ExamBro
ExamBro
MHT CET · Maths · Continuity and Differentiability

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

  1. A \(\frac{-7}{6}\)
  2. B \(\frac{4}{7}\)
  3. C \(\frac{-4}{7}\)
  4. D \(\frac{7}{6}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{-7}{6}\)

Step-by-step Solution

Detailed explanation

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