AIEEE 2012 | Limits, Continuity and Differentiability Question 180 | Mathematics

AIEEE 2012

MCQ (Single Correct Answer)

-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)

Statement - 1 is false, statement - 2 is true

Statement - 1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1

Statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1

Statement - 1 is true, statement - 2 is false

AIEEE 2011

MCQ (Single Correct Answer)

-1

$$\mathop {\lim }\limits_{x \to 2} \left( {{{\sqrt {1 - \cos \left\{ {2(x - 2)} \right\}} } \over {x - 2}}} \right)$$

Equals $$\sqrt 2 $$

Equals $$-\sqrt 2 $$

Equals $${1 \over {\sqrt 2 }}$$

does not exist

AIEEE 2011

MCQ (Single Correct Answer)

-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

$$p =$$ $${5 \over 2}$$, $$q = $$ $${1 \over 2}$$

$$p =$$ $$-{3 \over 2}$$, $$q = $$ $${1 \over 2}$$

$$p =$$ $${1 \over 2}$$, $$q = $$ $${3 \over 2}$$

$$p =$$ $${1 \over 2}$$, $$q = $$ $$-{3 \over 2}$$

AIEEE 2010

MCQ (Single Correct Answer)

-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 $$\mathop {\lim }\limits_{x \to \infty } {{f(2x)} \over {f(x)}} = $$

$${2 \over 3}$$

$${3 \over 2}$$

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