Question

63. \( f(x)=e^{-\sqrt{x}}+e^{\frac{-1}{x^{2}}} \) If \( f ^{\prime \prime}(x)=\alpha \cdot \frac{e^{-\sqrt{x}}}{x}\left(1+\frac{1}{\sqrt{x}}\right)+\beta \frac{e^{\frac{-1}{x^{2}}}}{x^{4}}\left(3-\frac{2}{x^{2}}\right) \) then \( (\alpha, \beta)= \) 1) \( \left(\frac{1}{4}, 2\right) \) 2) \( \left(\frac{1}{4},-2\right) \) 3) \( \left(-\frac{1}{4}, 2\right) \) 4) \( \left(-\frac{1}{4},-2\right) \)

Answer 1

Correct Answer: Option (2) \(\left({1\over4},-2\right)\)

Now, 

\(f'(x) = {d(e^{-{\sqrt x}})\over dx} + {d(e^{-1\over x^2})\over dx}\)    

\(f'(x) = {-e^{-{\sqrt x}}\over 2\sqrt{x}} + {2e^{-1\over x^2}\over x^3} \)   

And so,

\(f''(x) = {1\over 4}\cdot {e^{-\sqrt{x}}\over x}\left({1\over \sqrt x} +1\right) + (-2)\cdot {e^{-1\over x^2}\over x^4}\left(3 - {2\over x^2}\right)\)    

Comparing the coefficients of respectively terms, 

we get the ordered pair (\(\alpha,\beta\)) as \(\left({1\over4},-2\right)\).