If fx-i-12020cos-1-i-0- then -i-14-i-

Given : ∑_i=1^2020cos^ 1α_i=0 We know, 0≤cos^ 1x≤π So, given equation will be true only if cos^ 1α_i=0 for i=1,2,⋯,2020 ⇒α_i=1 for i=1,2,⋯,2020 ∴∑_i=1^4α_i=1+1+ ...

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Standard XII

Mathematics

Derivative of Standard Inverse Trigonometric Functions

If fx=∑i=1202...

Question

If $f (x) = 2020 \sum i = 1 {cos}^{- 1} α_{i} = 0$ , then $4 \sum i = 1 α_{i} =$

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Solution

Given : $2020 \sum i = 1 {cos}^{- 1} α_{i} = 0$
We know, $0 \leq {cos}^{- 1} x \leq π$
So, given equation will be true only if ${cos}^{- 1} α_{i} = 0 for i = 1, 2, \dots, 2020$
$\Rightarrow α_{i} = 1 for i = 1, 2, \dots, 2020$
$∴ 4 \sum i = 1 α_{i} = 1 + 1 + 1 + 1 = 4$

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Derivative of Standard Inverse Trigonometric Functions

Standard XII Mathematics

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If f(x)=2020∑i=1cos−1αi=0, then 4∑i=1αi=

Given : 2020∑i=1cos−1αi=0 We know, 0≤cos−1x≤π So, given equation will be true only if cos−1αi=0 for i=1,2,⋯,2020 ⇒αi=1 for i=1,2,⋯,2020 ∴4∑i=1αi=1+1+1+1=4

If $f (x) = 2020 \sum i = 1 {cos}^{- 1} α_{i} = 0$ , then $4 \sum i = 1 α_{i} =$

Given : $2020 \sum i = 1 {cos}^{- 1} α_{i} = 0$
We know, $0 \leq {cos}^{- 1} x \leq π$
So, given equation will be true only if ${cos}^{- 1} α_{i} = 0 for i = 1, 2, \dots, 2020$
$\Rightarrow α_{i} = 1 for i = 1, 2, \dots, 2020$
$∴ 4 \sum i = 1 α_{i} = 1 + 1 + 1 + 1 = 4$