Question

Which one of the following options represents the given graph? \begin{center} \includegraphics[width=0.5\textwidth]{02.jpeg} \end{center}

• A. \(f(x)=x^{2} 2^{-|x|}\)
• B. \(f(x)=x\,2^{-|x|}\)
• C. \(f(x)=|x|\,2^{-x}\)
• D. \(f(x)=x\,2^{-x}\)

Hint:

To identify a function from a sketch, check: (i) sign on each side of 0 (odd/even clues), (ii) behavior as \(x\to\pm\infty\), and (iii) local maxima/minima. Exponential factors like \(2^{-|x|}\) give rapid decay to 0 on both sides.

Answer 1

The Correct Option is B

Step 1: Read qualitative features of the curve.
- \(f(0)=0\) (graph passes through the origin).
- For \(x>0\): \(f(x)>0\), rises to a peak, then decays towards \(0\) as \(x\to+\infty\).
- For \(x<0\): \(f(x)<0\), has a minimum (most negative) for some \(x<0\), and approaches \(0^{-}\) as \(x\to-\infty\).
- The left and right sides look like mirror images with opposite sign \(\Rightarrow\) function is odd-like in sign behavior.

Step 2: Match to candidates.
(A) \(x^{2}2^{-|x|}\) is always \(\ge 0\) (even), so it cannot be negative for \(x<0\). ✗
(C) \(|x|2^{-x}\) is \(\ge 0\) for all \(x\). ✗
(D) \(x2^{-x}\): as \(x\to-\infty\), \(2^{-x}=2^{|x|}\to\infty\) so \(x2^{-x}\to-\infty\), not \(0^{-}\). ✗
(B) \(x2^{-|x|}\): for \(x>0\), \(2^{-|x|}=2^{-x}\Rightarrow f(x)=x2^{-x}>0\) with a single maximum and decay to \(0^{+}\); for \(x<0\), \(2^{-|x|}=2^{x}\Rightarrow f(x)=x2^{x}<0\) with a single minimum and approach to \(0^{-}\). This matches all features. ✓

Final Answer:\; \[ \boxed{f(x)=x\,2^{-|x|}} \]