117456
The domain of the function \(f(x)=\frac{1}{\sqrt{9-x^2}}\) is
1 \(-3 \leq \mathrm{x} \leq 3\)
2 \(-3\lt x\lt 3\)
3 \(-9 \leq x \leq 9\)
4 \(-9\lt x\lt 9\)
5 \(-\infty\lt x\lt \infty\)
Explanation:
\(: f(x)=\frac{1}{\sqrt{9-x^2}}\) For existence of \(f(x)\), \(9-x^2>0\) \(\mathrm{x}^2-9\lt 0\) \((x-3)(x+3)\lt 0\) So, \(x \in(-3,3)\) or \(-3\lt x\lt 3\)
Kerala CEE-2011
Sets, Relation and Function
117457
The range function \(f(x)=\log _e\left(3 x^2+4\right)\) is equal to
1 \(\left[\log _e 2, \infty\right)\)
2 \(\left[\log _{\mathrm{e}} 3, \infty\right)\)
3 \(\left[2 \log _e 3, \infty\right)\)
4 \([0, \infty)\)
5 \(\left(2 \log _e 2, \infty\right)\)
Explanation:
E Given function, \(f(x)=\log _e\left(3 x^2+4\right)\) Let, \(y=\log _e\left(3 x^2+4\right)\) \(\Rightarrow 3 \mathrm{x}^2+4=\mathrm{e}^{\mathrm{y}}\) \(\mathrm{x}=\sqrt{\frac{\mathrm{e}^{\mathrm{y}}-4}{3}}\) Hence, \(\frac{\mathrm{e}^{\mathrm{y}}-4}{3}>0\) \(\Rightarrow \mathrm{e}^{\mathrm{y}}>4\) \(\Rightarrow \mathrm{y}>2 \log _{\mathrm{e}} 2\) Hence, the range of the function is \(\left(2 \log _{\mathrm{e}} 2, \infty\right)\)
Kerala CEE-2011
Sets, Relation and Function
117458
The domain of \(\sin ^{-1}\left[\log _2\left(\frac{x}{12}\right)\right]\) is
1 \([2,12]\)
2 \([-1,1]\)
3 \(\left[\frac{1}{3}, 24\right]\)
4 \(\left[\frac{2}{3}, 24\right]\)
5 \([6,24]\)
Explanation:
E From question, \(\begin{aligned} & -1 \leq \sin ^{-1} x \leq 1 \\ & =-1 \leq \log _2 \frac{x}{12} \leq 1 \\ & =2^{-1} \leq \frac{x}{12} \leq 2^1 \\ & =6 \leq x \leq 24 \end{aligned}\) Required domain \(=[6,24]\)
Kerala CEE-2010
Sets, Relation and Function
117459
The range of the function \(f(x)=\frac{x^2-x+1}{x^2+x+1}\) where \(x \in R\), is
117456
The domain of the function \(f(x)=\frac{1}{\sqrt{9-x^2}}\) is
1 \(-3 \leq \mathrm{x} \leq 3\)
2 \(-3\lt x\lt 3\)
3 \(-9 \leq x \leq 9\)
4 \(-9\lt x\lt 9\)
5 \(-\infty\lt x\lt \infty\)
Explanation:
\(: f(x)=\frac{1}{\sqrt{9-x^2}}\) For existence of \(f(x)\), \(9-x^2>0\) \(\mathrm{x}^2-9\lt 0\) \((x-3)(x+3)\lt 0\) So, \(x \in(-3,3)\) or \(-3\lt x\lt 3\)
Kerala CEE-2011
Sets, Relation and Function
117457
The range function \(f(x)=\log _e\left(3 x^2+4\right)\) is equal to
1 \(\left[\log _e 2, \infty\right)\)
2 \(\left[\log _{\mathrm{e}} 3, \infty\right)\)
3 \(\left[2 \log _e 3, \infty\right)\)
4 \([0, \infty)\)
5 \(\left(2 \log _e 2, \infty\right)\)
Explanation:
E Given function, \(f(x)=\log _e\left(3 x^2+4\right)\) Let, \(y=\log _e\left(3 x^2+4\right)\) \(\Rightarrow 3 \mathrm{x}^2+4=\mathrm{e}^{\mathrm{y}}\) \(\mathrm{x}=\sqrt{\frac{\mathrm{e}^{\mathrm{y}}-4}{3}}\) Hence, \(\frac{\mathrm{e}^{\mathrm{y}}-4}{3}>0\) \(\Rightarrow \mathrm{e}^{\mathrm{y}}>4\) \(\Rightarrow \mathrm{y}>2 \log _{\mathrm{e}} 2\) Hence, the range of the function is \(\left(2 \log _{\mathrm{e}} 2, \infty\right)\)
Kerala CEE-2011
