C Given that, \(f(x)= \begin{cases}\frac{x^2-9}{x-3} & , x \neq 3 \\ 6, & x=3\end{cases}\) At, \(x=3\) \(\text { LHL }=\lim _{x \rightarrow 3^{-}} \frac{x^2-9}{x-3}=\lim _{x \rightarrow 3^{-}}(x+3)\) \(\Rightarrow \lim (3-h+3)=6\) \(R H L=\lim _{x \rightarrow 3^{+}} \frac{x^2-9}{x-3}=\lim _{x \rightarrow 3^{+}}(x+3)\) \(=\lim _{h \rightarrow 0}(3+h+3)=6\) \(\mathrm{f}(3)=6\) \(\mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(3)\) Hence, \(f(x)\) is continuous at \(x=3\) So, domain of \(f(x)=(-\infty, \infty)\)
Kerala CEE-2016
Sets, Relation and Function
117449
If \(\cos ^{-1} x>\sin ^{-1} x\), then \(x\) lies in the interval
1 \(\left(\frac{1}{2}, 1\right]\)
2 \((0,1]\)
3 \(\left[-1, \frac{1}{\sqrt{2}}\right)\)
4 \([-1,1]\)
5 \([0,1]\)
Explanation:
: We know that, $\cos ^{-1} x$ and $\sin ^{-1} x$ exist for $x \in[-1,1]$ Now, $\cos ^{-1} x>\sin ^{-1} x$ $\begin{aligned} & \cos ^{-1} x>\frac{\pi}{2}-\cos ^{-1} x \\ & 2 \cos ^{-1} x>\frac{\pi}{2} \\ & \cos ^{-1} x>\frac{\pi}{4} \\ & x \in\left[-1, \frac{1}{\sqrt{2}}\right) \end{aligned}$
Kerala CEE-2015
Sets, Relation and Function
117450
The range of the function $f(x)=\frac{1}{2-\cos 3 x}$ is
1 $(-2, \infty)$
2 $[-2,3]$
3 $\left(\frac{1}{3}, 2\right)$
4 $\left(\frac{1}{2}, 1\right)$
5 $\left[\frac{1}{3}, 1\right]$
Explanation:
E Given, function $\begin{aligned} & y=\frac{1}{2-\cos 3 x} \\ & 2-\cos 3 x=\frac{1}{y} \\ & \cos 3 x=2-\frac{1}{y} \\ & \cos 3 x=\frac{2 y-1}{y} \\ & 3 x=\cos ^{-1}\left(\frac{2 y-1}{y}\right) \\ & x=\frac{1}{3} \cos ^{-1}\left(\frac{2 y-1}{y}\right) \end{aligned}$ For x to be real, $\begin{aligned} & -1 \leq\left|\frac{2 y-1}{y}\right| \leq 1 \\ & -y \leq 2 y-1 \leq y \\ & 2 y-1 \geq-y \\ & 3 y \geq 1 \\ & y \geq \frac{1}{3} \end{aligned}$ $\begin{array}{ll} \text { or } & 2 \mathrm{y}-1 \leq \mathrm{y} \\ \text { or } & \mathrm{y} \leq 1 \end{array}$ So, range $=\left[\frac{1}{3}, 1\right]$
Kerala CEE-2015
Sets, Relation and Function
117451
The domain of the function $f(x)=\sqrt{7-3 x}+\log _e x$ is
1 $0\lt x\lt \infty$
2 $\frac{7}{3} \leq x\lt \infty$
3 $0\lt \mathrm{x} \leq \frac{7}{3}$
4 $-\infty\lt $ x $\lt 0$
5 $-\infty\lt x \leq \frac{7}{3}$
Explanation:
C Given function, $f(x)=\sqrt{7-3 x}+\log _e x$ For, $\sqrt{7-3 x}, 7-3 x \geq 0$ $7 \geq 3 \mathrm{x}$ Or $\begin{aligned} & x \leq \frac{7}{3} \\ & x>0 \\ & x \in\left(0, \frac{7}{3}\right] \\ & 0\lt x \leq \frac{7}{3} \end{aligned}$ Or $\quad 0\lt \mathrm{x} \leq \frac{7}{3}$
