JEE MAIN - Mathematics (2021 - 31st August Morning Shift - No. 1)
The number of real roots of the equation
$${e^{4x}} + 2{e^{3x}} - {e^x} - 6 = 0$$ is :
$${e^{4x}} + 2{e^{3x}} - {e^x} - 6 = 0$$ is :
2
4
1
0
Explanation
Let $${e^x} = t > 0$$
$$f(t) = {t^4} + 2{t^3} - t - 6 = 0$$
$$f'(t) = 4{t^3} + 6{t^2} - 1$$
_31st_August_Morning_Shift_en_1_1.png)
$$f''(t) = 12{t^2} + 12t > 0$$
_31st_August_Morning_Shift_en_1_2.png)
$$f(0) = - 6,f(1) = - 4,f(2) = 24$$
$$\Rightarrow$$ Number of real roots = 1
$$f(t) = {t^4} + 2{t^3} - t - 6 = 0$$
$$f'(t) = 4{t^3} + 6{t^2} - 1$$
$$f''(t) = 12{t^2} + 12t > 0$$
$$f(0) = - 6,f(1) = - 4,f(2) = 24$$
$$\Rightarrow$$ Number of real roots = 1
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