Suppose f- f- and f- are continuous on -0- e- and that f-e - fe - f1 - 1 and -1efxx2 dx - 12- then of -1ef-xlnxdx equals

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 g(0)-f(0) equals

Q. If for a continuous function

f, f (0) = f (1) = 0, f^{'} (1) = 2

and

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

, then

g^{'} (0)

is equal to

Q. If f (x) = (x + 1)^{cot x} be continuous at x = 0, then f (0) is equal to
(a) 0
(b) 1/e
(c) e
(d) none of these

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Suppose f,f′ and f′′ are continuous on [0,e] and that f′(e)=f(e)=f(1)=1 and ∫e1f(x)x2dx=12, then of ∫e1f′′(x)ln(x)dx equals

The correct option is A 0We have given that f′(e)=f(e)=f(1)=1Now,⇒∫e1f(x)x2dx=12⇒∫e1f′′(x)logxdx=0Hence the option (A) is correct.

Suppose $f, f^{'}$ and $f^{''}$ are continuous on $[0, e]$ and that $f^{'} (e) = f (e) = f (1) = 1$ and

$\int_{1}^{e} \frac{f (x)}{x^{2}} d x = \frac{1}{2}$ , then of $\int_{1}^{e} f^{''} (x) l n (x) d x$ equals

The correct option is A $0$
We have given that
$f^{'} (e) = f (e) = f (1) = 1$

Now,

$\Rightarrow \int_{1}^{e} \frac{f (x)}{x^{2}} d x = \frac{1}{2}$

$\Rightarrow \int_{1}^{e} f^{''} (x) log x d x = 0$

Hence the option $(A)$ is correct.