1
IIT-JEE 2011 Paper 1 Offline
+4
-1
Let the straight line $$x=b$$ divide the area enclosed by
$$y = {\left( {1 - x} \right)^2},y = 0,$$ and $$x=0$$ into two parts $${R_1}\left( {0 \le x \le b} \right)$$ and
$${R_2}\left( {b \le x \le 1} \right)$$ such that $${R_1} - {R_2} = {1 \over 4}.$$ Then $$b$$ equals
A
$${3 \over 4}$$
B
$${ 1\over 2}$$
C
$${1 \over 3}$$
D
$${1 \over 4}$$
2
IIT-JEE 2011 Paper 2 Offline
+4
-1
Let f $$:$$$$\left[ { - 1,2} \right] \to \left[ {0,\infty } \right]$$ be a continuous function such that
$$f\left( x \right) = f\left( {1 - x} \right)$$ for all $$x \in \left[ { - 1,2} \right]$$

Let $${R_1} = \int\limits_{ - 1}^2 {xf\left( x \right)dx,}$$ and $${R_2}$$ be the area of the region bounded by $$y=f(x),$$ $$x=-1,$$ $$x=2,$$ and the $$x$$-axis. Then

A
$${R_1} = 2{R_2}$$
B
$${R_1} = 3{R_2}$$
C
$${2R_1} = {R_2}$$
D
$${3R_1} = {R_2}$$
3
IIT-JEE 2010 Paper 1 Offline
+4
-1
The value of $$\mathop {\lim }\limits_{x \to 0} {1 \over {{x^3}}}\int\limits_0^x {{{t\ln \left( {1 + t} \right)} \over {{t^4} + 4}}} dt$$ is
A
$$0$$
B
$${1 \over 12}$$
C
$${1 \over 24}$$
D
$${1 \over 64}$$
4
IIT-JEE 2010 Paper 1 Offline
+4
-1
The value of $$\int\limits_0^1 {{{{x^4}{{\left( {1 - x} \right)}^4}} \over {1 + {x^2}}}dx}$$ is (are)
A
$${{22} \over 7} - \pi$$
B
$${2 \over {105}}$$
C
$$0$$
D
$${{71} \over {15}} - {{3\pi } \over 2}$$
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