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Standard XII

Mathematics

Theorems for Differentiability

If y=fx is ...

Question

If $y = f (x)$ is continuous on $[0, 6]$ . differentiable on $(0, 6)$ $f (0) = - 2$ and $f (6) = 16$ , then at some point between $x = 0$ and $x = 6$ , $f^{'} (x)$ must be equal to

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Solution

The correct option is B $3$
$f^{'} (x) = \frac{f (b) - f (a)}{b - a} = \frac{16 - (- 2)}{6 - 0} = 3$
(Using Lagrange's mean value theorem)

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Similar questions

Q. If y=f(x) is continuous on

[0, 6]

, differentiable on

(0, 6), f (0) = - 2

and

f (6) = 16

, then at some point between

x = 0

and

x = 6

, f'(x) must be equal to?

Q. Let

f

be a non-negative continuous and twice differentiable function defined on

R

such that

f (x) + f (x + \frac{3}{2}) = 6, \forall x \in R .

Then

Q. Let

f : [0, 2] \to

R be a function which is continuous on [0, 2] and is differentiable on (0, 2) with

f (0) = 1

Let

F (x) = \int_{0}^{x^{2}} f (\sqrt{t}) d t

for

x \in [0, 2]

. If

F^{'} (x) = f^{'} (x)

for all x

\in (0, 2)

, then F(2) equals:

Q. Let

f : [0, 2] \to R

a function which is continuous on

[0, 2]

and is differentiable on

(0, 2)

with

f (0) = 1

. Let

F (x) = x^{2} \int 0 f (\sqrt{t}) d t,

for

x \in [0, 2]

, if

F^{'} (x) = f^{'} (x), \forall x \in (0, 2),

then

F (2)

equals to

Q. If

f (x) = \frac{x^{2} - b x + 36}{x^{2} - 9 x + 18}

for

x \neq 6

is continuous at

x = 6

, then value of

f (6)

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If y=f(x) is continuous on [0,6]. differentiable on (0,6) f(0)=−2 and f(6)=16, then at some point between x=0 and x=6, f′(x) must be equal to

The correct option is B 3f′(x)=f(b)−f(a)b−a=16−(−2)6−0=3(Using Lagrange's mean value theorem)

If $y = f (x)$ is continuous on $[0, 6]$ . differentiable on $(0, 6)$ $f (0) = - 2$ and $f (6) = 16$ , then at some point between $x = 0$ and $x = 6$ , $f^{'} (x)$ must be equal to

The correct option is B $3$
$f^{'} (x) = \frac{f (b) - f (a)}{b - a} = \frac{16 - (- 2)}{6 - 0} = 3$
(Using Lagrange's mean value theorem)