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1

### IIT-JEE 2008

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

If $$f\left( { - 10\sqrt 2 } \right) = 2\sqrt 2 ,$$ then $$f''\left( { - 10\sqrt 2 } \right) =$$

A
$${{4\sqrt 2 } \over {{7^3}{3^2}}}$$
B
$$-{{4\sqrt 2 } \over {{7^3}{3^2}}}$$
C
$${{4\sqrt 2 } \over {{7^3}3}}$$
D
$$-{{4\sqrt 2 } \over {{7^3}3}}$$
2

### IIT-JEE 2008

The area of the region between the curves $$y = \sqrt {{{1 + \sin x} \over {\cos x}}}$$
and $$y = \sqrt {{{1 - \sin x} \over {\cos x}}}$$ bounded by the lines $$x=0$$ and $$x = {\pi \over 4}$$ is
A
$$\int\limits_0^{\sqrt 2 - 1} {{t \over {\left( {1 + {t^2}} \right)\sqrt {1 - {t^2}} }}dt}$$
B
$$\int\limits_0^{\sqrt 2 - 1} {{4t \over {\left( {1 + {t^2}} \right)\sqrt {1 - {t^2}} }}dt}$$
C
$$\int\limits_0^{\sqrt 2 + 1} {{4t \over {\left( {1 + {t^2}} \right)\sqrt {1 - {t^2}} }}dt}$$
D
$$\int\limits_0^{\sqrt 2 + 1} {{t \over {\left( {1 + {t^2}} \right)\sqrt {1 - {t^2}} }}dt}$$
3

### IIT-JEE 2006

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 2006

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
$$1$$

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