85274
The function \(f(x)=\log x-\frac{2 x}{x+2}\) is increasing for all
1 \(x \in(0, \infty)\)
2 \(x \in(-\infty, 1)\)
3 \(x \in(-1, \infty)\)
4 \(x \in(-\infty, 0)\)
Explanation:
(A) : Given, \(f(x)=\log x-\frac{2 x}{x+2}\) On differentiating both sides w. r. t. to \(x\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}-\left[\frac{2 \mathrm{x}-(\mathrm{x}+2) \cdot 2}{(\mathrm{x}+2)^{2}}\right]\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}-\left[\frac{2 \mathrm{x}-2 \mathrm{x}-4}{(\mathrm{x}+2)^{2}}\right]\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}+\frac{4}{(\mathrm{x}+2)^{2}}\) \(\because \quad f^{\prime}(x) \geq 0\) \(\frac{1}{x}+\frac{4}{(x+2)^{2}} \geq 0=\frac{(x+2)^{2}+4 x}{x(x+2)^{2}} \geq 0\) \(\frac{\mathrm{x}^{2}+8 \mathrm{x}+4}{\mathrm{x}(\mathrm{x}+2)^{2}} \geq 0 \Rightarrow \quad \therefore \quad \mathrm{x} \in(0, \infty)\)
MHT CET-2020
Application of Derivatives
85275
For every value of \(x\), the function \(f(x)=\frac{1}{a^{x}}, a>0\) is
1 constant
2 increasing
3 neither increasing nor decreasing
4 decreasing
Explanation:
(D) : Given function, \(f(x)=\frac{1}{a^{x}}\) \(f(x)=a^{-x}, \quad a>0\) \(f^{\prime}(x)=-a^{-x} \log a\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{-\log \mathrm{a}}{\mathrm{a}^{\mathrm{x}}}\) \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0 \quad \forall \mathrm{x} \in \mathrm{R}, \mathrm{a}>0\) \(\therefore \mathrm{f}(\mathrm{x})\) is decreasing for every value of \(\mathrm{x}\).
MHT CET-2020
Application of Derivatives
85276
The function \(f(x)=(x+2) e^{-x}\) is
1 decreasing in \((-\infty,-1)\) and increasing in \((-1, \infty)\)
2 decreasing for all \(x\)
3 increasing in \((-\infty,-1)\) and decreasing in \((-1, \infty)\)
4 increasing for all \(\mathrm{x}\)
Explanation:
(C) : Given, \(f(x)=(x+2) e^{-x} \Rightarrow f^{\prime}(x)=e^{-x}-e^{-x}(x+2)\) \(f^{\prime}(x)=-e^{-x}[x+1]\) For increasing function- \(-\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)>0\) or \(\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)\lt 0\) \(\mathrm{e}^{-\mathrm{x}}>0\) or \((\mathrm{x}+1)\lt 0\) \(\mathrm{x} \in(-\infty, \infty)\) and \(\mathrm{x} \in(-\infty,-1)\) Therefore, \(\mathrm{x} \in(-\infty,-1)\) The function is increasing in \((-\infty,-1)\) For the decreasing function- \(-\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)\lt 0\) or \(\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)>0, \mathrm{x} \in(-1, \infty)\) Hence, The function is decreasing in \((-1, \infty)\)
MHT CET-2020
Application of Derivatives
85277
The function \(f(x)=x^{3}-3 x\) is
1 decreasing in \((-\infty,-1) \cup(1, \infty)\) and increasing in \((-1,1)\)
2 decreasing in \((0, \infty)\) and increasing in \((-\infty, 0)\)
3 increasing in \((-\infty,-1) \cup(1, \infty)\) and decreasing in \((-1,1)\)
4 increasing in \((0, \infty)\) and decreasing in \((-\infty, 0)\)
Explanation:
(C): Given function, \(f(x)=x^{3}-3 x\) \(f^{\prime}(x)=3 x^{2}-3\) \(\mathrm{f}^{\prime}(\mathrm{x})=3(\mathrm{x}+1)(\mathrm{x}-1)\) Here, \(\mathrm{f}^{\prime}(\mathrm{x})>0\) in \((-\infty,-1) \cup(1, \infty)\) and \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) in \((-1,1)\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing in \((-\infty,-1) \cup(1, \infty)\) and decreasing in \((-1,1)\).
