Tangent & Normal – JEE Main Previous Year Question with Solutions
JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects. Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more. Download eSaral app for free study material and video tutorials.
Q. The equation of the tangent to the curve $y=x+\frac{4}{x^{2}},$ that is parallel to the x-axis, is :- (1) y = 0                (2) y = 1                 (3) y = 2                    (4) y = 3 [AIEEE-2010]

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Sol. (4) Q. The intercepts on x-axis made by tangents to the curve, $\mathrm{y}=\int_{0}^{\mathrm{x}}|\mathrm{t}| \mathrm{dt}, \mathrm{x} \in \mathrm{R}$ which are parallel to the line y = 2x, are equal to $(1) \pm 1$ $(2) \pm 2$ (3) $\pm 3$ (4) $\pm 4$ [JEE-MAIN 2013]

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Sol. (1) Q. The normal to the curve, $x^{2}+2 x y-3 y^{2}=0,$ at $(1,1):$ (1)meets the curve again in the third quadrant (2) meets the curve again in the fourth quadrant (3) does not meet the curve again (4) meets the curve again in the second quadrant [JEE-MAIN 2015]

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Sol. (2) Q. Consider $f(x)=\tan ^{-1}(\sqrt{\frac{1+\sin x}{1-\sin x}}), x \in\left(0, \frac{\pi}{2}\right) .$ A normal to $y=f(x)$ at $x=\frac{\pi}{6}$ also passes through the point : ( 1)$\left(\frac{\pi}{4}, 0\right)$b (2) (0, 0) (3) $\left(0, \frac{2 \pi}{3}\right)$ (4) $\left(\frac{\pi}{6}, 0\right)$ [JEE-MAIN 2016]

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Sol. (3)   Q. If the curves $\mathrm{y}^{2}=6 \mathrm{x}, 9 \mathrm{x}^{2}+\mathrm{by}^{2}=16$ intersect each other at right angles, then the value of $\mathrm{b}$ is : (1) $\frac{7}{2}$             (2) 4              (3) $\frac{9}{2}$                 (4) 6 [JEE-MAIN 2016]

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Sol. (3)  • April 15, 2021 at 11:12 pm

Oh bhai 2016 ke aage bhi badho 2021 ke bhi 2 session ho chuke tum itne pichde kyu ho

1
• December 1, 2020 at 5:19 pm

Op questions

1
• October 29, 2020 at 10:19 pm

nice

2
• October 20, 2020 at 5:05 pm

thanx

7
• October 14, 2020 at 9:49 pm

Good to see all questions here

3
• October 11, 2020 at 2:39 pm

it is a good platform but no of questions are quite less …

17
• September 4, 2020 at 3:55 pm

2 nd last question

1
• July 21, 2020 at 4:25 am

THANKS…FOR ….QUESTIONS

0
• October 20, 2020 at 3:14 pm

👍👍

0
• December 1, 2020 at 5:20 pm

Op questions

1
• July 10, 2020 at 12:08 am

plz can u explain second last question more properly??

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• March 4, 2021 at 12:35 pm

U just differentiate given f(x) u will get slope of tangent we known (slopeoftangent*slopeofnormal=-1) from here u get slope of normal they gave x value find y value by sub in eqn of curve then apply normal eqn then sub option one by one u will get answer !

1
• May 20, 2020 at 4:21 pm

good but need some more problems..

0
• October 20, 2020 at 3:14 pm

Yes

0
• February 22, 2021 at 12:54 pm

Good but some more prblms

0
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• May 2, 2020 at 1:46 pm

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0
• April 28, 2020 at 5:29 pm

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0
• April 17, 2020 at 3:47 pm

Some more

0
• April 14, 2020 at 5:51 pm

Thank you

1