NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117110
If \(f(x)\) is a real function defined on \([-1,1]\), then the function \(g(x)=f(5 x+4)\) is defined on the interval.
1 \([-4,9]\)
2 \([-1,9]\)
3 \([-2,9]\)
4 \([-3,9]\)
Explanation:
Exp: (B) : According to question - \(f(x)\) is define on the interval \([-1,1]\) and \(g(x)=f(5 x+4)\) Therefore, \(-1 \leq \mathrm{x} \leq 1\) On multiplying by 5 on all the terms, \(-5+4 \leq 5 \mathrm{x}+4 \leq 5+4\) \(-1 \leq 5 \mathrm{x}+4 \leq 9\)Hence, \(g(x)\) is define in the interval \([-1,9]\)
[TS EAMCET-2015]
Sets, Relation and Function
117111
If \(f(x)=\log _e\left(\frac{1-x}{1+x}\right),|x|\lt 1\), then \(f\left(\frac{2 x}{1+x^2}\right)\) is equal to
117112
Let \(f: R \rightarrow R\) be such that for all \(x \in R\left(2^{1+x}+2^{1-}\right.\) \(\left.{ }^x\right), f(x)\) and \(\left(3^x+3^{-x}\right)\) are in \(A P\), then the minimum value of \(f(x)\) is
1 4
2 3
3 2
4 0
Explanation:
Exp: (B) : Given that, \(\left(2^{1+x}+2^{1-x}\right) f(x)\) and \(3^x+3^{-x}\) are in A.P Therefore, \(f(x)=\frac{2^{1+x}+2^{1-x}+3^x+3^{-x}}{2}\) \(=\frac{2\left(2^x+2^{-x}\right)+\left(3^x+3^{-x}\right)}{2}\) \(=\left(2^x+2^{-x}\right)+\frac{\left(3^x+3^{-x}\right)}{2}\) We known that, \(\mathrm{AM} \geq \mathrm{GM}\) \(\frac{2^x+2^{-x}}{2} \geq\left(2^x \cdot 2^{-x}\right)^{\frac{1}{2}}=1\) \(=2^x+2^{-x} \geq 2\) Similarly, \(\frac{3^x+3^{-x}}{2} \geq\left(3^x \cdot 3^{-x}\right)^{1 / 2}=1\) \(=3^x+3^{-x} \geq 2\) Hence, the minimum value of \(f(x)\) is, \(=2+\frac{2}{2}=\frac{4+2}{2}=\frac{6}{2}=3\)
[JEE Main 06.09.2020 Shift-II]
Sets, Relation and Function
117113
A function \(f(x)\) is given by \(f(x)=\frac{5^x}{5^x+5}\), then the sum of the series \(\mathbf{f}\left(\frac{1}{20}\right)+\mathbf{f}\left(\frac{2}{20}\right)+\mathbf{f}\left(\frac{3}{20}\right)+\ldots . . f\left(\frac{39}{20}\right)\) is equal to
117110
If \(f(x)\) is a real function defined on \([-1,1]\), then the function \(g(x)=f(5 x+4)\) is defined on the interval.
