JEE Main 2019 (Online) 9th April Morning Slot | Application of Derivatives Question 146 | Mathematics

If the tangent to the curve, y = x³ + ax – b at the point (1, –5) is perpendicular to the line, –x + y + 4 = 0, then which one of the following points lies on the curve ?

(2, –2)

(2, –1)

(–2, 2)

(–2, 1)

JEE Main 2019 (Online) 8th April Evening Slot

MCQ (Single Correct Answer)

-1

Out of Syllabus

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 Evening Slot

MCQ (Single Correct Answer)

-1

The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is

$$\sqrt 3 $$

$$2\sqrt 3 $$

$$\sqrt 6 $$

$${2 \over 3} {\sqrt 3} $$

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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