Limit Continuity and Differentiability 4 Question 2
2. The function $f(x)=[x]^{2}-\left[x^{2}\right]$ (where, $[x]$ is the greatest integer less than or equal to $x$ ), is discontinuous at
(a) all integers
$(1999,2 M)$
(b) all integers except 0 and 1
(c) all integers except 0
(d) all integers except 1 Objective Questions II
(One more than one correct option)
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Answer:
Correct Answer: 2. (b)
Solution:
- Given function
$$ f(x)=\begin{array}{cc} \frac{\sin (p+1) x+\sin x}{x} & , x<0 \\ q & , x=0 \\ \frac{\sqrt{x+x^{2}}-\sqrt{x}}{x^{3 / 2}} & , x>0 \end{array} $$
is continuous at $x=0$, then
$$ \begin{aligned} & f(0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \ldots \text { (i) } \\ & \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{\sin (p+1) x+\sin x}{x} \\ & =p+1+1=p+2 \quad \because \lim _{x \rightarrow 0} \frac{\sin (a x)}{x}=a \\ & \text { and } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x+x^{2}}-\sqrt{x}}{x^{3 / 2}} \\ & =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}\left[(1+x)^{1 / 2}-1\right]}{x \sqrt{x}} \\ & 1+\frac{1}{2} x+\frac{\frac{1}{2} \frac{1}{2}-1}{2 !} x^{2}+\ldots .-1 \\ & =\lim _{x \rightarrow 0^{+}} \frac{}{x} \\ & {\left[\because(1+x)^{n}\right.} \\ & \left.=1+n x+\frac{n(n-1)}{1 \cdot 2} x^{2}+\frac{n(n-1(n-2))}{1 \cdot 2 \cdot 3} x^{3}+\ldots,|x|<1\right] \end{aligned} $$
$=\lim _{x \rightarrow 0^{+}} \frac{1}{2}+\frac{\frac{1}{2} \frac{1}{2}-1}{2 !} x+\ldots=\frac{1}{2}$
From Eq. (i), we get
$$ \begin{aligned} & & f(0) & =q=\frac{1}{2} \text { and } \lim _{x \rightarrow 0^{-}} f(x)=p+2=\frac{1}{2} \\ \Rightarrow & & p & =-\frac{3}{2} \\ \text { So, } & & (p, q) & =-\frac{3}{2}, \frac{1}{2} \end{aligned} $$
