85641
At \(t=0\), the function \(f(t)=\frac{\sin t}{t}\) has
1 a minimum
2 a discontinuity
3 a point of inflexion
4 a maximum
Explanation:
(D) : We have \(f(t)=\frac{\sin t}{t}\) At \(t=0\), we will check continuity of the function. \(\mathrm{L} \cdot \mathrm{H} \cdot \mathrm{L}=\mathrm{f}(0-\mathrm{h})\) \(\lim _{h \rightarrow 0} \frac{\sin (0-h)}{(0-h)} \Rightarrow=\lim _{h \rightarrow 0} \frac{-\sin h}{-h}=1\) \(\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{L}=\mathrm{f}(0+\mathrm{h})\) \(\lim _{h \rightarrow 0} \frac{\sin (0+h)}{(0+h)} \Rightarrow \lim _{h \rightarrow 0} \frac{\sin h}{h}=1\) and \(\mathrm{f}(0)=1\) \(\mathrm{L} \cdot \mathrm{H} \cdot \mathrm{L}=\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{L}=\mathrm{f}(0)\) So, the function is continuous at \(t=0\) Now, we check the function is maximum or minimum. \(\mathrm{f}^{\prime}(\mathrm{t})=\frac{1}{\mathrm{t}} \cos \mathrm{t}-\frac{1}{\mathrm{t}^{2}} \sin \mathrm{t}\) and \(\quad f^{\prime \prime}(t)=\frac{-1}{t} \sin t-\frac{1}{t^{2}} \cos \mathrm{t}-\frac{1}{t^{2}} \cos \mathrm{t}+\frac{2}{\mathrm{t}^{3}} \sin \mathrm{t}\) \(=\frac{-\sin t}{t}-\frac{2 \cos t}{t^{2}}+\frac{2 \sin t}{t^{3}}\) For maximum or minimum value of \(f(x)\), put \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\Rightarrow \frac{\cos \mathrm{t}}{\mathrm{t}} \frac{-\sin \mathrm{t}}{\mathrm{t}^{2}}=0 \Rightarrow \frac{\tan \mathrm{t}}{\mathrm{t}}=1\) Now \(\lim _{\mathrm{t} \rightarrow 0} \mathrm{f}^{\prime \prime}(\mathrm{t})\) \(=\lim _{t \rightarrow 0}\left(-\frac{\sin t}{t}\right)-2 \lim _{t \rightarrow 0}\left(\frac{t \cos t-\sin t}{t^{3}}\right) \quad \because\left[\frac{0}{0}\right.\) form \(]\) \(\because\) Using L-Hospital rule \(=-1-2 \lim _{t \rightarrow 0}\left(\frac{\cos t-t \sin t-\cos t}{3 t^{3}}\right)\) \(=-1+\frac{2}{3} \lim _{t \rightarrow 0} \frac{\sin t}{t}=-1+\frac{2}{3} \times 1=\frac{-1}{3}\lt 0\) So, function \(\mathrm{f}(\mathrm{t})\) is maximum at \(\mathrm{t}=0\)
VITEEE-2015
Application of Derivatives
85642
On the interval \([0,1]\), the function \(x^{25}(1-x)^{75}\) takes its maximum value at the point
1 0
2 \(\frac{1}{4}\)
3 \(\frac{1}{2}\)
4 \(\frac{1}{3}\)
Explanation:
