Trắc nghiệm Phương trình lượng giác cơ bản Toán Lớp 11
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Câu 1:
Đạo hàm của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaaikdaciGGZbGaaiyAaiaac6gadaahaaWcbeqaaiaaikdaaaGc % caWG4bGaeyOeI0Iaci4yaiaac+gacaGGZbGaaGOmaiaadIhacqGHRa % WkcaWG4baaaa!44D3! y = 2{\sin ^2}x - \cos 2x + x\) là
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGinaiGacohacaGGPbGaaiOBaiaadIhacqGHRaWkciGG % ZbGaaiyAaiaac6gacaaIYaGaamiEaiabgUcaRiaaigdacaGGUaaaaa!4458! y' = 4\sin x + \sin 2x + 1.\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGinaiGacohacaGGPbGaaiOBaiaaikdacaWG4bGaey4k % aSIaaGymaiaac6caaaa!3FA1! y' = 4\sin 2x + 1.\)
C. y'=1
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGinaiGacohacaGGPbGaaiOBaiaadIhacqGHsislcaaI % YaGaci4CaiaacMgacaGGUbGaaGOmaiaadIhacqGHRaWkcaaIXaGaai % Olaaaa!451F! y' = 4\sin x - 2\sin 2x + 1.\)
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Câu 2:
Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacogacaGGVbGaaiiDaiaaiodacaWG4bGaeyOeI0YaaSaaaeaa % caaIXaaabaGaaGOmaaaaciGG0bGaaiyyaiaac6gacaaIYaGaamiEaa % aa!4384! y = \cot 3x - \frac{1}{2}\tan 2x\) có đạo hàm là
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaaIZaaabaGaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaI % YaaaaOGaaG4maiaadIhaaaGaey4kaSYaaSaaaeaacaaIXaaabaGaci % 4yaiaac+gacaGGZbWaaWbaaSqabeaacaaIYaaaaOGaaGOmaiaadIha % aaGaeyyXICnaaa!46A9! \frac{{ - 3}}{{{{\sin }^2}3x}} + \frac{1}{{{{\cos }^2}2x}} \cdot \)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaaIZaaabaGaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaI % YaaaaOGaaG4maiaadIhaaaGaey4kaSYaaSaaaeaacaaIXaaabaGaci % 4yaiaac+gacaGGZbWaaWbaaSqabeaacaaIYaaaaOGaaGOmaiaadIha % aaGaeyyXICnaaa!46A9! \frac{{ - 3}}{{{{\sin }^2}3x}} - \frac{1}{{{{\cos }^2}2x}} \cdot \)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaaIZaaabaGaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaI % YaaaaOGaaG4maiaadIhaaaGaeyOeI0YaaSaaaeaacaWG4baabaGaci % 4yaiaac+gacaGGZbWaaWbaaSqabeaacaaIYaaaaOGaaGOmaiaadIha % aaGaeyyXICnaaa!46F6! \frac{{ - 3}}{{{{\sin }^2}3x}} - \frac{x}{{{{\cos }^2}2x}} \cdot \)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHsislcaaIXaaabaGaci4CaiaacMgacaGGUbWaaWbaaSqabeaacaaI % YaaaaOGaamiEaaaacqGHsisldaWcaaqaaiaaigdaaeaaciGGJbGaai % 4BaiaacohadaahaaWcbeqaaiaaikdaaaGccaaIYaGaamiEaaaacqGH % flY1aaa!45F5! \frac{{ - 1}}{{{{\sin }^2}x}} - \frac{1}{{{{\cos }^2}2x}} \cdot \)
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Câu 3:
Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacogacaGGVbGaai4CamaabmaabaWaaSaaaeaacaaIYaGaeqiW % dahabaGaaG4maaaacqGHRaWkcaaIYaGaamiEaaGaayjkaiaawMcaaa % aa!4235! y = \cos \left( {\frac{{2\pi }}{3} + 2x} \right)\). Khi đó phương trình y'=0 có nghiệm là:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iabgkHiTmaalaaabaGaeqiWdahabaGaaG4maaaacqGHRaWkcaWG % RbGaaGOmaiabec8aWbaa!3FB9! x = - \frac{\pi }{3} + k2\pi \)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaeqiWdahabaGaaG4maaaacqGHRaWkdaWcaaqaaiaa % dUgacqaHapaCaeaacaaIYaaaaaaa!3EDC! x = \frac{\pi }{3} + \frac{{k\pi }}{2}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iabgkHiTmaalaaabaGaeqiWdahabaGaaG4maaaacqGHRaWkcaWG % RbGaeqiWdahaaa!3EFD! x = - \frac{\pi }{3} + k\pi \)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iabgkHiTmaalaaabaGaeqiWdahabaGaaG4maaaacqGHRaWkdaWc % aaqaaiaadUgacqaHapaCaeaacaaIYaaaaaaa!3FC9! x = - \frac{\pi }{3} + \frac{{k\pi }}{2}\)
