NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
116889
The least number among \(\sqrt[3]{4}, \sqrt[4]{5}, \sqrt[4]{7}\) and \(\sqrt[3]{8}\) is
1 \(\sqrt[3]{8}\)
2 \(\sqrt[4]{7}\)
3 \(\sqrt[3]{4}\)
4 \(\sqrt[4]{5}\)
Explanation:
D Given, the numbers \(4^{\frac{1}{3}}, 5^{\frac{1}{4}}, 7^{\frac{1}{4}}, 8^{\frac{1}{3}}\) Then, \(\operatorname{LCM}\) of \((3,4)=12\) Now, \(\left(4^{\frac{1}{3}}\right)^{12}=4^4=256\) \(\left(5^{\frac{1}{4}}\right)^{12}=5^3=125\) \(\left(7^{\frac{1}{4}}\right)^{12}=7^3=343\) \(\left(8^{\frac{1}{3}}\right)^{12}=8^4=4096\) So, we see that 125 is least number. Hence, \(\sqrt[4]{5}\) is least number.
AP EAMCET-2002
Sets, Relation and Function
116890
If \(\log 2=a, \log 3=b, \log 7=c\) and \(6^x=7^{x+4}\) then \(x\) is equal to
1 \(\frac{4 b}{c+a-b}\)
2 \(\frac{4 c}{a+b-c}\)
3 \(\frac{4 b}{c-a-b}\)
4 \(\frac{4 a}{a+b-c}\)
Explanation:
B Given, \(\log 2=\mathrm{a}, \log 3=\mathrm{b}, \log 7=\mathrm{c} \text { and } 6^{\mathrm{x}}=7^{\mathrm{x}+4}\) Taking \(\log\) both the sides of the above condition- \(\log 6^x=\log 7^{x+4}\) \(\Rightarrow x \log (2 \times 3)=(x+4) \log 7\) \(\Rightarrow x[\log 2+\log 3]=(x+4) \cdot \log 7\) \(\Rightarrow x[a+b]=(x+4) \cdot c\) \(x[a+b]=x c+4 c\) Or \(x[a+b-c]=4 c\) \(x=\frac{4 c}{a+b-c}\)
AP EAMCET-2002
Sets, Relation and Function
116891
If \(x>0\), then \(\frac{x}{1+x}-\log (1+x)\)
1 is less than zero
2 is greater than zero
3 is equal to zero
4 takes all the real values
Explanation:
A Given, \(x>0\) then \(\frac{\mathrm{x}}{1+\mathrm{x}}-\log (1+\mathrm{x})\) \(=\left[1-\frac{1}{1+\mathrm{x}}\right]-\log (1+\mathrm{x})\) Then, Let, \(1+\mathrm{x}=\mathrm{x}\). \(\left(1-\frac{1}{x}\right)-\log x, x>0\) We will know by the graph- So, for \(\mathrm{x}>0\) \(\log \mathrm{x}>\left(1-\frac{1}{\mathrm{x}}\right)\) \(\therefore 1-\frac{1}{\mathrm{x}}-\log \mathrm{x}\lt 0\) \(\text { Or } \quad \frac{\mathrm{x}}{1+\mathrm{x}}-\log (1+\mathrm{x})\lt 0\)So, \(\frac{x}{1+x}-\log (1+x)\) is less than zero.
Shift-II
Sets, Relation and Function
116892
The value of \(x\) satisfying \(\log _2(3 x-2)=\log _{1 / 2} x\) is
1 1
2 \(-\frac{1}{3}\)
3 -1
4 \(\frac{1}{3}\)
Explanation:
\(\log _2(3 x-2)=\log _{\frac{1}{2}} x\) \(\log _2(3 x-2)=\frac{1}{\log _x\left(\frac{1}{2}\right)}=\frac{1}{-\log _x 2}\) \(\log _2(3 x-2)=-\log _2 x\) \(\log _2(3 x-2)+\log _2 x=0\) \(\log _2\{x(3 x-2)\}=0\) \(x(3 x-2)=2^0\) \(x(3 x-2)=1\) \(3 x^2-2 x-1=0\) \(\lvert\, (x-1)(3 x+1)=0\) \(x=1, \frac{-1}{3}.\) But \(x=\frac{-1}{3}\) is not possible. So \(x=1\) is the only one solution.
