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 $$\int\limits_a^b {f\left( x \right)dx = {{b - a} \over 4}\left( {f\left( a \right) + f\left( b \right) + 2f\left( c \right)} \right).} $$

$$\int\limits_0^{\pi /2} {\sin x\,dx = } $$

A
$${\pi \over 8}\left( {1 + \sqrt 2 } \right)$$
B
$${\pi \over 4}\left( {1 + \sqrt 2 } \right)$$
C
$${\pi \over {8\sqrt 2 }}$$
D
$${\pi \over {4\sqrt 2 }}$$
2
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 $$\mathop {\lim }\limits_{x \to a} {{\int\limits_a^x {f\left( x \right)dx - \left( {{{x - a} \over 2}} \right)\left( {f\left( x \right) + f\left( a \right)} \right)} } \over {{{\left( {x - a} \right)}^3}}} = 0,\,\,$$ then $$f(x)$$ is
of maximum degree

A
$$4$$
B
$$3$$
C
$$2$$
D
$$1$$
3
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
$$0$$
4
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
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