The function f(x) = p ,[x + 1] + q[x - 1], where [x]is the greate… - Vidyadip

MCQ

The function $f(x) = p\,[x + 1] + q[x - 1],$ where $[x]$is the greatest integer function is continuous at $x = 1$, if

A
$p - q = 0$
✓
$p + q = 0$
C
$p = 0$
D
$q = 0$

✓

Answer

Correct option: B.

$p + q = 0$

b
(b) $f(x) = p[x + 1] + q[x - 1]$ and $f(1) = p[1 + 1] + q[0] = 2p$
This function will be continuous at $x = 1$, then $L\mathop {\lim }\limits_{x \to 1} f(x) = R\mathop {\lim }\limits_{x \to 1} f(x) = f(1)$
==> $\mathop {\lim }\limits_{h \to 0} f(1 - h) = \mathop {\lim }\limits_{h \to 0} f(1 + h) = f(1)$
==> $\mathop {\lim }\limits_{h \to 0} p\,[1 - h + 1] + q\,[1 - h - 1\,]$
$ = \mathop {\lim }\limits_{h \to 0} p\,[1 + h + 1] + q[1 + h - 1] = f(1)$
==> $\mathop {\lim }\limits_{h \to 0} p\,[2 - h] + q\,[ - h] = \mathop {\lim }\limits_{h \to 0} p\,[2 + h] + q\,[h] = f(1)$
==> $\mathop {\lim }\limits_{h \to 0} \,[p(1 - h) + q( - h - 1)]$$ = \mathop {\lim }\limits_{h \to 0} \,[p(1 + h) + q(h - 1)] = 2p$
==> $p - q = 2p\,\, \Rightarrow p + q = 0.$

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