1
JEE Main 2021 (Online) 27th August Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
If $${U_n} = \left( {1 + {1 \over {{n^2}}}} \right)\left( {1 + {{{2^2}} \over {{n^2}}}} \right)^2.....\left( {1 + {{{n^2}} \over {{n^2}}}} \right)^n$$, then $$\mathop {\lim }\limits_{n \to \infty } {({U_n})^{{{ - 4} \over {{n^2}}}}}$$ is equal to :
A
$${{{e^2}} \over {16}}$$
B
$${4 \over e}$$
C
$${{16} \over {{e^2}}}$$
D
$${4 \over {{e^2}}}$$
2
JEE Main 2021 (Online) 27th August Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
The statement (p $$ \wedge $$ (p $$\to$$ q) $$\wedge$$ (q $$\to$$ r)) $$\to$$ r is :
A
a tautology
B
equivalent to p $$\to$$ $$\sim$$ r
C
a fallacy
D
equivalent to q $$\to$$ $$\sim$$ r
3
JEE Main 2021 (Online) 27th August Morning Shift
MCQ (Single Correct Answer)
+4
-1
Let us consider a curve, y = f(x) passing through the point ($$-$$2, 2) and the slope of the tangent to the curve at any point (x, f(x)) is given by f(x) + xf'(x) = x2. Then :
A
$${x^2} + 2xf(x) - 12 = 0$$
B
$${x^3} + xf(x) + 12 = 0$$
C
$${x^3} - 3xf(x) - 4 = 0$$
D
$${x^2} + 2xf(x) + 4 = 0$$
4
JEE Main 2021 (Online) 27th August Morning Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
$$\sum\limits_{k = 0}^{20} {{{\left( {{}^{20}{C_k}} \right)}^2}} $$ is equal to :
A
$${}^{40}{C_{21}}$$
B
$${}^{40}{C_{19}}$$
C
$${}^{40}{C_{20}}$$
D
$${}^{41}{C_{20}}$$
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