Let a function \( f(x)=\alpha+(\beta^2+5\beta+6)|x|+\gamma|x|^4 \), where \( \alpha,\beta \) and \( \gamma \) are real number. Then the function \( f(x) \) is differentiable at \( x=0 \) if
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \beta=-2 \) or \( \beta=-3 \) or both
At x = 0 the three terms of f behave as follows:
\( \displaystyle \alpha \text{ is a constant: differentiable} \)
\( \displaystyle \gamma|x|^4=\gamma x^4 \text{ is a polynomial: differentiable} \)
\( \displaystyle |x| \text{ is not differentiable at } x=0 \text{ (left derivative } -1\text{, right derivative } 1\text{)} \)
So f is differentiable at x = 0 only if the coefficient of |x| is 0, whatever α and γ are:
\( \displaystyle \beta^2+5\beta+6=0 \)
\( \displaystyle (\beta+2)(\beta+3)=0 \)
\( \displaystyle \beta=-2 \text{ or } \beta=-3 \)