85667
The maximum value of the function \(y=x(x-1)^{2}\), is
1 0
2 \(\frac{4}{27}\)
3 -4
4 None of these
Explanation:
(B) : Given, \(y=x(x-1)^{2}\) \(=x\left(x^{2}+1-2 x\right)=x^{3}+x-2 x^{2}\) \(\frac{d y}{d x}=3 x^{2}+1-4 x\) For maxima or minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(3 \mathrm{x}^{2}-4 \mathrm{x}+1=0\) \((3 \mathrm{x}-1)(\mathrm{x}-1)=0\) \(\mathrm{x}=1, \frac{1}{3}\) Now, \(\quad \frac{d^{2} y}{d x^{2}}=6 x-4\) At \(x=1, \frac{d^{2} y}{d^{2}}=6-4=2\) At \(\mathrm{x}=\frac{1}{3}, \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=6 \cdot \frac{1}{3}-4=-2\) Function has maxima at point \(\mathrm{x}=\frac{1}{3}\), then maximum value, \(\mathrm{y}=\frac{1}{3}\left(\frac{1}{3}-1\right)^{2} \Rightarrow \mathrm{y}=\frac{4}{27}\)
[JCECE-2010]
Application of Derivatives
85668
The function \(\mathrm{x}^{5}-5 \mathrm{x}^{4}+5 \mathrm{x}^{3}-1\) is
1 Neither maximum nor minimum at \(x=0\)
2 maximum at \(\mathrm{x}=0\)
3 maximum at \(\mathrm{x}=1\) and minimum at \(\mathrm{x}=3\)
4 Minimum at \(\mathrm{x}=0\)
Explanation:
(C) : \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{5}-5 \mathrm{x}^{4}+5 \mathrm{x}^{3}-1\) Differentiating both side with respect to \(x\) \(f^{\prime}(\mathrm{x})=5 \mathrm{x}^{4}-20 \mathrm{x}^{3}+15 \mathrm{x}^{2}\) For maxima or minima, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(5 \mathrm{x}^{4}-20 \mathrm{x}^{3}+15 \mathrm{x}^{2}=0\) \(5 \mathrm{x}^{2}\left(\mathrm{x}^{2}-4 \mathrm{x}+3\right)=0\) \((\mathrm{x}-1)(\mathrm{x}-3)=0\) \(\mathrm{x}=1,3\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=20 \mathrm{x}^{3}-60 \mathrm{x}^{2}+30 \mathrm{x}\) At \(x=1, \quad f^{\prime \prime}(1)=20(1)^{3}-60(1)^{2}+30(1)\) At \(\mathrm{x}=3, \quad \mathrm{f} \quad \mathrm{f}(3)=20(3)^{3}-60(3)^{2}+30(3)\) Hence, The function is maximum at \(\mathrm{x}=1\) and minimum at \(\mathrm{x}=3\).
UPSEE-2007
Application of Derivatives
85669
The maximum value of \(x^{3}-3 x\) in the interval \([0,2]\) is
1 -2
2 0
3 2
4 1
Explanation:
(A) : \(f(x)=x^{3}-3 x \quad x \in[0,2]\) \(f^{\prime}(x)=3 x^{2}-3\) For critical point, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(3 x^{2}-3=0\) \(3\left(\mathrm{x}^{2}-1\right)=0\) \(\mathrm{x}= \pm 1\) \(x=1 \in[0,2]\) Or \(\quad \mathrm{x}=1 \in[0,2]\) Function has maximum value, at \(\mathrm{x}=1\) \(f(1)=(1)^{3}-3(1)\) \(=-2\)
[JCECE-2007]
Application of Derivatives
85670
The function \(f(\mathrm{x})=4 \mathrm{x}^{3}-18 \mathrm{x}^{2}+27 \mathrm{x}-7\) has
1 one maximum
2 one minimum
3 one maximum and one minimum
4 neither a maximum nor minimum
Explanation:
(D) : \(\mathrm{f}(\mathrm{x})=4 \mathrm{x}^{3}-18 \mathrm{x}^{2}+27 \mathrm{x}-7\) \(f^{\prime}(x)=12 x^{2}-36 x+27\) For maxima and minima, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(12 x^{2}-36 x+27=0\) \(4 \mathrm{x}^{2}-12 \mathrm{x}+9=0\) \((2 \mathrm{x}-3)(2 \mathrm{x}-3)=0\) \(x=\frac{3}{2}\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=24 \mathrm{x}-36\) At \(\quad \begin{aligned} \mathrm{x}=\frac{3}{2}, \mathrm{f}^{\prime \prime}\left(\frac{3}{2}\right) =24 \times \frac{3}{2}-36 \\ =0\end{aligned}\) Hence, The function has neither a maximum nor minimum.
