Let `f(x) = min [ tan x, cot x, 1sqrt3]` for all `x in (0, pi/2)`Then the area bounded by y = ...

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Let `f(x)=min.[tanx, cotx, 1/sqrt(3)], x in [0, pi/2]`. If the area bounded by `y=f(x)` and x-axisLet `f(x)=min.[tanx, cotx, 1/sqrt(3)], x in [0, pi/2]`. If the area bounded by `y=f(x)` and x-axis
`f(x)`=Minimum`{tanx,cotx}AA x in(0,pi/2)` Then `int_0^(pi/3)f(x)dx` is equal to`f(x)`=Minimum`{tanx,cotx}AA x in(0,pi/2)` Then `int_0^(pi/3)f(x)dx` is equal to
f(x)= Min (tanx, cotx) ∀ x ϵ (0,π/2) Then ∫_0^(π/3)▒f(x)dx is equal to: a)ln(√3/2) b) ln(√(3/2))f(x)= Min (tanx, cotx) ∀ x ϵ (0,π/2) Then ∫_0^(π/3)▒f(x)dx is equal to: a)ln(√3/2) b) ln(√(3/2))
Area lying between the curves `y=tanx, y=cotx` and x-axis, `x in [0, pi/2]` is (A) `1/2log2`Area lying between the curves `y=tanx, y=cotx` and x-axis, `x in [0, pi/2]` is (A) `1/2log2`
Let `f(x)=min {tanx ,cotx}AAx in R`, findLet `f(x)=min {tanx ,cotx}AAx in R`, find
Q) For 𝑥∈ℝ, let tan^(−1) (𝑥)∈(−𝜋/2,𝜋/2). Then the minimum value of the function 𝑓:ℝ→ℝ defined byQ) For 𝑥∈ℝ, let tan^(−1) (𝑥)∈(−𝜋/2,𝜋/2). Then the minimum value of the function 𝑓:ℝ→ℝ defined by
Sketch and find area bounded by region of graph y = tan x, x = pi/3, y =0. Definite IntegralSketch and find area bounded by region of graph y = tan x, x = pi/3, y =0. Definite Integral
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