SATHEE: Functions 2 Question 6

Functions 2 Question 6

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

(2011)

(a) $\pm \sqrt{n π}, n \in 0, 1, 2, \dots$

(b) $\pm \sqrt{n π}, n \in 1, 2, \dots$

(d) $2 n π, n \in \dots, - 2, - 1, 0, 1, 2, \dots$

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Answer:

Correct Answer: 7. (b)

Solution:

$f (x) = x^{2}, g (x) = \sin x$

$\begin{aligned} (g \circ f) (x) & = \sin x^{2} \\ g o (g \circ f) (x) & = \sin (\sin x^{2}) \\ (fogogof) (x) & = {(\sin (\sin x^{2}))}^{2} \end{aligned}$

Again, $(g \circ f) (x) = \sin x^{2}$

$(gogof) (x) = \sin (\sin x^{2})$

Given, $($ fogogof $) (x) = (gogof) (x)$

$\Rightarrow {(\sin (\sin x^{2}))}^{2} = \sin (\sin x^{2})$

$\Rightarrow \sin (\sin x^{2}) \sin (\sin x^{2}) - 1 = 0$

$\Rightarrow \sin (\sin x^{2}) = 0$ or $\sin (\sin x^{2}) = 1$

$\Rightarrow \sin x^{2} = 0 or \sin x^{2} = \frac{π}{2}$

$∴ x^{2} = n π$

$\begin{aligned} [\sin x^{2} = \frac{π}{2} is not possible as - 1 \leq \sin θ \leq 1] \\ x = \pm \sqrt{n π} \end{aligned}$

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Functions 2 Question 6

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

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Functions 2 Question 6

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

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चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

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