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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Draw the graph of 𝑓(π‘₯)=𝑠𝑖𝑛^(βˆ’1) π‘₯, π‘₯∈[βˆ’1/√2, 1/√2]. Also, write range of f(x). #trigonometry
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Draw the graph of 𝑓(π‘₯)=𝑠𝑖𝑛^(βˆ’1) π‘₯, π‘₯∈[βˆ’1/√2, 1/√2]. Also, write range of f(x). #trigonometry
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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) Simplify : tan^(βˆ’1) {√((1βˆ’cosπ‘₯)/(1+cosπ‘₯))}}#class12maths #shivangmathsacademy #cbsemaths
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Q) Simplify : tan^(βˆ’1) {√((1βˆ’cosπ‘₯)/(1+cosπ‘₯))}}#class12maths #shivangmathsacademy #cbsemaths
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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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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) Find the domain of sec^(βˆ’1) (2π‘₯+1). #maths #shivangmathsacademy #shivangmathsacademy #cbsemaths
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Q) Find the domain of sec^(βˆ’1) (2π‘₯+1). #maths #shivangmathsacademy #shivangmathsacademy #cbsemaths
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Q) Evaluate sin[cos^(βˆ’1) cos(7πœ‹/6)] #shivangmathsacademy #cbsemaths #class12maths #maths
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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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The simplest form of tan^(βˆ’1) [(√(1+x)βˆ’βˆš(1βˆ’x))/(√(1+x)+√(1βˆ’x))] is (a Ο€/4βˆ’x/ #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) 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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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) 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 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) Express 𝐭𝐚𝐧^(βˆ’πŸ) (𝐜𝐨𝐬𝐱/(πŸβˆ’π¬π’π§π±)), (βˆ’πŸ‘π›‘)/𝟐=𝐱=𝛑/𝟐 in the simplest form.#class12
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Q) Simplify : .cos^(βˆ’1) ((sinπ‘₯+cosπ‘₯)/√2),πœ‹/4=π‘₯=5πœ‹/4 #shivangmathsacademy #maths #cbsemaths
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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) Simplify : tan^(βˆ’1) {√((1βˆ’cosπ‘₯)/(1+cosπ‘₯))}}#class12maths #shivangmathsacademy
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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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The simplest form of tan^(βˆ’1) [(√(1+x)βˆ’βˆš(1βˆ’x))/(√(1+x)+√(1βˆ’x))] is (a Ο€/4βˆ’x/ #cbsemaths #math
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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) 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 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) Express 𝐭𝐚𝐧^(βˆ’πŸ) (𝐜𝐨𝐬𝐱/(πŸβˆ’π¬π’π§π±)), (βˆ’πŸ‘π›‘)/𝟐=𝐱=𝛑/𝟐 in the simplest form.#class12
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