116989
If \(\mathrm{e}^{\mathrm{f}(\mathrm{x})}=\frac{10+\mathrm{x}}{10-\mathrm{x}}, \mathrm{x} \in(-10,10)\) and \(f(x)=k f\left(\frac{200 x}{100+x^2}\right)\), then \(k\) is equal to
1 0.5
2 0.6
3 0.7
4 0.8
Explanation:
A We have given, \(\mathrm{e}^{\mathrm{f}(\mathrm{x})}=\frac{10+\mathrm{x}}{10-\mathrm{x}}\) \(x \in(-10,10)\) Taking log on both side \(\log e^{f(x)}=\log \left(\frac{10+x}{10-x}\right)\) \(f(x)=\log \left(\frac{10+x}{10-x}\right)\) Given \(f(x)=k f\left(\frac{200 x}{100+x^2}\right)\) \(\log \left(\frac{10+x}{10-x}\right)=k \log \left(\frac{10+\frac{200 x}{100+x^2}}{10-\frac{200 x}{100+x^2}}\right)\) \(\log \left(\frac{10+x}{10-x}\right)=k \log \left(\frac{1000+10 x^2+200 x}{1000+10 x^2-200 x}\right)\) \(=k \log \left(\frac{100+x^2+20 x}{100+x^2-20 x}\right)\) \(=k \log \left(\frac{10+x}{10-x}\right)^2\) \(\log \left(\frac{10+x}{10-x}\right)=2 k \log \left(\frac{10+x}{10-x}\right)\) \(1=2 K\) \(k=\frac{1}{2}\) \(k=0.5\)
EAMCET-2003
Sets, Relation and Function
116990
If \(\frac{\log x}{\log 5}=\frac{\log 36}{\log 6}=\frac{\log 64}{\log y}\), what are the values of \(x\) and \(y\) respectively?
1 8,25
2 25,8
3 8,8
4 25,25
Explanation:
B We have \(\frac{\log x}{\log 5}=\frac{\log 36}{\log 6}=\frac{\log 64}{\log y}\) Now we take \(\frac{\log x}{\log 5}=\frac{\log 36}{\log 6}\) \(\frac{\log x}{\log 5}=\frac{\log 6^2}{\log 6}\) \(\frac{\log x}{\log 5}=\frac{2 \log 6}{\log 6}\) \(\log x=2 \log 5\) \(\log x=\log 25\) \(x=25\) Again we take \(\frac{\log 36}{\log 6}=\frac{\log 64}{\log y}\) \(\frac{2 \log 6}{\log 6}=\frac{3 \log 4}{\log y}\) \(\frac{3 \log 4}{\log y}=2\) \(3 \log 4=2 \log y\) \(\log 4^3=\log \mathrm{y}^2\) \(y^2=64\) \(y= \pm 8\) Hence, \(\mathrm{x}=25\) \(\mathrm{y}=8\)
Jamia Millia Islamia-2011
Sets, Relation and Function
116991
If \(f(x)=\log _e\{\log x\}\), then \(f(x)\) at \(x=e\), is
116992
Solve the equation, \(3^{x^2-x}=25-4^{x^2-x}\)
1 -1 only
2 2 only
3 Both -1 and 2
4 No solution
Explanation:
C Given, \(3^{x^2-x}=25-4^{x^2-x}\) \(3^{x^2-x}+4^{x^2-x}=25\) \(3^{x^2-x}+4^{x^2-x}=3^2+4^2\) Now comparing the coefficient \(3^{x^2-x}=3^2\) \(x^2-x=2\) \(x^2-x-2=0\) \((x-2)(x+1)=0\) \(x=2,-1\) And \(4^{x^2-x}=4^2\) \(x^2-x=2\) \(x^2-x-2=0\) \((x-2)(x+1)=0\) \(x=2,-1\)Hence, there are two solutions.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
116989
If \(\mathrm{e}^{\mathrm{f}(\mathrm{x})}=\frac{10+\mathrm{x}}{10-\mathrm{x}}, \mathrm{x} \in(-10,10)\) and \(f(x)=k f\left(\frac{200 x}{100+x^2}\right)\), then \(k\) is equal to
1 0.5
2 0.6
3 0.7
4 0.8
Explanation:
A We have given, \(\mathrm{e}^{\mathrm{f}(\mathrm{x})}=\frac{10+\mathrm{x}}{10-\mathrm{x}}\) \(x \in(-10,10)\) Taking log on both side \(\log e^{f(x)}=\log \left(\frac{10+x}{10-x}\right)\) \(f(x)=\log \left(\frac{10+x}{10-x}\right)\) Given \(f(x)=k f\left(\frac{200 x}{100+x^2}\right)\) \(\log \left(\frac{10+x}{10-x}\right)=k \log \left(\frac{10+\frac{200 x}{100+x^2}}{10-\frac{200 x}{100+x^2}}\right)\) \(\log \left(\frac{10+x}{10-x}\right)=k \log \left(\frac{1000+10 x^2+200 x}{1000+10 x^2-200 x}\right)\) \(=k \log \left(\frac{100+x^2+20 x}{100+x^2-20 x}\right)\) \(=k \log \left(\frac{10+x}{10-x}\right)^2\) \(\log \left(\frac{10+x}{10-x}\right)=2 k \log \left(\frac{10+x}{10-x}\right)\) \(1=2 K\) \(k=\frac{1}{2}\) \(k=0.5\)
EAMCET-2003
Sets, Relation and Function
116990
If \(\frac{\log x}{\log 5}=\frac{\log 36}{\log 6}=\frac{\log 64}{\log y}\), what are the values of \(x\) and \(y\) respectively?
