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Let beta(m,n)=integration 0to1 x^(m-1)(1-x)^(n-1)dx,if integration 0to1(1-x10)20dx=a beta(b,c),100(

Let I(x)=integration 6/sin^2x(1-cotx)^2 dx .if I(0)=3, then I(pi/12)is equal to

The value of k€N for which integral In=integration from 0 to1(1-x^k)dx n€N,satisfies147 I 20=148 I21

Let integration from alpha to log4 dx/sq root over(e^x -1)=pi/6 then e^(alpha)&e^(-alpha) are roots

Let rk = integration from 0 to 1(1-x^7)^k / integration 0 to 1 (1-x^7)^k+1 dx,k€N then the value of

The value of integral from-1 to 2 log(x+square root(x^2+1))dx is #iitjee_mains2024

Integration of 1/(x-1)^4/5 (x+3)^6 dx =A(alpha x -1/beta x+3)^B +C then the value of alpha+beta+20AB

Let integration of (2- tanx)/(3+tanx) dx = 1/2(alpha x+ log|beta sinx+ gamma cosx|)+C then alpha +

The integral from 1/4 to 3/4 cos(2 cot^-1 (1-x/1+x)^1/2) dx is equal to #iitjee_mains2024

The value of integral from 0 to pi/4 xdx/sin^4(2x) + cos^4(2x) equals #IITJEEMAINS 2024

If(1/ (alpha +1 )+1/(alpha+2)…+1/(alpha+1012))-(1/2.1+1/4.3+1/6.5+ …+1/2024.2023)=1/2024 then alpha

Let alpha &beta be roots of equation x2-root2 x-root 3=0 let Pn=alpha^n -beta^n.(11root3-10root2)P10

Let integration from 0 to x (1-(y’(t))^2 )^1/2dy=integration from 0 to x y(t)dt,x€[0,3],y(0)=0 then

#iitjeemains2024# Lim x tends to 0 (e- (1+2x)^1/2x ) / x is equal to

Let[x] be G.I.F. Then number of points in interval (-2,1), where f(x)=|[x]|+sq root (x-[x]) is disco

If local max value of function f(x)=((3e)^1/2 /2sinx)^sin^2x,x€(0,pi/2)is k/e then (k/8)^8 +k8/e5+k8

BEST PROBLEM OF IITJEE ADVANCED-2024,let function f:[1,infinity)-R be defined by f(t)= (-1)^(n+1)2

f(x)= sinx(x^2023+2024x+2025)/e^pi x (x^2-x+3) +2(x^2023+2024x+2025)/e^pi x (x^2-x+3) then number of

Considering only the principal values of the ITF,the value of tan(sin^-1(3/5) -2cos^-1(2/root5)) is

IITJEE ADVANCED2024, let f:R-R be a function defined by f(x)= x2 sin(pi/x2),ifx doesn’t 0&0 if x=0

IITJEE ADVANCED2024 Let k€R. If lim x -0+ (sin(sinkx)+cosx +x )^2/x =e^6, then the value of k is

The value of 2sin(pi/8)sin(2pi/8)sin(3pi/8)sin(5pi/8)sin(6pi/8)sin(7pi/8) #short method

If tanA= 1/(x(x^2+x+1))^1/2, tanB=x^1/2/(x2+x+1)^1/2 and tanC=(x^-3 +x^-2 + x^-1)1/2, then A+B is =

IITJEE ADVANCED 2024 PAPER SOLUTION- the value of 16/pi^3 integration from 0 to pi/2 f(x)g(x) dx is