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JEE Mains · Maths · STD 11 - 12. limits

\(f\left( x \right) = 5 - \left| {x - 2} \right|\) અને \(g\left( x \right) = \left| {x + 1} \right|,x \in R\). જો \(f(x)\) એ \(\alpha \) આગળ મહત્તમ અને \(g(x)\) એ \(\beta \) આગળ મહત્તમ થાય તો \(\mathop {\lim }\limits_{x \to \alpha \beta } \frac{{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)}}{{{x^2} - 6x + 8}}\) = 

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

Answer & Solution

Correct Answer

(C) \(\frac{1}{2}\)

Step-by-step Solution

Detailed explanation

\(f\left( x \right) = 5 - \left| {x - 2} \right|\) \(f\left( x \right)\) attains maximum value when \(\left| {x - 2} \right| = 0 \Rightarrow x = 2 = \alpha \) \(g\left( x \right) = \left| {x + 1} \right|\) \(g\left( x \right)\) attins minimum value of \(x = - 1 = \beta \)…
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