117256
If for all values of \(\mathrm{x}\) and \(\mathrm{y}, f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) f(\mathrm{y})\) and \(f(5)=2, f^{\prime}(0)=3\), then \(f^{\prime}(5)\) is
117257
If \(\boldsymbol{f}: \mathbf{R} \rightarrow \mathbf{R}\) is defined as \(f(x)=\left(2020-x^{2019}\right)^{1 / 2019}, \forall x \in \mathrm{R}\),find (fo fo fof) \(\left(\frac{2019}{2020}\right)\)
1 1
2 0
3 \(\frac{2019}{2020}\)
4 \(\frac{2020}{2019}\)
Explanation:
C The given real function \(\mathrm{f}(\mathrm{x})=\left(2020-\mathrm{x}^{2019}\right)^{\frac{1}{2019}}\) \(\therefore \quad \text { fof }(\mathrm{x})=\left[2020-\mathrm{f}(\mathrm{x})^{2019}\right]^{\frac{1}{2019}}\) \(=\left[2020-\left\{\left(2020-\mathrm{x}^{2019}\right)^{\frac{1}{2019}}\right\}^{2019}\right]^{\frac{1}{2019}}\) \(=\left[2020-2020+\mathrm{x}^{2019}\right]^{\frac{1}{2019}}=\left[\mathrm{x}^{2019}\right]^{\frac{1}{2019}}=\mathrm{x}\) Hence, \((\) fo fo fof \()\left(\frac{2019}{2020}\right)=\frac{2019}{2020}\)
Shift-I
Sets, Relation and Function
117258
If \(f(x)=x-\frac{1}{x}, x \neq 0\), then \(3 f(x)=\)
C Given that, \(f(x)=x-\frac{1}{x}\) \(\therefore(\mathrm{f}(\mathrm{x}))^3=\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^3=\mathrm{x}^3-\frac{1}{\mathrm{x}^3}-3(\mathrm{x})\left(\frac{1}{\mathrm{x}}\right)\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)\) \((\mathrm{f}(\mathrm{x}))^3=\mathrm{f}\left(\mathrm{x}^3\right)-3\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)\) \((\mathrm{f}(\mathrm{x}))^3=\mathrm{f}\left(\mathrm{x}^3\right)-3 \mathrm{f}(\mathrm{x})\) \(3 \mathrm{f}(\mathrm{x})=\mathrm{f}\left(\mathrm{x}^3\right)-[\mathrm{f}(\mathrm{x})]^3\)
Shift-II
Sets, Relation and Function
117259
For \(x \in\left(0, \frac{3}{2}\right)\) let \(f(x)=\sqrt{x} g(x)=\tan x\) and \(h(x)=\frac{1-x^2}{1+x^2}\). If \(\phi(x)=((h o f) o g)(x)\), then \(\phi\) \(\left(\frac{\pi}{3}\right)\) is equal to
117256
If for all values of \(\mathrm{x}\) and \(\mathrm{y}, f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) f(\mathrm{y})\) and \(f(5)=2, f^{\prime}(0)=3\), then \(f^{\prime}(5)\) is
117257
If \(\boldsymbol{f}: \mathbf{R} \rightarrow \mathbf{R}\) is defined as \(f(x)=\left(2020-x^{2019}\right)^{1 / 2019}, \forall x \in \mathrm{R}\),find (fo fo fof) \(\left(\frac{2019}{2020}\right)\)
1 1
2 0
3 \(\frac{2019}{2020}\)
4 \(\frac{2020}{2019}\)
Explanation:
C The given real function \(\mathrm{f}(\mathrm{x})=\left(2020-\mathrm{x}^{2019}\right)^{\frac{1}{2019}}\) \(\therefore \quad \text { fof }(\mathrm{x})=\left[2020-\mathrm{f}(\mathrm{x})^{2019}\right]^{\frac{1}{2019}}\) \(=\left[2020-\left\{\left(2020-\mathrm{x}^{2019}\right)^{\frac{1}{2019}}\right\}^{2019}\right]^{\frac{1}{2019}}\) \(=\left[2020-2020+\mathrm{x}^{2019}\right]^{\frac{1}{2019}}=\left[\mathrm{x}^{2019}\right]^{\frac{1}{2019}}=\mathrm{x}\) Hence, \((\) fo fo fof \()\left(\frac{2019}{2020}\right)=\frac{2019}{2020}\)
Shift-I
Sets, Relation and Function
117258
If \(f(x)=x-\frac{1}{x}, x \neq 0\), then \(3 f(x)=\)
C Given that, \(f(x)=x-\frac{1}{x}\) \(\therefore(\mathrm{f}(\mathrm{x}))^3=\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^3=\mathrm{x}^3-\frac{1}{\mathrm{x}^3}-3(\mathrm{x})\left(\frac{1}{\mathrm{x}}\right)\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)\) \((\mathrm{f}(\mathrm{x}))^3=\mathrm{f}\left(\mathrm{x}^3\right)-3\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)\) \((\mathrm{f}(\mathrm{x}))^3=\mathrm{f}\left(\mathrm{x}^3\right)-3 \mathrm{f}(\mathrm{x})\) \(3 \mathrm{f}(\mathrm{x})=\mathrm{f}\left(\mathrm{x}^3\right)-[\mathrm{f}(\mathrm{x})]^3\)
Shift-II
Sets, Relation and Function
117259
For \(x \in\left(0, \frac{3}{2}\right)\) let \(f(x)=\sqrt{x} g(x)=\tan x\) and \(h(x)=\frac{1-x^2}{1+x^2}\). If \(\phi(x)=((h o f) o g)(x)\), then \(\phi\) \(\left(\frac{\pi}{3}\right)\) is equal to
