Verified Solution • 2025–26 Session moderate

QJEEAdv11Ma75ec: JEE Advanced 2011 Mathematics - Class 11 JEEAdv11Ma75ec Solution: Let f(x) = x 2 and g(x) = sin x for all x R. Then the set of…

Let f(x) = x2 and g(x) = sin x for all x ∈\in R. Then the set of all x satisfying (f∘g∘g∘f)(x)=(g∘g∘f)(x)(f \circ g \circ g \circ f)(x) = (g \circ g \circ f)(x) , where (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x)) , is

Multiple Choice Options

  • A ±nπ, n∈{0,1,2,....}\pm \sqrt {n\pi } ,\,n \in \{ 0,1,2,....\}
  • B ±nπ, n∈{1,2,....}\pm \sqrt {n\pi } ,\,n \in \{ 1,2,....\}
  • C π2+2nπ, n∈{.....,−2,−1,0,1,2,....}{\pi \over 2} + 2n\pi ,\,n \in \{ ....., - 2, - 1,0,1,2,....\}
  • D 2nπ,n∈{.....,−2,−1,0,1,2,....}2n\pi ,n \in \{ ....., - 2, - 1,0,1,2,....\}

Step-by-Step Verified Solution

f(x)=x2f(x) = {x^2} , g(x)=sin⁡xg(x) = \sin x

(g∘f)(x)=sin⁡x2(g \circ f)(x) = \sin {x^2}

g∘(g∘f)(x)=sin⁡(sin⁡x2)g \circ (g \circ f)(x) = \sin (\sin {x^2})

(f∘g∘g∘f)(x)=(sin⁡(sin⁡x2))2(f \circ g \circ g \circ f)(x) = {(\sin (\sin {x^2}))^2} ..... (i)

Again, (g∘f)(x)=sin⁡x2(g \circ f)(x) = \sin {x^2}

(g∘g∘f)(x)=sin⁡(sin⁡x2)(g \circ g \circ f)(x) = \sin (\sin {x^2}) ..... (ii)

Given, (f∘g∘g∘f)(x)=(g∘g∘f)(x)(f \circ g \circ g \circ f)(x) = (g \circ g \circ f)(x)

⇒(sin⁡(sin⁡x2))2=sin⁡(sin⁡x2)\Rightarrow {(\sin (\sin {x^2}))^2} = \sin (\sin {x^2})

⇒sin⁡(sin⁡x2){sin⁡(sin⁡x2)−1}=0\Rightarrow \sin (\sin {x^2})\{ \sin (\sin {x^2}) - 1\} = 0

⇒sin⁡(sin⁡x2)=0\Rightarrow \sin (\sin {x^2}) = 0 or sin⁡(sin⁡x2)=1\sin (\sin {x^2}) = 1

⇒sin⁡x2=0\Rightarrow \sin {x^2} = 0 or sin⁡x2=π2\sin {x^2} = {\pi \over 2}

∴\therefore x2=nπ{x^2} = n\pi (i.e. not possible as −1≤sin⁡θ≤1- 1 \le \sin \theta \le 1 )

x=±nπx = \pm \sqrt {n\pi }

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