Let α, β ∈ ℝ be such that the function f(x) = { 2α(x² − 2) + 2βx , x < 1 (α + 3)x + (α − β) , x ≥ 1 } be differentiable at all x ∈ ℝ. Then 34(α + β) is equal to

Let α, β ∈ ℝ be such that the function f(x) = { 2α(x² − 2) + 2βx , x < 1 ; (α + 3)x + (α − β) , x ≥ 1 } be differentiable at all x ∈ ℝ. Then 34(α + β) is equal to

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

(A) 48

(B) 84

(D) 24

Correct Answer: 48

Explanation

For differentiability at $x=1$, the function must be continuous and have equal derivatives from left and right at $x=1$.

Continuity at $x=1$:

Left hand value at $x=1$:

$$ 2\alpha(1^2-2)+2\beta(1)=-2\alpha+2\beta $$

Right hand value at $x=1$:

$$ (\alpha+3)(1)+(\alpha-\beta)=2\alpha+3-\beta $$

Equating both:

$$ -2\alpha+2\beta=2\alpha+3-\beta $$

$$ 4\alpha-3\beta+3=0 \quad (1) $$

Differentiability at $x=1$:

Derivative for $x<1$:

$$ f'(x)=4\alpha x+2\beta $$

Left hand derivative at $x=1$:

$$ 4\alpha+2\beta $$

Derivative for $x\ge1$:

$$ f'(x)=\alpha+3 $$

Equating derivatives:

$$ 4\alpha+2\beta=\alpha+3 $$

$$ 3\alpha+2\beta=3 \quad (2) $$

Solving equations (1) and (2):

From (2),

$$ 2\beta=3-3\alpha \Rightarrow \beta=\frac{3-3\alpha}{2} $$

Substitute in (1):

$$ 4\alpha-3\left(\frac{3-3\alpha}{2}\right)+3=0 $$

$$ 8\alpha-9+9\alpha+6=0 $$

$$ 17\alpha=3 \Rightarrow \alpha=\frac{3}{17} $$

$$ \beta=\frac{21}{17} $$

$$ \alpha+\beta=\frac{24}{17} $$

$$ 34(\alpha+\beta)=34\times\frac{24}{17}=48 $$

Explanation

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