1
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
$${\log _e}\left( {{4 \over 3}} \right)$$
2
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)
3
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
$$\int\limits_0^1 {\left( {{{{y^2}} \over 2} + \sqrt {1 - {y^2}} + 1} \right)dy}$$
4
JEE Main 2022 (Online) 29th June Morning Shift
+4
-1

Let $$f:R \to R$$ be a function defined by :

$$f(x) = \left\{ {\matrix{ {\max \,\{ {t^3} - 3t\} \,t \le x} & ; & {x \le 2} \cr {{x^2} + 2x - 6} & ; & {2 < x < 3} \cr {[x - 3] + 9} & ; & {3 \le x \le 5} \cr {2x + 1} & ; & {x > 5} \cr } } \right.$$

where [t] is the greatest integer less than or equal to t. Let m be the number of points where f is not differentiable and $$I = \int\limits_{ - 2}^2 {f(x)\,dx}$$. Then the ordered pair (m, I) is equal to :

A
$$\left( {3,\,{{27} \over 4}} \right)$$
B
$$\left( {3,\,{{23} \over 4}} \right)$$
C
$$\left( {4,\,{{27} \over 4}} \right)$$
D
$$\left( {4,\,{{23} \over 4}} \right)$$
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