C Given that, \(f(x)= \begin{cases}\frac{x^2-9}{x-3} & , x \neq 3 \\ 6, & x=3\end{cases}\) At, \(x=3\) \(\text { LHL }=\lim _{x \rightarrow 3^{-}} \frac{x^2-9}{x-3}=\lim _{x \rightarrow 3^{-}}(x+3)\) \(\Rightarrow \lim (3-h+3)=6\) \(R H L=\lim _{x \rightarrow 3^{+}} \frac{x^2-9}{x-3}=\lim _{x \rightarrow 3^{+}}(x+3)\) \(=\lim _{h \rightarrow 0}(3+h+3)=6\) \(\mathrm{f}(3)=6\) \(\mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(3)\) Hence, \(f(x)\) is continuous at \(x=3\) So, domain of \(f(x)=(-\infty, \infty)\)
Kerala CEE-2016
Sets, Relation and Function
117449
If \(\cos ^{-1} x>\sin ^{-1} x\), then \(x\) lies in the interval
1 \(\left(\frac{1}{2}, 1\right]\)
2 \((0,1]\)
3 \(\left[-1, \frac{1}{\sqrt{2}}\right)\)
4 \([-1,1]\)
5 \([0,1]\)
Explanation:
: We know that, $\cos ^{-1} x$ and $\sin ^{-1} x$ exist for $x \in[-1,1]$ Now, $\cos ^{-1} x>\sin ^{-1} x$ $\begin{aligned} & \cos ^{-1} x>\frac{\pi}{2}-\cos ^{-1} x \\ & 2 \cos ^{-1} x>\frac{\pi}{2} \\ & \cos ^{-1} x>\frac{\pi}{4} \\ & x \in\left[-1, \frac{1}{\sqrt{2}}\right) \end{aligned}$
Kerala CEE-2015
Sets, Relation and Function
117450
The range of the function $f(x)=\frac{1}{2-\cos 3 x}$ is
1 $(-2, \infty)$
2 $[-2,3]$
3 $\left(\frac{1}{3}, 2\right)$
4 $\left(\frac{1}{2}, 1\right)$
5 $\left[\frac{1}{3}, 1\right]$
Explanation:
E Given, function $\begin{aligned} & y=\frac{1}{2-\cos 3 x} \\ & 2-\cos 3 x=\frac{1}{y} \\ & \cos 3 x=2-\frac{1}{y} \\ & \cos 3 x=\frac{2 y-1}{y} \\ & 3 x=\cos ^{-1}\left(\frac{2 y-1}{y}\right) \\ & x=\frac{1}{3} \cos ^{-1}\left(\frac{2 y-1}{y}\right) \end{aligned}$ For x to be real, $\begin{aligned} & -1 \leq\left|\frac{2 y-1}{y}\right| \leq 1 \\ & -y \leq 2 y-1 \leq y \\ & 2 y-1 \geq-y \\ & 3 y \geq 1 \\ & y \geq \frac{1}{3} \end{aligned}$ $\begin{array}{ll} \text { or } & 2 \mathrm{y}-1 \leq \mathrm{y} \\ \text { or } & \mathrm{y} \leq 1 \end{array}$ So, range $=\left[\frac{1}{3}, 1\right]$
Kerala CEE-2015
Sets, Relation and Function
117451
The domain of the function $f(x)=\sqrt{7-3 x}+\log _e x$ is
1 $0\lt x\lt \infty$
2 $\frac{7}{3} \leq x\lt \infty$
3 $0\lt \mathrm{x} \leq \frac{7}{3}$
4 $-\infty\lt $ x $\lt 0$
5 $-\infty\lt x \leq \frac{7}{3}$
Explanation:
C Given function, $f(x)=\sqrt{7-3 x}+\log _e x$ For, $\sqrt{7-3 x}, 7-3 x \geq 0$ $7 \geq 3 \mathrm{x}$ Or $\begin{aligned} & x \leq \frac{7}{3} \\ & x>0 \\ & x \in\left(0, \frac{7}{3}\right] \\ & 0\lt x \leq \frac{7}{3} \end{aligned}$ Or $\quad 0\lt \mathrm{x} \leq \frac{7}{3}$
