1
WB JEE 2022
+1
-0.25

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

A
$${{11} \over 9}$$ sq. unit
B
$${{32} \over 3}\sqrt 6$$ sq. unit
C
$${{16} \over 3}$$ sq. unit
D
$${{24} \over 5}$$ sq. unit
2
WB JEE 2022
+2
-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

A
$$f\left( {{1 \over 2}} \right) < {1 \over 2}$$ and $$f\left( {{1 \over 3}} \right) < {1 \over 3}$$
B
$$f\left( {{1 \over 2}} \right) > {1 \over 2}$$ and $$f\left( {{1 \over 3}} \right) < {1 \over 3}$$
C
$$f\left( {{4 \over 3}} \right) < {4 \over 3}$$ and $$f\left( {{2 \over 3}} \right) < {2 \over 3}$$
D
$$f\left( {{4 \over 3}} \right) > {4 \over 3}$$ and $$f\left( {{2 \over 3}} \right) > {2 \over 3}$$
3
WB JEE 2021
+1
-0.25
Let f : R $$\to$$ R be such that f(0) = 0 and $$\left| {f'(x)} \right| \le 5$$ for all x. Then f(1) is in
A
(5, 6)
B
[$$-$$5, 5]
C
($$-$$ $$\infty$$, $$-$$5) $$\cup$$ (5, $$\infty$$)
D
[$$-$$4, 4]
4
WB JEE 2021
+1
-0.25
The straight the through the origin which divides the area formed by the curves y = 2x $$-$$ x2, y = 0 and x = 1 into two equal halves is
A
y = x
B
y = 2x
C
y = $${3 \over 2}$$ x
D
y = $${2 \over 3}$$ x
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Physics
Mechanics
Electricity
Optics
Modern Physics
Chemistry
Physical Chemistry
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Organic Chemistry
Mathematics
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