1
JEE Main 2020 (Online) 8th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let y = y(x) be a solution of the differential equation,

$$\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0$$, |x| < 1.

If $$y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}$$, then $$y\left( { - {1 \over {\sqrt 2 }}} \right)$$ is equal to :
A
$$ - {{\sqrt 3 } \over 2}$$
B
None of those
C
$${{1 \over {\sqrt 2 }}}$$
D
$$-{{1 \over {\sqrt 2 }}}$$
2
JEE Main 2020 (Online) 8th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
For a > 0, let the curves C1 : y2 = ax and C2 : x2 = ay intersect at origin O and a point P. Let the line x = b (0 < b < a) intersect the chord OP and the x-axis at points Q and R, respectively. If the line x = b bisects the area bounded by the curves, C1 and C2, and the area of
$$\Delta $$OQR = $${1 \over 2}$$, then 'a' satisfies the equation :
A
x6 – 12x3 + 4 = 0
B
x6 – 12x3 – 4 = 0
C
x6 + 6x3 – 4 = 0
D
x6 – 6x3 + 4 = 0
3
JEE Main 2020 (Online) 8th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 resepectively. Each of these 10 observations is multiplied by p and then reduced by q, where p $$ \ne $$ 0 and q $$ \ne $$ 0. If the new mean and new s.d. become half of their original values, then q is equal to
A
10
B
-20
C
-10
D
-5
4
JEE Main 2020 (Online) 8th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let two points be A(1, –1) and B(0, 2). If a point P(x', y') be such that the area of $$\Delta $$PAB = 5 sq. units and it lies on the line, 3x + y – 4$$\lambda $$ = 0, then a value of $$\lambda $$ is :
A
4
B
1
C
-3
D
3
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