FunctionHard
Question
Let f(x) = x2 and g(x) = sinx for all xε R. Then the set of all x satisfying (f o g o g o f ) (x) = (g o g o f ) (x), where (f o g) (x) = f (g(x)), is
Options
A.
{0,1, 2, ....}
{0,1, 2, ....} B. {1, 2, ....}
{1, 2, ....}C.
+ 2nπ , n ε {...,-2, -1,0,1, 2, ....}
+ 2nπ , n ε {...,-2, -1,0,1, 2, ....}D.2nπ , n ε{...,-2, -1,0,1, 2, ....}
Solution
(fogogof) (x) = sin2 (sin x2)
(gogof) (x) = sin (sin x2)
∴ sin2 (sin x2) = sin (sin x2)
⇒ sin (sin x2) [sin (sin x2) - 1] = 0
⇒ sin (sin x2) = 0 or 1
⇒ sin x2 = nπ or 2mπ + π/2, where m, n ε I
⇒ sin x2 = 0
⇒ x2 = nπ ⇒ =
, n ε {0, 1, 2, ....}.
(gogof) (x) = sin (sin x2)
∴ sin2 (sin x2) = sin (sin x2)
⇒ sin (sin x2) [sin (sin x2) - 1] = 0
⇒ sin (sin x2) = 0 or 1
⇒ sin x2 = nπ or 2mπ + π/2, where m, n ε I
⇒ sin x2 = 0
⇒ x2 = nπ ⇒ =
, n ε {0, 1, 2, ....}.Create a free account to view solution
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