85274
The function \(f(x)=\log x-\frac{2 x}{x+2}\) is increasing for all
1 \(x \in(0, \infty)\)
2 \(x \in(-\infty, 1)\)
3 \(x \in(-1, \infty)\)
4 \(x \in(-\infty, 0)\)
Explanation:
(A) : Given, \(f(x)=\log x-\frac{2 x}{x+2}\) On differentiating both sides w. r. t. to \(x\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}-\left[\frac{2 \mathrm{x}-(\mathrm{x}+2) \cdot 2}{(\mathrm{x}+2)^{2}}\right]\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}-\left[\frac{2 \mathrm{x}-2 \mathrm{x}-4}{(\mathrm{x}+2)^{2}}\right]\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}+\frac{4}{(\mathrm{x}+2)^{2}}\) \(\because \quad f^{\prime}(x) \geq 0\) \(\frac{1}{x}+\frac{4}{(x+2)^{2}} \geq 0=\frac{(x+2)^{2}+4 x}{x(x+2)^{2}} \geq 0\) \(\frac{\mathrm{x}^{2}+8 \mathrm{x}+4}{\mathrm{x}(\mathrm{x}+2)^{2}} \geq 0 \Rightarrow \quad \therefore \quad \mathrm{x} \in(0, \infty)\)
MHT CET-2020
Application of Derivatives
85275
For every value of \(x\), the function \(f(x)=\frac{1}{a^{x}}, a>0\) is
1 constant
2 increasing
3 neither increasing nor decreasing
4 decreasing
Explanation:
(D) : Given function, \(f(x)=\frac{1}{a^{x}}\) \(f(x)=a^{-x}, \quad a>0\) \(f^{\prime}(x)=-a^{-x} \log a\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{-\log \mathrm{a}}{\mathrm{a}^{\mathrm{x}}}\) \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0 \quad \forall \mathrm{x} \in \mathrm{R}, \mathrm{a}>0\) \(\therefore \mathrm{f}(\mathrm{x})\) is decreasing for every value of \(\mathrm{x}\).
MHT CET-2020
Application of Derivatives
85276
The function \(f(x)=(x+2) e^{-x}\) is
1 decreasing in \((-\infty,-1)\) and increasing in \((-1, \infty)\)
2 decreasing for all \(x\)
3 increasing in \((-\infty,-1)\) and decreasing in \((-1, \infty)\)
4 increasing for all \(\mathrm{x}\)
Explanation:
(C) : Given, \(f(x)=(x+2) e^{-x} \Rightarrow f^{\prime}(x)=e^{-x}-e^{-x}(x+2)\) \(f^{\prime}(x)=-e^{-x}[x+1]\) For increasing function- \(-\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)>0\) or \(\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)\lt 0\) \(\mathrm{e}^{-\mathrm{x}}>0\) or \((\mathrm{x}+1)\lt 0\) \(\mathrm{x} \in(-\infty, \infty)\) and \(\mathrm{x} \in(-\infty,-1)\) Therefore, \(\mathrm{x} \in(-\infty,-1)\) The function is increasing in \((-\infty,-1)\) For the decreasing function- \(-\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)\lt 0\) or \(\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)>0, \mathrm{x} \in(-1, \infty)\) Hence, The function is decreasing in \((-1, \infty)\)
MHT CET-2020
Application of Derivatives
85277
The function \(f(x)=x^{3}-3 x\) is
1 decreasing in \((-\infty,-1) \cup(1, \infty)\) and increasing in \((-1,1)\)
2 decreasing in \((0, \infty)\) and increasing in \((-\infty, 0)\)
3 increasing in \((-\infty,-1) \cup(1, \infty)\) and decreasing in \((-1,1)\)
4 increasing in \((0, \infty)\) and decreasing in \((-\infty, 0)\)
Explanation:
(C): Given function, \(f(x)=x^{3}-3 x\) \(f^{\prime}(x)=3 x^{2}-3\) \(\mathrm{f}^{\prime}(\mathrm{x})=3(\mathrm{x}+1)(\mathrm{x}-1)\) Here, \(\mathrm{f}^{\prime}(\mathrm{x})>0\) in \((-\infty,-1) \cup(1, \infty)\) and \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) in \((-1,1)\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing in \((-\infty,-1) \cup(1, \infty)\) and decreasing in \((-1,1)\).