C Given that, \(f(x)= \begin{cases}\frac{x^2-9}{x-3} & , x \neq 3 \\ 6, & x=3\end{cases}\) At, \(x=3\) \(\text { LHL }=\lim _{x \rightarrow 3^{-}} \frac{x^2-9}{x-3}=\lim _{x \rightarrow 3^{-}}(x+3)\) \(\Rightarrow \lim (3-h+3)=6\) \(R H L=\lim _{x \rightarrow 3^{+}} \frac{x^2-9}{x-3}=\lim _{x \rightarrow 3^{+}}(x+3)\) \(=\lim _{h \rightarrow 0}(3+h+3)=6\) \(\mathrm{f}(3)=6\) \(\mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(3)\) Hence, \(f(x)\) is continuous at \(x=3\) So, domain of \(f(x)=(-\infty, \infty)\)
Kerala CEE-2016
Sets, Relation and Function
117449
If \(\cos ^{-1} x>\sin ^{-1} x\), then \(x\) lies in the interval
1 \(\left(\frac{1}{2}, 1\right]\)
2 \((0,1]\)
3 \(\left[-1, \frac{1}{\sqrt{2}}\right)\)
4 \([-1,1]\)
5 \([0,1]\)
Explanation:
: We know that, $\cos ^{-1} x$ and $\sin ^{-1} x$ exist for $x \in[-1,1]$ Now, $\cos ^{-1} x>\sin ^{-1} x$ $\begin{aligned} & \cos ^{-1} x>\frac{\pi}{2}-\cos ^{-1} x \\ & 2 \cos ^{-1} x>\frac{\pi}{2} \\ & \cos ^{-1} x>\frac{\pi}{4} \\ & x \in\left[-1, \frac{1}{\sqrt{2}}\right) \end{aligned}$
Kerala CEE-2015
Sets, Relation and Function
117450
The range of the function $f(x)=\frac{1}{2-\cos 3 x}$ is
1 $(-2, \infty)$
2 $[-2,3]$
3 $\left(\frac{1}{3}, 2\right)$
4 $\left(\frac{1}{2}, 1\right)$
5 $\left[\frac{1}{3}, 1\right]$
Explanation:
E Given, function $\begin{aligned} & y=\frac{1}{2-\cos 3 x} \\ & 2-\cos 3 x=\frac{1}{y} \\ & \cos 3 x=2-\frac{1}{y} \\ & \cos 3 x=\frac{2 y-1}{y} \\ & 3 x=\cos ^{-1}\left(\frac{2 y-1}{y}\right) \\ & x=\frac{1}{3} \cos ^{-1}\left(\frac{2 y-1}{y}\right) \end{aligned}$ For x to be real, $\begin{aligned} & -1 \leq\left|\frac{2 y-1}{y}\right| \leq 1 \\ & -y \leq 2 y-1 \leq y \\ & 2 y-1 \geq-y \\ & 3 y \geq 1 \\ & y \geq \frac{1}{3} \end{aligned}$ $\begin{array}{ll} \text { or } & 2 \mathrm{y}-1 \leq \mathrm{y} \\ \text { or } & \mathrm{y} \leq 1 \end{array}$ So, range $=\left[\frac{1}{3}, 1\right]$
Kerala CEE-2015
Sets, Relation and Function
117451
The domain of the function $f(x)=\sqrt{7-3 x}+\log _e x$ is
1 $0\lt x\lt \infty$
2 $\frac{7}{3} \leq x\lt \infty$
3 $0\lt \mathrm{x} \leq \frac{7}{3}$
4 $-\infty\lt $ x $\lt 0$
5 $-\infty\lt x \leq \frac{7}{3}$
Explanation:
C Given function, $f(x)=\sqrt{7-3 x}+\log _e x$ For, $\sqrt{7-3 x}, 7-3 x \geq 0$ $7 \geq 3 \mathrm{x}$ Or $\begin{aligned} & x \leq \frac{7}{3} \\ & x>0 \\ & x \in\left(0, \frac{7}{3}\right] \\ & 0\lt x \leq \frac{7}{3} \end{aligned}$ Or $\quad 0\lt \mathrm{x} \leq \frac{7}{3}$
