85660
The maximum value of \(f(x)=\frac{x}{4+x+x^{2}}\) on \([-1,1]\) i
1 \(-\frac{1}{3}\)
2 \(-\frac{1}{4}\)
3 \(\frac{1}{5}\)
4 \(\frac{1}{6}\)
Explanation:
(D) : \(f(x)=\frac{x}{4+x+x^{2}}, x \in[-1,1]\) On differentiating w.r.t \(x\), we get- \(f^{\prime}(x)= \frac{\left(4+x+x^{2}\right) \frac{d}{d x} x-x \frac{d}{d x}\left(4+x+x^{2}\right)}{\left(4+x+x^{2}\right)^{2}}\) \(=\frac{4+x+x^{2}-x(1+2 x)}{\left(4+x+x^{2}\right)^{2}}=\frac{4-x^{2}}{\left(4+x+x^{2}\right)^{2}}\) For maxima or minima, \(f^{\prime}(x)=0\) \(\frac{4-x^{2}}{\left(4+x+x^{2}\right)^{2}}=0\) \(4-x^{2}=0\) \(x= \pm 2[-1,1]\) The value of \(f(x)\) at extreme points are given by \(f(1)=\frac{1}{4+1+1^{2}}=\frac{1}{6}\) \(f(-1)=\frac{-1}{4-1+(-1)^{2}}=\frac{-1}{4}\) Thus, \(\frac{1}{6}\) is the maximum value.
UPSEE-2007
Application of Derivatives
85661
If \(y=3^{x-1}+3^{-x-1}\) (x real), then the least value of \(y\) is:
1 2
2 6
3 \(2 / 3\)
4 None of these
Explanation:
(C) : \(y=3^{x-1}+3^{-x-1}\) We know that, Arithmetic mean (A.M) \(\geq\) Geometric mean (G.M) \(\frac{3^{\mathrm{x}-1}+3^{-\mathrm{x}-1}}{2} \geq \sqrt{3^{\mathrm{x}-1} \times 3^{-\mathrm{x}-1}}\) \(\frac{\mathrm{y}}{2} \geq \sqrt{3^{-2}}\) \(\frac{\mathrm{y}}{2} \geq \frac{1}{3} \Rightarrow \mathrm{y} \geq \frac{2}{3}\) So, the least value of \(\mathrm{y}\) is \(\frac{2}{3}\).
UPSEE-2005
Application of Derivatives
85662
The maximum value of \(x^{1 / x}\) is
1 \(1 / \mathrm{e}^{\mathrm{e}}\)
2 e
3 \(\mathrm{e}^{1 / \mathrm{e}}\)
4 \(1 / \mathrm{e}\)
Explanation:
(C) : \(y=x^{1 / x}\) Taking \(\log\) on both side \(\log \mathrm{y}=\frac{1}{\mathrm{x}} \log \mathrm{x}\) Differentiate both side with respect to \(\mathrm{x}\), \(\frac{1}{y} \frac{d y}{d x}=\frac{1}{x^{2}}-\frac{\log x}{x^{2}}\) \(\frac{d y}{d x}=y\left(\frac{1}{x^{2}}-\frac{\log x}{x^{2}}\right)\) Now, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(\mathrm{y}\left(\frac{1-\log \mathrm{x}}{\mathrm{x}^{2}}\right)=0\) \(\log \mathrm{x}=1\) \(\mathrm{x}=\mathrm{e}\) \(y=e^{1 / e}\) Maximum value at \(\mathrm{x}=\mathrm{e}\)
85660
The maximum value of \(f(x)=\frac{x}{4+x+x^{2}}\) on \([-1,1]\) i
1 \(-\frac{1}{3}\)
2 \(-\frac{1}{4}\)
3 \(\frac{1}{5}\)
4 \(\frac{1}{6}\)
Explanation:
(D) : \(f(x)=\frac{x}{4+x+x^{2}}, x \in[-1,1]\) On differentiating w.r.t \(x\), we get- \(f^{\prime}(x)= \frac{\left(4+x+x^{2}\right) \frac{d}{d x} x-x \frac{d}{d x}\left(4+x+x^{2}\right)}{\left(4+x+x^{2}\right)^{2}}\) \(=\frac{4+x+x^{2}-x(1+2 x)}{\left(4+x+x^{2}\right)^{2}}=\frac{4-x^{2}}{\left(4+x+x^{2}\right)^{2}}\) For maxima or minima, \(f^{\prime}(x)=0\) \(\frac{4-x^{2}}{\left(4+x+x^{2}\right)^{2}}=0\) \(4-x^{2}=0\) \(x= \pm 2[-1,1]\) The value of \(f(x)\) at extreme points are given by \(f(1)=\frac{1}{4+1+1^{2}}=\frac{1}{6}\) \(f(-1)=\frac{-1}{4-1+(-1)^{2}}=\frac{-1}{4}\) Thus, \(\frac{1}{6}\) is the maximum value.
