If 0≤x≤π3 then range of  f(x)=sec⁡π6−x+sec⁡π6+x is

if a,b>0Using A.M. ≥ G.M., we get 1a+1b≥2ab⇒ f(x)≥2cos⁡π6−xcos⁡π6+x=2cos2⁡π6−sin2⁡x=234−1−cos⁡2x2=214+cos⁡2x2 Now for 0≤x≤π3,−12≤cos⁡2x≤1⇒ 0≤14+cos⁡2x2≤32⇒ f(x)≥43Since 'f is continuous, range of f′ is 43,∞.