Let the Range of f(x) = 1/(2 + sin 3x + cos 3x), x ∈ R, Be [a, b]. If α and β Are Respectively the A.M. and the G.M. of a and b, Then α/β Is Equal To

JEE Main 2024 — 9 April, Evening Shift. Previous Year Question.

Problem

Let the range of the function $f(x)=\dfrac{1}{2+\sin3x+\cos3x}$, $x\in\mathbb R$ be $[a,b]$. If $\alpha$ and $\beta$ are respectively the A.M. and the G.M. of $a$ and $b$, then $\dfrac{\alpha}{\beta}$ is equal to:

Key insight. $\sin3x+\cos3x$ is really just $\sqrt2\sin\left(3x+\frac{\pi}4\right)$ in disguise — a single sinusoid with amplitude $\sqrt2$. Once that’s recognised, the denominator’s whole range is pinned down immediately, and since $f$ is a decreasing function of its denominator, the smallest denominator gives the largest $f$-value and vice versa.

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Approach

Rewrite $\sin3x+\cos3x$ as a single sinusoid of amplitude $\sqrt2$, so it ranges over $[-\sqrt2,\sqrt2]$. This pins down the range of the denominator $2+\sin3x+\cos3x$, and hence the range of $f(x)=1/(\text{denominator})$ — with the largest denominator giving the smallest $f$-value and vice versa. Rationalise both endpoints to get clean forms for $a$ and $b$, then compute their AM and GM.

Solution

Step 1 — Rewrite sin3x + cos3x as a single sinusoid

$$\sin3x+\cos3x = \sqrt2\sin\left(3x+\frac{\pi}{4}\right)$$

This ranges over $[-\sqrt2,\sqrt2]$ as $x$ varies over $\mathbb R$.

Step 2 — Find the range of the denominator

$$2+\sin3x+\cos3x \in \left[2-\sqrt2,\ 2+\sqrt2\right]$$

Both endpoints are positive since $2-\sqrt2\approx0.586>0$.

Step 3 — Find the range of f(x)

Since $f(x)=\dfrac1{\text{denominator}}$ is a decreasing function of a positive denominator, the smallest denominator gives the largest $f$, and vice versa:

$$a = \frac{1}{2+\sqrt2}, \qquad b = \frac{1}{2-\sqrt2}$$

Step 4 — Rationalise a and b

$$a = \frac{1}{2+\sqrt2}\cdot\frac{2-\sqrt2}{2-\sqrt2} = \frac{2-\sqrt2}{4-2} = \frac{2-\sqrt2}{2}$$

$$b = \frac{1}{2-\sqrt2}\cdot\frac{2+\sqrt2}{2+\sqrt2} = \frac{2+\sqrt2}{2}$$

Step 5 — Find the arithmetic mean α

$$\alpha = \frac{a+b}{2} = \frac{1}{2}\left(\frac{2-\sqrt2}{2}+\frac{2+\sqrt2}{2}\right) = \frac{1}{2}\cdot\frac{4}{2} = 1$$

Step 6 — Find the geometric mean β

$$\beta = \sqrt{ab} = \sqrt{\frac{(2-\sqrt2)(2+\sqrt2)}{4}} = \sqrt{\frac{4-2}{4}} = \sqrt{\frac12} = \frac{1}{\sqrt2}$$

Step 7 — Compute α/β

$$\frac{\alpha}{\beta} = \frac{1}{1/\sqrt2} = \sqrt2$$

Answer

Common mistakes

• Assuming $f$’s maximum occurs where $\sin3x+\cos3x$ is maximum. Since $f$ is the reciprocal of the denominator, $f$ is largest exactly when the denominator is smallest — the relationship is inverted, and it’s easy to mix up which endpoint of the sinusoid’s range gives which endpoint of $f$’s range.
• Not rationalising $a$ and $b$ before computing the AM and GM. Working with $\frac1{2+\sqrt2}$ and $\frac1{2-\sqrt2}$ directly makes both the sum and product messier — rationalising first turns both computations into simple arithmetic.

Practise next

• Let the range of $g(x)=\dfrac{1}{3+\sin2x+\cos2x}$ be $[p,q]$. Find the ratio of the arithmetic mean to the geometric mean of $p$ and $q$, using the same amplitude-and-rationalise approach.