B Given, \(\mathrm{f}(\mathrm{x})=\log \mathrm{x}^2\) We know that \(\sqrt{\mathrm{x}^2}=|\mathrm{x}|\) Then \(f(\mathrm{x})=2 \log |\mathrm{x}|\) Hence equivalent formation of \(\log x^2=2 \log |x|\)
CG PET-2021
Sets, Relation and Function
116994
If \(f(x)=\left(a-x^n\right)^{1 / n}\) where \(a>0\) and \(n\) is a positive integer, then \(f[f(x)]\) is equal to
1 \(x^3\)
2 \(x^2\)
3 \(x\)
4 None of these
Explanation:
C Given that- \(f(x)=\left(a-x^n\right)^{1 / n}\) \(\therefore f[f(x)]=\left[a-\left\{f(x)^n\right\}\right]^{1 / n}\) \(=\left[a-\left(a-x^n\right)\right]^{1 / n}\) \(=\left[x^n\right]^{1 / n}\) \(=x\)
Manipal UGET-2019
Sets, Relation and Function
116995
If \(3^x+2^{2 x} \geq 5^x\), then the solution set for \(x\) is
1 \((-\infty, 2]\)
2 \([2, \infty)\)
3 \([0,2]\)
4 \(\{2\}\)
Explanation:
A Given that- \(3^{\mathrm{x}}+2^{2 \mathrm{x}} \geq 5^{\mathrm{x}}\) \(\left(\frac{3}{5}\right)^{\mathrm{x}}+\left(\frac{4}{5}\right)^{\mathrm{x}} \geq 1\) \((\sin \theta)^{\mathrm{x}}+(\cos \theta)^{\mathrm{x}} \geq 1\) \(\mathrm{x} \leq 2\) \(\therefore \text { solution set is }(-\infty, 2] .\)(by triangle inquality)
Manipal UGET-2019
Sets, Relation and Function
116996
The period of \(f(x)=\sin \left(\frac{\pi x}{n-1}\right)+\cos \left(\frac{\pi x}{n}\right), n \in Z, n>2 \text { is }\)
1 \(2 \pi n(n-1)\)
2 \(4 \mathrm{n}(\mathrm{n}-1)\)
3 \(2 \mathrm{n}(\mathrm{n}-1)\)
4 None of the above
Explanation:
C We have- \(f(x)=\sin \left(\frac{\pi x}{n-1}\right)+\cos \left(\frac{\pi x}{n}\right), n \varepsilon z, n>2\) Let, \(\mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{x})+\mathrm{p}(\mathrm{x}) \quad \forall \mathrm{n}>2\) \(\therefore\) Period of \(\mathrm{g}(\mathrm{x})=\frac{2 \pi(\mathrm{n}-1)}{\pi}=2(\mathrm{n}-1)\) and period of \(\mathrm{p}(\mathrm{x})=\frac{2 \pi \mathrm{n}}{\pi}=2 \mathrm{n}\) Period of \(f(x)=L C M\) of \([p(x)\) and \(g(x)]\) \(=2 \mathrm{n}(\mathrm{n}-1)\)
Manipal UGET-2019
Sets, Relation and Function
116997
The number of real solutions of the equation \(1+\left|e^x-1\right|=e^x\left(e^x-2\right)\) is
1 1
2 2
3 4
4 8
Explanation:
A Given that:- \(1+\left|e^x-1\right|=e^x\left(e^x-2\right)\) Adding 1 both sides:- \(1+1+\left|\mathrm{e}^{\mathrm{x}}-1\right|=\mathrm{e}^{\mathrm{x}}\left(\mathrm{e}^{\mathrm{x}}-2\right)+1\) \(1+1+\left|\mathrm{e}^{\mathrm{x}}-1\right|=\mathrm{e}^{2 \mathrm{x}}-2 \mathrm{e}^{\mathrm{x}}+1\) \(2+\left|\mathrm{e}^{\mathrm{x}}-1\right|=\left(\mathrm{e}^{\mathrm{x}}-1\right)^2\) \(\Rightarrow \left(\mathrm{e}^{\mathrm{x}}-1\right)^2-\left|\mathrm{e}^{\mathrm{x}}-1\right|-2=0\) Let \(\left(e^x-1\right)=y\) Then- \(y^2-y-2=0\) \((y-2)(y+1)=0\) \(y=2,-1\) Now, \(y=2\) \(\left|e^x-1\right|=2\) \(e^x-1= \pm 2\) \(e^x=1 \pm 2\) \(e^x=3,-1\) \(e^x=3\) \(x=\log _e 3\)\(\therefore\) There is one real solution of the equation.
