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Functions — JEE Maths practice question

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

Question

0\le x\le\dfrac{\pi}{3}

, then range of

f(x)=\sec\left(\dfrac{\pi}{6}-x\right)+\sec\left(\dfrac{\pi}{6}+x\right)

is:

\left(\dfrac{4}{\sqrt{3}},\infty\right)

\left[\dfrac{4}{\sqrt{3}},\infty\right)

correct

\left(0,\dfrac{4}{\sqrt{3}}\right]

\left(0,\dfrac{4}{\sqrt{3}}\right)

Solution

Step 1: Since

a,b>0\;\Rightarrow\;\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{2}{\sqrt{ab}}

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\!\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\le x\le\dfrac{\pi}{3}

\cos 2x\in\left[\cos\dfrac{2\pi}{3},\cos 0\right]=\left[-\dfrac{1}{2},1\right]

. So

\dfrac{1}{4}+\dfrac{\cos 2x}{2}\in\left[0,\dfrac{3}{4}\right]

. Step 4: Therefore

\sqrt{\dfrac{1}{4}+\dfrac{\cos 2x}{2}}\le\dfrac{\sqrt{3}}{2}

, and:

f(x)\ge\dfrac{2}{\sqrt{3}/2}=\dfrac{4}{\sqrt{3}}

Step 5: Equality holds when

\cos\!\left(\dfrac{\pi}{6}-x\right)=\cos\!\left(\dfrac{\pi}{6}+x\right)

, i.e.,

x=0

. At

x=0

f(0)=2\sec\dfrac{\pi}{6}=\dfrac{2}{{\sqrt{3}}/2}=\dfrac{4}{\sqrt{3}}

. Since

f

is continuous and equals

\dfrac{4}{\sqrt{3}}

x=0

, the range is

\left[\dfrac{4}{\sqrt{3}},\infty\right)

. Answer: (2)

\left[\dfrac{4}{\sqrt{3}},\infty\right)

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