(B) : Let \(f(x)=x^{25}(1-x)^{75}, x \in[0,1]\) \(\Rightarrow f^{\prime}(x)=25 x^{24}(1-x)^{75}-75 x^{25}(1-x)^{74}\) \(=25 x^{24}(1-x)^{74} \times\{(1-x)-3 x\}\) \(=25 x^{24}(1-x)^{74}(1-4 x)\) \(+\quad+\quad+\quad+\quad+\quad+\) We can see that \(\mathrm{f}^{\prime}(\mathrm{x})\) is positive for \(\mathrm{x}\lt \frac{1}{4}\) And \(\mathrm{f}^{\prime}(\mathrm{x})\) is negative for \(\mathrm{x}>\frac{1}{4}\) Hence, \(\mathrm{f}(\mathrm{x})\) attains maximum at \(\mathrm{x}=\frac{1}{4}\)
VITEEE-2015
Application of Derivatives
85643
The minimum value of
1 e
2 \(\frac{1}{\mathrm{e}}\)
3 \(\mathrm{e}^{2}\)
4 \(\mathrm{e}^{3}\)
Explanation:
(A) : Let \(f(x)=\frac{x}{\log x}\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\log \mathrm{x}-1}{(\log \mathrm{x})^{2}}\) For maxima and minima, put \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\frac{\log x-1}{(\log x)^{2}}=0\) \(\log x-1=0\) \(\log x=1\) \(\mathrm{x}=\mathrm{e}^{1} \quad \because \log _{\mathrm{a}} \mathrm{x}=\mathrm{y}\) \(\mathrm{x}=\mathrm{e} \quad \mathrm{x}=\mathrm{a}^{\mathrm{y}}\) Now \(f^{\prime \prime}(x)=\frac{(\log x)^{2} \cdot \frac{1}{x}-(\log x-1) \cdot \frac{2 \log x}{x}}{(\log x)^{4}}\) \(\mathrm{f}^{\prime \prime}(\mathrm{e})=\frac{\frac{1}{\mathrm{e}}-0}{1} \Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{e})=\frac{1}{\mathrm{e}}\) \(\mathrm{f}^{\prime \prime}(\mathrm{e})>0\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is minimum at \(\mathrm{x}=\mathrm{e}\). Hence, minimum value of \(f(x)\) at \(\mathrm{x}=\mathrm{e}\) is \(\mathrm{f}(\mathrm{e})=\frac{\mathrm{e}}{\log \mathrm{e}}=\mathrm{e} \quad \because \log _{\mathrm{e}} \mathrm{e}=1\) \(\mathrm{f}(\mathrm{e})=\mathrm{e}\)
VITEEE-2014
Application of Derivatives
85644
If \(f(x)=\left\{\begin{array}{cc}x^{2} x \leq 0 \\ 2 \sin x, x>0\end{array}\right.\), then \(x=0\) is
1 point of minima
2 point of maxima
3 point of discontinuity
4 None of the above
Explanation:
(A) : Given \(f(x)=\left\{\begin{array}{cc}x^{2}, x \leq 0 \\ 2 \sin x, x>0,\end{array}\right.\) \(f^{\prime}(x)=\left\{\begin{array}{cc}2 x, x\lt 0 \\ \text { non differentiable, } x=0 \\ 2 \cos x, x>0\end{array}\right.\) So, \(\mathrm{x}=0\) is a critical point \(\mathrm{f}\left(0^{-}\right)>0\) as well as \(\mathrm{f}\left(0^{+}\right)>0\) and \(\mathrm{f}(0)=0\) Hence, it is a point of minima.
