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1

### JEE Main 2021 (Online) 27th August Morning Shift

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 :
A
$${5 \over {2 + \sqrt 3 }}$$
B
$${{10} \over {2 + 3\sqrt 3 }}$$
C
$${5 \over {3 + \sqrt 3 }}$$
D
$${{10} \over {3 + 2\sqrt 3 }}$$

## Explanation

Let the wire is cut into two pieces of length x and 20 $$-$$ x. Area of square = $${\left( {{x \over 4}} \right)^2}$$

Area of regular hexagon = $$6 \times {{\sqrt 3 } \over 4}{\left( {{{20 - x} \over 6}} \right)^2}$$

Total area = $$A(x) = {{{x^2}} \over {16}} + {{3\sqrt 3 } \over 2}{{{{(20 - x)}^2}} \over {36}}$$

$$A'(x) = {{2x} \over {16}} + {{3\sqrt 3 \times 2} \over {2 \times 36}}(20 - x)( - 1)$$

A'(x) = 0 at $$x = {{40\sqrt 3 } \over {3 + 2\sqrt 3 }}$$

Length of side of regular Hexagon $$= {1 \over 6}(20 - x)$$

$$= {1 \over 6}\left( {20 - {{4.\sqrt 3 } \over {3 + 2\sqrt 3 }}} \right)$$

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

### JEE Main 2021 (Online) 20th July Morning Shift

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 :
A
local maximum at x = $$-$$ $${{3 \over 4}}$$
B
local minimum at x = $$-$$$${{3 \over 4}}$$
C
local maximum at x = $${{3 \over 4}}$$
D
local minimum at x = $${{3 \over 4}}$$

## Explanation

$${{ - B} \over {2A}} = {3 \over 4}$$

$$\Rightarrow {{ - (6)} \over {2a}} = {3 \over 4}$$

$$\Rightarrow a = {{ - 6 \times 4} \over 6} \Rightarrow a = - 4$$

$$\therefore$$ $$g(x) = 4{x^2} - 6x + 15$$

Local max. at $$x = {{ - B} \over {2A}} = - {{( - 6)} \over {2( - 4)}}$$

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

### JEE Main 2021 (Online) 16th March Evening Shift

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 :
A
$$\sqrt 5$$
B
$${3 \over 4}$$
C
5
D
$$\sqrt 7$$

## Explanation

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

$${C_1} \to {C_1} + {C_2}$$

= $$\left| {\matrix{ 2 & {1 + {{\cos }^2}x} & {\cos 2x} \cr 2 & {{{\cos }^2}x} & {\cos 2x} \cr 1 & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|$$

$${R_1} \to {R_1} - {R_2}$$

= $$\left| {\matrix{ 0 & 1 & 0 \cr 2 & {{{\cos }^2}x} & {\cos 2x} \cr 1 & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|$$

$$= ( - 1)[2\sin 2x - \cos 2x] = \cos 2x - 2\sin 2x$$

We know, maximum value of acosx $$\pm$$ bsinx

= $$\sqrt {{a^2} + {b^2}}$$

$$\therefore$$ Here maximum value = $$\sqrt {{1^2} + {{\left( { - 2} \right)}^2}}$$$$= \sqrt 5$$
4

### JEE Main 2021 (Online) 26th February Evening Shift

Let slope of the tangent line to a curve at any point P(x, y) be given by $${{x{y^2} + y} \over x}$$. If the curve intersects the line x + 2y = 4 at x = $$-$$2, then the value of y, for which the point (3, y) lies on the curve, is :
A
$$- {{18} \over {19}}$$
B
$$- {{4} \over {3}}$$
C
$${{18} \over {35}}$$
D
$$- {{18} \over {11}}$$

## Explanation

$${{dy} \over {dx}} = {{x{y^2} + y} \over x}$$

$$\Rightarrow {{xdy - ydx} \over {{y^2}}} = xdx$$
$$\Rightarrow - d\left( {{x \over y}} \right) = d\left( {{{{x^2}} \over 2}} \right)$$

$$\Rightarrow {{ - x} \over y} = {{{x^2}} \over 2} + C$$

Curve intersect the line x + 2y = 4 at x = $$-$$ 2

So, $$-$$ 2 + 2y = 4 $$\Rightarrow$$ y = 3

So the curve passes through ($$-$$2, 3)

$$\Rightarrow {2 \over 3} = 2 + C$$

$$\Rightarrow C = {{ - 4} \over 3}$$

$$\therefore$$ curve is $${{ - x} \over y} = {{{x^2}} \over 2} - {4 \over 3}$$

It also passes through (3, y)

$${{ - 3} \over y} = {9 \over 2} - {4 \over 3}$$

$$\Rightarrow {{ - 3} \over y} = {{19} \over 6}$$

$$\Rightarrow y = - {{18} \over {19}}$$

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