B Given, \(\mathrm{f}(\mathrm{x})=\log \mathrm{x}^2\) We know that \(\sqrt{\mathrm{x}^2}=|\mathrm{x}|\) Then \(f(\mathrm{x})=2 \log |\mathrm{x}|\) Hence equivalent formation of \(\log x^2=2 \log |x|\)
CG PET-2021
Sets, Relation and Function
116994
If \(f(x)=\left(a-x^n\right)^{1 / n}\) where \(a>0\) and \(n\) is a positive integer, then \(f[f(x)]\) is equal to
1 \(x^3\)
2 \(x^2\)
3 \(x\)
4 None of these
Explanation:
C Given that- \(f(x)=\left(a-x^n\right)^{1 / n}\) \(\therefore f[f(x)]=\left[a-\left\{f(x)^n\right\}\right]^{1 / n}\) \(=\left[a-\left(a-x^n\right)\right]^{1 / n}\) \(=\left[x^n\right]^{1 / n}\) \(=x\)
Manipal UGET-2019
Sets, Relation and Function
116995
If \(3^x+2^{2 x} \geq 5^x\), then the solution set for \(x\) is
1 \((-\infty, 2]\)
2 \([2, \infty)\)
3 \([0,2]\)
4 \(\{2\}\)
Explanation:
A Given that- \(3^{\mathrm{x}}+2^{2 \mathrm{x}} \geq 5^{\mathrm{x}}\) \(\left(\frac{3}{5}\right)^{\mathrm{x}}+\left(\frac{4}{5}\right)^{\mathrm{x}} \geq 1\) \((\sin \theta)^{\mathrm{x}}+(\cos \theta)^{\mathrm{x}} \geq 1\) \(\mathrm{x} \leq 2\) \(\therefore \text { solution set is }(-\infty, 2] .\)(by triangle inquality)
Manipal UGET-2019
Sets, Relation and Function
116996
The period of \(f(x)=\sin \left(\frac{\pi x}{n-1}\right)+\cos \left(\frac{\pi x}{n}\right), n \in Z, n>2 \text { is }\)
1 \(2 \pi n(n-1)\)
2 \(4 \mathrm{n}(\mathrm{n}-1)\)
3 \(2 \mathrm{n}(\mathrm{n}-1)\)
4 None of the above
Explanation:
C We have- \(f(x)=\sin \left(\frac{\pi x}{n-1}\right)+\cos \left(\frac{\pi x}{n}\right), n \varepsilon z, n>2\) Let, \(\mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{x})+\mathrm{p}(\mathrm{x}) \quad \forall \mathrm{n}>2\) \(\therefore\) Period of \(\mathrm{g}(\mathrm{x})=\frac{2 \pi(\mathrm{n}-1)}{\pi}=2(\mathrm{n}-1)\) and period of \(\mathrm{p}(\mathrm{x})=\frac{2 \pi \mathrm{n}}{\pi}=2 \mathrm{n}\) Period of \(f(x)=L C M\) of \([p(x)\) and \(g(x)]\) \(=2 \mathrm{n}(\mathrm{n}-1)\)
Manipal UGET-2019
Sets, Relation and Function
116997
The number of real solutions of the equation \(1+\left|e^x-1\right|=e^x\left(e^x-2\right)\) is
1 1
2 2
3 4
4 8
Explanation:
