Limit Continuity And Differentiability Ques 87
- $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\int _2^{\sec ^{2} x} f(t) d t}{x^{2}-\frac{\pi^{2}}{16}}$ equals
(a) $\frac{8}{\pi} f(2)$
(b) $\frac{2}{\pi} f(2)$
(c) $\frac{2}{\pi} f \frac{1}{2}$
(d) $4 f(2)$
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Answer:
Correct Answer: 87.(1)
Solution:
- NOTE All integers are critical points for greatest integer function. Case I When $x \in \mathbb{Z}$
$$ f(x)=[x]^{2}-\left\lfloor x^{2} \right\rfloor=x^{2}-\left\lfloor x^{2} \right\rfloor $$
Case II When $x \notin I$ If $0<x<1$, then $[x]=0$
$$ \text { and } \quad 0<x^{2}<1 \text {, then }\left[x^{2}\right]=0 $$
Next, if $\quad 1 \leq x^{2}<2 \Rightarrow -\sqrt{2} < x \leq -1 \text{ or } 1 \leq x < \sqrt{2}$
$$ \Rightarrow \quad[x]=1 \quad \text { and } \quad\left[x^{2}\right]=1 $$
Therefore, $f(x)=[x]^{2}-\left\lfloor x^{2} \right\rfloor=0$, if $1 \leq x<\sqrt{2}$
Therefore, $\quad f(x)=0$, if $0 \leq x<\sqrt{2}$
This shows that $f(x)$ is continuous at $x=1$.
Therefore, $f(x)$ is discontinuous in $(-\infty, 0) \cup (\sqrt{2}, \infty)$ on many other points. Therefore, (b) is the answer.
BETA
