Suppose f x and g x are two continuous functions defined for 0 - x - 1 Given fx-01 ex-t- ft d t and gx-01 ex-t- gt d t-xThe value of f 1 equalsA- 0B- eC- 1 D- e-1

The correct option is A 0 [ f(x)= amp; e^x∫_0^1 e^t.f(t) dt; amp; A say ] f(x)= A e^x ... ...

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Standard XII

Mathematics

Definition of Function

Suppose f x a...

Question

Suppose f(x) and g(x) are two continuous functions defined for $0 \leq x \leq 1$ .
Given $f (x) = \int_{0}^{1} e^{x + t} . f (t) d t$ and $g (x) = \int_{0}^{1} e^{x + t} . g (t) d t + x$ .

The value of f(1) equals

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$e^{- 1}$

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Solution

The correct option is A
0

$⎛ ⎝ \begin{matrix} f (x) = & e^{x} \int_{0}^{1} e^{t} . f (t) d t      A say \end{matrix} ⎞ ⎠$

$f (x) = A e^{x}$ ...(1) $\Rightarrow f (t) = A e^{t}$

where, $A = \int_{0}^{1} e^{t} . f (t) d t \Rightarrow A = \int_{0}^{1} . A e^{t} d t; A = A \int_{0}^{1} e^{2 t} d t$

now $A [\int_{0}^{1} e^{2 t} d t - 1] = 0 \Rightarrow A = 0$ as $\int_{0}^{1} e^{2 t} d t \neq 0$

Hence $f (x) = 0 \Rightarrow f (1) = 0$

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Suppose f(x) and g(x) are two continuous functions defined for $0 \leq x \leq 1$ .
Given $f (x) = \int_{0}^{1} e^{x + t} . f (t) d t$ and $g (x) = \int_{0}^{1} e^{x + t} . g (t) d t + x$ .

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Suppose f(x) and g(x) are two continuous functions defined for 0≤x≤1. Given f(x)=∫10ex+t.f(t) dt and g(x)=∫10ex+t.g(t) dt+x. The value of f(1) equals

The correct option is A 0 ⎛⎝f(x)=ex∫10et.f(t) dtA say⎞⎠ f(x)= A ex ...(1) ⇒f(t)=Aet where, A=∫10et.f(t)dt⇒A=∫10.Aetdt; A=A∫10e2tdt now A[∫10e2tdt−1]=0⇒A=0 as ∫10e2tdt≠0 Hence f(x)=0⇒f(1)=0

Suppose f(x) and g(x) are two continuous functions defined for $0 \leq x \leq 1$ .
Given $f (x) = \int_{0}^{1} e^{x + t} . f (t) d t$ and $g (x) = \int_{0}^{1} e^{x + t} . g (t) d t + x$ .

The value of f(1) equals