A Given that:- \(1+\left|e^x-1\right|=e^x\left(e^x-2\right)\) Adding 1 both sides:- \(1+1+\left|\mathrm{e}^{\mathrm{x}}-1\right|=\mathrm{e}^{\mathrm{x}}\left(\mathrm{e}^{\mathrm{x}}-2\right)+1\) \(1+1+\left|\mathrm{e}^{\mathrm{x}}-1\right|=\mathrm{e}^{2 \mathrm{x}}-2 \mathrm{e}^{\mathrm{x}}+1\) \(2+\left|\mathrm{e}^{\mathrm{x}}-1\right|=\left(\mathrm{e}^{\mathrm{x}}-1\right)^2\) \(\Rightarrow \left(\mathrm{e}^{\mathrm{x}}-1\right)^2-\left|\mathrm{e}^{\mathrm{x}}-1\right|-2=0\) Let \(\left(e^x-1\right)=y\) Then- \(y^2-y-2=0\) \((y-2)(y+1)=0\) \(y=2,-1\) Now, \(y=2\) \(\left|e^x-1\right|=2\) \(e^x-1= \pm 2\) \(e^x=1 \pm 2\) \(e^x=3,-1\) \(e^x=3\) \(x=\log _e 3\)\(\therefore\) There is one real solution of the equation.
B Given, \(\mathrm{f}(\mathrm{x})=\log \mathrm{x}^2\) We know that \(\sqrt{\mathrm{x}^2}=|\mathrm{x}|\) Then \(f(\mathrm{x})=2 \log |\mathrm{x}|\) Hence equivalent formation of \(\log x^2=2 \log |x|\)
CG PET-2021
Sets, Relation and Function
116994
If \(f(x)=\left(a-x^n\right)^{1 / n}\) where \(a>0\) and \(n\) is a positive integer, then \(f[f(x)]\) is equal to
1 \(x^3\)
2 \(x^2\)
3 \(x\)
4 None of these
Explanation:
C Given that- \(f(x)=\left(a-x^n\right)^{1 / n}\) \(\therefore f[f(x)]=\left[a-\left\{f(x)^n\right\}\right]^{1 / n}\) \(=\left[a-\left(a-x^n\right)\right]^{1 / n}\) \(=\left[x^n\right]^{1 / n}\) \(=x\)
Manipal UGET-2019
Sets, Relation and Function
116995
If \(3^x+2^{2 x} \geq 5^x\), then the solution set for \(x\) is
1 \((-\infty, 2]\)
2 \([2, \infty)\)
3 \([0,2]\)
4 \(\{2\}\)
Explanation:
A Given that- \(3^{\mathrm{x}}+2^{2 \mathrm{x}} \geq 5^{\mathrm{x}}\) \(\left(\frac{3}{5}\right)^{\mathrm{x}}+\left(\frac{4}{5}\right)^{\mathrm{x}} \geq 1\) \((\sin \theta)^{\mathrm{x}}+(\cos \theta)^{\mathrm{x}} \geq 1\) \(\mathrm{x} \leq 2\) \(\therefore \text { solution set is }(-\infty, 2] .\)(by triangle inquality)
Manipal UGET-2019
Sets, Relation and Function
116996
The period of \(f(x)=\sin \left(\frac{\pi x}{n-1}\right)+\cos \left(\frac{\pi x}{n}\right), n \in Z, n>2 \text { is }\)
1 \(2 \pi n(n-1)\)
2 \(4 \mathrm{n}(\mathrm{n}-1)\)
3 \(2 \mathrm{n}(\mathrm{n}-1)\)
4 None of the above
Explanation:
