1

### JEE Main 2021 (Online) 31st August Morning Shift

The number of real roots of the equation

${e^{4x}} + 2{e^{3x}} - {e^x} - 6 = 0$ is :
A
2
B
4
C
1
D
0

## Explanation

Let ${e^x} = t > 0$

$f(t) = {t^4} + 2{t^3} - t - 6 = 0$

$f'(t) = 4{t^3} + 6{t^2} - 1$

$f''(t) = 12{t^2} + 12t > 0$

$f(0) = - 6,f(1) = - 4,f(2) = 24$

$\Rightarrow$ Number of real roots = 1
2

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

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
${{a + b - \sqrt {{a^2} + {b^2} - ab} } \over {12}}$
B
${{a + b - \sqrt {{a^2} + {b^2} + ab} } \over 6}$
C
${{a + b - \sqrt {{a^2} + {b^2} - ab} } \over 6}$
D
${{a + b + \sqrt {{a^2} + {b^2} + ab} } \over 6}$

## Explanation

V = l . b . h = (a $-$ 2x)(b $-$ 2x) x

$\Rightarrow$ V(x) = (2x $-$ a)(2x $-$ b) x

$\Rightarrow$ V(x) = 4x3 $-$ 2(a + b)x2 + abx

$\Rightarrow {d \over {dx}}v(x) = 12{x^2} - 4(a + b)x + ab$

${d \over {dx}}(v(x)) = 0 \Rightarrow 12{x^2} - 4(a + b)x + ab = 0 < _\beta ^\alpha$

$\Rightarrow x = {{4(a + b) \pm \sqrt {16{{(a + b)}^2} - 48ab} } \over {2(12)}}$

$= {{(a + b) \pm \sqrt {{a^2} + {b^2} - ab} } \over 6}$

Let $x = \alpha = {{(a + b) + \sqrt {{a^2} + {b^2} - ab} } \over 6}$

$\beta = {{(a + b) - \sqrt {{a^2} + {b^2} - ab} } \over 6}$

Now, $12(x - \alpha )(x - \beta ) = 0$

$\therefore$ x = $\beta$ $= {{a + b - \sqrt {{a^2} + {b^2} - ab} } \over b}$
3

### 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 $= {1 \over 6}(20 - x)$

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

$= {{10} \over {2 + 2\sqrt 3 }}$
4

### 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 $x = {{ - B} \over {2A}} = - {{( - 6)} \over {2( - 4)}}$

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