NIMCET 2023 Question Paper with solutions

© examsiri.com

Question : 19 of 120

Marks: +1, -0

If

∏_{i=1}^n \tan (α_i) =1

, where

a_i ∈ [0, \;{π}/{2}] ∀ i=

1,2, . . . n

, then the maximum value of

∏_{i=1}^n sin \; (α_i)

is:

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

Solution:

sin \;α_1 sin \;α_2 sin \;α_3 . . . sin \;α_n

Multiplying both sides by

\; sin \;α_1 sin \;α_2 sin \;α_3, . . . sin \;α_n

\;⇒ sin \;^2 α_1 sin \;^2 α_2 sin \;^2 α_3 . . . sin \;^2 α_n

\;= ( sin \;α_1 \cos α_1) ( sin \;α_2 \cos α_2)

\;⇒ sin \;^2 α_1 sin \;^2 α_2 sin \;^2 α_3 . . . sin \;^2 α_n \cos α_n)

\;=\;{1}/{2^n} ( sin \;2 α_1) ( sin \;2 α_2) . . . ( sin \;2 α_n)

As we know maximum value of

sin \;θ

is 1

⇒ sin \;^2 α_1 sin \;^2 α_2 sin \;^2 α_3 . . . sin \;^2 α_n ≤ \;{1}/{2^n}

⇒ sin \;α_1 sin \;α_2 sin \;α_3 . . . sin \;α_n ≤ \;{1}/{2^{n \/ 2}}

Hence, maximum value of

∏_{i=1}^n sin \;α_1=\;{1}/{2^{n \/ 2}}

.

© examsiri.com

Go to Question:

Prev Question

Next Question