The values of A and B such that the function f(x) = * 20 c - 2 x,… - Vidyadip

MCQ

The values of $A$ and $B$ such that the function $f(x) = \left\{ {\begin{array}{*{20}{c}}{ - 2\sin x,}&{x \le - \frac{\pi }{2}}\\{A\sin x + B,}&{ - \frac{\pi }{2} < x < \frac{\pi }{2}}\\{\cos x,}&{x \ge \frac{\pi }{2}}\end{array}} \right.$, is continuous everywhere are

A
$A = 0,\,B = 1$
B
$A = 1,\,B = 1$
✓
$A = - 1,\,B = 1$
D
$A = - 1,\,B = 0$

✓

Answer

Correct option: C.

$A = - 1,\,B = 1$

c
(c) For continuity at all $x \in R,$ we must have

$f\left( { - \frac{\pi }{2}} \right) = \mathop {{\rm{lim}}}\limits_{x \to {{( - \pi /2)}^ - }} ( - 2\sin x)$

$ = \mathop {{\rm{lim}}}\limits_{x \to {{( - \pi /2)}^ + }} (A\sin x + B)$

==> $2 = - A + B$ .....$(i)$

and $f\left( {\frac{\pi }{2}} \right) = \mathop {\lim }\limits_{x \to {{(\pi /2)}^ - }} (A\sin x + B)$

$ = \mathop {{\rm{lim}}}\limits_{x \to {{(\pi /2)}^ + }} (\cos x)$

==> $0 = A + B$.....$(ii)$

From $(i)$ and $(ii),$ $A = - 1$ and $B = 1$.

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