1
IIT-JEE 2006
MCQ (Single Correct Answer)
+5
-1.25
Let the definite integral be defined by the formula
$$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 2}\left( {f\left( a \right) + f\left( b \right)} \right).}$$ For more accurate result for
$$c \in \left( {a,b} \right),$$ we can use $$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^c {f\left( x \right)dx + \int\limits_c^b {f\left( x \right)dx = F\left( c \right)} } }$$ so
that for $$c = {{a + b} \over 2},$$ we get $$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 4}\left( {f\left( a \right) + f\left( b \right) + 2f\left( c \right)} \right).}$$

If $$f''\left( x \right) < 0\,\forall x \in \left( {a,b} \right)$$ and $$c$$ is a point such that $$a < c < b,$$ and
$$\left( {c,f\left( c \right)} \right)$$ is the point lying on the curve for which $$F(c)$$ is
maximum, then $$f'(c)$$ is equal to

A
$${{f\left( b \right) - f\left( a \right)} \over {b - a}}$$
B
$${{2\left( {f\left( b \right)} \right) - f\left( a \right)} \over {b - a}}$$
C
$${{2f\left( b \right) - f\left( a \right)} \over {2b - a}}$$
D
$$0$$
2
IIT-JEE 2005 Screening
MCQ (Single Correct Answer)
+3
-0.75
$$\int\limits_{ - 2}^0 {\left\{ {{x^3} + 3{x^2} + 3x + 3 + \left( {x + 1} \right)\cos \left( {x + 1} \right)} \right\}\,\,dx}$$ is equal to
A
$$-4$$
B
$$0$$
C
$$4$$
D
$$6$$
3
IIT-JEE 2005 Screening
MCQ (Single Correct Answer)
+3
-0.75
The area bounded by the parabola $$y = {\left( {x + 1} \right)^2}$$ and
$$y = {\left( {x - 1} \right)^2}$$ and the line $$y=1/4$$ is
A
$$4$$ sq. units
B
$$1/6$$ sq. units
C
$$4/3$$ sq. units
D
$$1/3$$ sq. units
4
IIT-JEE 2004 Screening
MCQ (Single Correct Answer)
+3
-0.75
The value of the integral $$\int\limits_0^1 {\sqrt {{{1 - x} \over {1 + x}}} dx}$$ is
A
$${\pi \over 2} + 1$$
B
$${\pi \over 2} - 1$$
C
$$-1$$
D
$$1$$
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