116889
The least number among \(\sqrt[3]{4}, \sqrt[4]{5}, \sqrt[4]{7}\) and \(\sqrt[3]{8}\) is
1 \(\sqrt[3]{8}\)
2 \(\sqrt[4]{7}\)
3 \(\sqrt[3]{4}\)
4 \(\sqrt[4]{5}\)
Explanation:
D Given, the numbers \(4^{\frac{1}{3}}, 5^{\frac{1}{4}}, 7^{\frac{1}{4}}, 8^{\frac{1}{3}}\) Then, \(\operatorname{LCM}\) of \((3,4)=12\) Now, \(\left(4^{\frac{1}{3}}\right)^{12}=4^4=256\) \(\left(5^{\frac{1}{4}}\right)^{12}=5^3=125\) \(\left(7^{\frac{1}{4}}\right)^{12}=7^3=343\) \(\left(8^{\frac{1}{3}}\right)^{12}=8^4=4096\) So, we see that 125 is least number. Hence, \(\sqrt[4]{5}\) is least number.
AP EAMCET-2002
Sets, Relation and Function
116890
If \(\log 2=a, \log 3=b, \log 7=c\) and \(6^x=7^{x+4}\) then \(x\) is equal to
1 \(\frac{4 b}{c+a-b}\)
2 \(\frac{4 c}{a+b-c}\)
3 \(\frac{4 b}{c-a-b}\)
4 \(\frac{4 a}{a+b-c}\)
Explanation:
B Given, \(\log 2=\mathrm{a}, \log 3=\mathrm{b}, \log 7=\mathrm{c} \text { and } 6^{\mathrm{x}}=7^{\mathrm{x}+4}\) Taking \(\log\) both the sides of the above condition- \(\log 6^x=\log 7^{x+4}\) \(\Rightarrow x \log (2 \times 3)=(x+4) \log 7\) \(\Rightarrow x[\log 2+\log 3]=(x+4) \cdot \log 7\) \(\Rightarrow x[a+b]=(x+4) \cdot c\) \(x[a+b]=x c+4 c\) Or \(x[a+b-c]=4 c\) \(x=\frac{4 c}{a+b-c}\)
AP EAMCET-2002
Sets, Relation and Function
116891
If \(x>0\), then \(\frac{x}{1+x}-\log (1+x)\)
1 is less than zero
2 is greater than zero
3 is equal to zero
4 takes all the real values
Explanation:
A Given, \(x>0\) then \(\frac{\mathrm{x}}{1+\mathrm{x}}-\log (1+\mathrm{x})\) \(=\left[1-\frac{1}{1+\mathrm{x}}\right]-\log (1+\mathrm{x})\) Then, Let, \(1+\mathrm{x}=\mathrm{x}\). \(\left(1-\frac{1}{x}\right)-\log x, x>0\) We will know by the graph- So, for \(\mathrm{x}>0\) \(\log \mathrm{x}>\left(1-\frac{1}{\mathrm{x}}\right)\) \(\therefore 1-\frac{1}{\mathrm{x}}-\log \mathrm{x}\lt 0\) \(\text { Or } \quad \frac{\mathrm{x}}{1+\mathrm{x}}-\log (1+\mathrm{x})\lt 0\)So, \(\frac{x}{1+x}-\log (1+x)\) is less than zero.
Shift-II
Sets, Relation and Function
116892
The value of \(x\) satisfying \(\log _2(3 x-2)=\log _{1 / 2} x\) is
1 1
2 \(-\frac{1}{3}\)
3 -1
4 \(\frac{1}{3}\)
Explanation:
\(\log _2(3 x-2)=\log _{\frac{1}{2}} x\) \(\log _2(3 x-2)=\frac{1}{\log _x\left(\frac{1}{2}\right)}=\frac{1}{-\log _x 2}\) \(\log _2(3 x-2)=-\log _2 x\) \(\log _2(3 x-2)+\log _2 x=0\) \(\log _2\{x(3 x-2)\}=0\) \(x(3 x-2)=2^0\) \(x(3 x-2)=1\) \(3 x^2-2 x-1=0\) \(\lvert\, (x-1)(3 x+1)=0\) \(x=1, \frac{-1}{3}.\) But \(x=\frac{-1}{3}\) is not possible. So \(x=1\) is the only one solution.