85274
The function \(f(x)=\log x-\frac{2 x}{x+2}\) is increasing for all
1 \(x \in(0, \infty)\)
2 \(x \in(-\infty, 1)\)
3 \(x \in(-1, \infty)\)
4 \(x \in(-\infty, 0)\)
Explanation:
(A) : Given, \(f(x)=\log x-\frac{2 x}{x+2}\) On differentiating both sides w. r. t. to \(x\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}-\left[\frac{2 \mathrm{x}-(\mathrm{x}+2) \cdot 2}{(\mathrm{x}+2)^{2}}\right]\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}-\left[\frac{2 \mathrm{x}-2 \mathrm{x}-4}{(\mathrm{x}+2)^{2}}\right]\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}+\frac{4}{(\mathrm{x}+2)^{2}}\) \(\because \quad f^{\prime}(x) \geq 0\) \(\frac{1}{x}+\frac{4}{(x+2)^{2}} \geq 0=\frac{(x+2)^{2}+4 x}{x(x+2)^{2}} \geq 0\) \(\frac{\mathrm{x}^{2}+8 \mathrm{x}+4}{\mathrm{x}(\mathrm{x}+2)^{2}} \geq 0 \Rightarrow \quad \therefore \quad \mathrm{x} \in(0, \infty)\)
MHT CET-2020
Application of Derivatives
85275
For every value of \(x\), the function \(f(x)=\frac{1}{a^{x}}, a>0\) is
1 constant
2 increasing
3 neither increasing nor decreasing
4 decreasing
Explanation:
(D) : Given function, \(f(x)=\frac{1}{a^{x}}\) \(f(x)=a^{-x}, \quad a>0\) \(f^{\prime}(x)=-a^{-x} \log a\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{-\log \mathrm{a}}{\mathrm{a}^{\mathrm{x}}}\) \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0 \quad \forall \mathrm{x} \in \mathrm{R}, \mathrm{a}>0\) \(\therefore \mathrm{f}(\mathrm{x})\) is decreasing for every value of \(\mathrm{x}\).
MHT CET-2020
Application of Derivatives
85276
The function \(f(x)=(x+2) e^{-x}\) is
1 decreasing in \((-\infty,-1)\) and increasing in \((-1, \infty)\)
2 decreasing for all \(x\)
3 increasing in \((-\infty,-1)\) and decreasing in \((-1, \infty)\)
4 increasing for all \(\mathrm{x}\)
Explanation:
(C) : Given, \(f(x)=(x+2) e^{-x} \Rightarrow f^{\prime}(x)=e^{-x}-e^{-x}(x+2)\) \(f^{\prime}(x)=-e^{-x}[x+1]\) For increasing function- \(-\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)>0\) or \(\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)\lt 0\) \(\mathrm{e}^{-\mathrm{x}}>0\) or \((\mathrm{x}+1)\lt 0\) \(\mathrm{x} \in(-\infty, \infty)\) and \(\mathrm{x} \in(-\infty,-1)\) Therefore, \(\mathrm{x} \in(-\infty,-1)\) The function is increasing in \((-\infty,-1)\) For the decreasing function- \(-\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)\lt 0\) or \(\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)>0, \mathrm{x} \in(-1, \infty)\) Hence, The function is decreasing in \((-1, \infty)\)
MHT CET-2020
Application of Derivatives
85277
The function \(f(x)=x^{3}-3 x\) is
1 decreasing in \((-\infty,-1) \cup(1, \infty)\) and increasing in \((-1,1)\)
2 decreasing in \((0, \infty)\) and increasing in \((-\infty, 0)\)
3 increasing in \((-\infty,-1) \cup(1, \infty)\) and decreasing in \((-1,1)\)
4 increasing in \((0, \infty)\) and decreasing in \((-\infty, 0)\)
Explanation:
(C): Given function, \(f(x)=x^{3}-3 x\) \(f^{\prime}(x)=3 x^{2}-3\) \(\mathrm{f}^{\prime}(\mathrm{x})=3(\mathrm{x}+1)(\mathrm{x}-1)\) Here, \(\mathrm{f}^{\prime}(\mathrm{x})>0\) in \((-\infty,-1) \cup(1, \infty)\) and \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) in \((-1,1)\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing in \((-\infty,-1) \cup(1, \infty)\) and decreasing in \((-1,1)\).
