117226
Let \(f(x)=x^2\) and \(g(x)=2^x\) then the solution set of \(f(x)=\operatorname{gof}(x)\) is
1 \(\mathrm{R}\)
2 \(\{0\}\)
3 \(\{0,2\}\)
4 none of these
Explanation:
C We have, \(f(x)=x^2, g(x)=2^x\) And, fog \((x)=\operatorname{gof}(x)\) \(\mathrm{f}[\mathrm{g}(\mathrm{x})]=\mathrm{g}[\mathrm{f}(\mathrm{x})]\) \(\mathrm{f}\left(2^{\mathrm{x}}\right)=\mathrm{g}\left[\mathrm{x}^2\right]\) \(\left(2^{\mathrm{x}}\right)^2=2^{\mathrm{x}^2}\) \(2^{2^x}=2^{x^2}\) \(2 \mathrm{x}=\mathrm{x}^2\) \(2 \mathrm{x}-\mathrm{x}^2=0\) \(\mathrm{x}(2-\mathrm{x})=0\) \(\mathrm{x}=0, \mathrm{x}=2\) \(\{0,2\}\) Hence, fog \((\mathrm{x})=\operatorname{gof}(\mathrm{x})\) is \(\{0,2\}\)
SRMJEEE-2014
Sets, Relation and Function
117228
If \(f(x)\) and \(g(x)\) are two functions with \(g(x)=x\) \(-\frac{1}{x}\) and \(f o g(x)=x^3-\frac{1}{x^3}\), then \(f^{\prime}(x)\) is equal to
1 \(3 x^2+3\)
2 \(\mathrm{x}^2-\frac{1}{x^2}\)
3 \(1+\frac{1}{x^2}\)
4 \(3 x^2+\frac{3}{x^4}\)
Explanation:
A Given, \(\mathrm{f}(\mathrm{x})\) and \(\mathrm{g}(\mathrm{x})\) are two function With \(g(x)=x-\frac{1}{x}\) and fog \((x)=x^3-\frac{1}{x^3}\) We know that, \(f \circ(x)=f(g(x))\) \(x^3-\frac{1}{x^3}=f\left(x-\frac{1}{x}\right)\) \(f\left(x-\frac{1}{x}\right)=x^3-\frac{1}{x^3}\) \(f\left(x-\frac{1}{x}\right)=\left(x-\frac{1}{x}\right)^3+3 \cdot x \cdot \frac{1}{x}\left(x-\frac{1}{x}\right)\) \(f\left(x-\frac{1}{x}\right)=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)\) So, \(f(x)=x^3+3 x\) Then, \(f^{\prime}(x)=3 x^2+3\).
JCECE-2014
Sets, Relation and Function
117230
If \(f: R \rightarrow R\) and \(g: R \rightarrow R\) are defined by \(f(x)=2 x\) +3 and \(g(x)=x^2+7\), then the values of \(x\) for which \(f(g(x))=25\), are
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
117226
Let \(f(x)=x^2\) and \(g(x)=2^x\) then the solution set of \(f(x)=\operatorname{gof}(x)\) is
1 \(\mathrm{R}\)
2 \(\{0\}\)
3 \(\{0,2\}\)
4 none of these
Explanation:
C We have, \(f(x)=x^2, g(x)=2^x\) And, fog \((x)=\operatorname{gof}(x)\) \(\mathrm{f}[\mathrm{g}(\mathrm{x})]=\mathrm{g}[\mathrm{f}(\mathrm{x})]\) \(\mathrm{f}\left(2^{\mathrm{x}}\right)=\mathrm{g}\left[\mathrm{x}^2\right]\) \(\left(2^{\mathrm{x}}\right)^2=2^{\mathrm{x}^2}\) \(2^{2^x}=2^{x^2}\) \(2 \mathrm{x}=\mathrm{x}^2\) \(2 \mathrm{x}-\mathrm{x}^2=0\) \(\mathrm{x}(2-\mathrm{x})=0\) \(\mathrm{x}=0, \mathrm{x}=2\) \(\{0,2\}\) Hence, fog \((\mathrm{x})=\operatorname{gof}(\mathrm{x})\) is \(\{0,2\}\)
SRMJEEE-2014
Sets, Relation and Function
117228
