Q76 / 150 · MHT CET 2020 (12 Oct Shift 1) · Mathematics

If $\begin{aligned} f(x) &=6 \beta-3 \propto x, \text { if }-4 \leq x < -2 \\ &=4 x+1, \text { if }-2 \leq x \leq 2 \end{aligned}$ is continuous on $[-4,2]$, then $\propto+\beta=$

  • Correct answer
    A
    $\frac{-7}{6}$
  • B
    $\frac{4}{7}$
  • C
    $\frac{-4}{7}$
  • D
    $\frac{7}{6}$

Solution

$\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{-}}(6 \beta-3 \alpha x)$ $=6 \beta+6 \alpha$ ...(1) $\lim _{x \rightarrow-2^{+}} f(x)=\lim _{x \rightarrow-2^{+}}(4 x+1)=-7$ ...(2) Given $f(x)$ is continuous on $[-4,2]$ $\therefore 6 \beta+6 \alpha=-7$ $\therefore \alpha+\beta=\frac{-7}{6}$

Hence, the correct answer is option: 'A'.

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