116904
If \(x=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\), then \(x^2(x-4)^2\) is equal to:
1 7
2 4
3 2
4 1
Explanation:
D Given, \(x=\sqrt{\frac{(2+\sqrt{3})}{(2-\sqrt{3})}}\) On simplifying, we gets - \(x=\sqrt{\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}}\) \(x=\sqrt{\frac{(2+\sqrt{3})^2}{4-3}}\) \(x=2+\sqrt{3}\) Then, \(\quad x-4=-2+\sqrt{3}\) \(\therefore \mathrm{x}(\mathrm{x}-4) =(2+\sqrt{3})(-2+\sqrt{3})\) \(=3-4\) \(=-1\)So, \(\mathrm{x}^2(\mathrm{x}-4)^2=(-1)^2=1\)
AP EAMCET-2006
Sets, Relation and Function
116905
If \(f(x)=\log \left(x+\sqrt{x^2+1}\right)\), then \(f(x)\) is
1 even function
2 odd function
3 periodic function
4 none of these
Explanation:
B Given, \(f(x)=\log \left(x+\sqrt{x^2+1}\right)\) We know that, for odd function- \(\mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{f}(\mathrm{x})+\mathrm{f}(-\mathrm{x})=0\) For even function- \(f(-x)=f(x) \Rightarrow f(x)+f(-x)=2 f(x)\) We have, \(f(-x) =\log \left(-x+\sqrt{(-x)^2+1}\right)\) \(f(-x) =\log \left(-x+\sqrt{x^2+1}\right)\) \(f(x)+f(-x) =\log \left(x+\sqrt{x^2+1}\right)+\log \left(-x+\sqrt{x^2+1}\right)\) \(=\log \left(x+\sqrt{x^2+1}\right)\left(-x+\sqrt{x^2+1}\right)\) \(=\log \left(\left(\sqrt{x^2+1}\right)^2-x^2\right)\) \(=\log \left(x^2-1-x^2\right)\) \(=\log (1)\) \(=0\)Hence it is odd function
AIEEE-2003
Sets, Relation and Function
116908
If \(f(x)=\cos \left(\log _e x\right)\), then \(\mathbf{f}(\mathbf{x}) \mathbf{f}(\mathbf{y})-\frac{1}{2}\left[\mathbf{f}\left(\frac{\mathbf{y}}{\mathbf{x}}\right)+\mathbf{f}(\mathbf{x y})\right]\) has the value
116904
If \(x=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\), then \(x^2(x-4)^2\) is equal to:
1 7
2 4
3 2
4 1
Explanation:
D Given, \(x=\sqrt{\frac{(2+\sqrt{3})}{(2-\sqrt{3})}}\) On simplifying, we gets - \(x=\sqrt{\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}}\) \(x=\sqrt{\frac{(2+\sqrt{3})^2}{4-3}}\) \(x=2+\sqrt{3}\) Then, \(\quad x-4=-2+\sqrt{3}\) \(\therefore \mathrm{x}(\mathrm{x}-4) =(2+\sqrt{3})(-2+\sqrt{3})\) \(=3-4\) \(=-1\)So, \(\mathrm{x}^2(\mathrm{x}-4)^2=(-1)^2=1\)
AP EAMCET-2006
Sets, Relation and Function
116905
If \(f(x)=\log \left(x+\sqrt{x^2+1}\right)\), then \(f(x)\) is
1 even function
2 odd function
3 periodic function
4 none of these
Explanation:
B Given, \(f(x)=\log \left(x+\sqrt{x^2+1}\right)\) We know that, for odd function- \(\mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{f}(\mathrm{x})+\mathrm{f}(-\mathrm{x})=0\) For even function- \(f(-x)=f(x) \Rightarrow f(x)+f(-x)=2 f(x)\) We have, \(f(-x) =\log \left(-x+\sqrt{(-x)^2+1}\right)\) \(f(-x) =\log \left(-x+\sqrt{x^2+1}\right)\) \(f(x)+f(-x) =\log \left(x+\sqrt{x^2+1}\right)+\log \left(-x+\sqrt{x^2+1}\right)\) \(=\log \left(x+\sqrt{x^2+1}\right)\left(-x+\sqrt{x^2+1}\right)\) \(=\log \left(\left(\sqrt{x^2+1}\right)^2-x^2\right)\) \(=\log \left(x^2-1-x^2\right)\) \(=\log (1)\) \(=0\)Hence it is odd function
AIEEE-2003
Sets, Relation and Function
116908
If \(f(x)=\cos \left(\log _e x\right)\), then \(\mathbf{f}(\mathbf{x}) \mathbf{f}(\mathbf{y})-\frac{1}{2}\left[\mathbf{f}\left(\frac{\mathbf{y}}{\mathbf{x}}\right)+\mathbf{f}(\mathbf{x y})\right]\) has the value