C We have- \(f(x)=\sin \left(\frac{\pi x}{n-1}\right)+\cos \left(\frac{\pi x}{n}\right), n \varepsilon z, n>2\) Let, \(\mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{x})+\mathrm{p}(\mathrm{x}) \quad \forall \mathrm{n}>2\) \(\therefore\) Period of \(\mathrm{g}(\mathrm{x})=\frac{2 \pi(\mathrm{n}-1)}{\pi}=2(\mathrm{n}-1)\) and period of \(\mathrm{p}(\mathrm{x})=\frac{2 \pi \mathrm{n}}{\pi}=2 \mathrm{n}\) Period of \(f(x)=L C M\) of \([p(x)\) and \(g(x)]\) \(=2 \mathrm{n}(\mathrm{n}-1)\)
Manipal UGET-2019
Sets, Relation and Function
116997
The number of real solutions of the equation \(1+\left|e^x-1\right|=e^x\left(e^x-2\right)\) is
1 1
2 2
3 4
4 8
Explanation:
A Given that:- \(1+\left|e^x-1\right|=e^x\left(e^x-2\right)\) Adding 1 both sides:- \(1+1+\left|\mathrm{e}^{\mathrm{x}}-1\right|=\mathrm{e}^{\mathrm{x}}\left(\mathrm{e}^{\mathrm{x}}-2\right)+1\) \(1+1+\left|\mathrm{e}^{\mathrm{x}}-1\right|=\mathrm{e}^{2 \mathrm{x}}-2 \mathrm{e}^{\mathrm{x}}+1\) \(2+\left|\mathrm{e}^{\mathrm{x}}-1\right|=\left(\mathrm{e}^{\mathrm{x}}-1\right)^2\) \(\Rightarrow \left(\mathrm{e}^{\mathrm{x}}-1\right)^2-\left|\mathrm{e}^{\mathrm{x}}-1\right|-2=0\) Let \(\left(e^x-1\right)=y\) Then- \(y^2-y-2=0\) \((y-2)(y+1)=0\) \(y=2,-1\) Now, \(y=2\) \(\left|e^x-1\right|=2\) \(e^x-1= \pm 2\) \(e^x=1 \pm 2\) \(e^x=3,-1\) \(e^x=3\) \(x=\log _e 3\)\(\therefore\) There is one real solution of the equation.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
116993
The equivalent function of \(\log x^2\) is
1 \(2 \log \mathrm{x}\)
2 \(2 \log |x|\)
3 \(\left|\log x^2\right|\)
4 \((\log \mathrm{x})^2\)
Explanation:
B Given, \(\mathrm{f}(\mathrm{x})=\log \mathrm{x}^2\) We know that \(\sqrt{\mathrm{x}^2}=|\mathrm{x}|\) Then \(f(\mathrm{x})=2 \log |\mathrm{x}|\) Hence equivalent formation of \(\log x^2=2 \log |x|\)
CG PET-2021
Sets, Relation and Function
116994
If \(f(x)=\left(a-x^n\right)^{1 / n}\) where \(a>0\) and \(n\) is a positive integer, then \(f[f(x)]\) is equal to
1 \(x^3\)
2 \(x^2\)
3 \(x\)
4 None of these
Explanation:
C Given that- \(f(x)=\left(a-x^n\right)^{1 / n}\) \(\therefore f[f(x)]=\left[a-\left\{f(x)^n\right\}\right]^{1 / n}\) \(=\left[a-\left(a-x^n\right)\right]^{1 / n}\) \(=\left[x^n\right]^{1 / n}\) \(=x\)
Manipal UGET-2019
Sets, Relation and Function
