1
AIEEE 2007
+4
-1
The function $$f:R/\left\{ 0 \right\} \to R$$ given by

$$f\left( x \right) = {1 \over x} - {2 \over {{e^{2x}} - 1}}$$

can be made continuous at $$x$$ = 0 by defining $$f$$(0) as
A
0
B
1
C
2
D
$$-1$$
2
AIEEE 2006
+4
-1
The set of points where $$f\left( x \right) = {x \over {1 + \left| x \right|}}$$ is differentiable is
A
$$\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)$$
B
$$\left( { - \infty ,1} \right) \cup \left( { - 1,\infty } \right)$$
C
$$\left( { - \infty ,\infty } \right)$$
D
$$\left( {0,\infty } \right)$$
3
AIEEE 2005
+4
-1
Suppose $$f(x)$$ is differentiable at x = 1 and

$$\mathop {\lim }\limits_{h \to 0} {1 \over h}f\left( {1 + h} \right) = 5$$, then $$f'\left( 1 \right)$$ equals
A
3
B
4
C
5
D
6
4
AIEEE 2005
+4
-1
Let $$\alpha$$ and $$\beta$$ be the distinct roots of $$a{x^2} + bx + c = 0$$, then

$$\mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \left( {a{x^2} + bx + c} \right)} \over {{{\left( {x - \alpha } \right)}^2}}}$$ is equal to
A
$${{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}$$
B
0
C
$$- {{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}$$
D
$${{{{\left( {\alpha - \beta } \right)}^2}} \over 2}$$
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