(D) : Let, \(f(x)=\log (1+x)-\frac{x}{1+x}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{1+\mathrm{x}}-\frac{(1+\mathrm{x}) \times 1-\mathrm{x} \cdot 1}{(1+\mathrm{x})^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{1+\mathrm{x}}-\frac{1}{(1+\mathrm{x})^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x}}{(1+\mathrm{x})^{2}}\) Which is positive \(\quad[\because \mathrm{x}>0]\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is monotonic increasing, when \(\mathrm{x}>0\) \(\mathrm{f}(\mathrm{x})>\mathrm{f}(0)\) Now, \(\mathrm{f}(0)=\log 1-0=0\) \(\therefore \quad \mathrm{f}(\mathrm{x})>0\) \(\log (1+x)-\frac{x}{1+x}>0\) \(\frac{\mathrm{x}}{1+\mathrm{x}}\lt \log (1+\mathrm{x})\) Also, for \(\mathrm{x}>0\) \(\mathrm{x}^{2}>0\) \(\mathrm{x}^{2}+\mathrm{x}>\mathrm{x}\) \(\mathrm{x}(\mathrm{x}+1)>\mathrm{x}\) \(\mathrm{x}>\frac{\mathrm{x}}{\mathrm{x}+1}\) From equation (i) and (ii), we have \(\frac{\mathrm{x}}{\mathrm{x}+1}\lt \log (1+\mathrm{x})\lt \mathrm{x} \quad[\because \log (1+\mathrm{x})\lt \mathrm{x} \text { for } \mathrm{x}>0]\)
VITEEE-2015
Application of Derivatives
85295
The function \(y=\sqrt{2 x-x^{2}}\)
1 increase in \((0,1)\) but decreases in \((1,2)\)
2 decreases in \((0,2)\)
3 increases in \((1,2)\) but decreases in \((0,1)\)
4 increases in \((0,2)\)
Explanation:
(A) : Given, \(y=\sqrt{2 x-x^{2}}\) \(y=\sqrt{1-(x-1)^{2}}\) \(\frac{d y}{d x}=\frac{1-x}{\sqrt{1-(x-1)^{2}}} \quad\left\{\begin{array}{l}>0 \text { for } 0\lt x\lt 1 \\ \lt 0 \text { for } x \in(1,2)\end{array}\right.\) So, function increases in \((0,1)\) and decreases in \((1,2)\)
UPSEE-2017
Application of Derivatives
85296
For which interval, the given function \(f(x)=-\) \(2 x^{3}-9 x^{2}-12 x+1\) is decreasing?
1 \((-2, \infty)\)
2 \((-2,-1)\)
3 \((-\infty,-1)\)
4 \((-\infty,-2)\) and \((-1, \infty)\)
Explanation:
(D): Given, \(f(x)=-2 x^{3}-9 x^{2}-12 x+1\) On differentiating both sides w.r.t. to \(\mathrm{x}\) we get - \(f^{\prime}(x)=-6 x^{2}-18 x-12\) Since \(\mathrm{f}(\mathrm{x})\) is decreasing \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})\lt 0\) \(-6 x^{2}-18 x-12\lt 0\) \(x^{2}+3 x+2>0\) \((x+1)(x+2)>0\) \(\therefore\) The interval is \((-\infty,-2)\) and \((-1, \infty)\).
