Let the range of the function f(x)=\frac12+\sin 3 x+\cos 3 x, x \in \mathbbR be [a, b] . If \alpha and \beta ar respectively the A.M. and the G.M. of a and b , then \frac\alpha\beta is equal to

Functions PYQ — JEE Main 2024 · 9 Apr · Shift 2 · Q33 with Solution

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Correct answer: C

Given function

We need the range of

f(x)=\frac{1}{2+\sin 3x+\cos 3x}, \qquad x\in \mathbb R.

Let

t=\sin 3x+\cos 3x.

Using the standard identity,

\sin \theta+\cos \theta=\sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right),

t \in [-\sqrt{2},\sqrt{2}].

Range of the denominator

Thus,

2+\sin 3x+\cos 3x \in [2-\sqrt{2},\,2+\sqrt{2}].

Since both endpoints are positive, taking reciprocal reverses the order. Therefore the range of $f(x)$ is

\left[\frac{1}{2+\sqrt{2}},\,\frac{1}{2-\sqrt{2}}\right].

So,

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

Compute A.M. and G.M.

The arithmetic mean is

\alpha=\frac{a+b}{2}.

Now,

a+b=\frac{1}{2+\sqrt{2}}+\frac{1}{2-\sqrt{2}} =\frac{(2-\sqrt{2})+(2+\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})} =\frac{4}{4-2}=2.

Hence,

\alpha=\frac{2}{2}=1.

The geometric mean is

\beta=\sqrt{ab}.

Also,

ab=\frac{1}{(2+\sqrt{2})(2-\sqrt{2})}=\frac{1}{4-2}=\frac{1}{2}.

So,

\beta=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}.

Find the required ratio

\frac{\alpha}{\beta}=\frac{1}{1/\sqrt{2}}=\sqrt{2}.

Match with the options

Thus the correct option is

\boxed{\text{C: }\sqrt{2}}.

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View written solutionFree

Correct answer: C

Given function

We need the range of

f(x)=\frac{1}{2+\sin 3x+\cos 3x}, \qquad x\in \mathbb R.

Let

t=\sin 3x+\cos 3x.

Using the standard identity,

\sin \theta+\cos \theta=\sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right),

t \in [-\sqrt{2},\sqrt{2}].

Range of the denominator

Thus,

2+\sin 3x+\cos 3x \in [2-\sqrt{2},\,2+\sqrt{2}].

Since both endpoints are positive, taking reciprocal reverses the order. Therefore the range of $f(x)$ is

\left[\frac{1}{2+\sqrt{2}},\,\frac{1}{2-\sqrt{2}}\right].

So,

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

Compute A.M. and G.M.

The arithmetic mean is

\alpha=\frac{a+b}{2}.

Now,

a+b=\frac{1}{2+\sqrt{2}}+\frac{1}{2-\sqrt{2}} =\frac{(2-\sqrt{2})+(2+\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})} =\frac{4}{4-2}=2.

Hence,

\alpha=\frac{2}{2}=1.

The geometric mean is

\beta=\sqrt{ab}.

Also,

ab=\frac{1}{(2+\sqrt{2})(2-\sqrt{2})}=\frac{1}{4-2}=\frac{1}{2}.

So,

\beta=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}.

Find the required ratio

\frac{\alpha}{\beta}=\frac{1}{1/\sqrt{2}}=\sqrt{2}.

Match with the options

Thus the correct option is

\boxed{\text{C: }\sqrt{2}}.

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