1 8,25
2 25,8
3 8,8
4 25,25
Explanation:
B We have \(\frac{\log x}{\log 5}=\frac{\log 36}{\log 6}=\frac{\log 64}{\log y}\) Now we take \(\frac{\log x}{\log 5}=\frac{\log 36}{\log 6}\) \(\frac{\log x}{\log 5}=\frac{\log 6^2}{\log 6}\) \(\frac{\log x}{\log 5}=\frac{2 \log 6}{\log 6}\) \(\log x=2 \log 5\) \(\log x=\log 25\) \(x=25\) Again we take \(\frac{\log 36}{\log 6}=\frac{\log 64}{\log y}\) \(\frac{2 \log 6}{\log 6}=\frac{3 \log 4}{\log y}\) \(\frac{3 \log 4}{\log y}=2\) \(3 \log 4=2 \log y\) \(\log 4^3=\log \mathrm{y}^2\) \(y^2=64\) \(y= \pm 8\) Hence, \(\mathrm{x}=25\) \(\mathrm{y}=8\)
Jamia Millia Islamia-2011
Sets, Relation and Function
116991
If \(f(x)=\log _e\{\log x\}\), then \(f(x)\) at \(x=e\), is
116992
Solve the equation, \(3^{x^2-x}=25-4^{x^2-x}\)
1 -1 only
2 2 only
3 Both -1 and 2
4 No solution
Explanation:
C Given, \(3^{x^2-x}=25-4^{x^2-x}\) \(3^{x^2-x}+4^{x^2-x}=25\) \(3^{x^2-x}+4^{x^2-x}=3^2+4^2\) Now comparing the coefficient \(3^{x^2-x}=3^2\) \(x^2-x=2\) \(x^2-x-2=0\) \((x-2)(x+1)=0\) \(x=2,-1\) And \(4^{x^2-x}=4^2\) \(x^2-x=2\) \(x^2-x-2=0\) \((x-2)(x+1)=0\) \(x=2,-1\)Hence, there are two solutions.
116989
If \(\mathrm{e}^{\mathrm{f}(\mathrm{x})}=\frac{10+\mathrm{x}}{10-\mathrm{x}}, \mathrm{x} \in(-10,10)\) and \(f(x)=k f\left(\frac{200 x}{100+x^2}\right)\), then \(k\) is equal to
1 0.5
2 0.6
3 0.7
4 0.8
Explanation:
A We have given, \(\mathrm{e}^{\mathrm{f}(\mathrm{x})}=\frac{10+\mathrm{x}}{10-\mathrm{x}}\) \(x \in(-10,10)\) Taking log on both side \(\log e^{f(x)}=\log \left(\frac{10+x}{10-x}\right)\) \(f(x)=\log \left(\frac{10+x}{10-x}\right)\) Given \(f(x)=k f\left(\frac{200 x}{100+x^2}\right)\) \(\log \left(\frac{10+x}{10-x}\right)=k \log \left(\frac{10+\frac{200 x}{100+x^2}}{10-\frac{200 x}{100+x^2}}\right)\) \(\log \left(\frac{10+x}{10-x}\right)=k \log \left(\frac{1000+10 x^2+200 x}{1000+10 x^2-200 x}\right)\) \(=k \log \left(\frac{100+x^2+20 x}{100+x^2-20 x}\right)\) \(=k \log \left(\frac{10+x}{10-x}\right)^2\) \(\log \left(\frac{10+x}{10-x}\right)=2 k \log \left(\frac{10+x}{10-x}\right)\) \(1=2 K\) \(k=\frac{1}{2}\) \(k=0.5\)
EAMCET-2003
Sets, Relation and Function
116990
If \(\frac{\log x}{\log 5}=\frac{\log 36}{\log 6}=\frac{\log 64}{\log y}\), what are the values of \(x\) and \(y\) respectively?