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Câu 4:
Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaaikdaciGGJbGaai4BaiaacohacaWG4bWaaWbaaSqabeaacaaI % Yaaaaaaa!3D6D! y = 2\cos {x^2}\)có đạo hàm là
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % OmaiGacohacaGGPbGaaiOBaiaadIhadaahaaWcbeqaaiaaikdaaaaa % aa!3C5B! - 2\sin {x^2}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % inaiaadIhaciGGJbGaai4BaiaacohacaWG4bWaaWbaaSqabeaacaaI % Yaaaaaaa!3D55! - 4x\cos {x^2}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % OmaiaadIhaciGGZbGaaiyAaiaac6gacaWG4bWaaWbaaSqabeaacaaI % Yaaaaaaa!3D58! - 2x\sin {x^2}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % inaiaadIhaciGGZbGaaiyAaiaac6gacaWG4bWaaWbaaSqabeaacaaI % Yaaaaaaa!3D5A! - 4x\sin {x^2}\)
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Câu 5:
Đạo hàm của \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacohacaGGPbGaaiOBamaaCaaaleqabaGaaGOmaaaakiaaisda % caWG4baaaa!3D7D! y = {\sin ^2}4x\)là
A. 2sin8x
B. 8sin8x
C. sin8x
D. 4sin8x
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Câu 6:
Hàm số y = tan x – cot x có đạo hàm là:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaigdaaeaaciGGJbGaai4Baiaacohadaah % aaWcbeqaaiaaikdaaaGccaaIYaGaamiEaaaaaaa!3EED! y' = \frac{1}{{{{\cos }^2}2x}}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaisdaaeaaciGGZbGaaiyAaiaac6gadaah % aaWcbeqaaiaaikdaaaGccaaIYaGaamiEaaaaaaa!3EF5! y' = \frac{4}{{{{\sin }^2}2x}}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaisdaaeaaciGGJbGaai4Baiaacohadaah % aaWcbeqaaiaaikdaaaGccaaIYaGaamiEaaaaaaa!3EF0! y' = \frac{4}{{{{\cos }^2}2x}}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaigdaaeaaciGGZbGaaiyAaiaac6gadaah % aaWcbeqaaiaaikdaaaGccaaIYaGaamiEaaaaaaa!3EF2! y' = \frac{1}{{{{\sin }^2}2x}}\)
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Câu 7:
Hàm số y = cot x có đạo hàm là:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcqGHsisldaWcaaqaaiaaigdaaeaaciGGJbGaai4Baiaa % cohadaahaaWcbeqaaiaaikdaaaGccaWG4baaaaaa!3F1D! y' = - \frac{1}{{{{\sin }^2}x}}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcqGHsisldaWcaaqaaiaaigdaaeaaciGGJbGaai4Baiaa % cohadaahaaWcbeqaaiaaikdaaaGccaWG4baaaaaa!3F1D! y' = - \frac{1}{{{{\cos }^2}x}}\)
C. y' = tan x
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaaIXaGaey4kaSIaci4yaiaac+gacaGG0bWaaWbaaSqa % beaacaaIYaaaaOGaamiEaaaa!3F03! y' = 1 + {\cot ^2}x\)
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Câu 8:
Hàm số y = tan x có đạo hàm là:
A. y' = cot x