If \(f(x)\) and \(g(x)\) are two functions with \(g(x)=x\) \(-\frac{1}{x}\) and \(f o g(x)=x^3-\frac{1}{x^3}\), then \(f^{\prime}(x)\) is equal to
1 \(3 x^2+3\)
2 \(\mathrm{x}^2-\frac{1}{x^2}\)
3 \(1+\frac{1}{x^2}\)
4 \(3 x^2+\frac{3}{x^4}\)
Explanation:
A Given, \(\mathrm{f}(\mathrm{x})\) and \(\mathrm{g}(\mathrm{x})\) are two function With \(g(x)=x-\frac{1}{x}\) and fog \((x)=x^3-\frac{1}{x^3}\) We know that, \(f \circ(x)=f(g(x))\) \(x^3-\frac{1}{x^3}=f\left(x-\frac{1}{x}\right)\) \(f\left(x-\frac{1}{x}\right)=x^3-\frac{1}{x^3}\) \(f\left(x-\frac{1}{x}\right)=\left(x-\frac{1}{x}\right)^3+3 \cdot x \cdot \frac{1}{x}\left(x-\frac{1}{x}\right)\) \(f\left(x-\frac{1}{x}\right)=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)\) So, \(f(x)=x^3+3 x\) Then, \(f^{\prime}(x)=3 x^2+3\).
JCECE-2014
Sets, Relation and Function
117230
If \(f: R \rightarrow R\) and \(g: R \rightarrow R\) are defined by \(f(x)=2 x\) +3 and \(g(x)=x^2+7\), then the values of \(x\) for which \(f(g(x))=25\), are
117226
Let \(f(x)=x^2\) and \(g(x)=2^x\) then the solution set of \(f(x)=\operatorname{gof}(x)\) is
1 \(\mathrm{R}\)
2 \(\{0\}\)
3 \(\{0,2\}\)
4 none of these
Explanation:
C We have, \(f(x)=x^2, g(x)=2^x\) And, fog \((x)=\operatorname{gof}(x)\) \(\mathrm{f}[\mathrm{g}(\mathrm{x})]=\mathrm{g}[\mathrm{f}(\mathrm{x})]\) \(\mathrm{f}\left(2^{\mathrm{x}}\right)=\mathrm{g}\left[\mathrm{x}^2\right]\) \(\left(2^{\mathrm{x}}\right)^2=2^{\mathrm{x}^2}\) \(2^{2^x}=2^{x^2}\) \(2 \mathrm{x}=\mathrm{x}^2\) \(2 \mathrm{x}-\mathrm{x}^2=0\) \(\mathrm{x}(2-\mathrm{x})=0\) \(\mathrm{x}=0, \mathrm{x}=2\) \(\{0,2\}\) Hence, fog \((\mathrm{x})=\operatorname{gof}(\mathrm{x})\) is \(\{0,2\}\)
SRMJEEE-2014
Sets, Relation and Function
117228
If \(f(x)\) and \(g(x)\) are two functions with \(g(x)=x\) \(-\frac{1}{x}\) and \(f o g(x)=x^3-\frac{1}{x^3}\), then \(f^{\prime}(x)\) is equal to
1 \(3 x^2+3\)
2 \(\mathrm{x}^2-\frac{1}{x^2}\)
3 \(1+\frac{1}{x^2}\)
4 \(3 x^2+\frac{3}{x^4}\)
Explanation:
A Given, \(\mathrm{f}(\mathrm{x})\) and \(\mathrm{g}(\mathrm{x})\) are two function With \(g(x)=x-\frac{1}{x}\) and fog \((x)=x^3-\frac{1}{x^3}\) We know that, \(f \circ(x)=f(g(x))\) \(x^3-\frac{1}{x^3}=f\left(x-\frac{1}{x}\right)\) \(f\left(x-\frac{1}{x}\right)=x^3-\frac{1}{x^3}\) \(f\left(x-\frac{1}{x}\right)=\left(x-\frac{1}{x}\right)^3+3 \cdot x \cdot \frac{1}{x}\left(x-\frac{1}{x}\right)\) \(f\left(x-\frac{1}{x}\right)=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)\) So, \(f(x)=x^3+3 x\) Then, \(f^{\prime}(x)=3 x^2+3\).