1 \([-4,9]\)
2 \([-1,9]\)
3 \([-2,9]\)
4 \([-3,9]\)
Explanation:
Exp: (B) : According to question - \(f(x)\) is define on the interval \([-1,1]\) and \(g(x)=f(5 x+4)\) Therefore, \(-1 \leq \mathrm{x} \leq 1\) On multiplying by 5 on all the terms, \(-5+4 \leq 5 \mathrm{x}+4 \leq 5+4\) \(-1 \leq 5 \mathrm{x}+4 \leq 9\)Hence, \(g(x)\) is define in the interval \([-1,9]\)
[TS EAMCET-2015]
Sets, Relation and Function
117111
If \(f(x)=\log _e\left(\frac{1-x}{1+x}\right),|x|\lt 1\), then \(f\left(\frac{2 x}{1+x^2}\right)\) is equal to
117112
Let \(f: R \rightarrow R\) be such that for all \(x \in R\left(2^{1+x}+2^{1-}\right.\) \(\left.{ }^x\right), f(x)\) and \(\left(3^x+3^{-x}\right)\) are in \(A P\), then the minimum value of \(f(x)\) is
1 4
2 3
3 2
4 0
Explanation:
Exp: (B) : Given that, \(\left(2^{1+x}+2^{1-x}\right) f(x)\) and \(3^x+3^{-x}\) are in A.P Therefore, \(f(x)=\frac{2^{1+x}+2^{1-x}+3^x+3^{-x}}{2}\) \(=\frac{2\left(2^x+2^{-x}\right)+\left(3^x+3^{-x}\right)}{2}\) \(=\left(2^x+2^{-x}\right)+\frac{\left(3^x+3^{-x}\right)}{2}\) We known that, \(\mathrm{AM} \geq \mathrm{GM}\) \(\frac{2^x+2^{-x}}{2} \geq\left(2^x \cdot 2^{-x}\right)^{\frac{1}{2}}=1\) \(=2^x+2^{-x} \geq 2\) Similarly, \(\frac{3^x+3^{-x}}{2} \geq\left(3^x \cdot 3^{-x}\right)^{1 / 2}=1\) \(=3^x+3^{-x} \geq 2\) Hence, the minimum value of \(f(x)\) is, \(=2+\frac{2}{2}=\frac{4+2}{2}=\frac{6}{2}=3\)
[JEE Main 06.09.2020 Shift-II]
Sets, Relation and Function
117113
A function \(f(x)\) is given by \(f(x)=\frac{5^x}{5^x+5}\), then the sum of the series \(\mathbf{f}\left(\frac{1}{20}\right)+\mathbf{f}\left(\frac{2}{20}\right)+\mathbf{f}\left(\frac{3}{20}\right)+\ldots . . f\left(\frac{39}{20}\right)\) is equal to
117110
If \(f(x)\) is a real function defined on \([-1,1]\), then the function \(g(x)=f(5 x+4)\) is defined on the interval.
1 \([-4,9]\)
2 \([-1,9]\)
3 \([-2,9]\)
4 \([-3,9]\)
Explanation:
Exp: (B) : According to question - \(f(x)\) is define on the interval \([-1,1]\) and \(g(x)=f(5 x+4)\) Therefore, \(-1 \leq \mathrm{x} \leq 1\) On multiplying by 5 on all the terms, \(-5+4 \leq 5 \mathrm{x}+4 \leq 5+4\) \(-1 \leq 5 \mathrm{x}+4 \leq 9\)Hence, \(g(x)\) is define in the interval \([-1,9]\)
[TS EAMCET-2015]
Sets, Relation and Function
117111
If \(f(x)=\log _e\left(\frac{1-x}{1+x}\right),|x|\lt 1\), then \(f\left(\frac{2 x}{1+x^2}\right)\) is equal to
117112
Let \(f: R \rightarrow R\) be such that for all \(x \in R\left(2^{1+x}+2^{1-}\right.\) \(\left.{ }^x\right), f(x)\) and \(\left(3^x+3^{-x}\right)\) are in \(A P\), then the minimum value of \(f(x)\) is
1 4
2 3
3 2
4 0
Explanation:
Exp: (B) : Given that, \(\left(2^{1+x}+2^{1-x}\right) f(x)\) and \(3^x+3^{-x}\) are in A.P Therefore, \(f(x)=\frac{2^{1+x}+2^{1-x}+3^x+3^{-x}}{2}\) \(=\frac{2\left(2^x+2^{-x}\right)+\left(3^x+3^{-x}\right)}{2}\) \(=\left(2^x+2^{-x}\right)+\frac{\left(3^x+3^{-x}\right)}{2}\) We known that, \(\mathrm{AM} \geq \mathrm{GM}\) \(\frac{2^x+2^{-x}}{2} \geq\left(2^x \cdot 2^{-x}\right)^{\frac{1}{2}}=1\) \(=2^x+2^{-x} \geq 2\) Similarly, \(\frac{3^x+3^{-x}}{2} \geq\left(3^x \cdot 3^{-x}\right)^{1 / 2}=1\) \(=3^x+3^{-x} \geq 2\) Hence, the minimum value of \(f(x)\) is, \(=2+\frac{2}{2}=\frac{4+2}{2}=\frac{6}{2}=3\)
[JEE Main 06.09.2020 Shift-II]
Sets, Relation and Function
117113
A function \(f(x)\) is given by \(f(x)=\frac{5^x}{5^x+5}\), then the sum of the series \(\mathbf{f}\left(\frac{1}{20}\right)+\mathbf{f}\left(\frac{2}{20}\right)+\mathbf{f}\left(\frac{3}{20}\right)+\ldots . . f\left(\frac{39}{20}\right)\) is equal to
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
117110
If \(f(x)\) is a real function defined on \([-1,1]\), then the function \(g(x)=f(5 x+4)\) is defined on the interval.
