IIT-JEE 2003 Screening | Application of Derivatives Question 47 | Mathematics

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 2002 Screening

MCQ (Single Correct Answer)

-0.5

The point(s) in the curve $${y^3} + 3{x^2} = 12y$$ where the tangent is vertical, is (are)

$$\left( { \pm {4 \over {\sqrt 3 }}, - 2} \right)$$

$$\left( { \pm \sqrt {{{11} \over 3}} ,1} \right)$$

$$(0,0)$$

$$\left( { \pm {4 \over {\sqrt 3 }}, 2} \right)$$

IIT-JEE 2002 Screening

MCQ (Single Correct Answer)

-0.5

The length of a longest interval in which the function $$3\,\sin x - 4{\sin ^3}x$$ is increasing, is

$${\pi \over 3}$$

$${\pi \over 2}$$

$${3\pi \over 2}$$

$$\pi $$

IIT-JEE 2001 Screening

MCQ (Single Correct Answer)

-0.5

Let $$f\left( x \right) = \left( {1 + {b^2}} \right){x^2} + 2bx + 1$$ and let $$m(b)$$ be the minimum value of $$f(x)$$. As $$b$$ varies, the range of $$m(b)$$ is

$$\left[ {0,1} \right]$$

$$\left( {0,\,1/2} \right]$$

$$\left[ {1/2,\,1} \right]$$

$$\left( {0,\,1} \right]$$

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