1
AIEEE 2011
+4
-1
$$\mathop {\lim }\limits_{x \to 2} \left( {{{\sqrt {1 - \cos \left\{ {2(x - 2)} \right\}} } \over {x - 2}}} \right)$$
A
Equals $$\sqrt 2$$
B
Equals $$-\sqrt 2$$
C
Equals $${1 \over {\sqrt 2 }}$$
D
does not exist
2
AIEEE 2011
+4
-1
The value of $$p$$ and $$q$$ for which the function

$$f\left( x \right) = \left\{ {\matrix{ {{{\sin (p + 1)x + \sin x} \over x}} & {,x < 0} \cr q & {,x = 0} \cr {{{\sqrt {x + {x^2}} - \sqrt x } \over {{x^{3/2}}}}} & {,x > 0} \cr } } \right.$$

is continuous for all $$x$$ in R, are
A
$$p =$$ $${5 \over 2}$$, $$q =$$ $${1 \over 2}$$
B
$$p =$$ $$-{3 \over 2}$$, $$q =$$ $${1 \over 2}$$
C
$$p =$$ $${1 \over 2}$$, $$q =$$ $${3 \over 2}$$
D
$$p =$$ $${1 \over 2}$$, $$q =$$ $$-{3 \over 2}$$
3
AIEEE 2010
+4
-1
Let $$f:R \to R$$ be a positive increasing function with

$$\mathop {\lim }\limits_{x \to \infty } {{f(3x)} \over {f(x)}} = 1$$. Then $$\mathop {\lim }\limits_{x \to \infty } {{f(2x)} \over {f(x)}} =$$
A
$${2 \over 3}$$
B
$${3 \over 2}$$
C
3
D
1
4
AIEEE 2009
+4
-1
Let $$f\left( x \right) = x\left| x \right|$$ and $$g\left( x \right) = \sin x.$$
Statement-1: gof is differentiable at $$x=0$$ and its derivative is continuous at that point.
Statement-2: gof is twice differentiable at $$x=0$$.
A
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
B
Statement-1 is true, Statement-2 is false
C
Statement-1 is false, Statement-2 is true
D
Statement-1 is true, Statement-2 is true Statement-2 is a correct explanation for Statement-1
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