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If $f (x) = \frac{x^{2} - 1}{x^{2} + 1}$ for every real number $x$ , then the minimum value of $f (x)$

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Important Questions on Application of Derivatives

HARD

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

A solid hemisphere is mounted on a solid cylinder, both having equal radii. If the whole solid is to have a fixed surface area and the maximum possible volume, then the ratio of the height of the cylinder to the common radius is

HARD

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

If the function

f

given by

f (x) = x^{3} - 3 (a - 2) x^{2} + 3 a x + 7

, for some

a \in R

is increasing in

(0, 1]

and decreasing in

[1, 5)

, then a root of the equation,

\frac{f (x) - 14}{{(x - 1)}^{2}} = 0, (x \neq 1)

is :

MEDIUM

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

x = - 1

and

x = 2

are extreme points of

f (x) = α \log |x| + β x^{2} + x

, then

MEDIUM

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

From the top of a

64

metres high tower, a stone is thrown upwards vertically with the velocity of

48 m / s .

The greatest height (in metres) attained by the stone, assuming the value of the gravitational acceleration

g = 32 m / s^{2}

, is:

HARD

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

f (x)

is a non-zero polynomial of degree four, having local extreme points at

x = - 1, 0, 1;

then the set

S = {x \in R : f (x) = f (0)}

contains exactly

MEDIUM

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is:

HARD

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

Let

k

and

K

be the minimum and the maximum values of the function

f (x) = \frac{{(1 + x)}^{0.6}}{1 + x^{0.6}} in [0, 1]

, respectively, then the ordered pair

(k, K)

is equal to:

HARD

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

Among all sectors of a fixed perimeter, choose the one with maximum area. Then the angle at the center of this sector (i.e., the angle between the bounding radii) is-

MEDIUM

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

The least value of

α \in R

for which,

4 α x^{2} + \frac{1}{x} \geq 1,

for all

x > 0,

HARD

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

The maximum area (in sq. units) of a rectangle having its base on the

x -

axis and its other two vertices on the parabola,

y = 12 - x^{2}

such that the rectangle lies inside the parabola, is :

HARD

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

The maximum value of

f (x) = \frac{\log x}{x} (x \neq 0, x \neq 1)

HARD

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

The point on the curve

x^{2} = 2 y

which is nearest to the point

(0, 5)

EASY

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

The maximum value of the function

f (x) = 2 x^{3} - 15 x^{2} + 36 x + 4

is attained at

HARD

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

The maximum area of a rectangle that can be inscribed in a circle of radius

2

units is

HARD

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

Let

f (x) = α x^{2} - 2 + \frac{1}{x}

where

α

is a real constant. The smallest

α

for which

f (x) \geq 0

for all

x > 0

is-

HARD

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

Let

f (x) = x^{2} + \frac{1}{x^{2}}

and

g (x) = x - \frac{1}{x}, x \in R - \{- 1, 0, 1\} .

h (x) = \frac{f (x)}{g (x)}

, then the local minimum value of

h (x)

is:

MEDIUM

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

If non-zero real numbers

b

and

c

are such that

m i n f (x) > m a x g (x)

, where

f (x) = x^{2} + 2 b x + 2 c^{2}

and

g (x) = - x^{2} - 2 c x + b^{2}, (x \in R)

; then

|\frac{c}{b}|

lies in the interval

HARD

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

The maximum value of

f (x) = \frac{x}{4 + x + x^{2}}

[- 1, 1]

MEDIUM

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

S_{1}

and

S_{2}

are respectively the sets of local minimum and local maximum points of the function,

f (x) = 9 x^{4} + 12 x^{3} - 36 x^{2} + 25, x \in R,

then

EASY

Mathematics>Differential Calculus>Application of Derivatives>Maxima and Minima

Let

M

and

m

be respectively the absolute maximum and the absolute minimum values of the function,

f (x) = 2 x^{3} - 9 x^{2} + 12 x + 5

in the interval

[0, 3]

. Then

M - m

is equal to

If fx=x2-1x2+1 for every real number x, then the minimum value of f(x)

Important Questions on Application of Derivatives

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If $f (x) = \frac{x^{2} - 1}{x^{2} + 1}$ for every real number $x$ , then the minimum value of $f (x)$