Let f - -0- 2- - R be a function which is continuous on -0- 2- and is differentiable on 0- 2 with f0-1Let Fx - -0x2 f-t dt for x - -0- 2- If F-x - f-x for all x - 0- 2- then F2 equals-

The correct option is B e^4 1 F(0)=0 F' (x) = 2x #160;f (x) = f'(x) f(x) = e^x^2 + c f(x) = e^x^2 #160;(∵ f(0) =1) F(x) = ∫ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Variable Separable Method

Let f : [0,...

Question

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:

e^{2} - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e^{4} - 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

e - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e^{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $e^{4} - 1$
$F (0) = 0$
$F^{'} (x) = 2 x f (x) = f^{'} (x)$
$f (x) = e^{x^{2} + c}$
$f (x) = e^{x^{2}} (∵ f (0) = 1)$
$F (x) = \int_{0}^{x^{2}} e^{x} d x$
$F (x) = e^{x^{2}} - 1 (∵ F (0) = 0)$
$\Rightarrow F (2) = e^{4} - 1$

Suggest Corrections

Similar questions

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

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. Let

f (x)

be a differentiable function defined on

[0, 2]

such that

f^{'} (x) = f^{'} (2 - x)

for all

x \in (0, 2),

f (0) = 1

and

f (2) = e^{2} .

Then the value of

2 \int 0 f (x) d x

Q. Let

f (x)

be a differentiable function defined on

[0, 2]

such that

f^{'} (x) = f^{'} (2 - x)

for all

x \in (0, 2),

f (0) = 1

and

f (2) = e^{2} .

Then the value of

2 \int 0 f (x) d x

Q. Let

f (x)

satisfy the requirements of Lagrange's mean value theorem in

[0, 2]

. If

f (0) = 0

and

| f^{'} (x) | \leq \frac{1}{2}

for all

x \in [0, 2]

, then

Join BYJU'S Learning Program

Let f:[0,2]→ R be a function which is continuous on [0, 2] and is differentiable on (0, 2) with f(0)=1Let F(x)=∫x20f(√t)dt for x∈[0,2]. If F′(x)=f′(x) for all x ∈(0,2), then F(2) equals:

The correct option is B e4−1F(0)=0F′(x)=2xf(x)=f′(x)f(x)=ex2+cf(x)=ex2(∵f(0)=1)F(x)=∫x20exdxF(x)=ex2−1(∵F(0)=0)⇒F(2)=e4−1

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:

The correct option is B $e^{4} - 1$
$F (0) = 0$
$F^{'} (x) = 2 x f (x) = f^{'} (x)$
$f (x) = e^{x^{2} + c}$
$f (x) = e^{x^{2}} (∵ f (0) = 1)$
$F (x) = \int_{0}^{x^{2}} e^{x} d x$
$F (x) = e^{x^{2}} - 1 (∵ F (0) = 0)$
$\Rightarrow F (2) = e^{4} - 1$