85667
The maximum value of the function \(y=x(x-1)^{2}\), is
1 0
2 \(\frac{4}{27}\)
3 -4
4 None of these
Explanation:
(B) : Given, \(y=x(x-1)^{2}\) \(=x\left(x^{2}+1-2 x\right)=x^{3}+x-2 x^{2}\) \(\frac{d y}{d x}=3 x^{2}+1-4 x\) For maxima or minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(3 \mathrm{x}^{2}-4 \mathrm{x}+1=0\) \((3 \mathrm{x}-1)(\mathrm{x}-1)=0\) \(\mathrm{x}=1, \frac{1}{3}\) Now, \(\quad \frac{d^{2} y}{d x^{2}}=6 x-4\) At \(x=1, \frac{d^{2} y}{d^{2}}=6-4=2\) At \(\mathrm{x}=\frac{1}{3}, \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=6 \cdot \frac{1}{3}-4=-2\) Function has maxima at point \(\mathrm{x}=\frac{1}{3}\), then maximum value, \(\mathrm{y}=\frac{1}{3}\left(\frac{1}{3}-1\right)^{2} \Rightarrow \mathrm{y}=\frac{4}{27}\)
[JCECE-2010]
Application of Derivatives
85668
The function \(\mathrm{x}^{5}-5 \mathrm{x}^{4}+5 \mathrm{x}^{3}-1\) is
1 Neither maximum nor minimum at \(x=0\)
2 maximum at \(\mathrm{x}=0\)
3 maximum at \(\mathrm{x}=1\) and minimum at \(\mathrm{x}=3\)
4 Minimum at \(\mathrm{x}=0\)
Explanation:
(C) : \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{5}-5 \mathrm{x}^{4}+5 \mathrm{x}^{3}-1\) Differentiating both side with respect to \(x\) \(f^{\prime}(\mathrm{x})=5 \mathrm{x}^{4}-20 \mathrm{x}^{3}+15 \mathrm{x}^{2}\) For maxima or minima, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(5 \mathrm{x}^{4}-20 \mathrm{x}^{3}+15 \mathrm{x}^{2}=0\) \(5 \mathrm{x}^{2}\left(\mathrm{x}^{2}-4 \mathrm{x}+3\right)=0\) \((\mathrm{x}-1)(\mathrm{x}-3)=0\) \(\mathrm{x}=1,3\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=20 \mathrm{x}^{3}-60 \mathrm{x}^{2}+30 \mathrm{x}\) At \(x=1, \quad f^{\prime \prime}(1)=20(1)^{3}-60(1)^{2}+30(1)\) At \(\mathrm{x}=3, \quad \mathrm{f} \quad \mathrm{f}(3)=20(3)^{3}-60(3)^{2}+30(3)\) Hence, The function is maximum at \(\mathrm{x}=1\) and minimum at \(\mathrm{x}=3\).
UPSEE-2007
Application of Derivatives
85669
The maximum value of \(x^{3}-3 x\) in the interval \([0,2]\) is
1 -2
2 0
3 2
4 1
Explanation:
(A) : \(f(x)=x^{3}-3 x \quad x \in[0,2]\) \(f^{\prime}(x)=3 x^{2}-3\) For critical point, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(3 x^{2}-3=0\) \(3\left(\mathrm{x}^{2}-1\right)=0\) \(\mathrm{x}= \pm 1\) \(x=1 \in[0,2]\) Or \(\quad \mathrm{x}=1 \in[0,2]\) Function has maximum value, at \(\mathrm{x}=1\) \(f(1)=(1)^{3}-3(1)\) \(=-2\)
[JCECE-2007]
Application of Derivatives
85670
The function \(f(\mathrm{x})=4 \mathrm{x}^{3}-18 \mathrm{x}^{2}+27 \mathrm{x}-7\) has
1 one maximum
2 one minimum
3 one maximum and one minimum
4 neither a maximum nor minimum
Explanation:
(D) : \(\mathrm{f}(\mathrm{x})=4 \mathrm{x}^{3}-18 \mathrm{x}^{2}+27 \mathrm{x}-7\) \(f^{\prime}(x)=12 x^{2}-36 x+27\) For maxima and minima, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(12 x^{2}-36 x+27=0\) \(4 \mathrm{x}^{2}-12 \mathrm{x}+9=0\) \((2 \mathrm{x}-3)(2 \mathrm{x}-3)=0\) \(x=\frac{3}{2}\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=24 \mathrm{x}-36\) At \(\quad \begin{aligned} \mathrm{x}=\frac{3}{2}, \mathrm{f}^{\prime \prime}\left(\frac{3}{2}\right) =24 \times \frac{3}{2}-36 \\ =0\end{aligned}\) Hence, The function has neither a maximum nor minimum.
