Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 38

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Mathematics

IMPORTANT

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A function $f$ is defined by $f (x) = \int_{0}^{π} \cos t \cos (x - t) d t, 0 \leq x \leq 2 π .$ Which of the following hold(s) good?

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Important Questions on Definite Integration

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Mathematics

IMPORTANT

Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 39

Let

f (x) = \int_{0}^{\sin^{2} x} \sin^{- 1} (\sqrt{t}) d t + \int_{0}^{\cos^{2} x} \cos^{- 1} (\sqrt{t}) d t,

then

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Mathematics

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Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 40

The definite integral

\int_{\tan^{- 1} a}^{\cot^{- 1} a} \frac{d x}{1 + \tan x}

cannot take the value(s)

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Mathematics

IMPORTANT

Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 41

A function

f

is defined by

f (x) = \int_{0}^{π} c o s t c o s (x - t) d t, 0 \leq x \leq 2 π

, then which of the following hold(s) good

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Mathematics

IMPORTANT

Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 42

A function

f

is defined by

f (x) = \int_{0}^{π} \cos t \cos (x - t) d t

,

0 \leq x \leq 2 π

. Then which of the following hold(s) good?

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Mathematics

IMPORTANT

Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 43

The value of

\int_{0}^{[x]} (x - [x]) d x

(where

[.]

denotes the greatest integer function), is

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Mathematics

IMPORTANT

Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 44

Let

f (x)

be an odd continuous function which is periodic with period

2

. If

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

, then

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Mathematics

IMPORTANT

Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 45

If

A (x + y) = A (x) A (y) \forall x, y

and

A (0) \neq 0

and

B (x) = \frac{A (x)}{1 + {(A (x))}^{2}}

, then

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Mathematics

IMPORTANT

Mathematics Crash Course JEE Advanced>Chapter 23 - Definite Integration>Exercise 1>Q 46

Let

f (x) = \int_{0}^{x} \frac{e^{t}}{t} d t (x > 0)

, then

e^{- a} [f (x + a) - f (1 + a)] =