For all twice differentiable functions f : R → R , with f 0 = f 1 = f ' 0 = 0 ,

f " x = 0 , for some x ∈ 0 , 1

f " x ≠ 0 at every point x ∈ 0 , 1

f " x = 0 , at every point x ∈ 0 , 1

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If $f (x) = x^{2} - 2 x + 4$ on $[1, 5]$ , then the value of a constant $c$ such that $\frac{f (5) - f (1)}{5 - 1} = f^{'} (c)$ , is

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Important Questions on Application of Derivatives

HARD

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

Let

f (x) = \sqrt{x}, 4 \leq x \leq 16

. If the point

c \in (4, 16)

is such that the tangent line to the graph of

f

x = c

is parallel to the chord joining

(16, 4)

and

(4, 2),

then the value of

c

MEDIUM

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

The value of

c

, in the Lagrange’s mean value theorem for the function

f (x) = x^{3} - 4 x^{2} + 8 x + 11,

when

x \in [0,1]

, is

MEDIUM

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

Rolle’s theorem is not applicable in which one of the following cases?

EASY

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

The value of

x

in the interval

[4, 9]

at which the function

f (x) = \sqrt{x}

satisfies the mean value theorem is

MEDIUM

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

f : R \to R

is a twice differentiable function such that

f^{''} (x) > 0

for all

x \in R

and

f (\frac{1}{2}) = \frac{1}{2}, f (1) = 1

then

HARD

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

For a polynomial

g (x)

with real coefficients, let

m_{g}

denote the number of distinct real roots of

g (x)

. Suppose

S

is the set of polynomials with real coefficients defined by

S = \{{(x^{2} - 1)}^{2} (a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3}) : a_{0}, a_{1}, a_{2}, a_{3} \in R\}

. For a polynomial

f

, let

f^{'}

and

f^{″}

denote its first and second order derivatives respectively. Then the minimum possible value of

(m_{f^{'}} + m_{f^{″}})

, where

f \in S

, is ____

EASY

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

For the function

f (x) = x + \frac{1}{x}, x \in [1, 3]

, the value of

c

for mean value theorem is

HARD

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

Let

f (x)

be continuous on

[0, 6]

and differentiable on

(0, 6)

Let

f (0) = 12

and

f (6) = - 4

. If

g (x) = \frac{f (x)}{x + 1}

, then for some Lagrange's constant

c \in (0, 6), g^{'} (c) =

HARD

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

Let the function

, f : [- 7, 0] \to R

be continuous on

[- 7, 0]

and differentiable on

(- 7, 0) .

f (- 7) = - 3

and

f^{'} (x) \leq 2

for all

x \in (- 7, 0),

then for all such functions

f, f (- 1) + f (0)

lies in the interval

MEDIUM

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

Let

f

be any function continuous on

[a, b]

and twice differentiable on

(a, b)

. If all

x \in (a, b), f^{'} (x) > 0

and

f^{"} (x) < 0

, then for any

c \in (a, b), \frac{f (c) - f (a)}{f (b) - f (c)}

MEDIUM

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

Rolle's theorem is not applicable for the function

f (x) = | x |

in the interval

[- 1, 1]

because

MEDIUM

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

f & g

are differentiable functions in

[0, 1]

satisfying

f (0) = 2 = g (1), g (0) = 0 & f (1) = 6,

then for some

c \in]0, 1[

MEDIUM

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

f

and

g

are differentiable function in

[0, 1]

satisfying

f (0) = 2 = g (1), g (0) = 0

and

f (1) = 6

, then for some

c \in (0, 1)

MEDIUM

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

For all twice differentiable functions

f : R \to R

, with

f (0) = f (1) = f' (0) = 0

MEDIUM

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

The number of admissible values of

C

obtained when the Lagrange's mean value theorem is applied for the function

f (x) = x

[2, 5]

HARD

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

f (x) = (x - 1) (x - 2) (x - 3)

for

x \in [0, 4],

then the value of

c \in (0, 4)

satisfying Lagrange's mean value theorem is

MEDIUM

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

f (x) = |x - 2|, x \in [0, 4]

, then the Rolle's theorem cannot be applied to the function because

EASY

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

c

is a point at which Rolle’s theorem holds for the function,

f (x) = \log_{e} (\frac{x^{2} + α}{7 x})

in the interval

[3, 4],

where

α \in R,

then

f^{"} (c)

is equal to

MEDIUM

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

If the Rolle's theorem holds for the function

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

in the interval

[- 1, 1]

for the point

c = \frac{1}{2}

, then the value of

2 a + b

is:

MEDIUM

Mathematics>Calculus>Application of Derivatives>Mean Value Theorem

If Rolle's theorem holds for the function

f (x) = 2 x^{3} + b x^{2} + c x, x \in [- 1, 1]

at the point

x = \frac{1}{2},

then

2 b + c

is equal to

If fx=x2-2x+4 on [1, 5], then the value of a constant c such that f5-f(1)5-1=f′(c), is

Important Questions on Application of Derivatives

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If $f (x) = x^{2} - 2 x + 4$ on $[1, 5]$ , then the value of a constant $c$ such that $\frac{f (5) - f (1)}{5 - 1} = f^{'} (c)$ , is