UPSEE-2007
Application of Derivatives
85661
If \(y=3^{x-1}+3^{-x-1}\) (x real), then the least value of \(y\) is:
1 2
2 6
3 \(2 / 3\)
4 None of these
Explanation:
(C) : \(y=3^{x-1}+3^{-x-1}\) We know that, Arithmetic mean (A.M) \(\geq\) Geometric mean (G.M) \(\frac{3^{\mathrm{x}-1}+3^{-\mathrm{x}-1}}{2} \geq \sqrt{3^{\mathrm{x}-1} \times 3^{-\mathrm{x}-1}}\) \(\frac{\mathrm{y}}{2} \geq \sqrt{3^{-2}}\) \(\frac{\mathrm{y}}{2} \geq \frac{1}{3} \Rightarrow \mathrm{y} \geq \frac{2}{3}\) So, the least value of \(\mathrm{y}\) is \(\frac{2}{3}\).
UPSEE-2005
Application of Derivatives
85662
The maximum value of \(x^{1 / x}\) is
1 \(1 / \mathrm{e}^{\mathrm{e}}\)
2 e
3 \(\mathrm{e}^{1 / \mathrm{e}}\)
4 \(1 / \mathrm{e}\)
Explanation:
(C) : \(y=x^{1 / x}\) Taking \(\log\) on both side \(\log \mathrm{y}=\frac{1}{\mathrm{x}} \log \mathrm{x}\) Differentiate both side with respect to \(\mathrm{x}\), \(\frac{1}{y} \frac{d y}{d x}=\frac{1}{x^{2}}-\frac{\log x}{x^{2}}\) \(\frac{d y}{d x}=y\left(\frac{1}{x^{2}}-\frac{\log x}{x^{2}}\right)\) Now, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(\mathrm{y}\left(\frac{1-\log \mathrm{x}}{\mathrm{x}^{2}}\right)=0\) \(\log \mathrm{x}=1\) \(\mathrm{x}=\mathrm{e}\) \(y=e^{1 / e}\) Maximum value at \(\mathrm{x}=\mathrm{e}\)
85660
The maximum value of \(f(x)=\frac{x}{4+x+x^{2}}\) on \([-1,1]\) i
1 \(-\frac{1}{3}\)
2 \(-\frac{1}{4}\)
3 \(\frac{1}{5}\)
4 \(\frac{1}{6}\)
Explanation:
(D) : \(f(x)=\frac{x}{4+x+x^{2}}, x \in[-1,1]\) On differentiating w.r.t \(x\), we get- \(f^{\prime}(x)= \frac{\left(4+x+x^{2}\right) \frac{d}{d x} x-x \frac{d}{d x}\left(4+x+x^{2}\right)}{\left(4+x+x^{2}\right)^{2}}\) \(=\frac{4+x+x^{2}-x(1+2 x)}{\left(4+x+x^{2}\right)^{2}}=\frac{4-x^{2}}{\left(4+x+x^{2}\right)^{2}}\) For maxima or minima, \(f^{\prime}(x)=0\) \(\frac{4-x^{2}}{\left(4+x+x^{2}\right)^{2}}=0\) \(4-x^{2}=0\) \(x= \pm 2[-1,1]\) The value of \(f(x)\) at extreme points are given by \(f(1)=\frac{1}{4+1+1^{2}}=\frac{1}{6}\) \(f(-1)=\frac{-1}{4-1+(-1)^{2}}=\frac{-1}{4}\) Thus, \(\frac{1}{6}\) is the maximum value.
