Question

Let \(f:\mathbb{R}\to\mathbb{R}\), \(f(x)=|x+1|e^{-x^2}\), then which of the following option is true?

• A. \(f\) has point of maxima in \((-2,-1)\)
• B. \(f\) has point of global maxima in \((1,2)\)
• C. \(f\) has point of local minima in \((0,1)\)
• D. \(f\) has point of global minima in \((0,1)\)

Hint:

Whenever a function contains \(|x-a|\), split the analysis into two regions: \(x\ge a\) and \(x<a\). Then differentiate each piece separately.

Answer 1

The Correct Option is A

Concept: Since the modulus function changes its form at \(x=-1\), we analyze the function separately on the intervals \(x\ge -1\) and \(x<-1\).

Step 1: Remove the modulus sign.
For \(x\ge -1\), \[ f(x)=(x+1)e^{-x^2}. \] For \(x<-1\), \[ f(x)=-(x+1)e^{-x^2}. \]

Step 2: Differentiate for \(x<-1\).
\[ f'(x) = -e^{-x^2} + 2x(x+1)e^{-x^2}. \] Thus, \[ f'(x) = e^{-x^2} \left(2x^2+2x-1\right). \] Critical points satisfy \[ 2x^2+2x-1=0. \] \[ x=\frac{-1\pm \sqrt3}{2}. \] Only \[ x=\frac{-1-\sqrt3}{2}\approx -1.366 \] lies in \((-2,-1)\).

Step 3: Check nature of critical point.
The derivative changes from positive to negative at \[ x=\frac{-1-\sqrt3}{2}. \] Hence this point is a local maximum. Since it lies in \((-2,-1)\), statement (A) is true.

Step 4: Reject remaining options.
At \(x=-1\), \[ f(-1)=0. \] Since \(f(x)\ge 0\) for all \(x\), this is the global minimum. Thus options (C) and (D) are false. The maximum point is not in \((1,2)\), hence (B) is false. \[ \boxed{\text{Option (A)}} \]