(D) : Let, \(f(x)=\log (1+x)-\frac{x}{1+x}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{1+\mathrm{x}}-\frac{(1+\mathrm{x}) \times 1-\mathrm{x} \cdot 1}{(1+\mathrm{x})^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{1+\mathrm{x}}-\frac{1}{(1+\mathrm{x})^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x}}{(1+\mathrm{x})^{2}}\) Which is positive \(\quad[\because \mathrm{x}>0]\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is monotonic increasing, when \(\mathrm{x}>0\) \(\mathrm{f}(\mathrm{x})>\mathrm{f}(0)\) Now, \(\mathrm{f}(0)=\log 1-0=0\) \(\therefore \quad \mathrm{f}(\mathrm{x})>0\) \(\log (1+x)-\frac{x}{1+x}>0\) \(\frac{\mathrm{x}}{1+\mathrm{x}}\lt \log (1+\mathrm{x})\) Also, for \(\mathrm{x}>0\) \(\mathrm{x}^{2}>0\) \(\mathrm{x}^{2}+\mathrm{x}>\mathrm{x}\) \(\mathrm{x}(\mathrm{x}+1)>\mathrm{x}\) \(\mathrm{x}>\frac{\mathrm{x}}{\mathrm{x}+1}\) From equation (i) and (ii), we have \(\frac{\mathrm{x}}{\mathrm{x}+1}\lt \log (1+\mathrm{x})\lt \mathrm{x} \quad[\because \log (1+\mathrm{x})\lt \mathrm{x} \text { for } \mathrm{x}>0]\)
VITEEE-2015
Application of Derivatives
85295
The function \(y=\sqrt{2 x-x^{2}}\)
1 increase in \((0,1)\) but decreases in \((1,2)\)
2 decreases in \((0,2)\)
3 increases in \((1,2)\) but decreases in \((0,1)\)
4 increases in \((0,2)\)
Explanation:
(A) : Given, \(y=\sqrt{2 x-x^{2}}\) \(y=\sqrt{1-(x-1)^{2}}\) \(\frac{d y}{d x}=\frac{1-x}{\sqrt{1-(x-1)^{2}}} \quad\left\{\begin{array}{l}>0 \text { for } 0\lt x\lt 1 \\ \lt 0 \text { for } x \in(1,2)\end{array}\right.\) So, function increases in \((0,1)\) and decreases in \((1,2)\)
UPSEE-2017
Application of Derivatives
85296
For which interval, the given function \(f(x)=-\) \(2 x^{3}-9 x^{2}-12 x+1\) is decreasing?
1 \((-2, \infty)\)
2 \((-2,-1)\)
3 \((-\infty,-1)\)
4 \((-\infty,-2)\) and \((-1, \infty)\)
Explanation:
(D): Given, \(f(x)=-2 x^{3}-9 x^{2}-12 x+1\) On differentiating both sides w.r.t. to \(\mathrm{x}\) we get - \(f^{\prime}(x)=-6 x^{2}-18 x-12\) Since \(\mathrm{f}(\mathrm{x})\) is decreasing \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})\lt 0\) \(-6 x^{2}-18 x-12\lt 0\) \(x^{2}+3 x+2>0\) \((x+1)(x+2)>0\) \(\therefore\) The interval is \((-\infty,-2)\) and \((-1, \infty)\).
(D) : Let, \(f(x)=\log (1+x)-\frac{x}{1+x}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{1+\mathrm{x}}-\frac{(1+\mathrm{x}) \times 1-\mathrm{x} \cdot 1}{(1+\mathrm{x})^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{1+\mathrm{x}}-\frac{1}{(1+\mathrm{x})^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x}}{(1+\mathrm{x})^{2}}\) Which is positive \(\quad[\because \mathrm{x}>0]\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is monotonic increasing, when \(\mathrm{x}>0\) \(\mathrm{f}(\mathrm{x})>\mathrm{f}(0)\) Now, \(\mathrm{f}(0)=\log 1-0=0\) \(\therefore \quad \mathrm{f}(\mathrm{x})>0\) \(\log (1+x)-\frac{x}{1+x}>0\) \(\frac{\mathrm{x}}{1+\mathrm{x}}\lt \log (1+\mathrm{x})\) Also, for \(\mathrm{x}>0\) \(\mathrm{x}^{2}>0\) \(\mathrm{x}^{2}+\mathrm{x}>\mathrm{x}\) \(\mathrm{x}(\mathrm{x}+1)>\mathrm{x}\) \(\mathrm{x}>\frac{\mathrm{x}}{\mathrm{x}+1}\) From equation (i) and (ii), we have \(\frac{\mathrm{x}}{\mathrm{x}+1}\lt \log (1+\mathrm{x})\lt \mathrm{x} \quad[\because \log (1+\mathrm{x})\lt \mathrm{x} \text { for } \mathrm{x}>0]\)
VITEEE-2015
Application of Derivatives
85295
The function \(y=\sqrt{2 x-x^{2}}\)
1 increase in \((0,1)\) but decreases in \((1,2)\)
2 decreases in \((0,2)\)
3 increases in \((1,2)\) but decreases in \((0,1)\)
4 increases in \((0,2)\)
Explanation:
(A) : Given, \(y=\sqrt{2 x-x^{2}}\) \(y=\sqrt{1-(x-1)^{2}}\) \(\frac{d y}{d x}=\frac{1-x}{\sqrt{1-(x-1)^{2}}} \quad\left\{\begin{array}{l}>0 \text { for } 0\lt x\lt 1 \\ \lt 0 \text { for } x \in(1,2)\end{array}\right.\) So, function increases in \((0,1)\) and decreases in \((1,2)\)
UPSEE-2017
Application of Derivatives
85296
For which interval, the given function \(f(x)=-\) \(2 x^{3}-9 x^{2}-12 x+1\) is decreasing?