UPSEE-2007
Application of Derivatives
85661
If \(y=3^{x-1}+3^{-x-1}\) (x real), then the least value of \(y\) is:
1 2
2 6
3 \(2 / 3\)
4 None of these
Explanation:
(C) : \(y=3^{x-1}+3^{-x-1}\) We know that, Arithmetic mean (A.M) \(\geq\) Geometric mean (G.M) \(\frac{3^{\mathrm{x}-1}+3^{-\mathrm{x}-1}}{2} \geq \sqrt{3^{\mathrm{x}-1} \times 3^{-\mathrm{x}-1}}\) \(\frac{\mathrm{y}}{2} \geq \sqrt{3^{-2}}\) \(\frac{\mathrm{y}}{2} \geq \frac{1}{3} \Rightarrow \mathrm{y} \geq \frac{2}{3}\) So, the least value of \(\mathrm{y}\) is \(\frac{2}{3}\).
UPSEE-2005
Application of Derivatives
85662
The maximum value of \(x^{1 / x}\) is
1 \(1 / \mathrm{e}^{\mathrm{e}}\)
2 e
3 \(\mathrm{e}^{1 / \mathrm{e}}\)
4 \(1 / \mathrm{e}\)
Explanation:
(C) : \(y=x^{1 / x}\) Taking \(\log\) on both side \(\log \mathrm{y}=\frac{1}{\mathrm{x}} \log \mathrm{x}\) Differentiate both side with respect to \(\mathrm{x}\), \(\frac{1}{y} \frac{d y}{d x}=\frac{1}{x^{2}}-\frac{\log x}{x^{2}}\) \(\frac{d y}{d x}=y\left(\frac{1}{x^{2}}-\frac{\log x}{x^{2}}\right)\) Now, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(\mathrm{y}\left(\frac{1-\log \mathrm{x}}{\mathrm{x}^{2}}\right)=0\) \(\log \mathrm{x}=1\) \(\mathrm{x}=\mathrm{e}\) \(y=e^{1 / e}\) Maximum value at \(\mathrm{x}=\mathrm{e}\)
85660
The maximum value of \(f(x)=\frac{x}{4+x+x^{2}}\) on \([-1,1]\) i
1 \(-\frac{1}{3}\)
2 \(-\frac{1}{4}\)
3 \(\frac{1}{5}\)
4 \(\frac{1}{6}\)
Explanation:
(D) : \(f(x)=\frac{x}{4+x+x^{2}}, x \in[-1,1]\) On differentiating w.r.t \(x\), we get- \(f^{\prime}(x)= \frac{\left(4+x+x^{2}\right) \frac{d}{d x} x-x \frac{d}{d x}\left(4+x+x^{2}\right)}{\left(4+x+x^{2}\right)^{2}}\) \(=\frac{4+x+x^{2}-x(1+2 x)}{\left(4+x+x^{2}\right)^{2}}=\frac{4-x^{2}}{\left(4+x+x^{2}\right)^{2}}\) For maxima or minima, \(f^{\prime}(x)=0\) \(\frac{4-x^{2}}{\left(4+x+x^{2}\right)^{2}}=0\) \(4-x^{2}=0\) \(x= \pm 2[-1,1]\) The value of \(f(x)\) at extreme points are given by \(f(1)=\frac{1}{4+1+1^{2}}=\frac{1}{6}\) \(f(-1)=\frac{-1}{4-1+(-1)^{2}}=\frac{-1}{4}\) Thus, \(\frac{1}{6}\) is the maximum value.
UPSEE-2007
Application of Derivatives
85661
If \(y=3^{x-1}+3^{-x-1}\) (x real), then the least value of \(y\) is:
1 2
2 6
3 \(2 / 3\)
4 None of these
Explanation:
(C) : \(y=3^{x-1}+3^{-x-1}\) We know that, Arithmetic mean (A.M) \(\geq\) Geometric mean (G.M) \(\frac{3^{\mathrm{x}-1}+3^{-\mathrm{x}-1}}{2} \geq \sqrt{3^{\mathrm{x}-1} \times 3^{-\mathrm{x}-1}}\) \(\frac{\mathrm{y}}{2} \geq \sqrt{3^{-2}}\) \(\frac{\mathrm{y}}{2} \geq \frac{1}{3} \Rightarrow \mathrm{y} \geq \frac{2}{3}\) So, the least value of \(\mathrm{y}\) is \(\frac{2}{3}\).
