1
AIEEE 2004
+4
-1
If $$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}},{I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}dx}$$
and $${I_2} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {g\left\{ {x\left( {1 - x} \right)} \right\}dx} ,$$ then the value of $${{{I_2}} \over {{I_1}}}$$ is
A
$$1$$
B
$$-3$$
C
$$-1$$
D
$$2$$
2
AIEEE 2003
+4
-1
If $$f\left( y \right) = {e^y},$$ $$g\left( y \right) = y;y > 0$$ and

$$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)dy,}$$ then :
A
$$F\left( t \right) = t{e^{ - t}}$$
B
$$F\left( t \right) = 1t - t{e^{ - 1}}\left( {1 + t} \right)$$
C
$$F\left( t \right) = {e^t} - \left( {1 + t} \right)$$
D
$$F\left( t \right) = t{e^t}$$.
3
AIEEE 2003
+4
-1
Let $$f(x)$$ be a function satisfying $$f'(x)=f(x)$$ with $$f(0)=1$$ and $$g(x)$$ be a function that satisfies $$f\left( x \right) + g\left( x \right) = {x^2}$$. Then the value of the integral $$\int\limits_0^1 {f\left( x \right)g\left( x \right)dx,}$$ is
A
$$e + {{{e^2}} \over 2} + {5 \over 2}$$
B
$$e - {{{e^2}} \over 2} - {5 \over 2}$$
C
$$e + {{{e^2}} \over 2} - {3 \over 2}$$
D
$$e - {{{e^2}} \over 2} - {3 \over 2}$$
4
AIEEE 2003
+4
-1
If $$f\left( {a + b - x} \right) = f\left( x \right)$$ then $$\int\limits_a^b {xf\left( x \right)dx}$$ is equal to
A
$${{a + b} \over 2}\int\limits_a^b {f\left( {a + b + x} \right)dx}$$
B
$${{a + b} \over 2}\int\limits_a^b {f\left( {b - x} \right)dx}$$
C
$${{a + b} \over 2}\int\limits_a^b {f\left( x \right)dx}$$
D
$$\,{{b - a} \over 2}\int\limits_a^b {f\left( x \right)dx}$$
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