1 \([-4,9]\)
2 \([-1,9]\)
3 \([-2,9]\)
4 \([-3,9]\)
Explanation:
Exp: (B) : According to question - \(f(x)\) is define on the interval \([-1,1]\) and \(g(x)=f(5 x+4)\) Therefore, \(-1 \leq \mathrm{x} \leq 1\) On multiplying by 5 on all the terms, \(-5+4 \leq 5 \mathrm{x}+4 \leq 5+4\) \(-1 \leq 5 \mathrm{x}+4 \leq 9\)Hence, \(g(x)\) is define in the interval \([-1,9]\)
[TS EAMCET-2015]
Sets, Relation and Function
117111
If \(f(x)=\log _e\left(\frac{1-x}{1+x}\right),|x|\lt 1\), then \(f\left(\frac{2 x}{1+x^2}\right)\) is equal to
117112
Let \(f: R \rightarrow R\) be such that for all \(x \in R\left(2^{1+x}+2^{1-}\right.\) \(\left.{ }^x\right), f(x)\) and \(\left(3^x+3^{-x}\right)\) are in \(A P\), then the minimum value of \(f(x)\) is
1 4
2 3
3 2
4 0
Explanation:
Exp: (B) : Given that, \(\left(2^{1+x}+2^{1-x}\right) f(x)\) and \(3^x+3^{-x}\) are in A.P Therefore, \(f(x)=\frac{2^{1+x}+2^{1-x}+3^x+3^{-x}}{2}\) \(=\frac{2\left(2^x+2^{-x}\right)+\left(3^x+3^{-x}\right)}{2}\) \(=\left(2^x+2^{-x}\right)+\frac{\left(3^x+3^{-x}\right)}{2}\) We known that, \(\mathrm{AM} \geq \mathrm{GM}\) \(\frac{2^x+2^{-x}}{2} \geq\left(2^x \cdot 2^{-x}\right)^{\frac{1}{2}}=1\) \(=2^x+2^{-x} \geq 2\) Similarly, \(\frac{3^x+3^{-x}}{2} \geq\left(3^x \cdot 3^{-x}\right)^{1 / 2}=1\) \(=3^x+3^{-x} \geq 2\) Hence, the minimum value of \(f(x)\) is, \(=2+\frac{2}{2}=\frac{4+2}{2}=\frac{6}{2}=3\)
[JEE Main 06.09.2020 Shift-II]
Sets, Relation and Function
117113
A function \(f(x)\) is given by \(f(x)=\frac{5^x}{5^x+5}\), then the sum of the series \(\mathbf{f}\left(\frac{1}{20}\right)+\mathbf{f}\left(\frac{2}{20}\right)+\mathbf{f}\left(\frac{3}{20}\right)+\ldots . . f\left(\frac{39}{20}\right)\) is equal to