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaigdaaeaaciGGJbGaai4Baiaacohadaah % aaWcbeqaaiaaikdaaaGccaWG4baaaaaa!3E30! y' = \frac{1}{{{{\cos }^2}x}}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpdaWcaaqaaiaaigdaaeaaciGGZbGaaiyAaiaac6gadaah % aaWcbeqaaiaaikdaaaGccaWG4baaaaaa!3E35! y' = \frac{1}{{{{\sin }^2}x}}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiaacE % cacqGH9aqpcaaIXaGaeyOeI0IaciiDaiaacggacaGGUbWaaWbaaSqa % beaacaaIYaaaaOGaamiEaaaa!3F0B! y' = 1 - {\tan ^2}x\)
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Câu 9:
Hàm số y = cos x có đạo hàm là:
A. y' = sin x
B. y' = -sin x
C. y' = -cos x
D. y' = tan x
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Câu 10:
Hàm số y = sin x có đạo hàm là:
A. y' = cos x
B. y' = -cos x
C. y' = -sin x
D. y' = cot x
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Câu 11:
Tính \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGMbGaai4jamaabmaabaGaaGymaaGaayjkaiaawMcaaaqaaiabeA8a % QjaacEcadaqadaqaaiaaicdaaiaawIcacaGLPaaaaaaaaa!3E89! \frac{{f'\left( 1 \right)}}{{\varphi '\left( 0 \right)}}\). Biết rằng : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacI % cacaWG4bGaaiykaiabg2da9iaadIhadaahaaWcbeqaaiaaikdaaaaa % aa!3C21! f(x) = {x^2}\) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqOXdOMaai % ikaiaadIhacaGGPaGaeyypa0JaaGinaiaadIhacqGHRaWkciGGZbGa % aiyAaiaac6gadaWcaaqaaiabec8aWjaadIhaaeaacaaIYaaaaaaa!4408! \varphi (x) = 4x + \sin \frac{{\pi x}}{2}\).
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGMbGaai4jaiaacIcacaaIXaGaaiykaaqaaiabeA8aQjaacEcacaGG % OaGaaGimaiaacMcaaaGaeyypa0ZaaSaaaeaacaaI0aaabaGaaGioai % abgkHiTiabec8aWbaaaaa!4369! \frac{{f'(1)}}{{\varphi '(0)}} = \frac{4}{{8 - \pi }}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGMbGaai4jaiaacIcacaaIXaGaaiykaaqaaiabeA8aQjaacEcacaGG % OaGaaGimaiaacMcaaaGaeyypa0ZaaSaaaeaacaaIYaaabaGaaGioai % abgUcaRiabec8aWbaaaaa!435C! \frac{{f'(1)}}{{\varphi '(0)}} = \frac{2}{{8 + \pi }}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGMbGaai4jaiaacIcacaaIXaGaaiykaaqaaiabeA8aQjaacEcacaGG % OaGaaGimaiaacMcaaaGaeyypa0ZaaSaaaeaacaaI0aaabaGaeqiWda % haaaaa!41BA! \frac{{f'(1)}}{{\varphi '(0)}} = \frac{4}{\pi }\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGMbGaai4jaiaacIcacaaIXaGaaiykaaqaaiabeA8aQjaacEcacaGG % OaGaaGimaiaacMcaaaGaeyypa0ZaaSaaaeaacaaI0aaabaGaaGioai % abgUcaRiabec8aWbaaaaa!435E! \frac{{f'(1)}}{{\varphi '(0)}} = \frac{4}{{8 + \pi }}\)
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Câu 12:
Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgacaGGOaGaamiEaiaacMcacqGH9aqpdaWcaaqaaiGacoga % caGGVbGaai4CaiaadIhaaeaacaaIXaGaeyOeI0Iaci4CaiaacMgaca % GGUbGaamiEaaaaaaa!459C! y = f(x) = \frac{{\cos x}}{{1 - \sin x}}\). Giá trị biểu thức \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaadaWcaaqaaiabec8aWbqaaiaaiAdaaaaacaGLOaGaayzk % aaGaeyOeI0IabmOzayaafaWaaeWaaeaacqGHsisldaWcaaqaaiabec % 8aWbqaaiaaiAdaaaaacaGLOaGaayzkaaaaaa!41E8! f'\left( {\frac{\pi }{6}} \right) - f'\left( { - \frac{\pi }{6}} \right)\) là