Sets, Relation and Function
117458
The domain of \(\sin ^{-1}\left[\log _2\left(\frac{x}{12}\right)\right]\) is
1 \([2,12]\)
2 \([-1,1]\)
3 \(\left[\frac{1}{3}, 24\right]\)
4 \(\left[\frac{2}{3}, 24\right]\)
5 \([6,24]\)
Explanation:
E From question, \(\begin{aligned} & -1 \leq \sin ^{-1} x \leq 1 \\ & =-1 \leq \log _2 \frac{x}{12} \leq 1 \\ & =2^{-1} \leq \frac{x}{12} \leq 2^1 \\ & =6 \leq x \leq 24 \end{aligned}\) Required domain \(=[6,24]\)
Kerala CEE-2010
Sets, Relation and Function
117459
The range of the function \(f(x)=\frac{x^2-x+1}{x^2+x+1}\) where \(x \in R\), is
117456
The domain of the function \(f(x)=\frac{1}{\sqrt{9-x^2}}\) is
1 \(-3 \leq \mathrm{x} \leq 3\)
2 \(-3\lt x\lt 3\)
3 \(-9 \leq x \leq 9\)
4 \(-9\lt x\lt 9\)
5 \(-\infty\lt x\lt \infty\)
Explanation:
\(: f(x)=\frac{1}{\sqrt{9-x^2}}\) For existence of \(f(x)\), \(9-x^2>0\) \(\mathrm{x}^2-9\lt 0\) \((x-3)(x+3)\lt 0\) So, \(x \in(-3,3)\) or \(-3\lt x\lt 3\)
Kerala CEE-2011
Sets, Relation and Function
117457
The range function \(f(x)=\log _e\left(3 x^2+4\right)\) is equal to
1 \(\left[\log _e 2, \infty\right)\)
2 \(\left[\log _{\mathrm{e}} 3, \infty\right)\)
3 \(\left[2 \log _e 3, \infty\right)\)
4 \([0, \infty)\)
5 \(\left(2 \log _e 2, \infty\right)\)
Explanation:
E Given function, \(f(x)=\log _e\left(3 x^2+4\right)\) Let, \(y=\log _e\left(3 x^2+4\right)\) \(\Rightarrow 3 \mathrm{x}^2+4=\mathrm{e}^{\mathrm{y}}\) \(\mathrm{x}=\sqrt{\frac{\mathrm{e}^{\mathrm{y}}-4}{3}}\) Hence, \(\frac{\mathrm{e}^{\mathrm{y}}-4}{3}>0\) \(\Rightarrow \mathrm{e}^{\mathrm{y}}>4\) \(\Rightarrow \mathrm{y}>2 \log _{\mathrm{e}} 2\) Hence, the range of the function is \(\left(2 \log _{\mathrm{e}} 2, \infty\right)\)
Kerala CEE-2011
Sets, Relation and Function
117458
The domain of \(\sin ^{-1}\left[\log _2\left(\frac{x}{12}\right)\right]\) is
1 \([2,12]\)
2 \([-1,1]\)
3 \(\left[\frac{1}{3}, 24\right]\)
4 \(\left[\frac{2}{3}, 24\right]\)
5 \([6,24]\)
Explanation:
E From question, \(\begin{aligned} & -1 \leq \sin ^{-1} x \leq 1 \\ & =-1 \leq \log _2 \frac{x}{12} \leq 1 \\ & =2^{-1} \leq \frac{x}{12} \leq 2^1 \\ & =6 \leq x \leq 24 \end{aligned}\) Required domain \(=[6,24]\)
Kerala CEE-2010
Sets, Relation and Function
117459
The range of the function \(f(x)=\frac{x^2-x+1}{x^2+x+1}\) where \(x \in R\), is
117456
The domain of the function \(f(x)=\frac{1}{\sqrt{9-x^2}}\) is
1 \(-3 \leq \mathrm{x} \leq 3\)
2 \(-3\lt x\lt 3\)
3 \(-9 \leq x \leq 9\)
4 \(-9\lt x\lt 9\)
5 \(-\infty\lt x\lt \infty\)
Explanation:
\(: f(x)=\frac{1}{\sqrt{9-x^2}}\) For existence of \(f(x)\), \(9-x^2>0\) \(\mathrm{x}^2-9\lt 0\) \((x-3)(x+3)\lt 0\) So, \(x \in(-3,3)\) or \(-3\lt x\lt 3\)
Kerala CEE-2011
Sets, Relation and Function
117457
The range function \(f(x)=\log _e\left(3 x^2+4\right)\) is equal to
1 \(\left[\log _e 2, \infty\right)\)
2 \(\left[\log _{\mathrm{e}} 3, \infty\right)\)
