117030
For all real values of \(x\), the minimum value of the function \(f(x)=\frac{1-x+x^2}{1+x+x^2}\) is
1 1
2 0
3 \(\frac{1}{3}\)
4 3
Explanation:
C Given, \(f(x)=\frac{1-x+x^2}{1+x+x^2}\) To, get the minimum value, we differentiate the above. So, we get \(\frac{d f(x)}{d x}=\frac{\left(1-x+x^2\right)(1+2 x)-\left(1+x+x^2\right)(-1+2 x)}{\left(1+x+x^2\right)^2}\) \(\frac{\left[1-x+x^2+2 x-2 x^2+2 x^3\right]-\left[-1-x-x^2+2 x+2 x^2+2 x^3\right]}{\left(1+x+x^2\right)^2}\) For minimum value of \(f(x)\) \(\frac{d f(x)}{d x}=0\) \(1-x+x^2+2 x-2 x^2+2 x^3\) \(+1+x+x^2-2 x-2 x^2-2 x^3=0\) \(2+2 x^2-4 x^2=0\) \(\text { or, } 2-2 x^2=0\) \(x^2=1\) \(x=1\) \(\therefore f(1)=\frac{1-1+1}{1+1+1}=1 / 3\)
MHT CET-2021
Sets, Relation and Function
116857
If \(f(x)=\sqrt{\log _{10} x^2}\). The set of all values of \(x\) for which \(f(x)\) is real, is
1 \([-1,1]\)
2 \([-1, \infty]\)
3 \((-\infty, 1]\)
4 \((-\infty,-1] \cup[1, \infty]\)
Explanation:
D \(\mathrm{f}(\mathrm{x})=\sqrt{\log _{10} \mathrm{x}^2}\) is real, if \(\log _{10} x^2 \geq 0\) \(x^2 \geq 1\) \(x\lt -1 \text { and } x>1\) \(x \in(-\infty,-1] \cup[1, \infty)\)
VITEEE-2010
Sets, Relation and Function
116865
If \(f(x)=\sqrt[n]{x^m}, n \in N\) is an even function, then \(m\) is
1 even integer
2 Odd integer
3 any integer
4 \(\mathrm{f}(\mathrm{x})\) even is not possible
Explanation:
A Given, \(f(x)=\sqrt[n]{x^m}, n \in N\) is an even function. We know for even function \(f(x)=f(-x)\) \(\sqrt[n]{x^m}=\sqrt[n]{(-x)^m}\) \(x^m=(-x)^m\)So, \(\mathrm{m}\) is even integer.
JCECE-2013
Sets, Relation and Function
116871
The function \(f(x)=\sin x+\cos x\) will be
1 an even function
2 an odd function
3 a constant function
4 None of these
Explanation:
D The function \(f(x)=\sin x+\cos x\) \(=\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right)\) \(=\sqrt{2}\left(\sin x \cos \frac{\pi}{4}+\cos x \sin \frac{\pi}{4}\right)\) \(=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)\)But, \(\sin \left(x+\frac{\pi}{4}\right)\) is a periodic function. So,option (d) is correct.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117030
For all real values of \(x\), the minimum value of the function \(f(x)=\frac{1-x+x^2}{1+x+x^2}\) is
1 1
2 0
3 \(\frac{1}{3}\)
4 3
Explanation:
C Given, \(f(x)=\frac{1-x+x^2}{1+x+x^2}\) To, get the minimum value, we differentiate the above. So, we get \(\frac{d f(x)}{d x}=\frac{\left(1-x+x^2\right)(1+2 x)-\left(1+x+x^2\right)(-1+2 x)}{\left(1+x+x^2\right)^2}\) \(\frac{\left[1-x+x^2+2 x-2 x^2+2 x^3\right]-\left[-1-x-x^2+2 x+2 x^2+2 x^3\right]}{\left(1+x+x^2\right)^2}\) For minimum value of \(f(x)\) \(\frac{d f(x)}{d x}=0\) \(1-x+x^2+2 x-2 x^2+2 x^3\) \(+1+x+x^2-2 x-2 x^2-2 x^3=0\) \(2+2 x^2-4 x^2=0\) \(\text { or, } 2-2 x^2=0\) \(x^2=1\) \(x=1\) \(\therefore f(1)=\frac{1-1+1}{1+1+1}=1 / 3\)
MHT CET-2021
Sets, Relation and Function
