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JEE Main 20205 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
5 September
Shift
Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2020 · 5 September · Evening Shift · Mathematics · Application of Derivatives

Which of the following points lies on the tangent to the curve x ^...

Which of the following points lies on the tangent to the curve x4ey+2y+1=3{x}^{4}{e}^{y} + 2\sqrt{y + 1} = 3 at the point (1,0)( 1, 0 )?

Options

  1. A(2,2)( 2, 2 )
  2. B(2,6)( 2, 6 )
  3. C(–2,6)( – 2, 6 )
  4. D(−2,4)( - 2, 4 )

Correct answer

C. (–2,6)( – 2, 6 )

Step-by-step solution

eyy′x4+4x3ey+2y′12y+1=0{e}^{y}{y}^{'}{x}^{4} + 4{x}^{3}{e}^{y} + 2{y}^{'}\dfrac{1}{2\sqrt{y + 1}} = 0

at (1,0)
y′+4+y′=0⇒y′=−2{y}^{'} + 4 + {y}^{'} = 0 ⇒ {y}^{'} = - 2
equation of tangent at (1,0) is 2x+y−2=02x + y - 2 = 0
so option (3) is correct

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