Q) If 2cos^(βˆ’1) π‘₯ = 𝑦,then(A) 0β‰€π‘¦β‰€πœ‹(B) βˆ’πœ‹β‰€π‘¦β‰€πœ‹(C) 0≀𝑦≀2πœ‹(D) βˆ’πœ‹β‰€π² ≀ 0 #shivangmathsacademy
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Q) If 2cos^(βˆ’1) π‘₯ = 𝑦,then(A) 0β‰€π‘¦β‰€πœ‹(B) βˆ’πœ‹β‰€π‘¦β‰€πœ‹(C) 0≀𝑦≀2πœ‹(D) βˆ’πœ‹β‰€π² ≀ 0 #shivangmathsacademy
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Q) Simplify : sin(co𝑠𝑒𝑐^(βˆ’1) 17/8)}#class12maths #shivangmathsacademy #cbsemaths #cbsemaths
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Q) Simplify : sin(co𝑠𝑒𝑐^(βˆ’1) 17/8)}#class12maths #shivangmathsacademy #cbsemaths #cbsemaths
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Q) The value of tan[1/2 cos^(βˆ’1) (√5/3)] is (a) (3+√5)/2 (b) (3βˆ’βˆš5)/2(c) (βˆ’3+√g5)/2(d) (βˆ’3βˆ’βˆš5)/2
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Q) The value of tan[1/2 cos^(βˆ’1) (√5/3)] is (a) (3+√5)/2 (b) (3βˆ’βˆš5)/2(c) (βˆ’3+√g5)/2(d) (βˆ’3βˆ’βˆš5)/2
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Q) Check whether function f(x) defined as f(π‘₯)={((|π‘₯βˆ’3|)/(2(π‘₯βˆ’3)), π‘₯=3 (π‘₯βˆ’6)/6,π‘₯β‰₯3) Is
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Q) Check whether function f(x) defined as f(π‘₯)={((|π‘₯βˆ’3|)/(2(π‘₯βˆ’3)), π‘₯=3 (π‘₯βˆ’6)/6,π‘₯β‰₯3) Is
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Q) If 𝑓(π‘₯)={(sinπ‘₯/π‘₯+cosπ‘₯, π‘₯β‰ 0 π‘˜, π‘₯=0) is continuous at π‘₯=0, then the value of π‘˜ is :
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Q) If 𝑓(π‘₯)={(sinπ‘₯/π‘₯+cosπ‘₯, π‘₯β‰ 0 π‘˜, π‘₯=0) is continuous at π‘₯=0, then the value of π‘˜ is :
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Q) Simplify : tan^(βˆ’1) {√((1βˆ’cosπ‘₯)/(1+cosπ‘₯))}}#class12maths #shivangmathsacademy #cbsemaths
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Q) Simplify : tan^(βˆ’1) ((cos2π‘₯βˆ’sin2π‘₯)/(cos2π‘₯+sin2π‘₯)),0=π‘₯=πœ‹/4..#class12maths #shivangmathsaca
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Q) The function 𝑓 defined by 𝑓(π‘₯)={ (π‘₯,f π‘₯≀1 5, if π‘₯= 1 is not continuous at(A) x=0(B) π‘₯
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Find k so that 𝑓(π‘₯)={((π‘₯^2βˆ’2π‘₯βˆ’3)/(π‘₯+1), π‘₯β‰ βˆ’1 π‘˜, π‘₯=βˆ’1) is continuous at x=βˆ’1.#cbse2027 #maths
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Q) Simplify :sin^(βˆ’1) (5/13 cosπ‘₯+12/13 sinπ‘₯)#class12maths #shivangmathsacademy #cbsemaths
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Q) If 𝑓(π‘₯)={(3π‘₯βˆ’2, 0≀π‘₯≀1 2π‘₯^2+π‘Žπ‘₯, 1≀π‘₯≀2) is continuous for π‘₯∈(0,2), then a is equal to
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Q) If 𝑓(π‘₯)={ ((π‘₯^2βˆ’4π‘₯βˆ’5)/(π‘₯+1), π‘₯β‰ βˆ’1 π‘˜,&π‘₯=βˆ’1 is continuous at π‘₯=βˆ’1, then the value of π‘˜ i
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Q) Simplify: 𝐜𝐨𝐬^(βˆ’πŸ) 𝐱+𝐜𝐨𝐬^(βˆ’πŸ) [𝐱/𝟐+√(πŸ‘βˆ’πŸ‘π±^𝟐 )/𝟐]; 𝟏/πŸβ‰€π±β‰€πŸ #cbsemaths #maths