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI0aaabaGaaG4maaaaaaa!377F! \frac{4}{3}\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI0aaabaGaaG4maaaaaaa!377F! \frac{4}{9}\)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI0aaabaGaaG4maaaaaaa!377F! \frac{4}{5}\)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI0aaabaGaaG4maaaaaaa!377F! \frac{2}{3}\)
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Câu 13:
Cho \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iGacogacaGGVbGaai4C % amaaCaaaleqabaGaaGOmaaaakiaadIhacqGHsislciGGZbGaaiyAai % aac6gadaahaaWcbeqaaiaaikdaaaGccaWG4baaaa!44E2! f\left( x \right) = {\cos ^2}x - {\sin ^2}x\). Giá trị \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaadaWcaaqaaiabec8aWbqaaiaaisdaaaaacaGLOaGaayzk % aaaaaa!3AFE! f'\left( {\frac{\pi }{4}} \right)\) bằng:
A. 2
B. 1
C. -2
D. 0
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Câu 14:
Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgacaGGOaGaamiEaiaacMcacqGH9aqpdaGcaaqaaiGacsha % caGGHbGaaiOBaiaadIhacqGHRaWkciGGJbGaai4BaiaacshacaWG4b % aaleqaaaaa!44DA! y = f(x) = \sqrt {\tan x + \cot x} \). Giá trị \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaadaWcaaqaaiabec8aWbqaaiaaisdaaaaacaGLOaGaayzk % aaaaaa!3AFE! f'\left( {\frac{\pi }{4}} \right)\) bằng
A. 0
B. 1
C. -2
D. 2
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Câu 15:
Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacI % cacaWG4bGaaiykaiabg2da9iaaikdaciGGZbGaaiyAaiaac6gadaqa % daqaamaalaaabaGaaGynaiabec8aWbqaaiaaiAdaaaGaey4kaSIaam % iEaaGaayjkaiaawMcaaaaa!4482! f(x) = 2\sin \left( {\frac{{5\pi }}{6} + x} \right)\). Giá trị \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaadaWcaaqaaiabec8aWbqaaiaaiAdaaaaacaGLOaGaayzk % aaaaaa!3B00! f'\left( {\frac{\pi }{6}} \right)\) bằng
A. -2
B. -1
C. 0
D. 2
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Câu 16:
Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpdaWc % aaqaaiaaikdaaeaaciGGJbGaai4Baiaacshadaqadaqaaiabec8aWj % aadIhaaiaawIcacaGLPaaaaaaaaa!4451! y = f\left( x \right) = \frac{2}{{\cot \left( {\pi x} \right)}}\) có f'(3) bằng
A. \(2\pi\)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aI0aWaaOaaaeaacaaIZaaaleqaaaGcbaGaaG4maaaacqGHflY1aaa!3AAA! \frac{{4\sqrt 3 }}{3} \cdot \)
C. 8
D. \(\pi\)
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Câu 17:
Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpciGG % ZbGaaiyAaiaac6gadaGcaaqaaiaadIhaaSqabaGccqGHRaWkciGGJb % Gaai4BaiaacohadaGcaaqaaiaadIhaaSqabaaaaa!4535! y = f\left( x \right) = \sin \sqrt x + \cos \sqrt x \). Giá trị \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaadaWcaaqaaiabec8aWnaaCaaaleqabaGaaGOmaaaaaOqa % aiaaigdacaaI2aaaaaGaayjkaiaawMcaaaaa!3CAE! f'\left( {\frac{{{\pi ^2}}}{{16}}} \right)\) bằng
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaOaaaeaaca % aIYaaaleqaaaaa!36CA! \sqrt 2 \)
B. 0
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaOaaaeaaca % aIYaaaleqaaaaa!36CA! -2\sqrt 2 \)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaOaaaeaaca % aIYaaaleqaaaaa!36CA! -\sqrt 2 \)
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Câu 18:
Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWG5bGaeyypa0JaamOza8aadaqadaqaa8qacaWG4baapaGaayjk % aiaawMcaa8qacqGH9aqpciGGZbGaaiyAaiaac6gapaGaaiika8qacq % aHapaCciGGZbGaaiyAaiaac6gacaWG4bWdaiaacMcaaaa!46BD! y = f\left( x \right) = \sin (\pi \sin x)\). Giá trị \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaadaWcaaqaaiabec8aWbqaaiaaiAdaaaaacaGLOaGaayzk % aaaaaa!3B00! f'\left( {\frac{\pi }{6}} \right)\) bằng:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % aHapaCdaGcaaqaaiaaiodaaSqabaaakeaacaaIYaaaaiabgwSixdaa % !3BA8! \frac{{\pi \sqrt 3 }}{2} \cdot \)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % aHapaCdaGcaaqaaiaaiodaaSqabaaakeaacaaIYaaaaiabgwSixdaa % !3BA8! \frac{{\pi }}{2} \cdot \)
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % aHapaCdaGcaaqaaiaaiodaaSqabaaakeaacaaIYaaaaiabgwSixdaa % !3BA8! -\frac{{\pi \sqrt 3 }}{2} \cdot \)
D. 0
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Câu 19:
Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaWaaOaaaeaacaaIYaaaleqaaaGcbaGaci4yaiaac+ga % caGGZbGaaG4maiaadIhaaaaaaa!3D76! y = \frac{{\sqrt 2 }}{{\cos 3x}}\). Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % WaaeWaaeaadaWcaaqaaiabec8aWbqaaiaaiodaaaaacaGLOaGaayzk % aaaaaa!3B11! y'\left( {\frac{\pi }{3}} \right)\) là:
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIZaWaaOaaaeaacaaIYaaaleqaaaGcbaGaaGOmaaaacqGHflY1aaa!3AA8! \frac{{3\sqrt 2 }}{2} \cdot \)
B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIZaWaaOaaaeaacaaIYaaaleqaaaGcbaGaaGOmaaaacqGHflY1aaa!3AA8! -\frac{{3\sqrt 2 }}{2} \cdot \)
C. 1
D. 0
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Câu 20:
Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpdaWc % aaqaaiGacogacaGGVbGaai4CaiaadIhaaeaacaaIXaGaey4kaSIaaG % OmaiGacohacaGGPbGaaiOBaiaadIhaaaaaaa!467D! y = f\left( x \right) = \frac{{\cos x}}{{1 + 2\sin x}}\). Chọn kết quả SAI
A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaadaWcaaqaaiabec8aWbqaaiaaiAdaaaaacaGLOaGaayzk % aaGaeyypa0JaeyOeI0YaaSaaaeaacaaI1aaabaGaaGinaaaacqGHfl % Y1aaa!40CB! f'\left( {\frac{\pi }{6}} \right) = - \frac{5}{4} \cdot \)
B. f'(0) = -2
C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaadaWcaaqaaiabec8aWbqaaiaaikdaaaaacaGLOaGaayzk % aaGaeyypa0JaeyOeI0YaaSaaaeaacaaIXaaabaGaaG4maaaacqGHfl % Y1aaa!40C2! f'\left( {\frac{\pi }{2}} \right) = - \frac{1}{3} \cdot \)
D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacqaHapaCaiaawIcacaGLPaaacqGH9aqpcqGHsislcaaI % Yaaaaa!3CE0! f'\left( \pi \right) = - 2\)
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Câu 21:
Phương trình \(\tan ^{2} x+5 \tan x-6=0\) có nghiệm là:
A. \(\begin{array}{l} x=\frac{\pi}{4}+k \pi ; x=\arctan (-6)+k \pi(k \in \mathbb{Z}) \end{array}\)
B. \(x=\frac{\pi}{4}+k 2 \pi ; x=\arctan (-6)+k 2 \pi(k \in \mathbb{Z}) \)
C. \(x=-\frac{\pi}{4}+k \pi ; x=\arctan (-6)+k 2 \pi(k \in \mathbb{Z}) \)
D. \(x=k \pi ; x=\arctan (-6)+k \pi(k \in \mathbb{Z})\)
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Câu 22:
Phương trình \(\sin ^{2} x+\sin ^{2} 2 x=1\) có nghiệm là:
