AIEEE 2003 | Definite Integration Question 294 | Mathematics

If $$f\left( y \right) = {e^y},$$ $$g\left( y \right) = y;y > 0$$ and

$$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)dy,} $$ then :

$$F\left( t \right) = t{e^{ - t}}$$

$$F\left( t \right) = 1t - t{e^{ - 1}}\left( {1 + t} \right)$$

$$F\left( t \right) = {e^t} - \left( {1 + t} \right)$$

$$F\left( t \right) = t{e^t}$$.

AIEEE 2002

MCQ (Single Correct Answer)

-1

Out of Syllabus

$$\mathop {\lim }\limits_{n \to \infty } {{{1^p} + {2^p} + {3^p} + ..... + {n^p}} \over {{n^{p + 1}}}}$$ is

$${1 \over {p + 1}}$$

$${1 \over {1 - p}}$$

$${1 \over p} - {1 \over {p - 1}}$$

$${1 \over {p + 2}}$$

AIEEE 2002

MCQ (Single Correct Answer)

-1

$$\int\limits_0^{10\pi } {\left| {\sin x} \right|dx} $$ is

$$20$$

$$8$$

$$10$$

$$18$$

AIEEE 2002

MCQ (Single Correct Answer)

-1

$${I_n} = \int\limits_0^{\pi /4} {{{\tan }^n}x\,dx} $$ then $$\,\mathop {\lim }\limits_{n \to \infty } \,n\left[ {{I_n} + {I_{n + 2}}} \right]$$ equals

$${1 \over 2}$$

$$1$$

$$\infty $$

zero

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