JEE Main 2021 (Online) 27th August Morning Shift | Application of Derivatives Question 33 | Mathematics

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 = $${{3 \over 4}}$$

JEE Main 2021 (Online) 16th March Evening Shift

MCQ (Single Correct Answer)

-1

The maximum value of

$$f(x) = \left| {\matrix{ {{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\cos 2x} \cr {1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr {{{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|,x \in R$$ is :

$$\sqrt 5 $$

$${3 \over 4}$$

$$\sqrt 7 $$

JEE Main 2021 (Online) 26th February Evening Shift

MCQ (Single Correct Answer)

-1

Let A₁ be the area of the region bounded by the curves y = sinx, y = cosx and y-axis in the first quadrant. Also, let A₂ be the area of the region bounded by the curves y = sinx, y = cosx, x-axis and x = $${\pi \over 2}$$ in the first quadrant. Then,

$${A_1}:{A_2} = 1:\sqrt 2 $$ and $${A_1} + {A_2} = 1$$

$${A_1} = {A_2}$$ and $${A_1} + {A_2} = \sqrt 2 $$

$$2{A_1} = {A_2}$$ and $${A_1} + {A_2} = 1 + \sqrt 2 $$

$${A_1}:{A_2} = 1:2$$ and $${A_1} + {A_2} = 1$$

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