WB JEE 2011 | Application of Integration Question 13 | Mathematics

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WB JEE 2011

MCQ (Single Correct Answer)

-0.25

The area bounded by y² = 4x and x² = 4y is

$${{20} \over 3}$$ sq. units

$${{16} \over 3}$$ sq. units

$${{14} \over 3}$$ sq. units

$${{10} \over 3}$$ sq. units

WB JEE 2022

MCQ (Single Correct Answer)

-0.25

If $$x{{dy} \over {dx}} + y = x{{f(xy)} \over {f'(xy)}}$$, then $$|f(xy)|$$ is equal to

$$C{e^{{{{x^2}} \over 2}}}$$ (where C is the constant of integration)

$$C{e^{{x^2}}}$$ (where C is the constant of integration)

$$C{e^{2{x^2}}}$$ (where C is the constant of integration)

$$C{e^{{{{x^2}} \over 3}}}$$ (where C is the constant of integration)

WB JEE 2022

MCQ (Single Correct Answer)

-0.25

Area of the figure bounded by the parabola $${y^2} + 8x = 16$$ and $${y^2} - 24x = 48$$ is

$${{11} \over 9}$$ sq. unit

$${{32} \over 3}\sqrt 6 $$ sq. unit

$${{16} \over 3}$$ sq. unit

$${{24} \over 5}$$ sq. unit

WB JEE 2022

MCQ (Single Correct Answer)

-0.5

Let f be a non-negative function defined in $$[0,\pi /2]$$, f' exists and be continuous for all x and $$\int\limits_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int\limits_0^x {f(t)dt} } $$ and f (0) = 0. Then

$$f\left( {{1 \over 2}} \right) < {1 \over 2}$$ and $$f\left( {{1 \over 3}} \right) < {1 \over 3}$$

$$f\left( {{1 \over 2}} \right) > {1 \over 2}$$ and $$f\left( {{1 \over 3}} \right) < {1 \over 3}$$

$$f\left( {{4 \over 3}} \right) < {4 \over 3}$$ and $$f\left( {{2 \over 3}} \right) < {2 \over 3}$$

$$f\left( {{4 \over 3}} \right) > {4 \over 3}$$ and $$f\left( {{2 \over 3}} \right) > {2 \over 3}$$

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