IIT-JEE 2004 Screening | Application of Derivatives Question 44 | Mathematics

IIT-JEE 2004 Screening

MCQ (Single Correct Answer)

-0.5

If $$f\left( x \right) = {x^3} + b{x^2} + cx + d$$ and $$0 < {b^2} < c,$$ then in $$\left( { - \infty ,\infty } \right)$$

$$f\left( x \right)$$ is a strictly increasing function

$$f\left( x \right)$$ has a local maxima

$$f\left( x \right)$$ is a strictly decreasing function

$$f\left( x \right)$$ is bounded

IIT-JEE 2004 Screening

MCQ (Single Correct Answer)

-0.5

If $$f\left( x \right) = {x^a}\log x$$ and $$f\left( 0 \right) = 0,$$ then the value of $$\alpha $$ for which Rolle's theorem can be applied in $$\left[ {0,1} \right]$$ is

$$-2$$

$$-1$$

$$0$$

$$1/2$$

IIT-JEE 2003 Screening

MCQ (Single Correct Answer)

-0.5

In $$\left[ {0,1} \right]$$ Languages Mean Value theorem is NOT applicable to

$$f\left( x \right) = \left\{ {\matrix{ {{1 \over 2} - x} & {x < {1 \over 2}} \cr {{{\left( {{1 \over 2} - x} \right)}^2}} & {x \ge {1 \over 2}} \cr } } \right.$$

$$f\left( x \right) = \left\{ {\matrix{ {\sin x,} & {x \ne 0} \cr {1,} & {x = 0} \cr } } \right.$$

$$f\left( x \right) = x\left| x \right|$$

$$f\left( x \right) = \left| x \right|$$

IIT-JEE 2003 Screening

MCQ (Single Correct Answer)

-0.5

Tangent is drawn to ellipse
$${{{x^2}} \over {27}} + {y^2} = 1\,\,\,at\,\left( {3\sqrt 3 \cos \theta ,\sin \theta } \right)\left( {where\,\,\theta \in \left( {0,\pi /2} \right)} \right)$$.

Then the value of $$\theta $$ such that sum of intercepts on axes made by this tangent is minimum, is

$$\pi /3$$

$$\pi /6$$

$$\pi /8$$

$$\pi /4$$

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