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

"x 4 e y + 2$\\sqrt {y + 1}$ = 3\n Differentiating w.r.t. x, we get\n x 4 e y y' + e y 4x 3 + ${{2y'} \\over {2\\sqrt {y + 1} }}$ = 0\n At P(1, 0)\n ${y{'_P}}$ + 4 + ${y{'_P}}$ = 0\n $\\Rightarrow$ ${y{'_P}}$ = -2\n Tangent at P(1, 0) is\n y – 0 = – 2 (x – 1)\n 2x + y = 2\n Only (–2, 6) lies on it."

Medium MCQ +4 / -1 PYQ · JEE Mains 2020

Which of the following points lies on the tangent to the curve

x⁴e^y + 2$\sqrt {y + 1}$ = 3 at the point (1, 0)?

A (2, 2)
B (–2, 4)
C (2, 6)
D (–2, 6) Correct answer

Solution

x4ey + 2$\sqrt {y + 1}$ = 3 Differentiating w.r.t. x, we get x4eyy' + ey4x3 + ${{2y'} \over {2\sqrt {y + 1} }}$ = 0 At P(1, 0) ${y{'_P}}$ + 4 + ${y{'_P}}$ = 0 $\Rightarrow$ ${y{'_P}}$ = -2 Tangent at P(1, 0) is y – 0 = – 2 (x – 1) 2x + y = 2 Only (–2, 6) lies on it.

About this question

Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals

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

Solution

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Which of the following points lies on the tangent to the curve

x⁴e^y + 2$\sqrt {y + 1}$ = 3 at the point (1, 0)?