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Question: If a function is given \(f(x) = x{e^{x(1 - x)}}\), then \(f(x)\) is \( {\text{A}}{\text{. inc...

If a function is given f(x)=xex(1−x)f(x) = x{e^{x(1 - x)}}, then f(x)f(x) is
A. increasing on [−12,1] B. decrease on R C. increase on R D. decrease on [−12,1]  {\text{A}}{\text{. increasing on }}\left[ { - \dfrac{1}{2},1} \right] \\\ {\text{B}}{\text{. decrease on R}} \\\ {\text{C}}{\text{. increase on R}} \\\ {\text{D}}{\text{. decrease on }}\left[ { - \dfrac{1}{2},1} \right] \\\

Explanation

Solution

Hint:- In this question first we need to find the derivative of xex(1−x)x{e^{x(1 - x)}} using derivative formulas. Then equate f′(x)=0f'(x) = 0 to find the local extremum points of given function. After this we have to check between these local extremum points f′(x)f'(x) is increasing or decreasing.

Complete step-by-step answer:
Given: f(x)=xex(1−x)f(x) = x{e^{x(1 - x)}} ---- eq.1
We know, the derivative formula of d(u.v)dx=u.dvdx+vdudx\dfrac{{d(u.v)}}{{dx}} = u.\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}. ---- eq.2
⇒dxdx=1\Rightarrow \dfrac{{dx}}{{dx}} = 1 ----eq.3
And
⇒dexdx=ex\Rightarrow \dfrac{{d{e^x}}}{{dx}} = {e^x} ---eq.4
Now, differentiate f(x)f(x)with respect to xx and use above derivative formulas, we get
⇒df(x)dx=dxex(1−x)dx\Rightarrow \dfrac{{df(x)}}{{dx}} = \dfrac{{d\\{ x{e^{x(1 - x)}}\\} }}{{dx}}
⇒f′(x)=ex(1−x)+xex(1−x)(1−2x)\Rightarrow f'(x) = {e^{x(1 - x)}} + x{e^{x(1 - x)}}(1 - 2x)
Now, take ex(1−x){e^{x(1 - x)}} common in RHS, we get
⇒f′(x)=ex(1−x)1+x(1−2x)\Rightarrow f'(x) = {e^{x(1 - x)}}\\{ 1 + x(1 - 2x)\\}
Now, we make a quadratic equation on RHS.
⇒f′(x)=ex(1−x)1+x−2x2\Rightarrow f'(x) = {e^{x(1 - x)}}\\{ 1 + x - 2{x^2}\\}
Now, take negative sign from 1+x−2x2\\{ 1 + x - 2{x^2}\\} to make coefficient of x2{x^2} positive, we get
⇒f′(x)=−ex(1−x)2x2−x−1\Rightarrow f'(x) = - {e^{x(1 - x)}}\\{ 2{x^2} - x - 1\\}
Now, factorise the quadratic equation 2x2−x−1\\{ 2{x^2} - x - 1\\} , we get
⇒f′(x)=−ex(1−x)2x2−2x+x−1 ⇒f′(x)=−ex(1−x)(x−1)(2x+1) —- eq.5 \Rightarrow f'(x) = - {e^{x(1 - x)}}\\{ 2{x^2} - 2x + x - 1\\} \\\ \Rightarrow f'(x) = - {e^{x(1 - x)}}\\{ (x - 1)(2x + 1)\\} {\text{ ---- eq}}{\text{.5}} \\\
Now, for finding the increasing or decreasing of function f(x)f(x).
We have to put f′(x)=0f'(x) = 0.
⇒−ex(1−x)(x−1)(2x+1)=0\Rightarrow - {e^{x(1 - x)}}\\{ (x - 1)(2x + 1)\\} = 0
We know, exponential function (ex(1−x){e^{x(1 - x)}}) can never be zero. So
⇒(x−1)(2x+1)=0\Rightarrow \\{ (x - 1)(2x + 1)\\} = 0
Above equation gives us the condition f′(x)=0f'(x) = 0. On equating each factor equal to zero , we get
⇒x=1,−12\Rightarrow x = 1,\dfrac{{ - 1}}{2}
Now, we know if f′(x)f'(x)is positive then f(x)f(x)is increasing and if f′(x)f'(x) is negative then f(x)f(x)is decreasing.
We know, exponential function ex(1−x){e^{x(1 - x)}} can never be negative. So we have to check for (x−1)(2x+1)(x - 1)(2x + 1) to when it is greater or less than to zero.
f’(x) > 0 i.e. f(x) is increasing −12⩽ x⩽1 f’(x) < 0 i.e. f(x) is decreasing otherwise   {\text{f'(x) > 0 i}}{\text{.e}}{\text{. f(x) is increasing }} - \dfrac{1}{2} \leqslant {\text{ }}x \leqslant 1 \\\ {\text{f'(x) < 0 i}}{\text{.e}}{\text{. f(x) is decreasing otherwise }} \\\
Therefore, f(x) is greater than zero i.e. increasing between the range −12⩽ x⩽1 - \dfrac{1}{2} \leqslant {\text{ }}x \leqslant 1 and f(x) is less than zero i.e. decreasing for all x other than range −12⩽ x⩽1 - \dfrac{1}{2} \leqslant {\text{ }}x \leqslant 1. So, f(x)=xex(1−x)f(x) = x{e^{x(1 - x)}} is increasing  [−12,1]{\text{ }}\left[ { - \dfrac{1}{2},1} \right].
Hence, option A is correct.

Note:-Whenever you get this type of question the key concept to solve this is to learn the all basics derivative formulas. Using these formulas find the derivative of a given function (f′(x)f'(x)). And then put f′(x)=0f'(x) = 0 to get local extreme points. Then check where it is increasing if f′(x)>0f'(x) > 0 and decreasing f′(x)<0f'(x) < 0. And remember one more thing that exponential function can never be negative.