1
JEE Main 2013 (Offline)
MCQ (Single Correct Answer)
+4
-1
$$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)} \over {x\tan 4x}}$$ is equal to
A
$$ - {1 \over 4}$$
B
$${1 \over 2}$$
C
1
D
2
2
AIEEE 2012
MCQ (Single Correct Answer)
+4
-1
Consider the function, $$f\left( x \right) = \left| {x - 2} \right| + \left| {x - 5} \right|,x \in R$$

Statement - 1 : $$f'\left( 4 \right) = 0$$

Statement - 2 : $$f$$ is continuous in [2, 5], differentiable in (2, 5) and $$f$$(2) = $$f$$(5)
A
Statement - 1 is false, statement - 2 is true
B
Statement - 1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1
C
Statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1
D
Statement - 1 is true, statement - 2 is false
3
AIEEE 2012
MCQ (Single Correct Answer)
+4
-1
If $$f:R \to R$$ is a function defined by

$$f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi $$,

where [x] denotes the greatest integer function, then $$f$$ is
A
continuous for every real $$x$$
B
discontinuous only at $$x=0$$
C
discontinuous only at non-zero integral values of $$x$$
D
continuous only at $$x=0$$
4
AIEEE 2011
MCQ (Single Correct Answer)
+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
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