First slide

Continuity

Question

$Value of A and B, so that the function f (x) defined by$ $f (x) = \{\begin{cases} x + A \sqrt{2} \sin x if 0 \leq x < \frac{π}{4}, \\ 2 x + \cot x + B if \frac{π}{4} \leq x < \frac{π}{2}, becomes continuous are \\ A \cos 2 x - B \sin x if \frac{π}{2} \leq x \leq π \end{cases}$

Moderate

A

$A = π / 6, B = - π / 12$
B

$A = - π / 6, B = π / 12$
C

$A = π / 6, B = π / 12$
D

$A = - π / 6, B = - π / 12$

Solution

$\begin{array}{l} Here we will check continuity at x = π / 4 a n d x = π / 2 \\ A t x = π / 4 . \\ \lim_{x \to \frac{π^{-}}{4}} f (x) = \lim_{x \to \frac{π^{-}}{4}} (x + A \sqrt{2} \sin x) = \frac{π}{4} + A \\ and \lim_{x \to \frac{π^{+}}{4}} f (x) = \lim_{x \to \frac{π^{+}}{4}} (2 x \cot x + B) = \frac{π}{2} + B \\ F o r c o n f i n u i t y a t x = π / 4 \\ \frac{π}{4} + A = \frac{π}{2} + B \Rightarrow A - B = \frac{π}{4} \\ Again, \lim_{x \to \frac{π^{-}}{2}} f (x) = \lim_{x \to \frac{π^{-}}{2}} (2 x \cot x + B) = B \\ and, \lim_{x \to \frac{π^{+}}{2}} f (x) = \lim_{x \to \frac{π^{+}}{2}} (2 A \cos 2 x - B \sin x) = - A - B, \\ so f is continuous at this point if \\ - A - B = B \Rightarrow A = - 2 B \\ Solving (i) and (ii), A=π/6 and B = - π / 12 \\ Hence (1)is the correct answer. \end{array}$

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