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Last updated at April 16, 2024 by Teachoo

Example 34 Show that the function f given by f (x) = tanβ1(sin x + cos x), x > 0 is always an strictly increasing function in (0,π/4) f(π₯)=tan^(β1)β‘(sinβ‘π₯+cosβ‘π₯ ) Finding fβ(π) fβ(π₯) = (π(tan^(β1)β‘γ(sinβ‘π₯ +cosβ‘π₯ ))γ)/ππ₯ = 1/(1 + (sinβ‘γπ₯ + cosβ‘π₯ γ )^2 ) Γ π(sinβ‘γπ₯ + cosβ‘π₯ γ )/ππ₯ = 1/(1 + (γγπ¬π’π§γ^π πγβ‘γ+γππ¨π¬γ^πβ‘γπ γ+ 2 sinβ‘π₯ cosβ‘π₯ γ ) ) Γ (cosβ‘π₯βsinβ‘π₯ ) = 1/(1 + (π + 2 sinβ‘π₯ cosβ‘π₯ ) ) Γ (cosβ‘π₯βsinβ‘π₯ ) = 1/(2 + 2 sinβ‘π₯ cosβ‘π₯ ) Γ (cosβ‘π₯βsinβ‘π₯ ) = ππ¨π¬β‘γπ βγ π¬π’π§γβ‘π γ/(π + π¬π’π§β‘ππ ) For increasing, fβ(x) > 0. β΄ Numerator and denominator both must be > 0 Checking sign for denominator Since Hence, denominator is always positive for 0 < x < π /π Checking sign for numerator cos π₯ β sin π₯ > 0 cos π₯ > sin π₯ 1 > sinβ‘π₯/cosβ‘π₯ 1 > tan π₯ πππ§β‘π<π This is possible only if π<π<π /π Thus, fβ(π) = ((+))/((+) ) > 0 in x β (0 , π/4) Hence, f is strictly increasing function in (π , π /π)