AIEEE 2003 | Definite Integration Question 266 | Mathematics

Let $$f(x)$$ be a function satisfying $$f'(x)=f(x)$$ with $$f(0)=1$$ and $$g(x)$$ be a function that satisfies $$f\left( x \right) + g\left( x \right) = {x^2}$$. Then the value of the integral $$\int\limits_0^1 {f\left( x \right)g\left( x \right)dx,} $$ is

$$e + {{{e^2}} \over 2} + {5 \over 2}$$

$$e - {{{e^2}} \over 2} - {5 \over 2}$$

$$e + {{{e^2}} \over 2} - {3 \over 2}$$

$$e - {{{e^2}} \over 2} - {3 \over 2}$$

AIEEE 2003

MCQ (Single Correct Answer)

-1

If $$f\left( {a + b - x} \right) = f\left( x \right)$$ then $$\int\limits_a^b {xf\left( x \right)dx} $$ is equal to

$${{a + b} \over 2}\int\limits_a^b {f\left( {a + b + x} \right)dx} $$

$${{a + b} \over 2}\int\limits_a^b {f\left( {b - x} \right)dx} $$

$${{a + b} \over 2}\int\limits_a^b {f\left( x \right)dx} $$

$$\,{{b - a} \over 2}\int\limits_a^b {f\left( x \right)dx} $$

AIEEE 2003

MCQ (Single Correct Answer)

-1

The value of the integral $$I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} $$ is

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

$${1 \over {n + 1}}$$

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

$${1 \over {n + 1}} - {1 \over {n + 2}}$$

AIEEE 2003

MCQ (Single Correct Answer)

-1

Out of Syllabus

$$\mathop {\lim }\limits_{n \to \infty } {{1 + {2^4} + {3^4} + .... + {n^4}} \over {{n^5}}}$$ - $$\mathop {\lim }\limits_{n \to \infty } {{1 + {2^3} + {3^3} + .... + {n^3}} \over {{n^5}}}$$

$${1 \over 5}$$

$${1 \over 30}$$

zero

$${1 \over 4}$$

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