Test Index

Functions Part 2

Section: Mathematics

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Question : 35 of 73

Marks: +1, -0

Let the range of the function

f(x)=\; \frac{1}{2 + \sin 3x + \cos 3x}, x \in \mathbb{R} \; \text{ be }\; [a, b].

\alpha

\beta

are respectively the A.M. and the G.M. of

a

b

\; \frac{\alpha}{\beta}

[9 Apr 2024 Shift 2]

$\sqrt{2}$
2
$\sqrt{\pi}$
$\pi$

Solution: 👈: Video Solution

Let

t=\sin 3x+\cos 3x

.

For

x\in\mathbb{R}

, we have

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

.

So

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

.

Since the denominator is positive, taking reciprocals reverses the interval endpoints.

Hence the range of

f(x)

is

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

.

Thus

a=\frac{1}{2+\sqrt{2}}

and

b=\frac{1}{2-\sqrt{2}}

.

Now

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

.

\alpha=\frac{1}{2}\left(\frac{1}{2+\sqrt{2}}+\frac{1}{2-\sqrt{2}}\right)

.

\alpha=\frac{1}{2}\cdot\frac{4}{(2+\sqrt{2})(2-\sqrt{2})}

.

\alpha=\frac{1}{2}\cdot\frac{4}{2}=1

.

Also

\beta=\sqrt{ab}

.

\beta=\sqrt{\frac{1}{(2+\sqrt{2})(2-\sqrt{2})}}

.

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

.

Therefore

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

.

So the correct option is

\sqrt{2}

.

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