Question

What should be the value of λ such that the function defined below is continuous at x = π/2?

  \({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{{\rm{λ }}\cos {\rm{x}}}}{{\frac{{\rm{π }}}{2} - {\rm{x}}}}}&{{\rm{x}} \ne \frac{{\rm{π }}}{2}}\\ 1&{{\rm{x}} = \frac{{\rm{π }}}{2}} \end{array}} \right.\) 

Answer 1

Concept:

A function f(x) is continuous at x = a if,

Left limit = Right limit = Function value = Real and finite

A function is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

Analysis:

\(\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {{\rm{x}} \to \frac{{\rm{\pi }}}{2}} \end{array}{\rm{\;f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\frac{{\rm{\pi }}}{2}} \right)\)  for continuity

\(\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {{\rm{x}} \to \frac{{\rm{\pi }}}{2}} \end{array}{\rm{\;}}\frac{{{\rm{λ }}\cos {\rm{x}}}}{{\frac{{\rm{\pi }}}{2} - {\rm{x}}}} = 1\)

To find the LHS value, we can apply the L-Hospital’s rule

\(\begin{array}{*{20}{c}} {{\rm{lt}}}\\ {{\rm{x}} \to \frac{{\rm{\pi }}}{2}} \end{array}{\rm{\;}}\frac{{ - {\rm{λ }}\sin {\rm{x}}}}{{ - 1}} = 1\)

∴ λ = 1