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信息安全工程師每日一練試題內(nèi)容(2024/3/6)
試題1
從安全屬性對(duì)各種網(wǎng)絡(luò)攻擊進(jìn)行分類,阻斷攻擊是針對(duì)()的攻擊查看答案
試題參考答案:B
試題解析與討論:www.pokkc.com/st/3271121583.html
試題2
Linux系統(tǒng)的運(yùn)行日志存儲(chǔ)的目錄是( )。查看答案
試題參考答案:A
試題解析與討論:www.pokkc.com/st/3894728881.html
試題3
以下關(guān)于VPN的敘述中,正確的是()
A、VPN指的是用戶通過公用網(wǎng)絡(luò)建立的臨時(shí)的、安全的連接
B、VPN指的是用戶自己租用線路,和公共網(wǎng)絡(luò)物理上完全隔離的、安全的線路
C、VPN不能做到信息認(rèn)證和身份認(rèn)證
D、VPN只能提供身份認(rèn)證,不能提供數(shù)據(jù)加密的功能
查看答案
試題參考答案:A
試題解析與討論:www.pokkc.com/st/285048692.html
試題4
計(jì)算機(jī)網(wǎng)絡(luò)為了實(shí)現(xiàn)資源共享,采用協(xié)議分層設(shè)計(jì)思想,每層網(wǎng)絡(luò)協(xié)議都有地址信息,如網(wǎng)卡(MAC)地址、IP地址、端口地址和域名地址,以下有關(guān)上述地址轉(zhuǎn)換的描述錯(cuò)誤的是 ( )。查看答案
試題參考答案:A
試題解析與討論:www.pokkc.com/st/522428592.html
試題5
外部網(wǎng)關(guān)協(xié)議BGP是不同自治系統(tǒng)的路由器之間交換路由信息的協(xié)議,BGP-4使用四種報(bào)文:打開報(bào)文、更新報(bào)文、?;顖?bào)文和通知報(bào)文。其中用來確認(rèn)打開報(bào)文和周期性地證實(shí)鄰站關(guān)系的是( )。查看答案
試題參考答案:C
試題解析與討論:www.pokkc.com/st/411084710.html
試題6
一個(gè)密碼系統(tǒng)如果用E表示加密運(yùn)算,D表示解密運(yùn)算,M表示明文,C表示密文,則下面描述必然成立的是( )。查看答案
試題參考答案:B
試題解析與討論:www.pokkc.com/st/389293064.html
試題7
安全策略表達(dá)模型是一種對(duì)安全需求與安全策略的抽象概念模型,一般分為自主訪問控制模型和強(qiáng)制訪問控制模型。以下屬于自主訪問控制模型的是()查看答案
試題參考答案:B
試題解析與討論:www.pokkc.com/st/3272020263.html
試題8
Kerberos 是一個(gè)網(wǎng)絡(luò)認(rèn)證協(xié)議,其目標(biāo)是使用密鑰加密為客戶端/服務(wù)器應(yīng)用程序提供強(qiáng)身份認(rèn)證。以下關(guān)于 Kerberos 的說法中,錯(cuò)誤的是( )。查看答案
試題參考答案:D
試題解析與討論:www.pokkc.com/st/5737026640.html
試題9
數(shù)字簽名是對(duì)以數(shù)字形式儲(chǔ)存的消息就行某種處理,產(chǎn)生一種類似于傳統(tǒng)手書簽名功效的消息處理過程,一個(gè)數(shù)字簽名體制通常包括兩個(gè)部分,()查看答案
試題參考答案:A
試題解析與討論:www.pokkc.com/st/3270223694.html
試題10
Trust is typically interpreted as a subjective belief in the reliability, honesty and security of an entity on which we depend ( )our welfare .In online environments we depend on a wide spectrun of things , ranging from computer hardware,software and data to people and organizations. A security solution always assumes certain entities function according to specific policies.To trust is precisely to make this sort of assumptions , hence , a trusted entity is the same as an entity that is assumed to function according to policy . A consequence of this is that a trust component of a system must work correctly in order for the security of that system to hold, meaning that when a trusted( )fails , then the sytems and applications that depend on it can( )be considered secure.An often cited articulation of this principle is:" a trusted system or component is one that can break your security policy” ( which happens when the trust system fails ). The same applies to a trusted party such as a service provider ( SP for short )that is , it must operate according to the agreed or assumed policy in order to ensure the expected level of securty and quality of services . A paradoxical conclusion to be drawn from this analysis is that security assurance may decrease when increasing the number of trusted components and parties that a service infrastructure depends on . This is because the security of an infrastructure consisting of many.查看答案
試題參考答案:D、C、A、B、A
試題解析與討論:www.pokkc.com/st/389944612.html
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