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Let a function f(x)=α+(β2+5β+6)∣x∣+γ∣x∣4f(x)=\alpha+(\beta^2+5\beta+6)|x|+\gamma|x|^4, where α,β\alpha, \beta and γ\gamma are real number. Then the function f(x)f(x) is differentiable at x=0x=0 if

Solution

✅ Correct Option: 1

The terms α\alpha and γ∣x∣4=γx4\gamma|x|^4=\gamma x^4 are differentiable everywhere, so only the ∣x∣|x| term can create a corner at x=0x=0.

We therefore need β2+5β+6=0\beta^2+5\beta+6=0, i.e. (β+2)(β+3)=0(\beta+2)(\beta+3)=0, giving β=−2\beta=-2 or β=−3\beta=-3.

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