JEE MainMathematicsDefinite IntegrationMCQ+4 / −1If f:R→Rf: \mathbb{R} \rightarrow \mathbb{R}f:R→R be a continuous function satisfying ∫0π2f(sin2x)sinxdx+α∫0π4f(cos2x)cosxdx=0\int_{0}^{\frac{\pi}{2}} f(\sin 2 x) \sin x d x+\alpha \int_{0}^{\frac{\pi}{4}} f(\cos 2 x) \cos x d x=0∫02πf(sin2x)sinxdx+α∫04πf(cos2x)cosxdx=0, then the value of α\alphaα is :A−3-\sqrt{3}−3B2\sqrt{2}2C−2-\sqrt{2}−2D3\sqrt{3}3Attempt in chatSharePreviousNextMore from Definite IntegrationLet the function f:[0,2]→R be defined as f(x)={emin{x2,x−[x]},e[x−logex],x∈[0,1)x∈[1,2] where [t] denotes the greatest…2023 · MCQIf ∫−0.150.15100x2−1dx=3000k, then k is equal to .2023 · Numerical∫0∞e3x+6e2x+11ex+66dx=2023 · MCQLet for x∈R,S0(x)=x,Sk(x)=Ckx+k∫0xSk−1(t)dt, where C0=1,Ck=1−∫01Sk−1(x)dx,k=1,2,3,… Then S2(3)+6C3 is equal to .2023 · NumericalThe value of 0∫4πe−x(tan49x+tan51x)dxe−4π+0∫4πe−xtan50xdx is2023 · MCQLet fn=∫02π(∑k=1nsink−1x)(∑k=1n(2k−1)sink−1x)cosxdx,n∈N. Then f21−f20 is equal to 2023 · NumericalIf 0∫1(5+2x−2x2)(1+e(2−4x))1dx=α1loge(βα+1),α,β>0, then α4−β4 is equal to :2023 · MCQThe value of 120∫3x2−3x+2dx is 2023 · Numerical