Minimum value of f(x)=Integration(e^(t-cost)/1+t^2023)dt 0 to xtaninverse x is | JEE Advanced 2023

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Then the minimum value of the function f(x)= limit 0 to x tan-1(x) integration e^(t-cost)/1+t^2023dtThen the minimum value of the function f(x)= limit 0 to x tan-1(x) integration e^(t-cost)/1+t^2023dt
JEE Advanced 2023 Question 1 Paper 2 Solution | 3Integration[f(t)dt] 1 to x =xf(x)-x^3/3JEE Advanced 2023 Question 1 Paper 2 Solution | 3Integration[f(t)dt] 1 to x =xf(x)-x^3/3
The limit of (1/x)integral from 0 to x of f(t)dtThe limit of (1/x)integral from 0 to x of f(t)dt
JEE Main 2023 𝒇(𝒙)+∫_𝟎^𝒙〖𝒇(𝒕)√(πŸβˆ’(π₯𝐧 𝒇(𝒕))^𝟐 ) 𝒅𝒕〗=𝒆 βˆ€π’™βˆˆ[𝟎,𝝅/𝟐] then (πŸ” π₯𝐧𝒇(𝝅/πŸ”))^𝟐 is equal toJEE Main 2023 𝒇(𝒙)+∫_𝟎^𝒙〖𝒇(𝒕)√(πŸβˆ’(π₯𝐧 𝒇(𝒕))^𝟐 ) 𝒅𝒕〗=𝒆 βˆ€π’™βˆˆ[𝟎,𝝅/𝟐] then (πŸ” π₯𝐧𝒇(𝝅/πŸ”))^𝟐 is equal to
Q) Let 𝑦=𝑦(π‘₯) be the solution of the differential equation (1+𝑦^2 ) 𝑒^tanπ‘₯ 𝑑π‘₯+cos^2 π‘₯(1+𝑒^Q) Let 𝑦=𝑦(π‘₯) be the solution of the differential equation (1+𝑦^2 ) 𝑒^tanπ‘₯ 𝑑π‘₯+cos^2 π‘₯(1+𝑒^
Q8 JEE Advanced 2023 Mathematics Paper 2 Solution | Mixed Concepts | Extreme Values | IntegrationQ8 JEE Advanced 2023 Mathematics Paper 2 Solution | Mixed Concepts | Extreme Values | Integration
Q) The minimum value of the function 𝑓(π‘₯)= Integration ∫ ( 0 to 2 ) 𝑒^(|π‘₯βˆ’π‘‘|) 𝑑𝑑 is #jee #maths #jeeQ) The minimum value of the function 𝑓(π‘₯)= Integration ∫ ( 0 to 2 ) 𝑒^(|π‘₯βˆ’π‘‘|) 𝑑𝑑 is #jee #maths #jee
How to calculate inverse tan in scientific calculator #scintific #calculator #studenthacksHow to calculate inverse tan in scientific calculator #scintific #calculator #studenthacks