116889
The least number among \(\sqrt[3]{4}, \sqrt[4]{5}, \sqrt[4]{7}\) and \(\sqrt[3]{8}\) is
1 \(\sqrt[3]{8}\)
2 \(\sqrt[4]{7}\)
3 \(\sqrt[3]{4}\)
4 \(\sqrt[4]{5}\)
Explanation:
D Given, the numbers \(4^{\frac{1}{3}}, 5^{\frac{1}{4}}, 7^{\frac{1}{4}}, 8^{\frac{1}{3}}\) Then, \(\operatorname{LCM}\) of \((3,4)=12\) Now, \(\left(4^{\frac{1}{3}}\right)^{12}=4^4=256\) \(\left(5^{\frac{1}{4}}\right)^{12}=5^3=125\) \(\left(7^{\frac{1}{4}}\right)^{12}=7^3=343\) \(\left(8^{\frac{1}{3}}\right)^{12}=8^4=4096\) So, we see that 125 is least number. Hence, \(\sqrt[4]{5}\) is least number.
AP EAMCET-2002
Sets, Relation and Function
116890
If \(\log 2=a, \log 3=b, \log 7=c\) and \(6^x=7^{x+4}\) then \(x\) is equal to
1 \(\frac{4 b}{c+a-b}\)
2 \(\frac{4 c}{a+b-c}\)
3 \(\frac{4 b}{c-a-b}\)
4 \(\frac{4 a}{a+b-c}\)
Explanation:
B Given, \(\log 2=\mathrm{a}, \log 3=\mathrm{b}, \log 7=\mathrm{c} \text { and } 6^{\mathrm{x}}=7^{\mathrm{x}+4}\) Taking \(\log\) both the sides of the above condition- \(\log 6^x=\log 7^{x+4}\) \(\Rightarrow x \log (2 \times 3)=(x+4) \log 7\) \(\Rightarrow x[\log 2+\log 3]=(x+4) \cdot \log 7\) \(\Rightarrow x[a+b]=(x+4) \cdot c\) \(x[a+b]=x c+4 c\) Or \(x[a+b-c]=4 c\) \(x=\frac{4 c}{a+b-c}\)
AP EAMCET-2002
Sets, Relation and Function
116891
If \(x>0\), then \(\frac{x}{1+x}-\log (1+x)\)
1 is less than zero
2 is greater than zero
3 is equal to zero
4 takes all the real values
Explanation:
A Given, \(x>0\) then \(\frac{\mathrm{x}}{1+\mathrm{x}}-\log (1+\mathrm{x})\) \(=\left[1-\frac{1}{1+\mathrm{x}}\right]-\log (1+\mathrm{x})\) Then, Let, \(1+\mathrm{x}=\mathrm{x}\). \(\left(1-\frac{1}{x}\right)-\log x, x>0\) We will know by the graph- So, for \(\mathrm{x}>0\) \(\log \mathrm{x}>\left(1-\frac{1}{\mathrm{x}}\right)\) \(\therefore 1-\frac{1}{\mathrm{x}}-\log \mathrm{x}\lt 0\) \(\text { Or } \quad \frac{\mathrm{x}}{1+\mathrm{x}}-\log (1+\mathrm{x})\lt 0\)So, \(\frac{x}{1+x}-\log (1+x)\) is less than zero.
Shift-II
Sets, Relation and Function
116892
The value of \(x\) satisfying \(\log _2(3 x-2)=\log _{1 / 2} x\) is
1 1
2 \(-\frac{1}{3}\)
3 -1
4 \(\frac{1}{3}\)
Explanation:
\(\log _2(3 x-2)=\log _{\frac{1}{2}} x\) \(\log _2(3 x-2)=\frac{1}{\log _x\left(\frac{1}{2}\right)}=\frac{1}{-\log _x 2}\) \(\log _2(3 x-2)=-\log _2 x\) \(\log _2(3 x-2)+\log _2 x=0\) \(\log _2\{x(3 x-2)\}=0\) \(x(3 x-2)=2^0\) \(x(3 x-2)=1\) \(3 x^2-2 x-1=0\) \(\lvert\, (x-1)(3 x+1)=0\) \(x=1, \frac{-1}{3}.\) But \(x=\frac{-1}{3}\) is not possible. So \(x=1\) is the only one solution.
