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JEE Mathematics 2024 Question with Solution

Let the range of the function f(x)=12+sin⁡(3x)+cos⁡(3x)f(x) = \frac{1}{2 + \sin(3x) + \cos(3x)}, where x∈Rx \in \mathbb{R} and x∈[a,b]x \in [a, b]. If α\alpha and β\beta are respectively the A.M. and G.M. of aa and bb, then α/β\alpha/\beta is equal to:

  • A

    2\sqrt{2}

  • B

    22

  • C

    π\sqrt{\pi}

  • D

    π\pi

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: f(x)=12+sin⁡3x+cos⁡3xf(x) = \frac{1}{2 + \sin 3x + \cos 3x} with range [a,b][a,b].

Find: αβ\frac{\alpha}{\beta}, where α\alpha and β\beta are respectively the A.M. and G.M. of aa and bb.

Use the identity

sin⁡3x+cos⁡3x=2sin⁡(3x+π4)\sin 3x + \cos 3x = \sqrt{2}\sin\left(3x + \frac{\pi}{4}\right)

So,

2+sin⁡3x+cos⁡3x=2+2sin⁡(3x+π4)2 + \sin 3x + \cos 3x = 2 + \sqrt{2}\sin\left(3x + \frac{\pi}{4}\right)

Since sin⁡(3x+π4)∈[−1,1]\sin\left(3x + \frac{\pi}{4}\right) \in [-1,1], we get

2+sin⁡3x+cos⁡3x∈[2−2, 2+2]2 + \sin 3x + \cos 3x \in [2-\sqrt{2},\,2+\sqrt{2}]

Hence,

f(x)∈[12+2, 12−2]f(x) \in \left[\frac{1}{2+\sqrt{2}},\,\frac{1}{2-\sqrt{2}}\right]

Therefore,

a=12+2,b=12−2a = \frac{1}{2+\sqrt{2}}, \qquad b = \frac{1}{2-\sqrt{2}}

Now calculate the arithmetic mean:

α=a+b2=12(12+2+12−2)\alpha = \frac{a+b}{2} = \frac{1}{2}\left(\frac{1}{2+\sqrt{2}} + \frac{1}{2-\sqrt{2}}\right) α=12((2−2)+(2+2)(2+2)(2−2))\alpha = \frac{1}{2}\left(\frac{(2-\sqrt{2})+(2+\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}\right) α=12(44−2)=1\alpha = \frac{1}{2}\left(\frac{4}{4-2}\right) = 1

Now calculate the geometric mean:

β=ab=12+2⋅12−2\beta = \sqrt{ab} = \sqrt{\frac{1}{2+\sqrt{2}}\cdot \frac{1}{2-\sqrt{2}}} β=1(2+2)(2−2)=12=12\beta = \sqrt{\frac{1}{(2+\sqrt{2})(2-\sqrt{2})}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}

Therefore,

αβ=11/2=2\frac{\alpha}{\beta} = \frac{1}{1/\sqrt{2}} = \sqrt{2}

So the correct option is A.

Direct Mean Relation

Given: The range endpoints are a=12+2a = \frac{1}{2+\sqrt{2}} and b=12−2b = \frac{1}{2-\sqrt{2}}.

Find: αβ\frac{\alpha}{\beta}.

First compute

a+b=12+2+12−2=42=2a+b = \frac{1}{2+\sqrt{2}} + \frac{1}{2-\sqrt{2}} = \frac{4}{2} = 2

and

ab=1(2+2)(2−2)=12ab = \frac{1}{(2+\sqrt{2})(2-\sqrt{2})} = \frac{1}{2}

Now use

α=a+b2=1,β=ab=12\alpha = \frac{a+b}{2} = 1, \qquad \beta = \sqrt{ab} = \sqrt{\frac{1}{2}}

Hence,

αβ=11/2=2\frac{\alpha}{\beta} = \frac{1}{\sqrt{1/2}} = \sqrt{2}

Therefore, the correct option is A.

Common mistakes

  • Taking a=2+2a = 2+\sqrt{2} and b=2−2b = 2-\sqrt{2} instead of the range endpoints of f(x)f(x) is incorrect because these are the extrema of the denominator, not of the function. First find the denominator range, then take reciprocals to get the range of f(x)f(x).

  • Forgetting that reciprocal reverses order for positive quantities can give the wrong interval. Since 2−2>02-\sqrt{2} > 0 and 2+2>02+\sqrt{2} > 0, the range becomes [12+2,12−2]\left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}\right].

  • Using sin⁡3x+cos⁡3x\sin 3x + \cos 3x range as [−2,2][-2,2] is wrong because the two terms are not independently maximized together. Rewrite it as 2sin⁡(3x+π4)\sqrt{2}\sin\left(3x+\frac{\pi}{4}\right), whose range is [−2,2][-\sqrt{2},\sqrt{2}].

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