JEE Main 2020 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 8 January 2020, Shift 1, Maths Q62

Question 62 of 75 in this shift, Maths question 12 of 25, Section A.

Let

y=y(x)

be a solution of the differential equation,

\sqrt{1-x^2}\,\dfrac{dy}{dx}+\sqrt{1-y^2}=0

|x|<1

. If

y\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}

, then

y\left(\frac{-1}{\sqrt{2}}\right)

is equal to :

(1) $-\frac{1}{\sqrt{2}}$
(2) $\frac{1}{\sqrt{2}}$ Official answer
(3) $\frac{\sqrt{3}}{2}$
(4) $-\frac{\sqrt{3}}{2}$

Official answer

Option 2

NTA final key (Jan 2020).

Chapter: Differential Equations
Topic: Variable Separable Form
Idea tested: Initial Value Problem

Same idea in other shifts

Asked 7× in all

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.