Range of the function

JEE Mains 2015 Questions › Calculus questions › Range of the function

Range of the function f(x)=ln[cosx^cosx+1],xE(0,pi/2)
(A)(ln(1+e^1e ),ln2)
(B)(ln(1+1e^1e)ln2)
(C)(0,ln2)
(D)None of these
Ans.(B) is the correct choice.

#Mathematics #Calculus

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Manish Shankar ·2011-02-03 21:23:29

range is same as ln(x^x+1) x lies in 0 to 1

x^x(1+ln x)

This will be zero when x=1/e

Minimum value is given by ln(1/e^1/e+1)

Maxima will be at one of the end points.. at 1 it is ln 2

At 0 you can find it by using limit.... I leave that as an exercise..

correct answer is B

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Vinay Arya ·2011-02-03 21:30:38

Thank you Sir!

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Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

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Range of the function

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