A. \(\left[\begin{array}{l}x=\frac{\pi}{2}+k \pi \\ x=\pm \frac{\pi}{6}+k \pi\end{array}(k \in \mathbb{Z})\right.\)
B. \(\left[\begin{array}{l}x=\frac{\pi}{3}+k \frac{\pi}{2} \\ x=-\frac{\pi}{4}+k \pi\end{array}\right.\)
C. \(\left[\begin{array}{l}x=\frac{\pi}{12}+k \frac{\pi}{3} \\ x=-\frac{\pi}{3}+k \pi\end{array}\right.\)
D. Vô nghiệm.
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Câu 23:
Giải phương trình \(3 \cos ^{2} x+2 \cos x-5=0\)
A. \(x=k \pi\)
B. \(x=-\frac{\pi}{2}+k \pi\)
C. \(x=\frac{\pi}{2}+k 2 \pi\)
D. \(x=k 2 \pi\)
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Câu 24:
Các họ nghiệm của phương trình \(3 \sin ^{2} 2 x+3 \cos 2 x-3=0\)
A. \(k \pi ; \frac{\pi}{4}+k \frac{\pi}{2}\)
B. \(k \pi ;-\frac{\pi}{4}+k \frac{\pi}{2}\)
C. \(k \pi ; \frac{\pi}{4}+k \pi\)
D. \(k \pi ;-\frac{\pi}{4}+k \pi\)
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Câu 25:
Họ nghiệm của phương trình \(3 \cos 4 x+2 \cos 2 x-5=0\) là:
A. \(k 2 \pi\)
B. \(\frac{\pi}{3}+k 2 \pi\)
C. \(k \pi\)
D. \(-\frac{\pi}{3}+k 2 \pi\)
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Câu 26:
Họ nghiệm của phương trình \(\cos ^{2} 2 x-\cos 2 x-2=0\) là:
A. \(\frac{\pi}{2}+k \pi\)
B. \(-\frac{\pi}{2}+\frac{k \pi}{2}\)
C. \(\frac{-\pi}{2}+k 2 \pi\)
D. \(\frac{\pi}{2}+k 2 \pi\)
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Câu 27:
Phương trình \(\sin ^{2} 2 x-2 \cos ^{2} x+\frac{3}{4}=0\) có nghiệm là
A. \(x=\pm \frac{\pi}{6}+k \pi, k \in \mathbb{Z}\)
B. \(x=\pm \frac{\pi}{4}+k \pi, k \in \mathbb{Z}\)
C. \(x=\pm \frac{\pi}{3}+k \pi,, k \in \mathbb{Z}\)
D. \(x=\pm \frac{2 \pi}{3}+k \pi, k \in \mathbb{Z}\)
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Câu 28:
Phương trình lượng giác: \(\cos ^{2} x+2 \cos x-3=0\) có nghiệm là
A. \(x=k 2 \pi, k \in \mathbb{Z}\)
B. \(x=0\)
C. \(x=\frac{\pi}{2}+k 2 \pi, k \in \mathbb{Z}\)
D. Vô nghiệm
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Câu 29:
Phương trình lượng giác: \(\sin ^{2} x-3 \cos x-4=0\) có nghiệm là
A. \(x=-\frac{\pi}{2}+k 2 \pi, k \in \mathbb{Z}\)
B. \(x=-\pi+k 2 \pi, k \in \mathbb{Z}\)
C. \(x=\frac{\pi}{6}+k \pi, k \in \mathbb{Z}\)
D. Vô nghiệm
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Câu 30:
Phương trình \(2 \cos ^{2} x+3 \cos x-2=0\) có nghiệm là
A. \(\begin{aligned} &\pm \frac{\pi}{6}+k 2 \pi, k \in \mathbb{Z} \end{aligned}\)
B. \(\pm \frac{\pi}{3}+k 2 \pi, k \in \mathbb{Z}\)
C. \(\begin{aligned} &\pm \frac{2 \pi}{3}+k 2 \pi, k \in \mathbb{Z} \end{aligned}\)
D. \(\frac{\pi}{3}+k 2 \pi, k \in \mathbb{Z}\)
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Câu 31:
Nghiệm của pt \(2 \cos 2 x+2 \cos x-\sqrt{2}=0\)
A. \(x=\pm \frac{\pi}{4}+k 2 \pi\)
B. \(x=\pm \frac{\pi}{4}+k \pi\)
C. \(x=\pm \frac{\pi}{3}+k 2 \pi\)
D. \(x=\pm \frac{\pi}{3}+k \pi\)
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Câu 32:
Nghiệm của phương trình \(3 \cos ^{2} x=-8 \cos x-5\)
