1
IIT-JEE 2008 Paper 1 Offline
+3
-1

Consider the functions defined implicitly by the equation $$y^3-3y+x=0$$ on various intervals in the real line. If $$x\in(-\infty,-2)\cup(2,\infty)$$, the equation implicitly defines a unique real valued differentiable function $$y=f(x)$$. If $$x\in(-2,2)$$, the equation implicitly defines a unique real valued differentiable function $$y=g(x)$$ satisfying $$g(0)=0$$

$$\int\limits_{ - 1}^1 {g'\left( x \right)dx = }$$

A
$$2g(-1)$$
B
$$0$$
C
$$-2g(1)$$
D
$$2g(1)$$
2
IIT-JEE 2006
+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
$${\pi \over {4\sqrt 2 }}$$
3
IIT-JEE 2006
+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
$$1$$
4
IIT-JEE 2006
+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
$$0$$
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