1
JEE Main 2022 (Online) 25th July Morning Shift
+4
-1

For any real number $$x$$, let $$[x]$$ denote the largest integer less than equal to $$x$$. Let $$f$$ be a real valued function defined on the interval $$[-10,10]$$ by $$f(x)=\left\{\begin{array}{l}x-[x], \text { if }[x] \text { is odd } \\ 1+[x]-x, \text { if }[x] \text { is even } .\end{array}\right.$$ Then the value of $$\frac{\pi^{2}}{10} \int_{-10}^{10} f(x) \cos \pi x \,d x$$ is :

A
4
B
2
C
1
D
0
2
JEE Main 2022 (Online) 30th June Morning Shift
+4
-1
Out of Syllabus

$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}}$$ is equal to :

A
$${\log _e}\left( {{{\sqrt 3 } \over 2}} \right)$$
B
$${\log _e}\left( {{{3\sqrt 3 } \over 4}} \right)$$
C
$${\log _e}\left( {{{27} \over 4}} \right)$$
D
$${\log _e}\left( {{4 \over 3}} \right)$$
3
JEE Main 2022 (Online) 29th June Evening Shift
+4
-1

Let f be a real valued continuous function on [0, 1] and $$f(x) = x + \int\limits_0^1 {(x - t)f(t)dt}$$.

Then, which of the following points (x, y) lies on the curve y = f(x) ?

A
(2, 4)
B
(1, 2)
C
(4, 17)
D
(6, 8)
4
JEE Main 2022 (Online) 29th June Evening Shift
+4
-1

If $$\int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } }$$, then I equals

A
$$\int\limits_0^1 {\left( {1 + \sqrt {1 - {y^2}} } \right)dy}$$
B
$$\int\limits_0^1 {\left( {{{{y^2}} \over 2} - \sqrt {1 - {y^2}} + 1} \right)dy}$$
C
$$\int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} } \right)dy}$$
D
$$\int\limits_0^1 {\left( {{{{y^2}} \over 2} + \sqrt {1 - {y^2}} + 1} \right)dy}$$
EXAM MAP
Medical
NEET