116995
If \(3^x+2^{2 x} \geq 5^x\), then the solution set for \(x\) is
1 \((-\infty, 2]\)
2 \([2, \infty)\)
3 \([0,2]\)
4 \(\{2\}\)
Explanation:
A Given that- \(3^{\mathrm{x}}+2^{2 \mathrm{x}} \geq 5^{\mathrm{x}}\) \(\left(\frac{3}{5}\right)^{\mathrm{x}}+\left(\frac{4}{5}\right)^{\mathrm{x}} \geq 1\) \((\sin \theta)^{\mathrm{x}}+(\cos \theta)^{\mathrm{x}} \geq 1\) \(\mathrm{x} \leq 2\) \(\therefore \text { solution set is }(-\infty, 2] .\)(by triangle inquality)
Manipal UGET-2019
Sets, Relation and Function
116996
The period of \(f(x)=\sin \left(\frac{\pi x}{n-1}\right)+\cos \left(\frac{\pi x}{n}\right), n \in Z, n>2 \text { is }\)
1 \(2 \pi n(n-1)\)
2 \(4 \mathrm{n}(\mathrm{n}-1)\)
3 \(2 \mathrm{n}(\mathrm{n}-1)\)
4 None of the above
Explanation:
C We have- \(f(x)=\sin \left(\frac{\pi x}{n-1}\right)+\cos \left(\frac{\pi x}{n}\right), n \varepsilon z, n>2\) Let, \(\mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{x})+\mathrm{p}(\mathrm{x}) \quad \forall \mathrm{n}>2\) \(\therefore\) Period of \(\mathrm{g}(\mathrm{x})=\frac{2 \pi(\mathrm{n}-1)}{\pi}=2(\mathrm{n}-1)\) and period of \(\mathrm{p}(\mathrm{x})=\frac{2 \pi \mathrm{n}}{\pi}=2 \mathrm{n}\) Period of \(f(x)=L C M\) of \([p(x)\) and \(g(x)]\) \(=2 \mathrm{n}(\mathrm{n}-1)\)
Manipal UGET-2019
Sets, Relation and Function
116997
The number of real solutions of the equation \(1+\left|e^x-1\right|=e^x\left(e^x-2\right)\) is
1 1
2 2
3 4
4 8
Explanation:
A Given that:- \(1+\left|e^x-1\right|=e^x\left(e^x-2\right)\) Adding 1 both sides:- \(1+1+\left|\mathrm{e}^{\mathrm{x}}-1\right|=\mathrm{e}^{\mathrm{x}}\left(\mathrm{e}^{\mathrm{x}}-2\right)+1\) \(1+1+\left|\mathrm{e}^{\mathrm{x}}-1\right|=\mathrm{e}^{2 \mathrm{x}}-2 \mathrm{e}^{\mathrm{x}}+1\) \(2+\left|\mathrm{e}^{\mathrm{x}}-1\right|=\left(\mathrm{e}^{\mathrm{x}}-1\right)^2\) \(\Rightarrow \left(\mathrm{e}^{\mathrm{x}}-1\right)^2-\left|\mathrm{e}^{\mathrm{x}}-1\right|-2=0\) Let \(\left(e^x-1\right)=y\) Then- \(y^2-y-2=0\) \((y-2)(y+1)=0\) \(y=2,-1\) Now, \(y=2\) \(\left|e^x-1\right|=2\) \(e^x-1= \pm 2\) \(e^x=1 \pm 2\) \(e^x=3,-1\) \(e^x=3\) \(x=\log _e 3\)\(\therefore\) There is one real solution of the equation.