85274
The function \(f(x)=\log x-\frac{2 x}{x+2}\) is increasing for all
1 \(x \in(0, \infty)\)
2 \(x \in(-\infty, 1)\)
3 \(x \in(-1, \infty)\)
4 \(x \in(-\infty, 0)\)
Explanation:
(A) : Given, \(f(x)=\log x-\frac{2 x}{x+2}\) On differentiating both sides w. r. t. to \(x\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}-\left[\frac{2 \mathrm{x}-(\mathrm{x}+2) \cdot 2}{(\mathrm{x}+2)^{2}}\right]\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}-\left[\frac{2 \mathrm{x}-2 \mathrm{x}-4}{(\mathrm{x}+2)^{2}}\right]\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}}+\frac{4}{(\mathrm{x}+2)^{2}}\) \(\because \quad f^{\prime}(x) \geq 0\) \(\frac{1}{x}+\frac{4}{(x+2)^{2}} \geq 0=\frac{(x+2)^{2}+4 x}{x(x+2)^{2}} \geq 0\) \(\frac{\mathrm{x}^{2}+8 \mathrm{x}+4}{\mathrm{x}(\mathrm{x}+2)^{2}} \geq 0 \Rightarrow \quad \therefore \quad \mathrm{x} \in(0, \infty)\)
MHT CET-2020
Application of Derivatives
85275
For every value of \(x\), the function \(f(x)=\frac{1}{a^{x}}, a>0\) is
1 constant
2 increasing
3 neither increasing nor decreasing
4 decreasing
Explanation:
(D) : Given function, \(f(x)=\frac{1}{a^{x}}\) \(f(x)=a^{-x}, \quad a>0\) \(f^{\prime}(x)=-a^{-x} \log a\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{-\log \mathrm{a}}{\mathrm{a}^{\mathrm{x}}}\) \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0 \quad \forall \mathrm{x} \in \mathrm{R}, \mathrm{a}>0\) \(\therefore \mathrm{f}(\mathrm{x})\) is decreasing for every value of \(\mathrm{x}\).
MHT CET-2020
Application of Derivatives
85276
The function \(f(x)=(x+2) e^{-x}\) is
1 decreasing in \((-\infty,-1)\) and increasing in \((-1, \infty)\)
2 decreasing for all \(x\)
3 increasing in \((-\infty,-1)\) and decreasing in \((-1, \infty)\)
4 increasing for all \(\mathrm{x}\)
Explanation:
(C) : Given, \(f(x)=(x+2) e^{-x} \Rightarrow f^{\prime}(x)=e^{-x}-e^{-x}(x+2)\) \(f^{\prime}(x)=-e^{-x}[x+1]\) For increasing function- \(-\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)>0\) or \(\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)\lt 0\) \(\mathrm{e}^{-\mathrm{x}}>0\) or \((\mathrm{x}+1)\lt 0\) \(\mathrm{x} \in(-\infty, \infty)\) and \(\mathrm{x} \in(-\infty,-1)\) Therefore, \(\mathrm{x} \in(-\infty,-1)\) The function is increasing in \((-\infty,-1)\) For the decreasing function- \(-\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)\lt 0\) or \(\mathrm{e}^{-\mathrm{x}}(\mathrm{x}+1)>0, \mathrm{x} \in(-1, \infty)\) Hence, The function is decreasing in \((-1, \infty)\)
MHT CET-2020
Application of Derivatives
85277
The function \(f(x)=x^{3}-3 x\) is
1 decreasing in \((-\infty,-1) \cup(1, \infty)\) and increasing in \((-1,1)\)
2 decreasing in \((0, \infty)\) and increasing in \((-\infty, 0)\)
3 increasing in \((-\infty,-1) \cup(1, \infty)\) and decreasing in \((-1,1)\)
4 increasing in \((0, \infty)\) and decreasing in \((-\infty, 0)\)
Explanation:
(C): Given function, \(f(x)=x^{3}-3 x\) \(f^{\prime}(x)=3 x^{2}-3\) \(\mathrm{f}^{\prime}(\mathrm{x})=3(\mathrm{x}+1)(\mathrm{x}-1)\) Here, \(\mathrm{f}^{\prime}(\mathrm{x})>0\) in \((-\infty,-1) \cup(1, \infty)\) and \(\mathrm{f}^{\prime}(\mathrm{x})\lt 0\) in \((-1,1)\) \(\therefore \mathrm{f}(\mathrm{x})\) is increasing in \((-\infty,-1) \cup(1, \infty)\) and decreasing in \((-1,1)\).
