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Manipal Entrance Test 2013 Math Paper

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Question : 21 of 80

Marks: +1, -0

The maximum value of

(\cos \alpha_{1})\cdot(\cos \alpha_{2})\cdots(\cos \alpha_{n})

Under the restrictions

0 \leq \alpha_{1}, \alpha_{2}, \dots, \alpha_{n} \leq \frac{\pi}{2}

(\cot \alpha_{1})\cdot(\cot \alpha_{2})\cdots(\cot \alpha_{n}) = 1 \text{ is}

$\frac{1}{2^{n/2}}$
$\frac{1}{2n}$
$1/{2 n }$
1

Solution:

We are given that,

(\cot \alpha_{1})\cdot(\cot \alpha_{2})\cdots(\cot \alpha_{n}) = 1

⇒

(\cos \alpha_{1})\cdot(\cos \alpha_{2})\cdots(\cos \alpha_{n}) = (\sin \alpha_{1})

\cdot (\sin \alpha_{2}) \cdots (\sin \alpha_{n})

...(i)

Let

y = (\cos \alpha_{1}) \cdot (\cos \alpha_{2}) \cdots (\cos \alpha_{n})

(to be max.)

Squarring both sides, we get

y^{2} = (\cos^{2} \alpha_{1}) \cdot (\cos^{2} \alpha_{2}) \cdots (\cos^{2} \alpha_{n})

= \cos \alpha_{1} \cdot \sin \alpha_{1} \cdot \cos \alpha_{2} \cdot

\sin \alpha_{2} \cdot \cos \alpha_{n} \cdot \sin \alpha_{n}

[using (i)]

= \frac{1}{2^{n}} [\sin 2\alpha_{1} \cdot \sin 2\alpha_{2} \cdots \sin 2\alpha_{n}]

As

0 \leq \alpha_{1}, \alpha_{2}, \dots, \alpha_{n} \leq \frac{\pi}{2}

∴

0 \leq 2\alpha_{1}, 2\alpha_{2}, \dots, 2\alpha_{n} \leq \pi

⇒

0 \leq \sin 2\alpha_{1}, \sin 2\alpha_{2}, \dots, \sin 2\alpha_{n} \leq 1

∴

y^{2} \leq \frac{1}{2^{n}}. 1 \qquad \Rightarrow y \leq \frac{1}{2^{n/2}}

∴ Maximum value of

y

is

\frac{1}{2^{n/2}}

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