116857
If \(f(x)=\sqrt{\log _{10} x^2}\). The set of all values of \(x\) for which \(f(x)\) is real, is
1 \([-1,1]\)
2 \([-1, \infty]\)
3 \((-\infty, 1]\)
4 \((-\infty,-1] \cup[1, \infty]\)
Explanation:
D \(\mathrm{f}(\mathrm{x})=\sqrt{\log _{10} \mathrm{x}^2}\) is real, if \(\log _{10} x^2 \geq 0\) \(x^2 \geq 1\) \(x\lt -1 \text { and } x>1\) \(x \in(-\infty,-1] \cup[1, \infty)\)
VITEEE-2010
Sets, Relation and Function
116865
If \(f(x)=\sqrt[n]{x^m}, n \in N\) is an even function, then \(m\) is
1 even integer
2 Odd integer
3 any integer
4 \(\mathrm{f}(\mathrm{x})\) even is not possible
Explanation:
A Given, \(f(x)=\sqrt[n]{x^m}, n \in N\) is an even function. We know for even function \(f(x)=f(-x)\) \(\sqrt[n]{x^m}=\sqrt[n]{(-x)^m}\) \(x^m=(-x)^m\)So, \(\mathrm{m}\) is even integer.
JCECE-2013
Sets, Relation and Function
116871
The function \(f(x)=\sin x+\cos x\) will be
1 an even function
2 an odd function
3 a constant function
4 None of these
Explanation:
D The function \(f(x)=\sin x+\cos x\) \(=\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right)\) \(=\sqrt{2}\left(\sin x \cos \frac{\pi}{4}+\cos x \sin \frac{\pi}{4}\right)\) \(=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)\)But, \(\sin \left(x+\frac{\pi}{4}\right)\) is a periodic function. So,option (d) is correct.
117030
For all real values of \(x\), the minimum value of the function \(f(x)=\frac{1-x+x^2}{1+x+x^2}\) is
1 1
2 0
3 \(\frac{1}{3}\)
4 3
Explanation:
C Given, \(f(x)=\frac{1-x+x^2}{1+x+x^2}\) To, get the minimum value, we differentiate the above. So, we get \(\frac{d f(x)}{d x}=\frac{\left(1-x+x^2\right)(1+2 x)-\left(1+x+x^2\right)(-1+2 x)}{\left(1+x+x^2\right)^2}\) \(\frac{\left[1-x+x^2+2 x-2 x^2+2 x^3\right]-\left[-1-x-x^2+2 x+2 x^2+2 x^3\right]}{\left(1+x+x^2\right)^2}\) For minimum value of \(f(x)\) \(\frac{d f(x)}{d x}=0\) \(1-x+x^2+2 x-2 x^2+2 x^3\) \(+1+x+x^2-2 x-2 x^2-2 x^3=0\) \(2+2 x^2-4 x^2=0\) \(\text { or, } 2-2 x^2=0\) \(x^2=1\) \(x=1\) \(\therefore f(1)=\frac{1-1+1}{1+1+1}=1 / 3\)
MHT CET-2021
Sets, Relation and Function
116857
If \(f(x)=\sqrt{\log _{10} x^2}\). The set of all values of \(x\) for which \(f(x)\) is real, is
1 \([-1,1]\)
2 \([-1, \infty]\)
3 \((-\infty, 1]\)
4 \((-\infty,-1] \cup[1, \infty]\)
Explanation:
D \(\mathrm{f}(\mathrm{x})=\sqrt{\log _{10} \mathrm{x}^2}\) is real, if \(\log _{10} x^2 \geq 0\) \(x^2 \geq 1\) \(x\lt -1 \text { and } x>1\) \(x \in(-\infty,-1] \cup[1, \infty)\)
VITEEE-2010
Sets, Relation and Function
116865
If \(f(x)=\sqrt[n]{x^m}, n \in N\) is an even function, then \(m\) is
1 even integer
2 Odd integer
3 any integer
4 \(\mathrm{f}(\mathrm{x})\) even is not possible
Explanation:
A Given, \(f(x)=\sqrt[n]{x^m}, n \in N\) is an even function. We know for even function \(f(x)=f(-x)\) \(\sqrt[n]{x^m}=\sqrt[n]{(-x)^m}\) \(x^m=(-x)^m\)So, \(\mathrm{m}\) is even integer.