B Given, \(\mathrm{f}(\mathrm{x})=\log \mathrm{x}^2\) We know that \(\sqrt{\mathrm{x}^2}=|\mathrm{x}|\) Then \(f(\mathrm{x})=2 \log |\mathrm{x}|\) Hence equivalent formation of \(\log x^2=2 \log |x|\)
CG PET-2021
Sets, Relation and Function
116994
If \(f(x)=\left(a-x^n\right)^{1 / n}\) where \(a>0\) and \(n\) is a positive integer, then \(f[f(x)]\) is equal to
1 \(x^3\)
2 \(x^2\)
3 \(x\)
4 None of these
Explanation:
C Given that- \(f(x)=\left(a-x^n\right)^{1 / n}\) \(\therefore f[f(x)]=\left[a-\left\{f(x)^n\right\}\right]^{1 / n}\) \(=\left[a-\left(a-x^n\right)\right]^{1 / n}\) \(=\left[x^n\right]^{1 / n}\) \(=x\)
Manipal UGET-2019
Sets, Relation and Function
116995
If \(3^x+2^{2 x} \geq 5^x\), then the solution set for \(x\) is
1 \((-\infty, 2]\)
2 \([2, \infty)\)
3 \([0,2]\)
4 \(\{2\}\)
Explanation:
A Given that- \(3^{\mathrm{x}}+2^{2 \mathrm{x}} \geq 5^{\mathrm{x}}\) \(\left(\frac{3}{5}\right)^{\mathrm{x}}+\left(\frac{4}{5}\right)^{\mathrm{x}} \geq 1\) \((\sin \theta)^{\mathrm{x}}+(\cos \theta)^{\mathrm{x}} \geq 1\) \(\mathrm{x} \leq 2\) \(\therefore \text { solution set is }(-\infty, 2] .\)(by triangle inquality)
Manipal UGET-2019
Sets, Relation and Function
116996
The period of \(f(x)=\sin \left(\frac{\pi x}{n-1}\right)+\cos \left(\frac{\pi x}{n}\right), n \in Z, n>2 \text { is }\)
1 \(2 \pi n(n-1)\)
2 \(4 \mathrm{n}(\mathrm{n}-1)\)
3 \(2 \mathrm{n}(\mathrm{n}-1)\)
4 None of the above
Explanation:
C We have- \(f(x)=\sin \left(\frac{\pi x}{n-1}\right)+\cos \left(\frac{\pi x}{n}\right), n \varepsilon z, n>2\) Let, \(\mathrm{f}(\mathrm{x})=\mathrm{g}(\mathrm{x})+\mathrm{p}(\mathrm{x}) \quad \forall \mathrm{n}>2\) \(\therefore\) Period of \(\mathrm{g}(\mathrm{x})=\frac{2 \pi(\mathrm{n}-1)}{\pi}=2(\mathrm{n}-1)\) and period of \(\mathrm{p}(\mathrm{x})=\frac{2 \pi \mathrm{n}}{\pi}=2 \mathrm{n}\) Period of \(f(x)=L C M\) of \([p(x)\) and \(g(x)]\) \(=2 \mathrm{n}(\mathrm{n}-1)\)
Manipal UGET-2019
Sets, Relation and Function
116997
The number of real solutions of the equation \(1+\left|e^x-1\right|=e^x\left(e^x-2\right)\) is
1 1
2 2
3 4
4 8
Explanation:
A Given that:- \(1+\left|e^x-1\right|=e^x\left(e^x-2\right)\) Adding 1 both sides:- \(1+1+\left|\mathrm{e}^{\mathrm{x}}-1\right|=\mathrm{e}^{\mathrm{x}}\left(\mathrm{e}^{\mathrm{x}}-2\right)+1\) \(1+1+\left|\mathrm{e}^{\mathrm{x}}-1\right|=\mathrm{e}^{2 \mathrm{x}}-2 \mathrm{e}^{\mathrm{x}}+1\) \(2+\left|\mathrm{e}^{\mathrm{x}}-1\right|=\left(\mathrm{e}^{\mathrm{x}}-1\right)^2\) \(\Rightarrow \left(\mathrm{e}^{\mathrm{x}}-1\right)^2-\left|\mathrm{e}^{\mathrm{x}}-1\right|-2=0\) Let \(\left(e^x-1\right)=y\) Then- \(y^2-y-2=0\) \((y-2)(y+1)=0\) \(y=2,-1\) Now, \(y=2\) \(\left|e^x-1\right|=2\) \(e^x-1= \pm 2\) \(e^x=1 \pm 2\) \(e^x=3,-1\) \(e^x=3\) \(x=\log _e 3\)\(\therefore\) There is one real solution of the equation.