JEE Main 2021 (Online) 31st August Morning Shift | Application of Derivatives Question 31 | Mathematics

JEE Main 2021 (Online) 31st August Morning Shift

MCQ (Single Correct Answer)

-1

The number of real roots of the equation

$${e^{4x}} + 2{e^{3x}} - {e^x} - 6 = 0$$ is :

JEE Main 2021 (Online) 27th August Evening Shift

MCQ (Single Correct Answer)

-1

A box open from top is made from a rectangular sheet of dimension a $$\times$$ b by cutting squares each of side x from each of the four corners and folding up the flaps. If the volume of the box is maximum, then x is equal to :

$${{a + b - \sqrt {{a^2} + {b^2} - ab} } \over {12}}$$

$${{a + b - \sqrt {{a^2} + {b^2} + ab} } \over 6}$$

$${{a + b - \sqrt {{a^2} + {b^2} - ab} } \over 6}$$

$${{a + b + \sqrt {{a^2} + {b^2} + ab} } \over 6}$$

JEE Main 2021 (Online) 27th August Morning Shift

MCQ (Single Correct Answer)

-1

A wire of length 20 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is :

$${5 \over {2 + \sqrt 3 }}$$

$${{10} \over {2 + 3\sqrt 3 }}$$

$${5 \over {3 + \sqrt 3 }}$$

$${{10} \over {3 + 2\sqrt 3 }}$$

JEE Main 2021 (Online) 20th July Morning Shift

MCQ (Single Correct Answer)

-1

Let 'a' be a real number such that the function f(x) = ax² + 6x $$-$$ 15, x $$\in$$ R is increasing in $$\left( { - \infty ,{3 \over 4}} \right)$$ and decreasing in $$\left( {{3 \over 4},\infty } \right)$$. Then the function g(x) = ax² $$-$$ 6x + 15, x$$\in$$R has a :

local maximum at x = $$-$$ $${{3 \over 4}}$$

local minimum at x = $$-$$$${{3 \over 4}}$$

local maximum at x = $${{3 \over 4}}$$

local minimum at x = $${{3 \over 4}}$$

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