Test Index

NIMCET 2023 Question Paper with solutions

© examsnet.com

Question : 19 of 120

Marks: +1, -0

If

\prod_{i=1}^n \tan(\alpha_i)=1

a_i \in \left[0, \;\frac{\pi}{2}\right] \forall i=

1,2,\ldots n

, then the maximum value of

\prod_{i=1}^n \sin(\alpha_i)

$2^{-n}$
$2^{-\frac{n}{2}}$
1
None of these

Solution:

\sin\alpha_1 \sin\alpha_2 \sin\alpha_3 \ldots \sin\alpha_n

Multiplying both sides by

\; \sin\alpha_1 \sin\alpha_2 \sin\alpha_3, \ldots \sin\alpha_n

\Rightarrow \sin^2\alpha_1 \sin^2\alpha_2 \sin^2\alpha_3 \ldots \sin^2\alpha_n

= (\sin\alpha_1 \cos\alpha_1)(\sin\alpha_2 \cos\alpha_2)

\Rightarrow \sin^2\alpha_1 \sin^2\alpha_2 \sin^2\alpha_3 \ldots \sin^2\alpha_n \cos\alpha_n)

= \frac{1}{2^n} (\sin 2\alpha_1)(\sin 2\alpha_2) \ldots (\sin 2\alpha_n)

As we know maximum value of

\sin \theta

is 1

\Rightarrow \sin^2\alpha_1 \sin^2\alpha_2 \sin^2\alpha_3 \ldots \sin^2\alpha_n \leq \frac{1}{2^n}

\Rightarrow \sin\alpha_1 \sin\alpha_2 \sin\alpha_3 \ldots \sin\alpha_n \leq \frac{1}{2^{n/2}}

Hence, maximum value of

\prod_{i=1}^n \sin\alpha_1 = \frac{1}{2^{n/2}}

.

© examsnet.com

Go to Question:

Prev Question

Next Question