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Application of Derivatives
85641
At \(t=0\), the function \(f(t)=\frac{\sin t}{t}\) has
1 a minimum
2 a discontinuity
3 a point of inflexion
4 a maximum
Explanation:
(D) : We have \(f(t)=\frac{\sin t}{t}\) At \(t=0\), we will check continuity of the function. \(\mathrm{L} \cdot \mathrm{H} \cdot \mathrm{L}=\mathrm{f}(0-\mathrm{h})\) \(\lim _{h \rightarrow 0} \frac{\sin (0-h)}{(0-h)} \Rightarrow=\lim _{h \rightarrow 0} \frac{-\sin h}{-h}=1\) \(\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{L}=\mathrm{f}(0+\mathrm{h})\) \(\lim _{h \rightarrow 0} \frac{\sin (0+h)}{(0+h)} \Rightarrow \lim _{h \rightarrow 0} \frac{\sin h}{h}=1\) and \(\mathrm{f}(0)=1\) \(\mathrm{L} \cdot \mathrm{H} \cdot \mathrm{L}=\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{L}=\mathrm{f}(0)\) So, the function is continuous at \(t=0\) Now, we check the function is maximum or minimum. \(\mathrm{f}^{\prime}(\mathrm{t})=\frac{1}{\mathrm{t}} \cos \mathrm{t}-\frac{1}{\mathrm{t}^{2}} \sin \mathrm{t}\) and \(\quad f^{\prime \prime}(t)=\frac{-1}{t} \sin t-\frac{1}{t^{2}} \cos \mathrm{t}-\frac{1}{t^{2}} \cos \mathrm{t}+\frac{2}{\mathrm{t}^{3}} \sin \mathrm{t}\) \(=\frac{-\sin t}{t}-\frac{2 \cos t}{t^{2}}+\frac{2 \sin t}{t^{3}}\) For maximum or minimum value of \(f(x)\), put \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\Rightarrow \frac{\cos \mathrm{t}}{\mathrm{t}} \frac{-\sin \mathrm{t}}{\mathrm{t}^{2}}=0 \Rightarrow \frac{\tan \mathrm{t}}{\mathrm{t}}=1\) Now \(\lim _{\mathrm{t} \rightarrow 0} \mathrm{f}^{\prime \prime}(\mathrm{t})\) \(=\lim _{t \rightarrow 0}\left(-\frac{\sin t}{t}\right)-2 \lim _{t \rightarrow 0}\left(\frac{t \cos t-\sin t}{t^{3}}\right) \quad \because\left[\frac{0}{0}\right.\) form \(]\) \(\because\) Using L-Hospital rule \(=-1-2 \lim _{t \rightarrow 0}\left(\frac{\cos t-t \sin t-\cos t}{3 t^{3}}\right)\) \(=-1+\frac{2}{3} \lim _{t \rightarrow 0} \frac{\sin t}{t}=-1+\frac{2}{3} \times 1=\frac{-1}{3}\lt 0\) So, function \(\mathrm{f}(\mathrm{t})\) is maximum at \(\mathrm{t}=0\)
VITEEE-2015
Application of Derivatives
85642
On the interval \([0,1]\), the function \(x^{25}(1-x)^{75}\) takes its maximum value at the point
1 0
2 \(\frac{1}{4}\)
3 \(\frac{1}{2}\)
4 \(\frac{1}{3}\)
Explanation:
(B) : Let \(f(x)=x^{25}(1-x)^{75}, x \in[0,1]\) \(\Rightarrow f^{\prime}(x)=25 x^{24}(1-x)^{75}-75 x^{25}(1-x)^{74}\) \(=25 x^{24}(1-x)^{74} \times\{(1-x)-3 x\}\) \(=25 x^{24}(1-x)^{74}(1-4 x)\) \(+\quad+\quad+\quad+\quad+\quad+\) We can see that \(\mathrm{f}^{\prime}(\mathrm{x})\) is positive for \(\mathrm{x}\lt \frac{1}{4}\) And \(\mathrm{f}^{\prime}(\mathrm{x})\) is negative for \(\mathrm{x}>\frac{1}{4}\) Hence, \(\mathrm{f}(\mathrm{x})\) attains maximum at \(\mathrm{x}=\frac{1}{4}\)
VITEEE-2015
Application of Derivatives
85643
The minimum value of
1 e
2 \(\frac{1}{\mathrm{e}}\)
3 \(\mathrm{e}^{2}\)
4 \(\mathrm{e}^{3}\)
Explanation:
(A) : Let \(f(x)=\frac{x}{\log x}\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\log \mathrm{x}-1}{(\log \mathrm{x})^{2}}\) For maxima and minima, put \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\frac{\log x-1}{(\log x)^{2}}=0\) \(\log x-1=0\) \(\log x=1\) \(\mathrm{x}=\mathrm{e}^{1} \quad \because \log _{\mathrm{a}} \mathrm{x}=\mathrm{y}\) \(\mathrm{x}=\mathrm{e} \quad \mathrm{x}=\mathrm{a}^{\mathrm{y}}\) Now \(f^{\prime \prime}(x)=\frac{(\log x)^{2} \cdot \frac{1}{x}-(\log x-1) \cdot \frac{2 \log x}{x}}{(\log x)^{4}}\) \(\mathrm{f}^{\prime \prime}(\mathrm{e})=\frac{\frac{1}{\mathrm{e}}-0}{1} \Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{e})=\frac{1}{\mathrm{e}}\) \(\mathrm{f}^{\prime \prime}(\mathrm{e})>0\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is minimum at \(\mathrm{x}=\mathrm{e}\). Hence, minimum value of \(f(x)\) at \(\mathrm{x}=\mathrm{e}\) is \(\mathrm{f}(\mathrm{e})=\frac{\mathrm{e}}{\log \mathrm{e}}=\mathrm{e} \quad \because \log _{\mathrm{e}} \mathrm{e}=1\) \(\mathrm{f}(\mathrm{e})=\mathrm{e}\)
VITEEE-2014
Application of Derivatives
85644
If \(f(x)=\left\{\begin{array}{cc}x^{2} x \leq 0 \\ 2 \sin x, x>0\end{array}\right.\), then \(x=0\) is
1 point of minima
2 point of maxima
3 point of discontinuity
4 None of the above
Explanation:
(A) : Given \(f(x)=\left\{\begin{array}{cc}x^{2}, x \leq 0 \\ 2 \sin x, x>0,\end{array}\right.\) \(f^{\prime}(x)=\left\{\begin{array}{cc}2 x, x\lt 0 \\ \text { non differentiable, } x=0 \\ 2 \cos x, x>0\end{array}\right.\) So, \(\mathrm{x}=0\) is a critical point \(\mathrm{f}\left(0^{-}\right)>0\) as well as \(\mathrm{f}\left(0^{+}\right)>0\) and \(\mathrm{f}(0)=0\) Hence, it is a point of minima.
85641
At \(t=0\), the function \(f(t)=\frac{\sin t}{t}\) has
1 a minimum
2 a discontinuity
3 a point of inflexion
4 a maximum
Explanation:
(D) : We have \(f(t)=\frac{\sin t}{t}\) At \(t=0\), we will check continuity of the function. \(\mathrm{L} \cdot \mathrm{H} \cdot \mathrm{L}=\mathrm{f}(0-\mathrm{h})\) \(\lim _{h \rightarrow 0} \frac{\sin (0-h)}{(0-h)} \Rightarrow=\lim _{h \rightarrow 0} \frac{-\sin h}{-h}=1\) \(\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{L}=\mathrm{f}(0+\mathrm{h})\) \(\lim _{h \rightarrow 0} \frac{\sin (0+h)}{(0+h)} \Rightarrow \lim _{h \rightarrow 0} \frac{\sin h}{h}=1\) and \(\mathrm{f}(0)=1\) \(\mathrm{L} \cdot \mathrm{H} \cdot \mathrm{L}=\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{L}=\mathrm{f}(0)\) So, the function is continuous at \(t=0\) Now, we check the function is maximum or minimum. \(\mathrm{f}^{\prime}(\mathrm{t})=\frac{1}{\mathrm{t}} \cos \mathrm{t}-\frac{1}{\mathrm{t}^{2}} \sin \mathrm{t}\) and \(\quad f^{\prime \prime}(t)=\frac{-1}{t} \sin t-\frac{1}{t^{2}} \cos \mathrm{t}-\frac{1}{t^{2}} \cos \mathrm{t}+\frac{2}{\mathrm{t}^{3}} \sin \mathrm{t}\) \(=\frac{-\sin t}{t}-\frac{2 \cos t}{t^{2}}+\frac{2 \sin t}{t^{3}}\) For maximum or minimum value of \(f(x)\), put \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\Rightarrow \frac{\cos \mathrm{t}}{\mathrm{t}} \frac{-\sin \mathrm{t}}{\mathrm{t}^{2}}=0 \Rightarrow \frac{\tan \mathrm{t}}{\mathrm{t}}=1\) Now \(\lim _{\mathrm{t} \rightarrow 0} \mathrm{f}^{\prime \prime}(\mathrm{t})\) \(=\lim _{t \rightarrow 0}\left(-\frac{\sin t}{t}\right)-2 \lim _{t \rightarrow 0}\left(\frac{t \cos t-\sin t}{t^{3}}\right) \quad \because\left[\frac{0}{0}\right.