Let f -0-2- - R be continuous on -0-2- and twice differentiable in 0-2- If f 0-0- f 1-1 and f 2-1- thenA- f - c -1-3 for at least one c -0-2B- 2 f - c -2 c -3 for at least one c -0-1C- 2 f - c -2 c -3 for at least one c -0-2D- f - c -1 for at least one c -0-2

The correct options are A f'(c) = 13 for at least one c ∈ nbsp;(0, 2) B 2f'(c) + 2c = 3 for at least one c ∈ nbsp;(0, 1) C 2f'(c) + 2c = 3 for at least one ...

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Byju's Answer

Standard XII

Mathematics

LaGrange's Mean Value theorem

Let f :[0,2] ...

Question

Let $f : [0, 2] \to R$ be continuous on $[0, 2]$ and twice differentiable in $(0, 2) .$ If $f (0) = 0,$ $f (1) = 1$ and $f (2) = 1,$ then

f^{'} (c) = \frac{1}{3}

for at least one

c \in (0, 2)

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2 f^{'} (c) + 2 c = 3

for at least one

c \in (0, 1)

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2 f^{'} (c) + 2 c = 3

for at least one

c \in (0, 2)

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f^{''} (c) = - 1

for at least one

c \in (0, 2)

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Solution

The correct options are
A $f^{'} (c) = \frac{1}{3}$ for at least one $c \in (0, 2)$
B $2 f^{'} (c) + 2 c = 3$ for at least one $c \in (0, 1)$
C $2 f^{'} (c) + 2 c = 3$ for at least one $c \in (0, 2)$
D $f^{''} (c) = - 1$ for at least one $c \in (0, 2)$
$\to$ By LMVT in the interval $[0, 1],$
$\frac{f (1) - f (0)}{1} = f^{'} (c_{1}) = 1$ for some $c_{1} \in (0, 1)$

Similarly, $\frac{f (2) - f (1)}{1} = f^{'} (c_{2}) = 0$ for some $c_{2} \in (1, 2)$
Since $\frac{1}{3} \in (f^{'} (c_{2}), f^{'} (c_{1}))$ and $f^{'}$ is continuous,
$\Rightarrow f^{'} (c) = \frac{1}{3}$ for at least one $c \in (0, 2)$ by IVT.

$\to$ Consider a function, $H (x) = 2 f (x) + x^{2}$
$H (0) = 0, H (1) = 3, H (2) = 6$
By LMVT in $[0, 1],$
$H^{'} (c) = 2 f^{'} (c) + 2 c = \frac{H (1) - H (0)}{1} = 3$ for some $c \in (0, 1)$

$\to$ By LMVT in $[0, 2],$
$H^{'} (c) = 2 f^{'} (c) + 2 c = \frac{H (2) - H (0)}{2} = 3$ for some $c \in (0, 2)$

$\to$ Applying Rolle's theorem to $H^{'} (x)$ in $[c_{1}, c_{2}]$ where $0 \leq c_{1} < c_{2} \leq 2$
$H^{'} (c_{1}) = 2 f^{'} (c_{1}) + 2 c_{1} = \frac{H (1) - H (0)}{1} = 3$
$H^{'} (c_{2}) = 2 f^{'} (c_{2}) + 2 c_{2} = \frac{H (2) - H (1)}{2 - 1} = 3$
$H^{''} (c) = 0$ for some $c \in (0, 2)$
$\Rightarrow 2 f^{''} (c) + 2 = 0$
$\Rightarrow f^{''} (c) = - 1$

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Similar questions

Q. Assertion :Let

f : [0, 2] \to [0, 2]

be a continuous function.

f (x) = x

for at least one

0 \leq x \leq 2

Reason:

f (x) = - x

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Q. Let

f : [0, 2] \to R

a function which is continuous on

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and is differentiable on

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with

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. 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

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equals to

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

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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, f o r x ϵ [0, 2], I f F^{'} (x) = f^{'} (x), \forall x ϵ (0, 2),

then F(2) equals

Q. Let

f : [0, 2] \to R

be a twice differentiable function such that

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Let f:[0,2]→R be continuous on [0,2] and twice differentiable in (0,2). If f(0)=0, f(1)=1 and f(2)=1, then

Let $f : [0, 2] \to R$ be continuous on $[0, 2]$ and twice differentiable in $(0, 2) .$ If $f (0) = 0,$ $f (1) = 1$ and $f (2) = 1,$ then