117256
If for all values of \(\mathrm{x}\) and \(\mathrm{y}, f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) f(\mathrm{y})\) and \(f(5)=2, f^{\prime}(0)=3\), then \(f^{\prime}(5)\) is
117257
If \(\boldsymbol{f}: \mathbf{R} \rightarrow \mathbf{R}\) is defined as \(f(x)=\left(2020-x^{2019}\right)^{1 / 2019}, \forall x \in \mathrm{R}\),find (fo fo fof) \(\left(\frac{2019}{2020}\right)\)
1 1
2 0
3 \(\frac{2019}{2020}\)
4 \(\frac{2020}{2019}\)
Explanation:
C The given real function \(\mathrm{f}(\mathrm{x})=\left(2020-\mathrm{x}^{2019}\right)^{\frac{1}{2019}}\) \(\therefore \quad \text { fof }(\mathrm{x})=\left[2020-\mathrm{f}(\mathrm{x})^{2019}\right]^{\frac{1}{2019}}\) \(=\left[2020-\left\{\left(2020-\mathrm{x}^{2019}\right)^{\frac{1}{2019}}\right\}^{2019}\right]^{\frac{1}{2019}}\) \(=\left[2020-2020+\mathrm{x}^{2019}\right]^{\frac{1}{2019}}=\left[\mathrm{x}^{2019}\right]^{\frac{1}{2019}}=\mathrm{x}\) Hence, \((\) fo fo fof \()\left(\frac{2019}{2020}\right)=\frac{2019}{2020}\)
Shift-I
Sets, Relation and Function
117258
If \(f(x)=x-\frac{1}{x}, x \neq 0\), then \(3 f(x)=\)
C Given that, \(f(x)=x-\frac{1}{x}\) \(\therefore(\mathrm{f}(\mathrm{x}))^3=\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^3=\mathrm{x}^3-\frac{1}{\mathrm{x}^3}-3(\mathrm{x})\left(\frac{1}{\mathrm{x}}\right)\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)\) \((\mathrm{f}(\mathrm{x}))^3=\mathrm{f}\left(\mathrm{x}^3\right)-3\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)\) \((\mathrm{f}(\mathrm{x}))^3=\mathrm{f}\left(\mathrm{x}^3\right)-3 \mathrm{f}(\mathrm{x})\) \(3 \mathrm{f}(\mathrm{x})=\mathrm{f}\left(\mathrm{x}^3\right)-[\mathrm{f}(\mathrm{x})]^3\)
Shift-II
Sets, Relation and Function
117259
For \(x \in\left(0, \frac{3}{2}\right)\) let \(f(x)=\sqrt{x} g(x)=\tan x\) and \(h(x)=\frac{1-x^2}{1+x^2}\). If \(\phi(x)=((h o f) o g)(x)\), then \(\phi\) \(\left(\frac{\pi}{3}\right)\) is equal to
117256
If for all values of \(\mathrm{x}\) and \(\mathrm{y}, f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) f(\mathrm{y})\) and \(f(5)=2, f^{\prime}(0)=3\), then \(f^{\prime}(5)\) is
117257
If \(\boldsymbol{f}: \mathbf{R} \rightarrow \mathbf{R}\) is defined as \(f(x)=\left(2020-x^{2019}\right)^{1 / 2019}, \forall x \in \mathrm{R}\),find (fo fo fof) \(\left(\frac{2019}{2020}\right)\)
1 1
2 0
3 \(\frac{2019}{2020}\)
4 \(\frac{2020}{2019}\)
Explanation:
C The given real function \(\mathrm{f}(\mathrm{x})=\left(2020-\mathrm{x}^{2019}\right)^{\frac{1}{2019}}\) \(\therefore \quad \text { fof }(\mathrm{x})=\left[2020-\mathrm{f}(\mathrm{x})^{2019}\right]^{\frac{1}{2019}}\) \(=\left[2020-\left\{\left(2020-\mathrm{x}^{2019}\right)^{\frac{1}{2019}}\right\}^{2019}\right]^{\frac{1}{2019}}\) \(=\left[2020-2020+\mathrm{x}^{2019}\right]^{\frac{1}{2019}}=\left[\mathrm{x}^{2019}\right]^{\frac{1}{2019}}=\mathrm{x}\) Hence, \((\) fo fo fof \()\left(\frac{2019}{2020}\right)=\frac{2019}{2020}\)
Shift-I
Sets, Relation and Function
117258
If \(f(x)=x-\frac{1}{x}, x \neq 0\), then \(3 f(x)=\)
C Given that, \(f(x)=x-\frac{1}{x}\) \(\therefore(\mathrm{f}(\mathrm{x}))^3=\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^3=\mathrm{x}^3-\frac{1}{\mathrm{x}^3}-3(\mathrm{x})\left(\frac{1}{\mathrm{x}}\right)\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)\) \((\mathrm{f}(\mathrm{x}))^3=\mathrm{f}\left(\mathrm{x}^3\right)-3\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)\) \((\mathrm{f}(\mathrm{x}))^3=\mathrm{f}\left(\mathrm{x}^3\right)-3 \mathrm{f}(\mathrm{x})\) \(3 \mathrm{f}(\mathrm{x})=\mathrm{f}\left(\mathrm{x}^3\right)-[\mathrm{f}(\mathrm{x})]^3\)
Shift-II
Sets, Relation and Function
117259
For \(x \in\left(0, \frac{3}{2}\right)\) let \(f(x)=\sqrt{x} g(x)=\tan x\) and \(h(x)=\frac{1-x^2}{1+x^2}\). If \(\phi(x)=((h o f) o g)(x)\), then \(\phi\) \(\left(\frac{\pi}{3}\right)\) is equal to