A. \(x=k \pi\)
B. \(x=\pi+k 2 \pi\)
C. \(x=k 2 \pi\)
D. \(x=\pm \frac{\pi}{2}+k 2 \pi\)
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Câu 33:
Nghiệm của phương trình \(\cos ^{2} x+\cos x=0\) thỏa điều kiện:\(\frac{\pi}{2}<x<\frac{3 \pi}{2}\) là
A. \(x=\pi\)
B. \(x=\frac{\pi}{3}\)
C. \(x=\frac{3 \pi}{2}\)
D. \(x=-\frac{3 \pi}{2}\)
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Câu 34:
Nghiệm của phương trình \(\cos ^{2} x-\cos x=0\) thỏa điều kiện \(0<x<\pi:\)
A. \(x=\frac{\pi}{6}\)
B. \(x=\frac{\pi}{2}\)
C. \(x=\frac{\pi}{4}\)
D. \(x=-\frac{\pi}{2}\)
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Câu 35:
Phương trình : \(\cos ^{2} 2 x+\cos 2 x-\frac{3}{4}=0\) có nghiệm là
A. \(x=\pm \frac{2 \pi}{3}+k \pi, k \in \mathbb{Z}\)
B. \(x=\pm \frac{\pi}{3}+k \pi, k \in \mathbb{Z}\)
C. \(x=\pm \frac{\pi}{6}+k \pi, k \in \mathbb{Z}\)
D. \(x=\pm \frac{\pi}{6}+k 2 \pi, k \in \mathbb{Z}\)
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Câu 36:
Phương trình: \(\sin ^{2} \frac{x}{3}-2 \cos \frac{x}{3}+2=0\) có nghiệm là:
A. \(x=k \pi, k \in \mathbb{Z}\)
B. \(x=k 3 \pi, k \in \mathbb{Z}\)
C. \(x=k 2 \pi, k \in \mathbb{Z}\)
D. \(x=k 6 \pi, k \in \mathbb{Z}\)
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Câu 37:
Phương trình nào sau đây vô nghiệm:
A. \(\sin x+3=0\)
B. \(2 \cos ^{2} x-\cos x-1=0\)
C. \(\tan x+3=0\)
D. \(3 \sin x-2=0\)
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Câu 38:
Phương trình \(\cos 2 x+2 \cos x-11=0\) có tập nghiệm là:
A. \(\begin{aligned} &x=\arccos (-3)+k 2 \pi, k \in \mathbb{Z}, x=\arccos (-2)+k 2 \pi, k \in \mathbb{Z} \end{aligned}\)
B. \(\varnothing\)
C. \(x=\arccos (-2)+k 2 \pi, k \in \mathbb{Z}\)
D. \(x=\arccos (-3)+k 2 \pi, k \in \mathbb{Z}\)
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Câu 39:
Giải phương trình \(2 \cos ^{2} x-3 \cos x+1=0\)
A. \(x=-\frac{\pi}{3}+k 2 \pi, k \in \mathbb{Z}\)
B. \(\left\{k 2 \pi, \pm \frac{\pi}{3}+k 2 \pi, k \in \mathbb{Z}\right\}\)
C. \(x=\frac{\pi}{3}+k 2 \pi, k \in \mathbb{Z}\)
D. \(x=k 2 \pi, k \in \mathbb{Z}\)
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Câu 40:
Tìm tất cả các họ nghiệm của phương trình \(\cos ^{2} x-4 \cos x+3=0\)
A. \(x=\pi+k 2 \pi(k \in \mathbb{Z})\)
B. \(x=\frac{\pi}{2}+k 2 \pi(k \in \mathbb{Z})\)
C. \(x=k 2 \pi(k \in \mathbb{Z})\)
D. \(x=k \pi(k \in \mathbb{Z})\)
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Câu 41:
Phương trình \(\cos 2\left(x+\frac{\pi}{3}\right)+4 \cos \left(\frac{\pi}{6}-x\right)=\frac{5}{2}\) có nghiệm là:
A. \(\left[\begin{array}{l} x=-\frac{\pi}{6}+k 2 \pi \\ x=\frac{\pi}{2}+k 2 \pi \end{array}\right.\)
B. \(\left[\begin{array}{l}x=\frac{\pi}{6}+k 2 \pi \\ x=\frac{3 \pi}{2}+k 2 \pi\end{array}\right.\)
C. \(\left[\begin{array}{l}x=-\frac{\pi}{3}+k 2 \pi \\ x=\frac{5 \pi}{6}+k 2 \pi\end{array}\right.\)
D. \(\left[\begin{array}{l} x=\frac{\pi}{3}+k 2 \pi \\ x=\frac{\pi}{4}+k 2 \pi \end{array}\right.\)