1 8,25
2 25,8
3 8,8
4 25,25
Explanation:
B We have \(\frac{\log x}{\log 5}=\frac{\log 36}{\log 6}=\frac{\log 64}{\log y}\) Now we take \(\frac{\log x}{\log 5}=\frac{\log 36}{\log 6}\) \(\frac{\log x}{\log 5}=\frac{\log 6^2}{\log 6}\) \(\frac{\log x}{\log 5}=\frac{2 \log 6}{\log 6}\) \(\log x=2 \log 5\) \(\log x=\log 25\) \(x=25\) Again we take \(\frac{\log 36}{\log 6}=\frac{\log 64}{\log y}\) \(\frac{2 \log 6}{\log 6}=\frac{3 \log 4}{\log y}\) \(\frac{3 \log 4}{\log y}=2\) \(3 \log 4=2 \log y\) \(\log 4^3=\log \mathrm{y}^2\) \(y^2=64\) \(y= \pm 8\) Hence, \(\mathrm{x}=25\) \(\mathrm{y}=8\)
Jamia Millia Islamia-2011
Sets, Relation and Function
116991
If \(f(x)=\log _e\{\log x\}\), then \(f(x)\) at \(x=e\), is
116992
Solve the equation, \(3^{x^2-x}=25-4^{x^2-x}\)
1 -1 only
2 2 only
3 Both -1 and 2
4 No solution
Explanation:
C Given, \(3^{x^2-x}=25-4^{x^2-x}\) \(3^{x^2-x}+4^{x^2-x}=25\) \(3^{x^2-x}+4^{x^2-x}=3^2+4^2\) Now comparing the coefficient \(3^{x^2-x}=3^2\) \(x^2-x=2\) \(x^2-x-2=0\) \((x-2)(x+1)=0\) \(x=2,-1\) And \(4^{x^2-x}=4^2\) \(x^2-x=2\) \(x^2-x-2=0\) \((x-2)(x+1)=0\) \(x=2,-1\)Hence, there are two solutions.
116989
If \(\mathrm{e}^{\mathrm{f}(\mathrm{x})}=\frac{10+\mathrm{x}}{10-\mathrm{x}}, \mathrm{x} \in(-10,10)\) and \(f(x)=k f\left(\frac{200 x}{100+x^2}\right)\), then \(k\) is equal to
1 0.5
2 0.6
3 0.7
4 0.8
Explanation:
A We have given, \(\mathrm{e}^{\mathrm{f}(\mathrm{x})}=\frac{10+\mathrm{x}}{10-\mathrm{x}}\) \(x \in(-10,10)\) Taking log on both side \(\log e^{f(x)}=\log \left(\frac{10+x}{10-x}\right)\) \(f(x)=\log \left(\frac{10+x}{10-x}\right)\) Given \(f(x)=k f\left(\frac{200 x}{100+x^2}\right)\) \(\log \left(\frac{10+x}{10-x}\right)=k \log \left(\frac{10+\frac{200 x}{100+x^2}}{10-\frac{200 x}{100+x^2}}\right)\) \(\log \left(\frac{10+x}{10-x}\right)=k \log \left(\frac{1000+10 x^2+200 x}{1000+10 x^2-200 x}\right)\) \(=k \log \left(\frac{100+x^2+20 x}{100+x^2-20 x}\right)\) \(=k \log \left(\frac{10+x}{10-x}\right)^2\) \(\log \left(\frac{10+x}{10-x}\right)=2 k \log \left(\frac{10+x}{10-x}\right)\) \(1=2 K\) \(k=\frac{1}{2}\) \(k=0.5\)
EAMCET-2003
Sets, Relation and Function
116990
If \(\frac{\log x}{\log 5}=\frac{\log 36}{\log 6}=\frac{\log 64}{\log y}\), what are the values of \(x\) and \(y\) respectively?
1 8,25
2 25,8
3 8,8
4 25,25
Explanation:
B We have \(\frac{\log x}{\log 5}=\frac{\log 36}{\log 6}=\frac{\log 64}{\log y}\) Now we take \(\frac{\log x}{\log 5}=\frac{\log 36}{\log 6}\) \(\frac{\log x}{\log 5}=\frac{\log 6^2}{\log 6}\) \(\frac{\log x}{\log 5}=\frac{2 \log 6}{\log 6}\) \(\log x=2 \log 5\) \(\log x=\log 25\) \(x=25\) Again we take \(\frac{\log 36}{\log 6}=\frac{\log 64}{\log y}\) \(\frac{2 \log 6}{\log 6}=\frac{3 \log 4}{\log y}\) \(\frac{3 \log 4}{\log y}=2\) \(3 \log 4=2 \log y\) \(\log 4^3=\log \mathrm{y}^2\) \(y^2=64\) \(y= \pm 8\) Hence, \(\mathrm{x}=25\) \(\mathrm{y}=8\)
Jamia Millia Islamia-2011
Sets, Relation and Function
116991
If \(f(x)=\log _e\{\log x\}\), then \(f(x)\) at \(x=e\), is
116992
Solve the equation, \(3^{x^2-x}=25-4^{x^2-x}\)
1 -1 only
2 2 only
3 Both -1 and 2
4 No solution
Explanation:
C Given, \(3^{x^2-x}=25-4^{x^2-x}\) \(3^{x^2-x}+4^{x^2-x}=25\) \(3^{x^2-x}+4^{x^2-x}=3^2+4^2\) Now comparing the coefficient \(3^{x^2-x}=3^2\) \(x^2-x=2\) \(x^2-x-2=0\) \((x-2)(x+1)=0\) \(x=2,-1\) And \(4^{x^2-x}=4^2\) \(x^2-x=2\) \(x^2-x-2=0\) \((x-2)(x+1)=0\) \(x=2,-1\)Hence, there are two solutions.
