Let [ t ] denote the greatest integer less than or equal to t . Let f :[0, ∞) rightarrow R be a function defined by f(x)= [x/2+3 ]-[√(x)] . Let S be the set of all points in the interval [0,8] at which f is not continuous. Then sum a in S a is equal to

Mathematics · Limits, Continuity and Differentiability

JEE Main 2024 — 6 April, Shift 2 — Question 21

Let [ $t$ ] denote the greatest integer less than or equal to t . Let $\mathrm{f}:[0, \infty) \rightarrow \mathrm{R}$ be a function defined by $f(x)=\left[\frac{x}{2}+3\right]-[\sqrt{x}]$ . Let $S$ be the set of all points in the interval $[0,8]$ at which f is not continuous. Then $\sum_{\mathrm{a} \in \mathrm{S}} \mathrm{a}$ is equal to $\qquad$

Answer: 17

Numerical answer — enter this value.

Step-by-step solution

$\left[\frac{x}{2}+3\right]$ is discontinuous at $x=2,4,6,8$

$\sqrt{\mathrm{x}}$ is discontinuous at $\mathrm{x}=1,4$

$\mathrm{F}(\mathrm{x})$ is discontinuous at $\mathrm{x}=1,2,6,8$

$\sum \mathrm{a}=1+2+6+8=17$

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Continuity

lim n arrow infty frac (1^2-1 )(n-1)+ (2^2-2 )(n-2)+ldots .+ ((n-1)^2-(n-1) ) × 1 (1^3+2^3+ldots .+n^3 )- (1^2+2^2+ldots .+n^2 ) is equal…

Answer: 17

Step-by-step solution

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