Consider a function f defined by f(x) = sin –1 sin 
, ∀ x ∈ [0, π ], which satisfies
f(x) + f(2 π – x) = π , ∀ x ∈ [ π , 2 π ] and f(x) = f(4 π – x) for all x ∈ [2 π , 4 π ], then
(i) If α is the length of the largest interval on which f(x) is increasing, then α =
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
i(c),ii(b),iii(a)
Let g(x) = 
, x ∈ [0, π ]. g(x) is increasing function of x.
∴ range of g(x) is 
∴ f(x) = 
, x ∈ [0, π ]
Now let π ≤ t ≤ 2 π , then
f(t) + f(2 π – t) = π
i.e f(t) + 
= π
i.e f(t) + π – 
– 
= π
i.e f(t) = 
∴ f(x) = 
for π ≤ x ≤ 2 π
Thus f(x) = 
for 0 ≤ x ≤ 2 π
Also f(x) = f(4 π – x) for all x ∈ [2 π , 4 π ]
⇒ f(x) is symmetric about x = 2 π
∴ from graph of f(x)
∴ α = 2 π – 0 = 2 π
∴ β = α
Maximum value is f(2 π ) = π = 
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