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  • Explain solution for RD Sharma Class 12 Chapter Function Exercise 2.3 Question 6 maths textbook solution.

Explain solution for RD Sharma Class 12 Chapter Function Exercise 2.3 Question 6 maths textbook solution.

Answers (1)

Given : Here given that

             f(x)=\sin x ; g(x)=2 x \& h(x)=\cos x

To prove : Here we have to prove that f \circ g=g \circ(f h).

Solution :

We know that

f: R \rightarrow[-1,1] \text { and } g: R \rightarrow R

Clearly, the range of g is a subset of the domain of f .

f \circ g: R \rightarrow R

Now,

       \begin{aligned} (f h)(x)=f(x) h(x) &=(\sin x)(\cos x) \\ &=\frac{1}{2} \sin (2 x) \end{aligned}

Domain of h is R.

Since range of \sin x \text { is }[-1,1] ;-1 \leq \sin \sin 2 x \leq 1

\Rightarrow \quad-\frac{1}{2} \leq \sin \sin \frac{x}{2} \leq \frac{1}{2}

Range of f h=\left[-\frac{1}{2}, \frac{1}{2}\right]

So, (f h): R \rightarrow\left[-\frac{1}{2}, \frac{1}{2}\right]

Clearly the range of fh is a subset of g

g \circ(f h): R \rightarrow R

Domain of f\; o\; g\; \text {and}\; g\; o\; (fh) are the same gain

So,

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

And again ,

           \begin{aligned} &g \circ(f h)(x)=g(f(x) \cdot h(x)) \\ &=g(\sin x \cos x) \\ &=2 \sin x \cos x \\ &=\sin 2 x \end{aligned}

\Rightarrow \quad f \circ g(x)=g o(f h)(x) ; \forall x \in R

Hence, f \circ g=g \circ(f h)

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