Let f(x) be a function satisfying $f^{^{'}} (x) = f (x)$ with f(0) = 1
and g(x) be the function satisfying $f (x) + g (x) = x^{2}$
The value of integral $\int_{0}^{1} f (x) g (x) d x$ is equal to [AIEEE 2003; DCE 2005]

Q. Let

f, g, g^{'}

be continuous and differentiable functions such that

f (x) = e^{x} g (x),

f (x + y) = f (x) + f (y),

g (0) = 0,

g^{'} (0) = 4.

Then

Q.
Let

f (x)

be a function satisfying

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

with

f (0) = 1

and

g (x)

be a function that satisfies

f (x) + g (x) = x^{2}

. Then, the value of the integral

\int_{0}^{1} f (x) g (x) d x

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Let f be the exponential function and g be the logarithmic function. Then, (f+g)(1) :

The correct option is B eWe have,f:R→R given by f(x)=ex and, g:R+→R given by g(x)=logex.Since Domain(f) ∩ Domain (g) = R∩R+=R+. Therefore,f+g:R+→R is given by(f+g)(x)=f(x)+g(x)=ex+logex for all x∈R+Therefore, (f+g)(1)=e1+loge1=e+0=e

Let $f$ be the exponential function and $g$ be the logarithmic function. Then, $(f + g) (1)$ :