WB JEE 2022 | Definite Integration Question 36 | Mathematics

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WB JEE 2022

MCQ (Single Correct Answer)

-0.25

Let $$f(x) = \int\limits_{\sin x}^{\cos x} {{e^{ - {t^2}}}dt} $$. Then $$f'\left( {{\pi \over 4}} \right)$$ equals

$$\sqrt {{1 \over e}} $$

$$ - \sqrt {{2 \over e}} $$

$$\sqrt {{2 \over e}} $$

$$ - \sqrt {{1 \over e}} $$

WB JEE 2022

MCQ (Single Correct Answer)

-0.5

If I is the greatest of $${I_1} = \int\limits_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} $$, $${I_2} = \int\limits_0^1 {{e^{ - {x^2}}}{{\cos }^2}x\,dx} $$, $${I_3} = \int\limits_0^1 {{e^{ - {x^2}}}dx} $$, $${I_4} = \int\limits_0^1 {{e^{ - {x^2}/2}}dx} $$, then

I = I₁

I = I₂

I = I₃

I = I₄

WB JEE 2021

MCQ (Single Correct Answer)

-0.25

$$\int\limits_1^3 {{{\left| {x - 1} \right|} \over {\left| {x - 2} \right| + \left| {x - 3} \right|}}dx} $$ is equal to

$$1 + {4 \over 3}{\log _e}3$$

$$1 + {3 \over 4}{\log _e}3$$

$$1 - {4 \over 3}{\log _e}3$$

$$1 - {3 \over 4}{\log _e}3$$

WB JEE 2021

MCQ (Single Correct Answer)

-0.25

The value of the integral $$\int\limits_{ - {1 \over 2}}^{{1 \over 2}} {{{\left\{ {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} \right\}}^{1/2}}} dx$$ is equal to

$${\log _e}\left( {{4 \over 3}} \right)$$

$$4\,{\log _e}\left( {{3 \over 4}} \right)$$

$$4\,{\log _e}\left( {{4 \over 3}} \right)$$

$${\log _e}\left( {{3 \over 4}} \right)$$

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