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If 𝑓(π‘₯)={ ((sin^2 π‘Žπ‘₯)/π‘₯^2 if π‘₯β‰ 0 1 , if π‘₯=0) is continuous at π‘₯=0, then the value of ' π‘Ž β€˜
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Q) Find the value of π‘˜, so that 𝑓(π‘₯)={(π‘˜cosπ‘₯/(πœ‹βˆ’2π‘₯), if π‘₯β‰ πœ‹/2@3, if π‘₯=πœ‹/2) is continu
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Q) Evaluate sin[cos^(βˆ’1) cos(7πœ‹/6)] #shivangmathsacademy #cbsemaths #class12maths #maths
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Q) Domain of 𝑠𝑖𝑛^(βˆ’1) (2π‘₯^2βˆ’3) is (A) (βˆ’1,0)βˆͺ(1,√2)(B) (βˆ’βˆš2,βˆ’1)βˆͺ(0,1)(C) [βˆ’βˆš2,βˆ’1]βˆͺ[1,√2](D) (βˆ’
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Q) If 𝑓(π‘₯)={((1βˆ’sin^3 π‘₯)/(3cos^2 π‘₯), for π‘₯β‰ πœ‹/2 π‘˜, for π‘₯=πœ‹/2) is continuous at π‘₯=πœ‹/2, t
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Q) If 2cos^(βˆ’1) π‘₯ = 𝑦,then(A) 0β‰€π‘¦β‰€πœ‹(B) βˆ’πœ‹β‰€π‘¦β‰€πœ‹(C) 0≀𝑦≀2πœ‹(D) βˆ’πœ‹β‰€π² ≀ 0 #shivangmathsacademy
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Q) Check whether function f(x) defined as f(π‘₯)={((|π‘₯βˆ’3|)/(2(π‘₯βˆ’3)), π‘₯=3 (π‘₯βˆ’6)/6,π‘₯β‰₯3) Is
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Q) If 𝑓(π‘₯)={(sinπ‘₯/π‘₯+cosπ‘₯, π‘₯β‰ 0 π‘˜, π‘₯=0) is continuous at π‘₯=0, then the value of π‘˜ is :
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Q) Simplify : tan^(βˆ’1) {√((1βˆ’cosπ‘₯)/(1+cosπ‘₯))}}#class12maths #shivangmathsacademy
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Q) Simplify : tan^(βˆ’1) ((cos2π‘₯βˆ’sin2π‘₯)/(cos2π‘₯+sin2π‘₯)),0=π‘₯=πœ‹/4..#class12maths #shivangmathsaca
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Find k so that 𝑓(π‘₯)={((π‘₯^2βˆ’2π‘₯βˆ’3)/(π‘₯+1), π‘₯β‰ βˆ’1 π‘˜, π‘₯=βˆ’1) is continuous at
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Q) If 𝑓(π‘₯)={(3π‘₯βˆ’2, 0≀π‘₯≀1 2π‘₯^2+π‘Žπ‘₯, 1≀π‘₯≀2) is continuous for π‘₯∈(0,2), then a is equal to
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Q) If 𝑓(π‘₯)={ ((π‘₯^2βˆ’4π‘₯βˆ’5)/(π‘₯+1), π‘₯β‰ βˆ’1 π‘˜,&π‘₯=βˆ’1 is continuous at π‘₯=βˆ’1, then the value of π‘˜ i
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Q) Simplify: 𝐜𝐨𝐬^(βˆ’πŸ) 𝐱+𝐜𝐨𝐬^(βˆ’πŸ) [𝐱/𝟐+√(πŸ‘βˆ’πŸ‘π±^𝟐 )/𝟐]; 𝟏/πŸβ‰€π±β‰€πŸ #cbsemaths #maths
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If 𝑓(π‘₯)={ ((sin^2 π‘Žπ‘₯)/π‘₯^2 if π‘₯β‰ 0 1 , if π‘₯=0) is continuous at π‘₯=0, then the value of ' π‘Ž β€˜
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Q) Find the value of π‘˜, so that 𝑓(π‘₯)={(π‘˜cosπ‘₯/(πœ‹βˆ’2π‘₯), if π‘₯β‰ πœ‹/2@3, if π‘₯=πœ‹/2) is continu
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Q) Domain of 𝑠𝑖𝑛^(βˆ’1) (2π‘₯^2βˆ’3) is (A) (βˆ’1,0)βˆͺ(1,√2)(B) (βˆ’βˆš2,βˆ’1)βˆͺ(0,1)(C)
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Q) If 𝑓(π‘₯)={((1βˆ’sin^3 π‘₯)/(3cos^2 π‘₯), for π‘₯β‰ πœ‹/2 π‘˜, for π‘₯=πœ‹/2) is continuous at π‘₯=πœ‹/2, t
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