Question

Let $\alpha,\beta\in\mathbb{R}$ be such that the function \[ f(x)= \begin{cases} 2\alpha(x^2-2)+2\beta x, & x<1 \\ (\alpha+3)x+(\alpha-\beta), & x\ge1 \end{cases} \] is differentiable at all $x\in\mathbb{R}$. Then $34(\alpha+\beta)$ is equal to}
 

• A. 48
• B. 84
• C. 24
• D. 36

Hint:

For differentiability, always check both continuity and equality of derivatives at the point.

Answer 1

The Correct Option is A

Step 1: Continuity at $x=1$.
Left limit: \[ \lim_{x\to1^-}f(x)=2\alpha(1-2)+2\beta(1)=-2\alpha+2\beta \] Right value: \[ f(1)=(\alpha+3)+(\alpha-\beta)=2\alpha+3-\beta \] Equating, \[ -2\alpha+2\beta=2\alpha+3-\beta \Rightarrow 4\alpha-3\beta=-3 \] Step 2: Differentiability at $x=1$.
Left derivative: \[ f'(x)=4\alpha x+2\beta \Rightarrow f'(1^-)=4\alpha+2\beta \] Right derivative: \[ f'(x)=\alpha+3 \Rightarrow f'(1^+)=\alpha+3 \] Equating, \[ 4\alpha+2\beta=\alpha+3 \Rightarrow 3\alpha+2\beta=3 \] Step 3: Solving the equations.
\[ \begin{cases} 4\alpha-3\beta=-3 \\ 3\alpha+2\beta=3 \end{cases} \] Solving, \[ \alpha=1,\quad \beta=\frac{1}{2} \] Step 4: Final calculation.
\[ 34(\alpha+\beta)=34\left(1+\frac{1}{2}\right)=48 \]