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Complex
Analysis Continuity
Continuity of
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Continuity of Functions
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1:53
Q) If π(π₯)={((1βsin^3 π₯)/(3cos^2 π₯), for π₯β π/2 π, for π₯=π/2) is continuous at π₯=π/2, t
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1:05
Q) Evaluate sin[cos^(β1) cos(7π/6)] #shivangmathsacademy #cbsemaths #class12maths #maths
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2:11
Q) Check whether function f(x) defined as f(π₯)={((|π₯β3|)/(2(π₯β3)), π₯=3 (π₯β6)/6,π₯β₯3) Is
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1:26
Q) The value of π for which the function π(π₯)={(π₯^2 sin 1/π₯, π₯β 0 π(π₯+1), π₯=0) is a continuo
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2:15
JThe number of points of discontinuity f(x) = {(|π±|+π, π’π π±β€βπ βππ±, π’π βπ=π±=π ππ±+π,
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1:18
Find the values of π if the function π(π±)={((γπ¬π’π§^π ππ±)/π±^π , π’π π±β π π, π’π π±=π)
360 views
1 week ago
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Shivang Maths Academy
1:44
Find the value of k for which the function π(π±)={((πβππ¨π¬π±)/(ππ±^π ), π’π π±β π π€, π’π
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1 week ago
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1:07
Q) If π(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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1:50
Q) If π ππ πππ‘^(β1) (π₯+1)]=πππ (tan^(β1) π₯), π‘βππ ππππ π₯.#class12maths #shivang
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Q) Simplify : cot^(β1) β((1+cos2π₯)/(1βcos2π₯)),π₯β(0,π/2).#maths #cbsemaths #class12 #cbse2026
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1 month ago
YouTube
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1:50
Q) The function π defined by π(π₯)={ (π₯,f π₯β€1 5, if π₯= 1 is not continuous at(A) x=0(B) π₯
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3 weeks ago
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2:33
The simplest form of tan^(β1) [(β(1+x)ββ(1βx))/(β(1+x)+β(1βx))] is (a Ο/4βx/ #cbsemaths #maths
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1:01
If a function defined by π(π₯)={(ππ₯+1, π₯β€π cosπ₯ ,π₯=π) is continuous at π₯=π, then the val
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1:46
Q) Find the value of π, so that π(π₯)={(πcosπ₯/(πβ2π₯), if π₯β π/2@3, if π₯=π/2) is continu
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0:55
Q) If sin^(β1) π₯ + pie =3π¦, thenCBSE Class 12 Maths PYQ 2026 | Continuity of Piecewise Function
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2:17
Q) Express πππ§^(βπ) (ππ¨π¬π±/(πβπ¬π’π§π±)), (βππ)/π=π±=π/π in the simplest form.#class12
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2:13
Q) If π(π₯)={ ((π₯^2β4π₯β5)/(π₯+1), π₯β β1 π,&π₯=β1 is continuous at π₯=β1, then the value of π i
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1:09
Q) If π(π₯)={(3π₯β2, 0β€π₯β€1 2π₯^2+ππ₯, 1β€π₯β€2) is continuous for π₯β(0,2), then a is equal to
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2:22
Find k so that π(π₯)={((π₯^2β2π₯β3)/(π₯+1), π₯β β1 π, π₯=β1) is continuous at x=β1.#cbse2027 #maths
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2:05
Q) Simplify: ππ¨π¬^(βπ) π±+ππ¨π¬^(βπ) [π±/π+β(πβππ±^π )/π]; π/πβ€π±β€π #cbsemaths #maths
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