AIEEE 2012 | Application of Derivatives Question 194 | Mathematics

A line is drawn through the point $$(1, 2)$$ to meet the coordinate axes at $$P$$ and $$Q$$ such that it forms a triangle $$OPQ,$$ where $$O$$ is the origin. If the area of the triangle $$OPQ$$ is least, then the slope of the line $$PQ$$ is :

$$-{1 \over 4}$$

$$-4$$

$$-2$$

$$-{1 \over 2}$$

AIEEE 2011

MCQ (Single Correct Answer)

-1

For $$x \in \left( {0,{{5\pi } \over 2}} \right),$$ define $$f\left( x \right) = \int\limits_0^x {\sqrt t \sin t\,dt.} $$ Then $$f$$ has

local minimum at $$\pi $$ and $$2\pi $$

local minimum at $$\pi $$ and local maximum at $$2\pi $$

local maximum at $$\pi $$ and local minimum at $$2\pi $$

local maximum at $$\pi $$ and $$2\pi $$

AIEEE 2011

MCQ (Single Correct Answer)

-1

Out of Syllabus

The shortest distance between line $$y-x=1$$ and curve $$x = {y^2}$$ is

$${{3\sqrt 2 } \over 8}$$

$${8 \over {3\sqrt 2 }}$$

$${4 \over {\sqrt 3 }}$$

$${{\sqrt 3 } \over 4}$$

AIEEE 2010

MCQ (Single Correct Answer)

-1

Let $$f:R \to R$$ be defined by $$$f\left( x \right) = \left\{ {\matrix{ {k - 2x,\,\,if} & {x \le - 1} \cr {2x + 3,\,\,if} & {x > - 1} \cr } } \right.$$$

If $$f$$has a local minimum at $$x=-1$$, then a possible value of $$k$$ is

$$0$$

$$ - {1 \over 2}$$

$$-1$$

$$1$$

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