1 \((-2, \infty)\)
2 \((-2,-1)\)
3 \((-\infty,-1)\)
4 \((-\infty,-2)\) and \((-1, \infty)\)
Explanation:
(D): Given, \(f(x)=-2 x^{3}-9 x^{2}-12 x+1\) On differentiating both sides w.r.t. to \(\mathrm{x}\) we get - \(f^{\prime}(x)=-6 x^{2}-18 x-12\) Since \(\mathrm{f}(\mathrm{x})\) is decreasing \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})\lt 0\) \(-6 x^{2}-18 x-12\lt 0\) \(x^{2}+3 x+2>0\) \((x+1)(x+2)>0\) \(\therefore\) The interval is \((-\infty,-2)\) and \((-1, \infty)\).
(D) : Let, \(f(x)=\log (1+x)-\frac{x}{1+x}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{1+\mathrm{x}}-\frac{(1+\mathrm{x}) \times 1-\mathrm{x} \cdot 1}{(1+\mathrm{x})^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{1+\mathrm{x}}-\frac{1}{(1+\mathrm{x})^{2}}\) \(\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x}}{(1+\mathrm{x})^{2}}\) Which is positive \(\quad[\because \mathrm{x}>0]\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is monotonic increasing, when \(\mathrm{x}>0\) \(\mathrm{f}(\mathrm{x})>\mathrm{f}(0)\) Now, \(\mathrm{f}(0)=\log 1-0=0\) \(\therefore \quad \mathrm{f}(\mathrm{x})>0\) \(\log (1+x)-\frac{x}{1+x}>0\) \(\frac{\mathrm{x}}{1+\mathrm{x}}\lt \log (1+\mathrm{x})\) Also, for \(\mathrm{x}>0\) \(\mathrm{x}^{2}>0\) \(\mathrm{x}^{2}+\mathrm{x}>\mathrm{x}\) \(\mathrm{x}(\mathrm{x}+1)>\mathrm{x}\) \(\mathrm{x}>\frac{\mathrm{x}}{\mathrm{x}+1}\) From equation (i) and (ii), we have \(\frac{\mathrm{x}}{\mathrm{x}+1}\lt \log (1+\mathrm{x})\lt \mathrm{x} \quad[\because \log (1+\mathrm{x})\lt \mathrm{x} \text { for } \mathrm{x}>0]\)
VITEEE-2015
Application of Derivatives
85295
The function \(y=\sqrt{2 x-x^{2}}\)
1 increase in \((0,1)\) but decreases in \((1,2)\)
2 decreases in \((0,2)\)
3 increases in \((1,2)\) but decreases in \((0,1)\)
4 increases in \((0,2)\)
Explanation:
(A) : Given, \(y=\sqrt{2 x-x^{2}}\) \(y=\sqrt{1-(x-1)^{2}}\) \(\frac{d y}{d x}=\frac{1-x}{\sqrt{1-(x-1)^{2}}} \quad\left\{\begin{array}{l}>0 \text { for } 0\lt x\lt 1 \\ \lt 0 \text { for } x \in(1,2)\end{array}\right.\) So, function increases in \((0,1)\) and decreases in \((1,2)\)
UPSEE-2017
Application of Derivatives
85296
For which interval, the given function \(f(x)=-\) \(2 x^{3}-9 x^{2}-12 x+1\) is decreasing?
1 \((-2, \infty)\)
2 \((-2,-1)\)
3 \((-\infty,-1)\)
4 \((-\infty,-2)\) and \((-1, \infty)\)
Explanation:
(D): Given, \(f(x)=-2 x^{3}-9 x^{2}-12 x+1\) On differentiating both sides w.r.t. to \(\mathrm{x}\) we get - \(f^{\prime}(x)=-6 x^{2}-18 x-12\) Since \(\mathrm{f}(\mathrm{x})\) is decreasing \(\therefore \quad \mathrm{f}^{\prime}(\mathrm{x})\lt 0\) \(-6 x^{2}-18 x-12\lt 0\) \(x^{2}+3 x+2>0\) \((x+1)(x+2)>0\) \(\therefore\) The interval is \((-\infty,-2)\) and \((-1, \infty)\).
