Maxima and Minima
« Previous Topic: Tangent and Normal Next Topic: Rolle's Theorem »
For More Updates join with Us
Application of Derivatives

85633 Maximum value of the function \(f(x)=\frac{x}{8}+\frac{2}{x}\) on the interval \([1,6]\) is

1 1
2 \(\frac{9}{8}\)
3 \(\frac{13}{12}\)
4 \(\frac{17}{8}\)
WB JEE-2012
Application of Derivatives

85634 Let \(f(x)=x^{3} e^{-3 x}, x>0\). Then the maximum value of \(f(x)\) is

1 \(\mathrm{e}^{-3}\)
2 \(3 \mathrm{e}^{-3}\)
3 \(27 \mathrm{e}^{-9}\)
4 \(\infty\)
WB JEE-2011
Application of Derivatives

85635 The displacement of a particle at time \(t\) is \(x\), where \(x=t^{4}-k t^{3}\). If the velocity of the particle at time \(\mathbf{t}=\mathbf{2}\) is minimum, the

1 \(\mathrm{k}=4\)
2 \(k=-4\)
3 \(\mathrm{k}=8\)
4 \(\mathrm{k}=-8\)
WB JEE-2010
Application of Derivatives

85636 The point in the interval \([0,2 \pi]\), where \(f(x)=e^{x}\) sin \(x\) has maximum slope, \(i\)

1 \(\pi / 4\)
2 \(\pi / 2\)
3 \(\pi\)
4 \(3 \pi / 2\)
WB JEE-2010
NEET DIGITAL LIBRARY
Application of Derivatives

85633 Maximum value of the function \(f(x)=\frac{x}{8}+\frac{2}{x}\) on the interval \([1,6]\) is

1 1
2 \(\frac{9}{8}\)
3 \(\frac{13}{12}\)
4 \(\frac{17}{8}\)
WB JEE-2012
Application of Derivatives

85634 Let \(f(x)=x^{3} e^{-3 x}, x>0\). Then the maximum value of \(f(x)\) is

1 \(\mathrm{e}^{-3}\)
2 \(3 \mathrm{e}^{-3}\)
3 \(27 \mathrm{e}^{-9}\)
4 \(\infty\)
WB JEE-2011
Application of Derivatives

85635 The displacement of a particle at time \(t\) is \(x\), where \(x=t^{4}-k t^{3}\). If the velocity of the particle at time \(\mathbf{t}=\mathbf{2}\) is minimum, the

1 \(\mathrm{k}=4\)
2 \(k=-4\)
3 \(\mathrm{k}=8\)
4 \(\mathrm{k}=-8\)
WB JEE-2010
Application of Derivatives

85636 The point in the interval \([0,2 \pi]\), where \(f(x)=e^{x}\) sin \(x\) has maximum slope, \(i\)

1 \(\pi / 4\)
2 \(\pi / 2\)
3 \(\pi\)
4 \(3 \pi / 2\)
WB JEE-2010
Join With Us for More Updates
Application of Derivatives

85633 Maximum value of the function \(f(x)=\frac{x}{8}+\frac{2}{x}\) on the interval \([1,6]\) is

1 1
2 \(\frac{9}{8}\)
3 \(\frac{13}{12}\)
4 \(\frac{17}{8}\)
WB JEE-2012
Application of Derivatives

85634 Let \(f(x)=x^{3} e^{-3 x}, x>0\). Then the maximum value of \(f(x)\) is

1 \(\mathrm{e}^{-3}\)
2 \(3 \mathrm{e}^{-3}\)
3 \(27 \mathrm{e}^{-9}\)
4 \(\infty\)
WB JEE-2011
Application of Derivatives

85635 The displacement of a particle at time \(t\) is \(x\), where \(x=t^{4}-k t^{3}\). If the velocity of the particle at time \(\mathbf{t}=\mathbf{2}\) is minimum, the

1 \(\mathrm{k}=4\)
2 \(k=-4\)
3 \(\mathrm{k}=8\)
4 \(\mathrm{k}=-8\)
WB JEE-2010
Application of Derivatives

85636 The point in the interval \([0,2 \pi]\), where \(f(x)=e^{x}\) sin \(x\) has maximum slope, \(i\)

1 \(\pi / 4\)
2 \(\pi / 2\)
3 \(\pi\)
4 \(3 \pi / 2\)
WB JEE-2010
For More Updates join with Us
Application of Derivatives

85633 Maximum value of the function \(f(x)=\frac{x}{8}+\frac{2}{x}\) on the interval \([1,6]\) is

1 1
2 \(\frac{9}{8}\)
3 \(\frac{13}{12}\)
4 \(\frac{17}{8}\)
WB JEE-2012
Application of Derivatives

85634 Let \(f(x)=x^{3} e^{-3 x}, x>0\). Then the maximum value of \(f(x)\) is

1 \(\mathrm{e}^{-3}\)
2 \(3 \mathrm{e}^{-3}\)
3 \(27 \mathrm{e}^{-9}\)
4 \(\infty\)
WB JEE-2011
Application of Derivatives

85635 The displacement of a particle at time \(t\) is \(x\), where \(x=t^{4}-k t^{3}\). If the velocity of the particle at time \(\mathbf{t}=\mathbf{2}\) is minimum, the

1 \(\mathrm{k}=4\)
2 \(k=-4\)
3 \(\mathrm{k}=8\)
4 \(\mathrm{k}=-8\)
WB JEE-2010
Application of Derivatives

85636 The point in the interval \([0,2 \pi]\), where \(f(x)=e^{x}\) sin \(x\) has maximum slope, \(i\)

1 \(\pi / 4\)
2 \(\pi / 2\)
3 \(\pi\)
4 \(3 \pi / 2\)
WB JEE-2010