116904
If \(x=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\), then \(x^2(x-4)^2\) is equal to:
1 7
2 4
3 2
4 1
Explanation:
D Given, \(x=\sqrt{\frac{(2+\sqrt{3})}{(2-\sqrt{3})}}\) On simplifying, we gets - \(x=\sqrt{\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}}\) \(x=\sqrt{\frac{(2+\sqrt{3})^2}{4-3}}\) \(x=2+\sqrt{3}\) Then, \(\quad x-4=-2+\sqrt{3}\) \(\therefore \mathrm{x}(\mathrm{x}-4) =(2+\sqrt{3})(-2+\sqrt{3})\) \(=3-4\) \(=-1\)So, \(\mathrm{x}^2(\mathrm{x}-4)^2=(-1)^2=1\)
AP EAMCET-2006
Sets, Relation and Function
116905
If \(f(x)=\log \left(x+\sqrt{x^2+1}\right)\), then \(f(x)\) is
1 even function
2 odd function
3 periodic function
4 none of these
Explanation:
B Given, \(f(x)=\log \left(x+\sqrt{x^2+1}\right)\) We know that, for odd function- \(\mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{f}(\mathrm{x})+\mathrm{f}(-\mathrm{x})=0\) For even function- \(f(-x)=f(x) \Rightarrow f(x)+f(-x)=2 f(x)\) We have, \(f(-x) =\log \left(-x+\sqrt{(-x)^2+1}\right)\) \(f(-x) =\log \left(-x+\sqrt{x^2+1}\right)\) \(f(x)+f(-x) =\log \left(x+\sqrt{x^2+1}\right)+\log \left(-x+\sqrt{x^2+1}\right)\) \(=\log \left(x+\sqrt{x^2+1}\right)\left(-x+\sqrt{x^2+1}\right)\) \(=\log \left(\left(\sqrt{x^2+1}\right)^2-x^2\right)\) \(=\log \left(x^2-1-x^2\right)\) \(=\log (1)\) \(=0\)Hence it is odd function
AIEEE-2003
Sets, Relation and Function
116908
If \(f(x)=\cos \left(\log _e x\right)\), then \(\mathbf{f}(\mathbf{x}) \mathbf{f}(\mathbf{y})-\frac{1}{2}\left[\mathbf{f}\left(\frac{\mathbf{y}}{\mathbf{x}}\right)+\mathbf{f}(\mathbf{x y})\right]\) has the value
116904
If \(x=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\), then \(x^2(x-4)^2\) is equal to:
1 7
2 4
3 2
4 1
Explanation:
D Given, \(x=\sqrt{\frac{(2+\sqrt{3})}{(2-\sqrt{3})}}\) On simplifying, we gets - \(x=\sqrt{\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}}\) \(x=\sqrt{\frac{(2+\sqrt{3})^2}{4-3}}\) \(x=2+\sqrt{3}\) Then, \(\quad x-4=-2+\sqrt{3}\) \(\therefore \mathrm{x}(\mathrm{x}-4) =(2+\sqrt{3})(-2+\sqrt{3})\) \(=3-4\) \(=-1\)So, \(\mathrm{x}^2(\mathrm{x}-4)^2=(-1)^2=1\)
AP EAMCET-2006
Sets, Relation and Function
116905
If \(f(x)=\log \left(x+\sqrt{x^2+1}\right)\), then \(f(x)\) is
1 even function
2 odd function
3 periodic function
4 none of these
Explanation:
B Given, \(f(x)=\log \left(x+\sqrt{x^2+1}\right)\) We know that, for odd function- \(\mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{f}(\mathrm{x})+\mathrm{f}(-\mathrm{x})=0\) For even function- \(f(-x)=f(x) \Rightarrow f(x)+f(-x)=2 f(x)\) We have, \(f(-x) =\log \left(-x+\sqrt{(-x)^2+1}\right)\) \(f(-x) =\log \left(-x+\sqrt{x^2+1}\right)\) \(f(x)+f(-x) =\log \left(x+\sqrt{x^2+1}\right)+\log \left(-x+\sqrt{x^2+1}\right)\) \(=\log \left(x+\sqrt{x^2+1}\right)\left(-x+\sqrt{x^2+1}\right)\) \(=\log \left(\left(\sqrt{x^2+1}\right)^2-x^2\right)\) \(=\log \left(x^2-1-x^2\right)\) \(=\log (1)\) \(=0\)Hence it is odd function
AIEEE-2003
Sets, Relation and Function
116908
If \(f(x)=\cos \left(\log _e x\right)\), then \(\mathbf{f}(\mathbf{x}) \mathbf{f}(\mathbf{y})-\frac{1}{2}\left[\mathbf{f}\left(\frac{\mathbf{y}}{\mathbf{x}}\right)+\mathbf{f}(\mathbf{x y})\right]\) has the value
