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Functions: Range

JEE Maths question with a full step-by-step solution.

Question
If 0≤x≤π30\le x\le\dfrac{\pi}{3}, then range of f(x)=sec⁡(π6−x)+sec⁡(π6+x)f(x)=\sec\left(\dfrac{\pi}{6}-x\right)+\sec\left(\dfrac{\pi}{6}+x\right) is:
A(43,∞)\left(\dfrac{4}{\sqrt{3}},\infty\right)
B[43,∞)\left[\dfrac{4}{\sqrt{3}},\infty\right)correct
C(0,43]\left(0,\dfrac{4}{\sqrt{3}}\right]
D(0,43)\left(0,\dfrac{4}{\sqrt{3}}\right)
Solution
Step 1: Since a,b>0  ⇒  1a+1b≥2aba,b>0\;\Rightarrow\;\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{2}{\sqrt{ab}}:
f(x)=sec⁡ ⁣(π6−x)+sec⁡ ⁣(π6+x)≥2cos⁡ ⁣(π6−x)cos⁡ ⁣(π6+x)f(x)=\sec\!\left(\dfrac{\pi}{6}-x\right)+\sec\!\left(\dfrac{\pi}{6}+x\right)\ge\dfrac{2}{\sqrt{\cos\!\left(\dfrac{\pi}{6}-x\right)\cos\!\left(\dfrac{\pi}{6}+x\right)}}
 Step 2: Use product-to-sum:
cos⁡ ⁣(π6−x)cos⁡ ⁣(π6+x)=12 ⁣[cos⁡ ⁣(π3)+cos⁡(2x)]=12 ⁣[12+cos⁡2x]=14+cos⁡2x2\cos\!\left(\dfrac{\pi}{6}-x\right)\cos\!\left(\dfrac{\pi}{6}+x\right)=\dfrac{1}{2}\!\left[\cos\!\left(\dfrac{\pi}{3}\right)+\cos(2x)\right]=\dfrac{1}{2}\!\left[\dfrac{1}{2}+\cos 2x\right]=\dfrac{1}{4}+\dfrac{\cos 2x}{2}
 Step 3: For 0≤x≤π30\le x\le\dfrac{\pi}{3}: cos⁡2x∈[cos⁡2π3,cos⁡0]=[−12,1]\cos 2x\in\left[\cos\dfrac{2\pi}{3},\cos 0\right]=\left[-\dfrac{1}{2},1\right]. So 14+cos⁡2x2∈[0,34]\dfrac{1}{4}+\dfrac{\cos 2x}{2}\in\left[0,\dfrac{3}{4}\right]. Step 4: Therefore 14+cos⁡2x2≤32\sqrt{\dfrac{1}{4}+\dfrac{\cos 2x}{2}}\le\dfrac{\sqrt{3}}{2}, and:
f(x)≥23/2=43f(x)\ge\dfrac{2}{\sqrt{3}/2}=\dfrac{4}{\sqrt{3}}
 Step 5: Equality holds when cos⁡ ⁣(π6−x)=cos⁡ ⁣(π6+x)\cos\!\left(\dfrac{\pi}{6}-x\right)=\cos\!\left(\dfrac{\pi}{6}+x\right), i.e., x=0x=0. At x=0x=0: f(0)=2sec⁡π6=23/2=43f(0)=2\sec\dfrac{\pi}{6}=\dfrac{2}{{\sqrt{3}}/2}=\dfrac{4}{\sqrt{3}}. Since ff is continuous and equals 43\dfrac{4}{\sqrt{3}} at x=0x=0, the range is [43,∞)\left[\dfrac{4}{\sqrt{3}},\infty\right). Answer: (2) [43,∞)\left[\dfrac{4}{\sqrt{3}},\infty\right)
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