf { f } \left( \frac { \mathbf { t x } } { 36 } \right) \mathbf { d t } = \mathbf { 4 \alpha f } ( \mathbf { x } )$
If the curve represented by $\mathrm { y } = \mathrm { f } ( \mathrm { x } )$ is a standard parabola passing through $( 2,1 )$ and $( - 4 , \beta )$ then find

Given the following statements about a function f

• $\mathrm { f } ^ { \prime \prime } ( x ) = a$ for all $x$
• $\mathrm { f } ( 0 ) = 1 , \mathrm { f } ( 1 ) = 2$
• $\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x = 1$

find the value of $a$.

Given that
$$\int _ { 0 } ^ { 1 } ( a x + b ) \mathrm { d } x = 1$$
and
$$\int _ { 0 } ^ { 1 } x ( a x + b ) \mathrm { d } x = 1$$
find the value of $a + b$.

Let $a$ and $b$ be real numbers. A function $f$ that is continuous on the set of real numbers is defined as
$$f ( x ) = \begin{cases} 6 - \frac { 3 x ^ { 2 } } { 2 } , & x < 2 \\ a x - b & x \geq 2 \end{cases}$$
$$\int _ { 0 } ^ { 4 } f ( x ) d x = \int _ { 2 } ^ { 6 } f ( x ) d x$$
Given that, what is the sum $a + b$?
A) 1
B) 2
C) 3
D) 4
E) 5