If f ( x )= x 11+ x 9- x 7+ x 3+1 and f ( sin -1( sin 8))=α, then f ( tan -1( tan 8)) is equal to

Question

If f ( x )= x 11+ x 9- x 7+ x 3+1 and f ( sin -1( sin 8))=α, then f ( tan -1( tan 8)) is equal to (A) α (B) α-2 (C) α+2 (D) 2-α

Tardigrade Admin · Accepted Answer

Θ sin -1( sin 8)= sin -1( sin (3 π-8))=3 π-8 text and tan -1( tan 8)= tan -1( tan (8-3 π))=8-3 π ∴ f ( x )+ f (- x )=2 ∴ f (3 π-8)+ f (8-3 π)=2 ⇒ f (8-3 π)= f ( tan -1( tan 8))=2-α .

Q. If $f (x) = x^{11} + x^{9} - x^{7} + x^{3} + 1$ and $f (sin^{- 1} (sin 8)) = α$ , then $f (tan^{- 1} (tan 8))$ is equal to

$α$

$α - 2$

$α + 2$

$2 - α$

$Θ sin^{- 1} (sin 8) = sin^{- 1} (sin (3 π - 8)) = 3 π - 8$
$and tan^{- 1} (tan 8) = tan^{- 1} (tan (8 - 3 π)) = 8 - 3 π$
$∴ f (x) + f (- x) = 2$
$∴ f (3 π - 8) + f (8 - 3 π) = 2$
$\Rightarrow f (8 - 3 π) = f (tan^{- 1} (tan 8)) = 2 - α .$

Q. If f(x)=x11+x9−x7+x3+1 and f(sin−1(sin8))=α, then f(tan−1(tan8)) is equal to

α

α−2

α+2

2−α

Θsin−1(sin8)=sin−1(sin(3π−8))=3π−8 and tan−1(tan8)=tan−1(tan(8−3π))=8−3π ∴f(x)+f(−x)=2 ∴f(3π−8)+f(8−3π)=2 ⇒f(8−3π)=f(tan−1(tan8))=2−α.

Q. If $f (x) = x^{11} + x^{9} - x^{7} + x^{3} + 1$ and $f (sin^{- 1} (sin 8)) = α$ , then $f (tan^{- 1} (tan 8))$ is equal to

$α$

$α - 2$

$α + 2$

$2 - α$

$Θ sin^{- 1} (sin 8) = sin^{- 1} (sin (3 π - 8)) = 3 π - 8$
$and tan^{- 1} (tan 8) = tan^{- 1} (tan (8 - 3 π)) = 8 - 3 π$
$∴ f (x) + f (- x) = 2$
$∴ f (3 π - 8) + f (8 - 3 π) = 2$
$\Rightarrow f (8 - 3 π) = f (tan^{- 1} (tan 8)) = 2 - α .$