Let f ( x ) = 1 + x 2 e x and g ( x ) = f ′ ( x ) , then

g ′ ( x ) = ( x 2 + 1 ) 3 ( x − 1 ) ( x 3 − 3 x 2 + 5 x + 1 ) e x Now, x 3 − 3 x 2 + 5 x + 1 is strictly increasing and has a root in ( − 1 , 0 ) .

Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMaxima-MinimaMedium2 minQB

MathematicsMediummultiple choice

Let $f (x) = \frac{e ^{x}}{1 + x ^{2}}$ and $g (x) = f^{'} (x)$ , then

Answer:

B, D

Solution:

$g^{'} (x) = \frac{( x - 1 ) ( x ^{3} - 3 x ^{2} + 5 x + 1 ) e ^{x}}{( x ^{2} + 1 ) ^{3}}$

Now, $x^{3} - 3 x^{2} + 5 x + 1$ is strictly increasing and has a root in $(- 1, 0)$ .

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Maxima-Minima

⏱ 2mℹ️ Source: QB

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