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1
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Given : $$f(x) = \left\{ {\matrix{ {x\,\,\,\,\,,} & {0 \le x < {1 \over 2}} \cr {{1 \over 2}\,\,\,\,,} & {x = {1 \over 2}} \cr {1 - x\,\,\,,} & {{1 \over 2} < x \le 1} \cr } } \right.$$

and $$g(x) = \left( {x - {1 \over 2}} \right)^2,x \in R$$

Then the area (in sq. units) of the region bounded by the curves, y = ƒ(x) and y = g(x) between the lines, 2x = 1 and 2x = $$\sqrt 3 $$, is :
A
$${1 \over 2} + {{\sqrt 3 } \over 4}$$
B
$${1 \over 2} - {{\sqrt 3 } \over 4}$$
C
$${1 \over 3} + {{\sqrt 3 } \over 4}$$
D
$${{\sqrt 3 } \over 4} - {1 \over 3}$$
2
JEE Main 2020 (Online) 9th January Evening Slot
Numerical
+4
-0
Change Language
If the curves, x2 – 6x + y2 + 8 = 0 and
x2 – 8y + y2 + 16 – k = 0, (k > 0) touch each other at a point, then the largest value of k is ______.
Your input ____
3
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If z be a complex number satisfying |Re(z)| + |Im(z)| = 4, then |z| cannot be :
A
$$\sqrt {10} $$
B
$$\sqrt {7} $$
C
$$\sqrt {{{17} \over 2}} $$
D
$$\sqrt {8} $$
4
JEE Main 2020 (Online) 9th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If $${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$$; y(1) = 1; then a value of x satisfying y(x) = e is :
A
$$\sqrt 2 e$$
B
$${1 \over 2}\sqrt 3 e$$
C
$${e \over {\sqrt 2 }}$$
D
$$\sqrt 3 e$$

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