Let fx-x2 and gx -sin x for all x - R- Then the set of all x satisfying fogogof x -gogof x- where fog x -fgx- is

(fogogof)(x)=sin^2 (sin x^2) (gogof)(x)=sin(sin x^2) ∴ sin^2(sin x^2)=sin (sin x^2) ⇒ sin(sin x^2)[sin(sin x^2) 1]=0 ⇒ sin(sin x^2)=0 or 1 ⇒ sin x^2 =n π or 2m ...

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Standard XIII

Mathematics

Solving Simultaneous Trigonometric Equations

Let fx=x2 and...

Question

Let $f (x) = x^{2}$ and g(x) =sin x for all $x \in R .$ Then the set of all x satisfying (fogogof) (x) =(gogof) (x), where (fog) (x) =f(g(x)), is

\pm \sqrt{n π}, n \in {0, 1, 2, . . . .}

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\pm \sqrt{n π}, n \in {1, 2, . . . .}

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\frac{π}{2} + 2 n π, n i n {. . . ., - 2, - 1, 0, 1, 2, . . . .}

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2 n π, n \in {. . . ., - 2, - 1, 0, 1, 2, . . . .}

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Solution

The correct option is A $\pm \sqrt{n π}, n \in {0, 1, 2, . . . .}$

$(f o g o g o f) (x) = s i n^{2} (s i n x^{2}) (g o g o f) (x) = s i n (s i n x^{2}) ∴ s i n^{2} (s i n x^{2}) = s i n (s i n x^{2}) \Rightarrow s i n (s i n x^{2}) [s i n (s i n x^{2}) - 1] = 0 \Rightarrow s i n (s i n x^{2}) = 0 o r 1$
$\Rightarrow s i n x^{2} = n π o r 2 m π + \frac{π}{2}$ , where $m, n \in I$
$\Rightarrow s i n x^{2} = 0 \Rightarrow x^{2} = n π \Rightarrow x = \pm \sqrt{n π}, n \in {0, 1, 2, . . . .}$

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Similar questions

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Let f(x)=x2 and g(x) =sin x for all x∈R. Then the set of all x satisfying (fogogof) (x) =(gogof) (x), where (fog) (x) =f(g(x)), is

The correct option is A ±√nπ,n∈{0,1,2,....} (fogogof)(x)=sin2(sinx2)(gogof)(x)=sin(sinx2)∴sin2(sinx2)=sin(sinx2)⇒sin(sinx2)[sin(sinx2)−1]=0⇒sin(sinx2)=0or1 ⇒sinx2=nπor2mπ+π2, where m,n∈I ⇒sinx2=0⇒x2=nπ⇒x=±√nπ,n∈{0,1,2,....}

Let $f (x) = x^{2}$ and g(x) =sin x for all $x \in R .$ Then the set of all x satisfying (fogogof) (x) =(gogof) (x), where (fog) (x) =f(g(x)), is