Let α = sin^2k (π/6). Let g : [0,1] → R be the function defined by g(x) =2^αx+2^(α(1-x)).Then,which

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Let α=∞∑k=1 sin^2k(π/6). Let g : [0,1] → R be the function defined by g(x) = 2^𝛼x + 2^𝛼(1-x).Then

Let f : [0,π/2]→[0,1] be the function defined by f(x) = 〖sin〗^2 x and let g : [0,π/2]→[0,∞) be the

α = Σ(sin π/6)^2k k = 1 to ∞ and g(x) = 2^αx + 2^α(1 - x) then maximum and minimum value of g(x)

Let f:[0,2]→R be the function defined byf(x)=(3-sin⁡(2πx))sin⁡(πx-π/4)-sin⁡(3πx+π/4) #jeeadvanced

Let f:[0,pi/2] to [0,1] be the function defined by f(x)=sin^2(x) …| part-1 &2 | jeeadvce2024

Let f : [ 0 , π 2 ] → [ 0 , 1 ] be the function defined by f ( x ) = sin 2 x and let g : [ 0 ,π/2

Let f:[0,2]⟶R be the function defined by f(x)=(3-sin⁡2πx ) sin⁡(πx-π/4)-sin⁡(3πx+π/4). If α,β∈[0,2]

Q) Let 𝑓:𝑅={0}→ℝ be a function such that 𝑓(𝑥)=6𝑓(1/𝑥)=35/3𝑥−5/2, If the lim(𝑥→0) (1/𝛼𝑥+𝑓(𝑥))=𝛽;