\) form \(]\) \(\because\) Using L-Hospital rule \(=-1-2 \lim _{t \rightarrow 0}\left(\frac{\cos t-t \sin t-\cos t}{3 t^{3}}\right)\) \(=-1+\frac{2}{3} \lim _{t \rightarrow 0} \frac{\sin t}{t}=-1+\frac{2}{3} \times 1=\frac{-1}{3}\lt 0\) So, function \(\mathrm{f}(\mathrm{t})\) is maximum at \(\mathrm{t}=0\)
VITEEE-2015
Application of Derivatives
85642
On the interval \([0,1]\), the function \(x^{25}(1-x)^{75}\) takes its maximum value at the point
1 0
2 \(\frac{1}{4}\)
3 \(\frac{1}{2}\)
4 \(\frac{1}{3}\)
Explanation:
(B) : Let \(f(x)=x^{25}(1-x)^{75}, x \in[0,1]\) \(\Rightarrow f^{\prime}(x)=25 x^{24}(1-x)^{75}-75 x^{25}(1-x)^{74}\) \(=25 x^{24}(1-x)^{74} \times\{(1-x)-3 x\}\) \(=25 x^{24}(1-x)^{74}(1-4 x)\) \(+\quad+\quad+\quad+\quad+\quad+\) We can see that \(\mathrm{f}^{\prime}(\mathrm{x})\) is positive for \(\mathrm{x}\lt \frac{1}{4}\) And \(\mathrm{f}^{\prime}(\mathrm{x})\) is negative for \(\mathrm{x}>\frac{1}{4}\) Hence, \(\mathrm{f}(\mathrm{x})\) attains maximum at \(\mathrm{x}=\frac{1}{4}\)
VITEEE-2015
Application of Derivatives
85643
The minimum value of
1 e
2 \(\frac{1}{\mathrm{e}}\)
3 \(\mathrm{e}^{2}\)
4 \(\mathrm{e}^{3}\)
Explanation:
(A) : Let \(f(x)=\frac{x}{\log x}\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\log \mathrm{x}-1}{(\log \mathrm{x})^{2}}\) For maxima and minima, put \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\frac{\log x-1}{(\log x)^{2}}=0\) \(\log x-1=0\) \(\log x=1\) \(\mathrm{x}=\mathrm{e}^{1} \quad \because \log _{\mathrm{a}} \mathrm{x}=\mathrm{y}\) \(\mathrm{x}=\mathrm{e} \quad \mathrm{x}=\mathrm{a}^{\mathrm{y}}\) Now \(f^{\prime \prime}(x)=\frac{(\log x)^{2} \cdot \frac{1}{x}-(\log x-1) \cdot \frac{2 \log x}{x}}{(\log x)^{4}}\) \(\mathrm{f}^{\prime \prime}(\mathrm{e})=\frac{\frac{1}{\mathrm{e}}-0}{1} \Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{e})=\frac{1}{\mathrm{e}}\) \(\mathrm{f}^{\prime \prime}(\mathrm{e})>0\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is minimum at \(\mathrm{x}=\mathrm{e}\). Hence, minimum value of \(f(x)\) at \(\mathrm{x}=\mathrm{e}\) is \(\mathrm{f}(\mathrm{e})=\frac{\mathrm{e}}{\log \mathrm{e}}=\mathrm{e} \quad \because \log _{\mathrm{e}} \mathrm{e}=1\) \(\mathrm{f}(\mathrm{e})=\mathrm{e}\)
VITEEE-2014
Application of Derivatives
85644
If \(f(x)=\left\{\begin{array}{cc}x^{2} x \leq 0 \\ 2 \sin x, x>0\end{array}\right.\), then \(x=0\) is
1 point of minima
2 point of maxima
3 point of discontinuity
4 None of the above
Explanation:
(A) : Given \(f(x)=\left\{\begin{array}{cc}x^{2}, x \leq 0 \\ 2 \sin x, x>0,\end{array}\right.\) \(f^{\prime}(x)=\left\{\begin{array}{cc}2 x, x\lt 0 \\ \text { non differentiable, } x=0 \\ 2 \cos x, x>0\end{array}\right.\) So, \(\mathrm{x}=0\) is a critical point \(\mathrm{f}\left(0^{-}\right)>0\) as well as \(\mathrm{f}\left(0^{+}\right)>0\) and \(\mathrm{f}(0)=0\) Hence, it is a point of minima.