C Given that, \(f(x)= \begin{cases}\frac{x^2-9}{x-3} & , x \neq 3 \\ 6, & x=3\end{cases}\) At, \(x=3\) \(\text { LHL }=\lim _{x \rightarrow 3^{-}} \frac{x^2-9}{x-3}=\lim _{x \rightarrow 3^{-}}(x+3)\) \(\Rightarrow \lim (3-h+3)=6\) \(R H L=\lim _{x \rightarrow 3^{+}} \frac{x^2-9}{x-3}=\lim _{x \rightarrow 3^{+}}(x+3)\) \(=\lim _{h \rightarrow 0}(3+h+3)=6\) \(\mathrm{f}(3)=6\) \(\mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(3)\) Hence, \(f(x)\) is continuous at \(x=3\) So, domain of \(f(x)=(-\infty, \infty)\)
Kerala CEE-2016
Sets, Relation and Function
117449
If \(\cos ^{-1} x>\sin ^{-1} x\), then \(x\) lies in the interval
1 \(\left(\frac{1}{2}, 1\right]\)
2 \((0,1]\)
3 \(\left[-1, \frac{1}{\sqrt{2}}\right)\)
4 \([-1,1]\)
5 \([0,1]\)
Explanation:
: We know that, $\cos ^{-1} x$ and $\sin ^{-1} x$ exist for $x \in[-1,1]$ Now, $\cos ^{-1} x>\sin ^{-1} x$ $\begin{aligned} & \cos ^{-1} x>\frac{\pi}{2}-\cos ^{-1} x \\ & 2 \cos ^{-1} x>\frac{\pi}{2} \\ & \cos ^{-1} x>\frac{\pi}{4} \\ & x \in\left[-1, \frac{1}{\sqrt{2}}\right) \end{aligned}$
Kerala CEE-2015
Sets, Relation and Function
117450
The range of the function $f(x)=\frac{1}{2-\cos 3 x}$ is
1 $(-2, \infty)$
2 $[-2,3]$
3 $\left(\frac{1}{3}, 2\right)$
4 $\left(\frac{1}{2}, 1\right)$
5 $\left[\frac{1}{3}, 1\right]$
Explanation:
E Given, function $\begin{aligned} & y=\frac{1}{2-\cos 3 x} \\ & 2-\cos 3 x=\frac{1}{y} \\ & \cos 3 x=2-\frac{1}{y} \\ & \cos 3 x=\frac{2 y-1}{y} \\ & 3 x=\cos ^{-1}\left(\frac{2 y-1}{y}\right) \\ & x=\frac{1}{3} \cos ^{-1}\left(\frac{2 y-1}{y}\right) \end{aligned}$ For x to be real, $\begin{aligned} & -1 \leq\left|\frac{2 y-1}{y}\right| \leq 1 \\ & -y \leq 2 y-1 \leq y \\ & 2 y-1 \geq-y \\ & 3 y \geq 1 \\ & y \geq \frac{1}{3} \end{aligned}$ $\begin{array}{ll} \text { or } & 2 \mathrm{y}-1 \leq \mathrm{y} \\ \text { or } & \mathrm{y} \leq 1 \end{array}$ So, range $=\left[\frac{1}{3}, 1\right]$
Kerala CEE-2015
Sets, Relation and Function
117451
The domain of the function $f(x)=\sqrt{7-3 x}+\log _e x$ is
1 $0\lt x\lt \infty$
2 $\frac{7}{3} \leq x\lt \infty$
3 $0\lt \mathrm{x} \leq \frac{7}{3}$
4 $-\infty\lt $ x $\lt 0$
5 $-\infty\lt x \leq \frac{7}{3}$
Explanation:
C Given function, $f(x)=\sqrt{7-3 x}+\log _e x$ For, $\sqrt{7-3 x}, 7-3 x \geq 0$ $7 \geq 3 \mathrm{x}$ Or $\begin{aligned} & x \leq \frac{7}{3} \\ & x>0 \\ & x \in\left(0, \frac{7}{3}\right] \\ & 0\lt x \leq \frac{7}{3} \end{aligned}$ Or $\quad 0\lt \mathrm{x} \leq \frac{7}{3}$
