WB JEE 2023 | Definite Integration Question 31 | Mathematics

WB JEE 2023

MCQ (More than One Correct Answer)

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Let f be a non-negative function defined on $$\left[ {0,{\pi \over 2}} \right]$$. If $$\int\limits_0^x {(f'(t) - \sin 2t)dt = \int\limits_x^0 {f(t)\tan t\,dt} } ,f(0) = 1$$ then $$\int\limits_0^{{\pi \over 2}} {f(x)dx} $$ is

$$3 - {\pi \over 2}$$

$$3 + {\pi \over 2}$$

$${\pi \over 2}$$

WB JEE 2023

MCQ (More than One Correct Answer)

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Which of the following statements are true?

If f(x) be continuous and periodic with periodicity T, then $$I = \int\limits_a^{a + T} {f(x)} ~dx$$ depend on 'a'.

If f(x) be continuous and periodic with periodicity T, then $$I = \int\limits_a^{a + T} {f(x)} ~dx$$ does not depend on 'a'.

Let $$\mathrm{f(x)} = \left\{ \matrix{ 1,\,\,\,\mathrm{if\,x\,is\,rational} \hfill \cr 0,\,\,\mathrm{if\,x\,is\,irrational} \hfill \cr} \right.$$, then f is periodic of the periodicity T only if T is rational.

f defined in (C) is periodic for all T.

WB JEE 2021

MCQ (More than One Correct Answer)

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Whichever of the following is/are correct?

To evaluate $${I_1} = \int\limits_{ - 2}^2 {{{dx} \over {4 + {x^2}}}} $$, it is possible to $$x = {1 \over t}$$

To evaluate $${I_2} = \int\limits_0^1 {\sqrt {({x^2} + 1)} dx} $$, it is possible to put $$x = \sec t$$

To evaluate $${I_2} = \int\limits_0^1 {\sqrt {({x^2} + 1)} dx} $$, it is not possible to put $$x = \cos ec\theta $$

To evaluate I₁, it is not possible to put $$x = {1 \over t}$$

WB JEE 2021

MCQ (More than One Correct Answer)

-0

Let $$f(x) = \left\{ {\matrix{ {0,} & {if} & { - 1 \le x \le 0} \cr {1,} & {if} & {x = 0} \cr {2,} & {if} & {0 < x \le 1} \cr } } \right.$$ and let $$F(x) = \int\limits_{ - 1}^x {f(t)dt} $$, $$-$$1 $$\le$$ x $$\le$$ 1, then

F is continuous function in [$$-$$1, 1]

F is discontinuous function in [$$-$$1, 1]

F'(x) exists at x = 0

F'(x) does not exist at x = 0