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Câu 42:
Giải phương trình lượng giác \(4 \sin ^{4} x+12 \cos ^{2} x-7=0\) có nghiệm là:
A. \(x=\pm \frac{\pi}{4}+k 2 \pi\)
B. \(x=\frac{\pi}{4}+k \frac{\pi}{2}\)
C. \(x=\frac{\pi}{4}+k \pi\)
D. \(x=-\frac{\pi}{4}+k \pi\)
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Câu 43:
Giải phương trình \(\sin ^{2} x+2 \sin x-3=0\)
A. \(k \pi\)
B. \(-\frac{\pi}{2}+k \pi\)
C. \(\frac{\pi}{2}+k 2 \pi\)
D. \(-\frac{\pi}{2}+k 2 \pi\)
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Câu 44:
Nghiệm của phương trình \(\sin ^{2} 2 x+2 \sin 2 x+1=0 \) trong khoảng \((-\pi ; \pi)\) là :
A. \(\left\{-\frac{\pi}{4} ;-\frac{3 \pi}{4}\right\}\)
B. \(\left\{-\frac{\pi}{4} ; \frac{3 \pi}{4}\right\}\)
C. \(\left\{\frac{\pi}{4} ; \frac{3 \pi}{4}\right\}\)
D. \(\left\{\frac{\pi}{4} ;-\frac{3 \pi}{4}\right\}\)
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Câu 45:
Một họ nghiệm của phương trình \(2 \cos 2 x+3 \sin x-1=0\)
A. \(\pi+\arcsin \left(-\frac{1}{4}\right)+k 2 \pi\)
B. \(\pi-\arcsin \left(-\frac{1}{4}\right)+k 2 \pi\)
C. \(\frac{\pi}{2}-\frac{1}{2} \arcsin \left(-\frac{1}{4}\right)+k \pi\)
D. \(\frac{\pi}{2}-\arcsin \left(-\frac{1}{4}\right)+k \pi\)
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Câu 46:
Một họ nghiệm của phương trình \(\cos ^{2} 2 x+\sin 2 x-1=0\) là:
A. \(\frac{\pi}{2}+k \pi\)
B. \(k \frac{\pi}{3}\)
C. \(-\frac{\pi}{2}+k \frac{\pi}{2}\)
D. \(k \frac{\pi}{2}\)
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Câu 47:
Họ nghiệm của phương trình \(\sin ^{2} 2 x-2 \sin 2 x+1=0\) là:
A. \(-\frac{\pi}{4}+k \pi\)
B. \(\frac{\pi}{4}+k \pi\)
C. \( \frac{\pi}{4}+k 2 \pi\)
D. \(-\frac{\pi}{4}+k 2 \pi\)
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Câu 48:
Nghiệm của phương trình \(5-5 \sin x-2 \cos ^{2} x=0\) là:
A. \(k \pi, k \in \mathbb{Z}\)
B. \(k 2 \pi, k \in \mathbb{Z}\)
C. \(\frac{\pi}{2}+k 2 \pi, k \in \mathbb{Z}\)
D. \(\frac{\pi}{6}+k 2 \pi, k \in \mathbb{Z}\)
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Câu 49:
Nghiệm của phương trình \(1-5 \sin x+2 \cos ^{2} x=0\) là:
A. \(\left[\begin{array}{c}x=\frac{\pi}{6}+k 2 \pi \\ x=-\frac{\pi}{6}+k 2 \pi\end{array}, k \in \mathbb{Z}\right.\)
B. \(\left[\begin{array}{c}x=\frac{\pi}{6}+k 2 \pi \\ x=\frac{5 \pi}{6}+k 2 \pi\end{array}, k \in \mathbb{Z}\right.\)
C. \(\left[\begin{array}{c}x=\frac{\pi}{3}+k 2 \pi \\ x=-\frac{\pi}{3}+k 2 \pi\end{array}, k \in \mathbb{Z}\right.\)
D. \(\left[\begin{array}{l}x=\frac{\pi}{3}+k 2 \pi \\ x=\frac{2 \pi}{3}+k 2 \pi\end{array}, k \in \mathbb{Z}\right.\)
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Câu 50:
Nghiệm của phương trình lượng giác:\(2 \cos ^{2} x+3 \sin x-3=0\) thõa điều kiện \(0<x<\frac{\pi}{2}\) là:
A. \(x=\frac{\pi}{3}\)
B. \(x=\frac{\pi}{2}\)
C. \(x=\frac{\pi}{6}\)
D. \(x=\frac{5 \pi}{6}\)
