1
JEE Main 2022 (Online) 28th June Morning Shift
MCQ (Single Correct Answer)
+4
-1

Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral $$\int\limits_0^1 {[ - 8{x^2} + 6x - 1]dx}$$ is equal to

A
$$-$$1
B
$${{ - 5} \over 4}$$
C
$${{\sqrt {17} - 13} \over 8}$$
D
$${{\sqrt {17} - 16} \over 8}$$
2
JEE Main 2022 (Online) 28th June Morning Shift
MCQ (Single Correct Answer)
+4
-1

The area of the region S = {(x, y) : y2 $$\le$$ 8x, y $$\ge$$ $$\sqrt2$$x, x $$\ge$$ 1} is

A
$${{13\sqrt 2 } \over 6}$$
B
$${{11\sqrt 2 } \over 6}$$
C
$${{5\sqrt 2 } \over 6}$$
D
$${{19\sqrt 2 } \over 6}$$
3
JEE Main 2022 (Online) 27th June Evening Shift
MCQ (Single Correct Answer)
+4
-1

If m and n respectively are the number of local maximum and local minimum points of the function $$f(x) = \int\limits_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt}$$, then the ordered pair (m, n) is equal to

A
(3, 2)
B
(2, 3)
C
(2, 2)
D
(3, 4)
4
JEE Main 2022 (Online) 27th June Evening Shift
MCQ (Single Correct Answer)
+4
-1

Let f be a differentiable function in $$\left( {0,{\pi \over 2}} \right)$$. If $$\int\limits_{\cos x}^1 {{t^2}\,f(t)dt = {{\sin }^3}x + \cos x}$$, then $${1 \over {\sqrt 3 }}f'\left( {{1 \over {\sqrt 3 }}} \right)$$ is equal to

A
$$6 - 9\sqrt 2$$
B
$$6 - {9 \over {\sqrt 2 }}$$
C
$${9 \over 2} - 6\sqrt 2$$
D
$${9 \over {\sqrt 2 }} - 6$$
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