Let f: [0, 2] → R be a function which is continuous on [0, 2] and is differentiable on (0, 2)f (0) = 1 . Let F(x) = ∫ _ 0 ^ x ^ 2 f( t )dt for x ∈ [0, 2] . If F ^ ' (x) = f ^ ' (x) for all x ∈ (0, 2), then F(2) equals

Correct answer: B. e ^ 4 - 1 F(0) = 0 F ^ ' (x) = 2xf(x) = f ^ ' (x) f(x) = e ^ x ^ 2 + c f(x) = e ^ x ^ 2 (∵ f(0) = 1) F(x) = ∫ _ 0 ^ x ^ 2 e ^ x dx F(x) = e ^ x ^ 2 - 1(∵ F(0) = 0) ⇒ F(2) = e ^ 4 - 1

JEE Advanced 2014Paper 2

Exam: JEE Advanced
Year: 2014
Paper: Paper 2
Subject: Mathematics
Chapter: Differential Equations

JEE Advanced 2014 · Paper 2 · Mathematics · Differential Equations

Let f: [0, 2] → R be a function which is continuous on [0, 2] and...

Let

f: \left[0, 2\right] \to R be a function which is continuous on \left[0, 2\right]

and is differentiable on \left(0, 2\right)f

\left(0\right) = 1

. Let

F\left(x\right) = \mathop{∫}_{0}^{{x}^{2}}f\left(\sqrt{t}\right)dt

for

x \in \left[0, 2\right] .

{F}^{'}\left(x\right) = {f}^{'}\left(x\right)

for all

x \in \left(0, 2\right),

then

F\left(2\right)

equals

Options

A ${e}^{2} - 1$
B ${e}^{4} - 1$
C $e - 1$
D ${e}^{4}$

Correct answer

B. ${e}^{4} - 1$

Step-by-step solution

$F\left(0\right) = 0$

${F}^{'}\left(x\right) = 2xf\left(x\right) = {f}^{'}\left(x\right)$

$f\left(x\right) = {e}^{{x}^{2} + c}$

$f\left(x\right) = {e}^{{x}^{2}}\left(∵ f\left(0\right) = 1\right)$

$F\left(x\right) = \mathop{∫}_{0}^{{x}^{2}}{e}^{x}dx$

$F\left(x\right) = {e}^{{x}^{2}} - 1\left(∵ F\left(0\right) = 0\right)$

$⇒ F\left(2\right) = {e}^{4} - 1$

Practice Differential Equations on Quantrex Academy →

More from Differential Equations