1
AIEEE 2004
+4
-1
The value of $$I = \int\limits_0^{\pi /2} {{{{{\left( {\sin x + \cos x} \right)}^2}} \over {\sqrt {1 + \sin 2x} }}dx}$$ is
A
$$3$$
B
$$1$$
C
$$2$$
D
$$0$$
2
AIEEE 2004
+4
-1
If $$\int\limits_0^\pi {xf\left( {\sin x} \right)dx = A\int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx,} }$$ then $$A$$ is
A
$$2\pi$$
B
$$\pi$$
C
$${\pi \over 4}$$
D
$$0$$
3
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
$$2$$
4
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
$$F\left( t \right) = t{e^t}$$.
EXAM MAP
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