Let fx=sin⁡(π)/6sin⁡(π)/2sin⁡x, for all x∈R and gx=(π)/2sin⁡x, for all x∈R. Let fog x denote fgx and gof x denote gfx. Then which of the following is…

Question

Let fx=sin⁡(π)/6sin⁡(π)/2sin⁡x, for all x∈R and gx=(π)/2sin⁡x, for all x∈R. Let fog x denote fgx and gof x denote gfx. Then which of the following is (are) true?

MARKS Content Team · Accepted Answer

Correct answer: (A) Range of f is -1/2,1/2; (B) Range of fog is -1/2,1/2; (C) limx→0⁡(f(x))/(g(x))=(π)/6. fx=sin⁡(π)/6sin⁡(π)/2sin⁡x (-π)/2≤(π)/2sin⁡x≤(π)/2 -1≤sin⁡(π)/2sin⁡x≤1 sin⁡(-π)/6≤sin⁡(π)/6sin⁡(π)/2sin⁡x≤sin⁡(π)/6 (-1)/2≤sin⁡(π)/6sin⁡(π)/2sin⁡x≤1/2 fgx=sin⁡(π)/6sin⁡(π)/2sin⁡(π)/2sin⁡x -1≤sin⁡(π)/2sin⁡x≤1 (-π)/2≤(π)/2sin⁡(π)/2sin⁡x≤(π)/2 -1≤sin⁡(π)/2sin⁡(π)/2sin⁡x≤1 -(π)/6≤(π)/6sin⁡(π)/2sin⁡(π)/2sin⁡x≤(π)/6 (-1)/2≤sin⁡(π)/6sin⁡(π)/2sin⁡(π)/2sin⁡x≤1/2 limx→0fx/gx=limx→0⁡(sin⁡π6sinπ2sinx)/(π2sin⁡x)×(π6sin⁡π2sin⁡x)/(π6sin⁡π2sin⁡x) =limx→0⁡(π)/6(sin⁡π2sin⁡x)/(π2sin⁡x)=(π)/6. Range of gfx is (π)/2sin⁡(-1)/2,(π)/2sin⁡1/2 =-(π)/2sin⁡1/2,(π)/2sin⁡1/2 Hence, 1, does not belong to this range.

Let $f (x) = \sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))$ , for all $x \in R$ and $g (x) = \frac{π}{2} \sin x$ , for all $x \in R$ . Let $(f o g) (x)$ denote $f (g (x))$ and $(g o f) (x)$ denote $g (f (x))$ . Then which of the following is (are) true?

Solution

More Functions Questions

Let fx=sin⁡π6sin⁡π2sin⁡x, for all x∈R and gx=π2sin⁡x, for all x∈R. Let fog x denote fgx and gof x denote gfx. Then which of the following is (are) true?

Solution

More Functions Questions

Let $f (x) = \sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))$ , for all $x \in R$ and $g (x) = \frac{π}{2} \sin x$ , for all $x \in R$ . Let $(f o g) (x)$ denote $f (g (x))$ and $(g o f) (x)$ denote $g (f (x))$ . Then which of the following is (are) true?