UPSEE-2005
Application of Derivatives
85662
The maximum value of \(x^{1 / x}\) is
1 \(1 / \mathrm{e}^{\mathrm{e}}\)
2 e
3 \(\mathrm{e}^{1 / \mathrm{e}}\)
4 \(1 / \mathrm{e}\)
Explanation:
(C) : \(y=x^{1 / x}\) Taking \(\log\) on both side \(\log \mathrm{y}=\frac{1}{\mathrm{x}} \log \mathrm{x}\) Differentiate both side with respect to \(\mathrm{x}\), \(\frac{1}{y} \frac{d y}{d x}=\frac{1}{x^{2}}-\frac{\log x}{x^{2}}\) \(\frac{d y}{d x}=y\left(\frac{1}{x^{2}}-\frac{\log x}{x^{2}}\right)\) Now, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(\mathrm{y}\left(\frac{1-\log \mathrm{x}}{\mathrm{x}^{2}}\right)=0\) \(\log \mathrm{x}=1\) \(\mathrm{x}=\mathrm{e}\) \(y=e^{1 / e}\) Maximum value at \(\mathrm{x}=\mathrm{e}\)
85660
The maximum value of \(f(x)=\frac{x}{4+x+x^{2}}\) on \([-1,1]\) i
1 \(-\frac{1}{3}\)
2 \(-\frac{1}{4}\)
3 \(\frac{1}{5}\)
4 \(\frac{1}{6}\)
Explanation:
(D) : \(f(x)=\frac{x}{4+x+x^{2}}, x \in[-1,1]\) On differentiating w.r.t \(x\), we get- \(f^{\prime}(x)= \frac{\left(4+x+x^{2}\right) \frac{d}{d x} x-x \frac{d}{d x}\left(4+x+x^{2}\right)}{\left(4+x+x^{2}\right)^{2}}\) \(=\frac{4+x+x^{2}-x(1+2 x)}{\left(4+x+x^{2}\right)^{2}}=\frac{4-x^{2}}{\left(4+x+x^{2}\right)^{2}}\) For maxima or minima, \(f^{\prime}(x)=0\) \(\frac{4-x^{2}}{\left(4+x+x^{2}\right)^{2}}=0\) \(4-x^{2}=0\) \(x= \pm 2[-1,1]\) The value of \(f(x)\) at extreme points are given by \(f(1)=\frac{1}{4+1+1^{2}}=\frac{1}{6}\) \(f(-1)=\frac{-1}{4-1+(-1)^{2}}=\frac{-1}{4}\) Thus, \(\frac{1}{6}\) is the maximum value.
UPSEE-2007
Application of Derivatives
85661
If \(y=3^{x-1}+3^{-x-1}\) (x real), then the least value of \(y\) is:
1 2
2 6
3 \(2 / 3\)
4 None of these
Explanation:
(C) : \(y=3^{x-1}+3^{-x-1}\) We know that, Arithmetic mean (A.M) \(\geq\) Geometric mean (G.M) \(\frac{3^{\mathrm{x}-1}+3^{-\mathrm{x}-1}}{2} \geq \sqrt{3^{\mathrm{x}-1} \times 3^{-\mathrm{x}-1}}\) \(\frac{\mathrm{y}}{2} \geq \sqrt{3^{-2}}\) \(\frac{\mathrm{y}}{2} \geq \frac{1}{3} \Rightarrow \mathrm{y} \geq \frac{2}{3}\) So, the least value of \(\mathrm{y}\) is \(\frac{2}{3}\).
UPSEE-2005
Application of Derivatives
85662
The maximum value of \(x^{1 / x}\) is
1 \(1 / \mathrm{e}^{\mathrm{e}}\)
2 e
3 \(\mathrm{e}^{1 / \mathrm{e}}\)
4 \(1 / \mathrm{e}\)
Explanation:
(C) : \(y=x^{1 / x}\) Taking \(\log\) on both side \(\log \mathrm{y}=\frac{1}{\mathrm{x}} \log \mathrm{x}\) Differentiate both side with respect to \(\mathrm{x}\), \(\frac{1}{y} \frac{d y}{d x}=\frac{1}{x^{2}}-\frac{\log x}{x^{2}}\) \(\frac{d y}{d x}=y\left(\frac{1}{x^{2}}-\frac{\log x}{x^{2}}\right)\) Now, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(\mathrm{y}\left(\frac{1-\log \mathrm{x}}{\mathrm{x}^{2}}\right)=0\) \(\log \mathrm{x}=1\) \(\mathrm{x}=\mathrm{e}\) \(y=e^{1 / e}\) Maximum value at \(\mathrm{x}=\mathrm{e}\)