JCECE-2013
Sets, Relation and Function
116871
The function \(f(x)=\sin x+\cos x\) will be
1 an even function
2 an odd function
3 a constant function
4 None of these
Explanation:
D The function \(f(x)=\sin x+\cos x\) \(=\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right)\) \(=\sqrt{2}\left(\sin x \cos \frac{\pi}{4}+\cos x \sin \frac{\pi}{4}\right)\) \(=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)\)But, \(\sin \left(x+\frac{\pi}{4}\right)\) is a periodic function. So,option (d) is correct.
117030
For all real values of \(x\), the minimum value of the function \(f(x)=\frac{1-x+x^2}{1+x+x^2}\) is
1 1
2 0
3 \(\frac{1}{3}\)
4 3
Explanation:
C Given, \(f(x)=\frac{1-x+x^2}{1+x+x^2}\) To, get the minimum value, we differentiate the above. So, we get \(\frac{d f(x)}{d x}=\frac{\left(1-x+x^2\right)(1+2 x)-\left(1+x+x^2\right)(-1+2 x)}{\left(1+x+x^2\right)^2}\) \(\frac{\left[1-x+x^2+2 x-2 x^2+2 x^3\right]-\left[-1-x-x^2+2 x+2 x^2+2 x^3\right]}{\left(1+x+x^2\right)^2}\) For minimum value of \(f(x)\) \(\frac{d f(x)}{d x}=0\) \(1-x+x^2+2 x-2 x^2+2 x^3\) \(+1+x+x^2-2 x-2 x^2-2 x^3=0\) \(2+2 x^2-4 x^2=0\) \(\text { or, } 2-2 x^2=0\) \(x^2=1\) \(x=1\) \(\therefore f(1)=\frac{1-1+1}{1+1+1}=1 / 3\)
MHT CET-2021
Sets, Relation and Function
116857
If \(f(x)=\sqrt{\log _{10} x^2}\). The set of all values of \(x\) for which \(f(x)\) is real, is
1 \([-1,1]\)
2 \([-1, \infty]\)
3 \((-\infty, 1]\)
4 \((-\infty,-1] \cup[1, \infty]\)
Explanation:
D \(\mathrm{f}(\mathrm{x})=\sqrt{\log _{10} \mathrm{x}^2}\) is real, if \(\log _{10} x^2 \geq 0\) \(x^2 \geq 1\) \(x\lt -1 \text { and } x>1\) \(x \in(-\infty,-1] \cup[1, \infty)\)
VITEEE-2010
Sets, Relation and Function
116865
If \(f(x)=\sqrt[n]{x^m}, n \in N\) is an even function, then \(m\) is
1 even integer
2 Odd integer
3 any integer
4 \(\mathrm{f}(\mathrm{x})\) even is not possible
Explanation:
A Given, \(f(x)=\sqrt[n]{x^m}, n \in N\) is an even function. We know for even function \(f(x)=f(-x)\) \(\sqrt[n]{x^m}=\sqrt[n]{(-x)^m}\) \(x^m=(-x)^m\)So, \(\mathrm{m}\) is even integer.
JCECE-2013
Sets, Relation and Function
116871
The function \(f(x)=\sin x+\cos x\) will be
1 an even function
2 an odd function
3 a constant function
4 None of these
Explanation:
D The function \(f(x)=\sin x+\cos x\) \(=\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right)\) \(=\sqrt{2}\left(\sin x \cos \frac{\pi}{4}+\cos x \sin \frac{\pi}{4}\right)\) \(=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right)\)But, \(\sin \left(x+\frac{\pi}{4}\right)\) is a periodic function. So,option (d) is correct.