85641
At \(t=0\), the function \(f(t)=\frac{\sin t}{t}\) has
1 a minimum
2 a discontinuity
3 a point of inflexion
4 a maximum
Explanation:
(D) : We have \(f(t)=\frac{\sin t}{t}\) At \(t=0\), we will check continuity of the function. \(\mathrm{L} \cdot \mathrm{H} \cdot \mathrm{L}=\mathrm{f}(0-\mathrm{h})\) \(\lim _{h \rightarrow 0} \frac{\sin (0-h)}{(0-h)} \Rightarrow=\lim _{h \rightarrow 0} \frac{-\sin h}{-h}=1\) \(\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{L}=\mathrm{f}(0+\mathrm{h})\) \(\lim _{h \rightarrow 0} \frac{\sin (0+h)}{(0+h)} \Rightarrow \lim _{h \rightarrow 0} \frac{\sin h}{h}=1\) and \(\mathrm{f}(0)=1\) \(\mathrm{L} \cdot \mathrm{H} \cdot \mathrm{L}=\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{L}=\mathrm{f}(0)\) So, the function is continuous at \(t=0\) Now, we check the function is maximum or minimum. \(\mathrm{f}^{\prime}(\mathrm{t})=\frac{1}{\mathrm{t}} \cos \mathrm{t}-\frac{1}{\mathrm{t}^{2}} \sin \mathrm{t}\) and \(\quad f^{\prime \prime}(t)=\frac{-1}{t} \sin t-\frac{1}{t^{2}} \cos \mathrm{t}-\frac{1}{t^{2}} \cos \mathrm{t}+\frac{2}{\mathrm{t}^{3}} \sin \mathrm{t}\) \(=\frac{-\sin t}{t}-\frac{2 \cos t}{t^{2}}+\frac{2 \sin t}{t^{3}}\) For maximum or minimum value of \(f(x)\), put \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\Rightarrow \frac{\cos \mathrm{t}}{\mathrm{t}} \frac{-\sin \mathrm{t}}{\mathrm{t}^{2}}=0 \Rightarrow \frac{\tan \mathrm{t}}{\mathrm{t}}=1\) Now \(\lim _{\mathrm{t} \rightarrow 0} \mathrm{f}^{\prime \prime}(\mathrm{t})\) \(=\lim _{t \rightarrow 0}\left(-\frac{\sin t}{t}\right)-2 \lim _{t \rightarrow 0}\left(\frac{t \cos t-\sin t}{t^{3}}\right) \quad \because\left[\frac{0}{0}\right.\) form \(]\) \(\because\) Using L-Hospital rule \(=-1-2 \lim _{t \rightarrow 0}\left(\frac{\cos t-t \sin t-\cos t}{3 t^{3}}\right)\) \(=-1+\frac{2}{3} \lim _{t \rightarrow 0} \frac{\sin t}{t}=-1+\frac{2}{3} \times 1=\frac{-1}{3}\lt 0\) So, function \(\mathrm{f}(\mathrm{t})\) is maximum at \(\mathrm{t}=0\)
VITEEE-2015
Application of Derivatives
85642
On the interval \([0,1]\), the function \(x^{25}(1-x)^{75}\) takes its maximum value at the point
1 0
2 \(\frac{1}{4}\)
3 \(\frac{1}{2}\)
4 \(\frac{1}{3}\)
Explanation:
(B) : Let \(f(x)=x^{25}(1-x)^{75}, x \in[0,1]\) \(\Rightarrow f^{\prime}(x)=25 x^{24}(1-x)^{75}-75 x^{25}(1-x)^{74}\) \(=25 x^{24}(1-x)^{74} \times\{(1-x)-3 x\}\) \(=25 x^{24}(1-x)^{74}(1-4 x)\) \(+\quad+\quad+\quad+\quad+\quad+\) We can see that \(\mathrm{f}^{\prime}(\mathrm{x})\) is positive for \(\mathrm{x}\lt \frac{1}{4}\) And \(\mathrm{f}^{\prime}(\mathrm{x})\) is negative for \(\mathrm{x}>\frac{1}{4}\) Hence, \(\mathrm{f}(\mathrm{x})\) attains maximum at \(\mathrm{x}=\frac{1}{4}\)
VITEEE-2015
Application of Derivatives
85643
The minimum value of
1 e
2 \(\frac{1}{\mathrm{e}}\)
3 \(\mathrm{e}^{2}\)
4 \(\mathrm{e}^{3}\)
Explanation:
(A) : Let \(f(x)=\frac{x}{\log x}\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\log \mathrm{x}-1}{(\log \mathrm{x})^{2}}\) For maxima and minima, put \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\frac{\log x-1}{(\log x)^{2}}=0\) \(\log x-1=0\) \(\log x=1\) \(\mathrm{x}=\mathrm{e}^{1} \quad \because \log _{\mathrm{a}} \mathrm{x}=\mathrm{y}\) \(\mathrm{x}=\mathrm{e} \quad \mathrm{x}=\mathrm{a}^{\mathrm{y}}\) Now \(f^{\prime \prime}(x)=\frac{(\log x)^{2} \cdot \frac{1}{x}-(\log x-1) \cdot \frac{2 \log x}{x}}{(\log x)^{4}}\) \(\mathrm{f}^{\prime \prime}(\mathrm{e})=\frac{\frac{1}{\mathrm{e}}-0}{1} \Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{e})=\frac{1}{\mathrm{e}}\) \(\mathrm{f}^{\prime \prime}(\mathrm{e})>0\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is minimum at \(\mathrm{x}=\mathrm{e}\). Hence, minimum value of \(f(x)\) at \(\mathrm{x}=\mathrm{e}\) is \(\mathrm{f}(\mathrm{e})=\frac{\mathrm{e}}{\log \mathrm{e}}=\mathrm{e} \quad \because \log _{\mathrm{e}} \mathrm{e}=1\) \(\mathrm{f}(\mathrm{e})=\mathrm{e}\)
VITEEE-2014
Application of Derivatives
85644
If \(f(x)=\left\{\begin{array}{cc}x^{2} x \leq 0 \\ 2 \sin x, x>0\end{array}\right.\), then \(x=0\) is
1 point of minima
2 point of maxima
3 point of discontinuity
4 None of the above
Explanation:
(A) : Given \(f(x)=\left\{\begin{array}{cc}x^{2}, x \leq 0 \\ 2 \sin x, x>0,\end{array}\right.\) \(f^{\prime}(x)=\left\{\begin{array}{cc}2 x, x\lt 0 \\ \text { non differentiable, } x=0 \\ 2 \cos x, x>0\end{array}\right.\) So, \(\mathrm{x}=0\) is a critical point \(\mathrm{f}\left(0^{-}\right)>0\) as well as \(\mathrm{f}\left(0^{+}\right)>0\) and \(\mathrm{f}(0)=0\) Hence, it is a point of minima.
