1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

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

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