JEE Main 2020 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 2 September 2020, Shift 2, Maths Q60

Question 60 of 75 in this shift, Maths question 10 of 25.

The equation of the normal to the curve

y=(1+x)^{2y}+\cos^2(\sin^{-1}x)

x=0

is :

(1) $x+4y=8$ Official answer
(2) $y=4x+2$
(3) $y+4x=2$
(4) $2y+x=4$

Official answer

Option 1

NTA final key (Sep 2020).

Chapter: Application of Derivatives
Topic: Tangents and Normals
Idea tested: Equation of Normal

Same idea in other shifts

Asked 6× in all

4 Aug 2021, Shift 1 · Q68The equation of the normal to the curve $y=e^{-|x|}$ at a point, where the curve cuts the line $x=-1$ isEasySingle correct
26 Feb 2021, Shift 2 · Q83Let the normals at all the points on a given curve pass through a fixed point $(a, b)$ . If the curve passes through $(3, -3)$ and…MediumNumerical value
8 Jan 2020, Shift 2 · Q60The length of the perpendicular from the origin, on the normal to the curve, $x^2 + 2xy - 3y^2 = 0$ at the point $(2, 2)$ is :MediumSingle correct
2 Sep 2020, Shift 1 · Q64Let P(h, k) be a point on the curve $y=x^{2}+7x+2$ , nearest to the line, $y=3x-3$ . Then the equation of the normal to the curve at P is :MediumSingle correct
2 Apr 2017 (offline) · Q72The normal to the curve $y(x-2)(x-3)=x+6$ at the point where the curve intersects the $y$ -axis passes through the point :MediumSingle correct

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 108 of the 174 JEE Main shifts held from 2017 to 2026.