Question

Find the domain and range of the following functions
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Answer 1

(i) $\smallf(x)=\cos[\ln(5x^2-8x+7)]$

$\small\Longrightarrow[\ln(5x^2-8x+7)]\in\mathbb{R\cap Z}$

$\small\longrightarrow[\ln(5x^2-8x+7)]\in\mathbb{Z}$

$\small\Longrightarrow\ln(5x^2-8x+7)\in\mathbb{R}$

$\small\Longrightarrow5x^2-8x+7\in(0,\ \infty)$

$\small\longrightarrow5\left(x-\dfrac{4}{5}\right)^2+\dfrac{19}{5}\in(0,\ \infty)$

$\small\longrightarrow\left(x-\dfrac{4}{5}\right)^2\in\bigg(-\dfrac{19}{25},\ \infty\bigg)\cap[0,\ \infty)$

$\small\longrightarrow\left(x-\dfrac{4}{5}\right)^2\in[0,\ \infty)$

$\small\text{ \Longrightarrow\underline{\underline{x\in\mathbb{R}}} }$

This is the domain.

$\small\longrightarrow x\in\mathbb{R}$

$\small\longrightarrow5\left(x-\dfrac{4}{5}\right)^2+\dfrac{19}{5}=5x^2-8x+7\in\left[\dfrac{19}{5},\ \infty\right)$

$\small\longrightarrow\ln(5x^2-8x+7)\in\left[\ln\left(\dfrac{19}{5}\right),\ \infty\right)$

$\small\longrightarrow[\ln(5x^2-8x+7)]\in\mathbb{Z^+}$

So $\small\cos[\ln(5x^2-8x+7)]$ gives cosine of positive integers. Therefore,

$\small\text{ \longrightarrow\underline{\underline{f(x)\in\{x:x=\cos\theta,\ \theta\in\mathbb{Z^+}\}}} }$

This is the range.

(ii)

$\small\longrightarrow f(x)=\log_2\left(\dfrac{\sin x-\cos x+7\sqrt2}{\sqrt2}\right)$

$\small\text{ \longrightarrow f(x)=\log_2\left(\dfrac{\sqrt2\sin\left(x-\dfrac{\pi}{4}\right)+7\sqrt2}{\sqrt2}\right) }$

$\small\longrightarrow f(x)=\log_2\left(\sin\left(x-\dfrac{\pi}{4}\right)+7\right)$

$\small\Longrightarrow\sin\left(x-\dfrac{\pi}{4}\right)+7\in(0,\ \infty)$

$\small\longrightarrow\sin\left(x-\dfrac{\pi}{4}\right)\in(-7,\ \infty)\cap[-1,\ 1]$

$\small\longrightarrow\sin\left(x-\dfrac{\pi}{4}\right)\in[-1,\ 1]$

$\small\text{ \Longrightarrow\underline{\underline{x\in\mathbb{R}}} }$

This is the domain.

$\small\longrightarrow x\in\mathbb{R}$

$\small\longrightarrow\sin\left(x-\dfrac{\pi}{4}\right)\in[-1,\ 1]$

$\small\longrightarrow\sin\left(x-\dfrac{\pi}{4}\right)+7=\dfrac{\sin x-\cos x+7\sqrt2}{\sqrt2}\in[6,\ 8]$

$\small\longrightarrow\log_2\left(\dfrac{\sin x-\cos x+7\sqrt2}{\sqrt2}\right)\in[\log_26,\ \log_28]$

$\small\text{ \displaystyle\longrightarrow\underline{\underline{f(x)\in[1+\log_23,\ 3]}} }$

This is the range.

(iii) $\small\text{ f(x)=\sqrt{1-\left(\ln\left[\sqrt{\left[x^3+1\right]}\right]\right)^2} }$

$\small\Longrightarrow 1-\left(\ln\left[\sqrt{\left[x^3+1\right]}\right]\right)^2\in[0,\ \infty)$

$\small\longrightarrow\left(\ln\left[\sqrt{\left[x^3+1\right]}\right]\right)^2\in(-\infty,\ 1]\cap[0,\ \infty)$

$\small\longrightarrow\left(\ln\left[\sqrt{\left[x^3+1\right]}\right]\right)^2\in[0,\ 1]$

$\small\longrightarrow\ln\left[\sqrt{\left[x^3+1\right]}\right]\in[-1,\ 1]$

$\small\longrightarrow\left[\sqrt{\left[x^3+1\right]}\right]\in\left[\dfrac{1}{e},\ e\right]\cap\mathbb{Z}$

$\small\longrightarrow\left[\sqrt{\left[x^3+1\right]}\right]\in\{1,\ 2\}$

$\small\Longrightarrow\sqrt{\left[x^3+1\right]}\in[1,\ 3)$

$\small\longrightarrow\left[x^3+1\right]\in[1,\ 9)\cap\mathbb{Z}$

$\small\longrightarrow\left[x^3+1\right]\in\{1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8\}$

$\small\Longrightarrow x^3+1\in[1,\ 9)$

$\small\longrightarrow x^3\in[0,\ 8)$

$\small\text{ \longrightarrow\underline{\underline{x\in[0,\ 2)}} }$

This is the domain.

$\small\longrightarrow x\in[0,\ 2)$

$\small\longrightarrow x^3+1\in[1,\ 9)$

$\small\longrightarrow\left[x^3+1\right]\in\{1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8\}$

$\small\longrightarrow\sqrt{\left[x^3+1\right]}\in\left\{1,\ \sqrt2,\ \sqrt3,\ 2,\ \sqrt5,\ \sqrt6,\ \sqrt7,\ \sqrt8\right\}$

$\small\longrightarrow\left[\sqrt{\left[x^3+1\right]}\right]\in\left\{1,\ 2\right\}$

$\small\longrightarrow\ln\left[\sqrt{\left[x^3+1\right]}\right]\in\left\{0,\ \ln2\right\}$

$\small\longrightarrow\left(\ln\left[\sqrt{\left[x^3+1\right]}\right]\right)^2\in\left\{0,\ (\ln2)^2\right\}$

$\small\longrightarrow1-\left(\ln\left[\sqrt{\left[x^3+1\right]}\right]\right)^2\in\left\{1,\ 1-(\ln2)^2\right\}$

$\small\text{ \longrightarrow\sqrt{1-\left(\ln\left[\sqrt{\left[x^3+1\right]}\right]\right)^2}\in\left\{1,\ \sqrt{1-(\ln2)^2}\right\} }$

$\small\text{ \longrightarrow\underline{\underline{f(x)\in\left\{1,\ \sqrt{1-(\ln2)^2}\right\}}} }$

This is the range.