1
JEE Main 2019 (Online) 9th January Evening Slot
+4
-1
The area of the region

A = {(x, y) : 0 $$\le$$ y $$\le$$x |x| + 1  and  $$-$$1 $$\le$$ x $$\le$$1} in sq. units, is :
A
$${2 \over 3}$$
B
2
C
$${4 \over 3}$$
D
$${1 \over 3}$$
2
JEE Main 2019 (Online) 9th January Morning Slot
+4
-1
The area (in sq. units) bounded by the parabolae y = x2 – 1, the tangent at the point (2, 3) to it and the y-axis is :
A
$$56\over3$$
B
$$32\over3$$
C
$$8\over3$$
D
$$14\over3$$
3
JEE Main 2018 (Online) 16th April Morning Slot
+4
-1
If the area of the region bounded by the curves, $$y = {x^2},y = {1 \over x}$$ and the lines y = 0 and x= t (t >1) is 1 sq. unit, then t is equal to :
A
$${e^{{3 \over 2}}}$$
B
$${4 \over 3}$$
C
$${3 \over 2}$$
D
$${e^{{2 \over 3}}}$$
4
JEE Main 2018 (Offline)
+4
-1
Let g(x) = cosx2, f(x) = $$\sqrt x$$ and $$\alpha ,\beta \left( {\alpha < \beta } \right)$$ be the roots of the quadratic equation 18x2 - 9$$\pi$$x + $${\pi ^2}$$ = 0. Then the area (in sq. units) bounded by the curve
y = (gof)(x) and the lines $$x = \alpha$$, $$x = \beta$$ and y = 0 is :
A
$${1 \over 2}\left( {\sqrt 2 - 1} \right)$$
B
$${1 \over 2}\left( {\sqrt 3 - 1} \right)$$
C
$${1 \over 2}\left( {\sqrt 3 + 1} \right)$$
D
$${1 \over 2}\left( {\sqrt 3 - \sqrt 2 } \right)$$
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