Question

If the local maximum 'M' and local minimum 'm' of the function $f(x)=x-\frac{x^{2}}{2}-xe^{2-x}$ exist at $x=\alpha$ and $x=\beta$ respectively, then $2\alpha m+\beta M=$

• A. -2e
• B. $\frac{1}{e}$
• C. -4e
• D. $\frac{1}{e^{2}}$

Hint:

Factor out common algebraic terms like $(1-x)$ from your derivative to easily find all critical points of a function.

Answer 1

The Correct Option is C

Step 1: Differentiate the function.
For $f(x)=x-\dfrac{x^2}{2}-xe^{2-x}$, the derivative is \[ f'(x)=1-x-\left(e^{2-x}-xe^{2-x}\right)=(1-x)(1-e^{2-x}). \]
Step 2: Find the critical points.
Set $f'(x)=0$. Either $1-x=0$ giving $x=1$, or $1-e^{2-x}=0$ giving $2-x=0$, that is $x=2$.
Step 3: Decide max or min.
Looking at the sign of $f'$, at $x=1$ the function has a local minimum, and at $x=2$ it has a local maximum. So $\beta=1$ (for $m$) and $\alpha=2$ (for $M$).
Step 4: Compute the minimum value $m$.
\[ f(1)=1-\tfrac12-1\cdot e^{1}=\tfrac12-e=m. \]
Step 5: Compute the maximum value $M$.
\[ f(2)=2-2-2\cdot e^{0}=-2=M. \]
Step 6: Plug into the expression.
\[ 2\alpha m+\beta M=2(2)\left(\tfrac12-e\right)+1(-2)=2-4e-2=-4e. \] \[ \boxed{-4e} \]