1
JEE Main 2018 (Online) 16th April Morning Slot
+4
-1
If $$f(x) = \int\limits_0^x {t\left( {\sin x - \sin t} \right)dt\,\,\,}$$ then :
A
f'''(x) + f''(x) = sinx
B
f'''(x) + f''(x) $$-$$ f'(x) = cosx
C
f'''(x) + f'(x) = cosx $$-$$ 2x sinx
D
f'''(x) $$-$$ f''(x) = cosx $$-$$ 2x sinx
2
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
$${e^{{2 \over 3}}}$$
3
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
$${1 \over 2}\left( {\sqrt 3 - \sqrt 2 } \right)$$
4
JEE Main 2018 (Offline)
+4
-1
The value of $$\int\limits_{ - \pi /2}^{\pi /2} {{{{{\sin }^2}x} \over {1 + {2^x}}}} dx$$ is
A
$${\pi \over 4}$$
B
$${\pi \over 8}$$
C
$${\pi \over 2}$$
D
$${4\pi }$$
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