JCECE-2014
Sets, Relation and Function
117230
If \(f: R \rightarrow R\) and \(g: R \rightarrow R\) are defined by \(f(x)=2 x\) +3 and \(g(x)=x^2+7\), then the values of \(x\) for which \(f(g(x))=25\), are
117226
Let \(f(x)=x^2\) and \(g(x)=2^x\) then the solution set of \(f(x)=\operatorname{gof}(x)\) is
1 \(\mathrm{R}\)
2 \(\{0\}\)
3 \(\{0,2\}\)
4 none of these
Explanation:
C We have, \(f(x)=x^2, g(x)=2^x\) And, fog \((x)=\operatorname{gof}(x)\) \(\mathrm{f}[\mathrm{g}(\mathrm{x})]=\mathrm{g}[\mathrm{f}(\mathrm{x})]\) \(\mathrm{f}\left(2^{\mathrm{x}}\right)=\mathrm{g}\left[\mathrm{x}^2\right]\) \(\left(2^{\mathrm{x}}\right)^2=2^{\mathrm{x}^2}\) \(2^{2^x}=2^{x^2}\) \(2 \mathrm{x}=\mathrm{x}^2\) \(2 \mathrm{x}-\mathrm{x}^2=0\) \(\mathrm{x}(2-\mathrm{x})=0\) \(\mathrm{x}=0, \mathrm{x}=2\) \(\{0,2\}\) Hence, fog \((\mathrm{x})=\operatorname{gof}(\mathrm{x})\) is \(\{0,2\}\)
SRMJEEE-2014
Sets, Relation and Function
117228
If \(f(x)\) and \(g(x)\) are two functions with \(g(x)=x\) \(-\frac{1}{x}\) and \(f o g(x)=x^3-\frac{1}{x^3}\), then \(f^{\prime}(x)\) is equal to
1 \(3 x^2+3\)
2 \(\mathrm{x}^2-\frac{1}{x^2}\)
3 \(1+\frac{1}{x^2}\)
4 \(3 x^2+\frac{3}{x^4}\)
Explanation:
A Given, \(\mathrm{f}(\mathrm{x})\) and \(\mathrm{g}(\mathrm{x})\) are two function With \(g(x)=x-\frac{1}{x}\) and fog \((x)=x^3-\frac{1}{x^3}\) We know that, \(f \circ(x)=f(g(x))\) \(x^3-\frac{1}{x^3}=f\left(x-\frac{1}{x}\right)\) \(f\left(x-\frac{1}{x}\right)=x^3-\frac{1}{x^3}\) \(f\left(x-\frac{1}{x}\right)=\left(x-\frac{1}{x}\right)^3+3 \cdot x \cdot \frac{1}{x}\left(x-\frac{1}{x}\right)\) \(f\left(x-\frac{1}{x}\right)=\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)\) So, \(f(x)=x^3+3 x\) Then, \(f^{\prime}(x)=3 x^2+3\).
JCECE-2014
Sets, Relation and Function
117230
If \(f: R \rightarrow R\) and \(g: R \rightarrow R\) are defined by \(f(x)=2 x\) +3 and \(g(x)=x^2+7\), then the values of \(x\) for which \(f(g(x))=25\), are