The Range of the function f(x)=log10 sin-1 (x-5)+3π2

Let f(x)=log10 sin-1 (x-5)+3π2, The function is defined if (i) x-5≥0 (ii) -1≤x-5≤1 and (iii) sin-1 (x-5)+3π2>0. Now (i) ⇒x≥5 (ii) ⇒0≤x-5≤1⇒5≤x≤6(iii) is satisfied by virtue of (ii).Hence, considering (i) and (ii), we find that the domain of the function viz. Df=[5,6]. Let y1=sin-1 (x-5) and y2=sin-1 (x-5)+3π2 so that y=log10 y2 where y2=y1+3π2Now for y1 since x∈[5,6],y1≥0 so that 0≤y1≤π2∵-π2≤sin-1 (z)≤π2 Consequently 0+3π2≤y1+3π2≤π2+3π2⇒3π2≤y2≤2π ⇒ log 3π2≤log y2≤log (2π)since u=log z is an increasing function ⇒ log 3π2≤log y2≤log (2π). Hence the range of f(x) is log 3π2,log 2π

The Range of the function f(x)=log10 sin-1 (x-5)+3π2

Let f(x)=log10 sin-1 (x-5)+3π2, The function is defined if (i) x-5≥0 (ii) -1≤x-5≤1 and (iii) sin-1 (x-5)+3π2>0. Now (i) ⇒x≥5 (ii) ⇒0≤x-5≤1⇒5≤x≤6(iii) is satisfied by virtue of (ii).Hence, considering (i) and (ii), we find that the domain of the function viz. Df=[5,6]. Let y1=sin-1 (x-5) and y2=sin-1 (x-5)+3π2 so that y=log10 y2 where y2=y1+3π2Now for y1 since x∈[5,6],y1≥0 so that 0≤y1≤π2∵-π2≤sin-1 (z)≤π2 Consequently 0+3π2≤y1+3π2≤π2+3π2⇒3π2≤y2≤2π ⇒ log 3π2≤log y2≤log (2π)since u=log z is an increasing function ⇒ log 3π2≤log y2≤log (2π). Hence the range of f(x) is log 3π2,log 2π

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The Range of the function $f (x) = \log_{10} \{\sin^{- 1} (\sqrt{x - 5}) + \frac{3 π}{2}\}$

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$[\log \frac{3 π}{2}, \log π]$

$[\log \frac{3 π}{2}, \log 2 π]$

$[\log \frac{3 π}{2}, \log 3 π]$

$[\log \frac{3 π}{4}, \log 2 π]$

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Detailed Solution

Let $f (x) = \log_{10} \{\sin^{- 1} (\sqrt{x - 5}) + \frac{3 π}{2}\}$ , The function is defined if (i) $x - 5 \geq 0$ (ii) $- 1 \leq \sqrt{x - 5} \leq 1$ and (iii) $\sin^{- 1} (\sqrt{x - 5}) + \frac{3 π}{2} > 0$ . Now (i) $\Rightarrow x \geq 5$ (ii) $\Rightarrow 0 \leq x - 5 \leq 1 \Rightarrow 5 \leq x \leq 6$

(iii) is satisfied by virtue of (ii).Hence, considering (i) and (ii), we find that the domain of the function viz. $D_{f} = [5,6]$ . Let $y_{1} = \sin^{- 1} (\sqrt{x - 5})$ and $y_{2} = \sin^{- 1} (\sqrt{x - 5}) + \frac{3 π}{2}$ so that $y = \log_{10} (y_{2})$ where $y_{2} = y_{1} + \frac{3 π}{2} Now$ for $y_{1}$ since $x \in [5,6], y_{1} \geq 0$ so that $0 \leq y_{1} \leq \frac{π}{2} (∵ - \frac{π}{2} \leq \sin^{- 1} (z) \leq \frac{π}{2})$ Consequently $0 + \frac{3 π}{2} \leq y_{1} + \frac{3 π}{2} \leq \frac{π}{2} + \frac{3 π}{2} \Rightarrow \frac{3 π}{2} \leq y_{2} \leq 2 π \Rightarrow \log (\frac{3 π}{2}) \leq \log (y_{2}) \leq \log (2 π)$

since $u = \log z$ is an increasing function $\Rightarrow \log (\frac{3 π}{2}) \leq \log (y_{2}) \leq \log (2 π)$ . Hence the range of $f (x) is [\log \frac{3 π}{2}, \log 2 π]$