116889
The least number among \(\sqrt[3]{4}, \sqrt[4]{5}, \sqrt[4]{7}\) and \(\sqrt[3]{8}\) is
1 \(\sqrt[3]{8}\)
2 \(\sqrt[4]{7}\)
3 \(\sqrt[3]{4}\)
4 \(\sqrt[4]{5}\)
Explanation:
D Given, the numbers \(4^{\frac{1}{3}}, 5^{\frac{1}{4}}, 7^{\frac{1}{4}}, 8^{\frac{1}{3}}\) Then, \(\operatorname{LCM}\) of \((3,4)=12\) Now, \(\left(4^{\frac{1}{3}}\right)^{12}=4^4=256\) \(\left(5^{\frac{1}{4}}\right)^{12}=5^3=125\) \(\left(7^{\frac{1}{4}}\right)^{12}=7^3=343\) \(\left(8^{\frac{1}{3}}\right)^{12}=8^4=4096\) So, we see that 125 is least number. Hence, \(\sqrt[4]{5}\) is least number.
AP EAMCET-2002
Sets, Relation and Function
116890
If \(\log 2=a, \log 3=b, \log 7=c\) and \(6^x=7^{x+4}\) then \(x\) is equal to
1 \(\frac{4 b}{c+a-b}\)
2 \(\frac{4 c}{a+b-c}\)
3 \(\frac{4 b}{c-a-b}\)
4 \(\frac{4 a}{a+b-c}\)
Explanation:
B Given, \(\log 2=\mathrm{a}, \log 3=\mathrm{b}, \log 7=\mathrm{c} \text { and } 6^{\mathrm{x}}=7^{\mathrm{x}+4}\) Taking \(\log\) both the sides of the above condition- \(\log 6^x=\log 7^{x+4}\) \(\Rightarrow x \log (2 \times 3)=(x+4) \log 7\) \(\Rightarrow x[\log 2+\log 3]=(x+4) \cdot \log 7\) \(\Rightarrow x[a+b]=(x+4) \cdot c\) \(x[a+b]=x c+4 c\) Or \(x[a+b-c]=4 c\) \(x=\frac{4 c}{a+b-c}\)
AP EAMCET-2002
Sets, Relation and Function
116891
If \(x>0\), then \(\frac{x}{1+x}-\log (1+x)\)
1 is less than zero
2 is greater than zero
3 is equal to zero
4 takes all the real values
Explanation:
A Given, \(x>0\) then \(\frac{\mathrm{x}}{1+\mathrm{x}}-\log (1+\mathrm{x})\) \(=\left[1-\frac{1}{1+\mathrm{x}}\right]-\log (1+\mathrm{x})\) Then, Let, \(1+\mathrm{x}=\mathrm{x}\). \(\left(1-\frac{1}{x}\right)-\log x, x>0\) We will know by the graph- So, for \(\mathrm{x}>0\) \(\log \mathrm{x}>\left(1-\frac{1}{\mathrm{x}}\right)\) \(\therefore 1-\frac{1}{\mathrm{x}}-\log \mathrm{x}\lt 0\) \(\text { Or } \quad \frac{\mathrm{x}}{1+\mathrm{x}}-\log (1+\mathrm{x})\lt 0\)So, \(\frac{x}{1+x}-\log (1+x)\) is less than zero.
Shift-II
Sets, Relation and Function
116892
The value of \(x\) satisfying \(\log _2(3 x-2)=\log _{1 / 2} x\) is
1 1
2 \(-\frac{1}{3}\)
3 -1
4 \(\frac{1}{3}\)
Explanation:
\(\log _2(3 x-2)=\log _{\frac{1}{2}} x\) \(\log _2(3 x-2)=\frac{1}{\log _x\left(\frac{1}{2}\right)}=\frac{1}{-\log _x 2}\) \(\log _2(3 x-2)=-\log _2 x\) \(\log _2(3 x-2)+\log _2 x=0\) \(\log _2\{x(3 x-2)\}=0\) \(x(3 x-2)=2^0\) \(x(3 x-2)=1\) \(3 x^2-2 x-1=0\) \(\lvert\, (x-1)(3 x+1)=0\) \(x=1, \frac{-1}{3}.\) But \(x=\frac{-1}{3}\) is not possible. So \(x=1\) is the only one solution.