3 \(\left[2 \log _e 3, \infty\right)\)
4 \([0, \infty)\)
5 \(\left(2 \log _e 2, \infty\right)\)
Explanation:
E Given function, \(f(x)=\log _e\left(3 x^2+4\right)\) Let, \(y=\log _e\left(3 x^2+4\right)\) \(\Rightarrow 3 \mathrm{x}^2+4=\mathrm{e}^{\mathrm{y}}\) \(\mathrm{x}=\sqrt{\frac{\mathrm{e}^{\mathrm{y}}-4}{3}}\) Hence, \(\frac{\mathrm{e}^{\mathrm{y}}-4}{3}>0\) \(\Rightarrow \mathrm{e}^{\mathrm{y}}>4\) \(\Rightarrow \mathrm{y}>2 \log _{\mathrm{e}} 2\) Hence, the range of the function is \(\left(2 \log _{\mathrm{e}} 2, \infty\right)\)
Kerala CEE-2011
Sets, Relation and Function
117458
The domain of \(\sin ^{-1}\left[\log _2\left(\frac{x}{12}\right)\right]\) is
1 \([2,12]\)
2 \([-1,1]\)
3 \(\left[\frac{1}{3}, 24\right]\)
4 \(\left[\frac{2}{3}, 24\right]\)
5 \([6,24]\)
Explanation:
E From question, \(\begin{aligned} & -1 \leq \sin ^{-1} x \leq 1 \\ & =-1 \leq \log _2 \frac{x}{12} \leq 1 \\ & =2^{-1} \leq \frac{x}{12} \leq 2^1 \\ & =6 \leq x \leq 24 \end{aligned}\) Required domain \(=[6,24]\)
Kerala CEE-2010
Sets, Relation and Function
117459
The range of the function \(f(x)=\frac{x^2-x+1}{x^2+x+1}\) where \(x \in R\), is
117456
The domain of the function \(f(x)=\frac{1}{\sqrt{9-x^2}}\) is
1 \(-3 \leq \mathrm{x} \leq 3\)
2 \(-3\lt x\lt 3\)
3 \(-9 \leq x \leq 9\)
4 \(-9\lt x\lt 9\)
5 \(-\infty\lt x\lt \infty\)
Explanation:
\(: f(x)=\frac{1}{\sqrt{9-x^2}}\) For existence of \(f(x)\), \(9-x^2>0\) \(\mathrm{x}^2-9\lt 0\) \((x-3)(x+3)\lt 0\) So, \(x \in(-3,3)\) or \(-3\lt x\lt 3\)
Kerala CEE-2011
Sets, Relation and Function
117457
The range function \(f(x)=\log _e\left(3 x^2+4\right)\) is equal to
1 \(\left[\log _e 2, \infty\right)\)
2 \(\left[\log _{\mathrm{e}} 3, \infty\right)\)
3 \(\left[2 \log _e 3, \infty\right)\)
4 \([0, \infty)\)
5 \(\left(2 \log _e 2, \infty\right)\)
Explanation:
E Given function, \(f(x)=\log _e\left(3 x^2+4\right)\) Let, \(y=\log _e\left(3 x^2+4\right)\) \(\Rightarrow 3 \mathrm{x}^2+4=\mathrm{e}^{\mathrm{y}}\) \(\mathrm{x}=\sqrt{\frac{\mathrm{e}^{\mathrm{y}}-4}{3}}\) Hence, \(\frac{\mathrm{e}^{\mathrm{y}}-4}{3}>0\) \(\Rightarrow \mathrm{e}^{\mathrm{y}}>4\) \(\Rightarrow \mathrm{y}>2 \log _{\mathrm{e}} 2\) Hence, the range of the function is \(\left(2 \log _{\mathrm{e}} 2, \infty\right)\)
Kerala CEE-2011
Sets, Relation and Function
117458
The domain of \(\sin ^{-1}\left[\log _2\left(\frac{x}{12}\right)\right]\) is
1 \([2,12]\)
2 \([-1,1]\)
3 \(\left[\frac{1}{3}, 24\right]\)
4 \(\left[\frac{2}{3}, 24\right]\)
5 \([6,24]\)
Explanation:
E From question, \(\begin{aligned} & -1 \leq \sin ^{-1} x \leq 1 \\ & =-1 \leq \log _2 \frac{x}{12} \leq 1 \\ & =2^{-1} \leq \frac{x}{12} \leq 2^1 \\ & =6 \leq x \leq 24 \end{aligned}\) Required domain \(=[6,24]\)
Kerala CEE-2010
Sets, Relation and Function
117459
The range of the function \(f(x)=\frac{x^2-x+1}{x^2+x+1}\) where \(x \in R\), is
