JEE Main 2019 (Online) 8th April Evening Slot | Application of Derivatives Question 144 | Mathematics

Given that the slope of the tangent to a curve y = y(x) at any point (x, y) is $$2y \over x^2$$. If the curve passes through the centre of the circle x² + y² – 2x – 2y = 0, then its equation is :

x log_e|y| = 2(x – 1)

x² log_e|y| = –2(x – 1)

x log_e|y| = x – 1

x log_e|y| = –2(x – 1)

JEE Main 2019 (Online) 8th April Morning Slot

MCQ (Single Correct Answer)

-1

Let ƒ : [0, 2] $$ \to $$ R be a twice differentiable function such that ƒ''(x) > 0, for all x $$ \in $$ (0, 2). If $$\phi $$(x) = ƒ(x) + ƒ(2 – x), then $$\phi $$ is :

decreasing on (0, 2)

decreasing on (0, 1) and increasing on (1, 2)

increasing on (0, 2)

increasing on (0, 1) and decreasing on (1, 2)

JEE Main 2019 (Online) 8th April Morning Slot

MCQ (Single Correct Answer)

-1

If S₁ and S₂ are respectively the sets of local minimum and local maximum points of the function,

ƒ(x) = 9x⁴ + 12x³ – 36x² + 25, x $$ \in $$ R, then :

S₁ = {–1}; S₂ = {0, 2}

S₁ = {–2}; S₂ = {0, 1}

S₁ = {–2, 0}; S₂ = {1}

S₁ = {–2, 1}; S₂ = {0}

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