85667
The maximum value of the function \(y=x(x-1)^{2}\), is
1 0
2 \(\frac{4}{27}\)
3 -4
4 None of these
Explanation:
(B) : Given, \(y=x(x-1)^{2}\) \(=x\left(x^{2}+1-2 x\right)=x^{3}+x-2 x^{2}\) \(\frac{d y}{d x}=3 x^{2}+1-4 x\) For maxima or minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(3 \mathrm{x}^{2}-4 \mathrm{x}+1=0\) \((3 \mathrm{x}-1)(\mathrm{x}-1)=0\) \(\mathrm{x}=1, \frac{1}{3}\) Now, \(\quad \frac{d^{2} y}{d x^{2}}=6 x-4\) At \(x=1, \frac{d^{2} y}{d^{2}}=6-4=2\) At \(\mathrm{x}=\frac{1}{3}, \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=6 \cdot \frac{1}{3}-4=-2\) Function has maxima at point \(\mathrm{x}=\frac{1}{3}\), then maximum value, \(\mathrm{y}=\frac{1}{3}\left(\frac{1}{3}-1\right)^{2} \Rightarrow \mathrm{y}=\frac{4}{27}\)
[JCECE-2010]
Application of Derivatives
85668
The function \(\mathrm{x}^{5}-5 \mathrm{x}^{4}+5 \mathrm{x}^{3}-1\) is
1 Neither maximum nor minimum at \(x=0\)
2 maximum at \(\mathrm{x}=0\)
3 maximum at \(\mathrm{x}=1\) and minimum at \(\mathrm{x}=3\)
4 Minimum at \(\mathrm{x}=0\)
Explanation:
(C) : \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{5}-5 \mathrm{x}^{4}+5 \mathrm{x}^{3}-1\) Differentiating both side with respect to \(x\) \(f^{\prime}(\mathrm{x})=5 \mathrm{x}^{4}-20 \mathrm{x}^{3}+15 \mathrm{x}^{2}\) For maxima or minima, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(5 \mathrm{x}^{4}-20 \mathrm{x}^{3}+15 \mathrm{x}^{2}=0\) \(5 \mathrm{x}^{2}\left(\mathrm{x}^{2}-4 \mathrm{x}+3\right)=0\) \((\mathrm{x}-1)(\mathrm{x}-3)=0\) \(\mathrm{x}=1,3\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=20 \mathrm{x}^{3}-60 \mathrm{x}^{2}+30 \mathrm{x}\) At \(x=1, \quad f^{\prime \prime}(1)=20(1)^{3}-60(1)^{2}+30(1)\) At \(\mathrm{x}=3, \quad \mathrm{f} \quad \mathrm{f}(3)=20(3)^{3}-60(3)^{2}+30(3)\) Hence, The function is maximum at \(\mathrm{x}=1\) and minimum at \(\mathrm{x}=3\).
UPSEE-2007
Application of Derivatives
85669
The maximum value of \(x^{3}-3 x\) in the interval \([0,2]\) is
1 -2
2 0
3 2
4 1
Explanation:
(A) : \(f(x)=x^{3}-3 x \quad x \in[0,2]\) \(f^{\prime}(x)=3 x^{2}-3\) For critical point, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(3 x^{2}-3=0\) \(3\left(\mathrm{x}^{2}-1\right)=0\) \(\mathrm{x}= \pm 1\) \(x=1 \in[0,2]\) Or \(\quad \mathrm{x}=1 \in[0,2]\) Function has maximum value, at \(\mathrm{x}=1\) \(f(1)=(1)^{3}-3(1)\) \(=-2\)
[JCECE-2007]
Application of Derivatives
85670
The function \(f(\mathrm{x})=4 \mathrm{x}^{3}-18 \mathrm{x}^{2}+27 \mathrm{x}-7\) has
1 one maximum
2 one minimum
3 one maximum and one minimum
4 neither a maximum nor minimum
Explanation:
(D) : \(\mathrm{f}(\mathrm{x})=4 \mathrm{x}^{3}-18 \mathrm{x}^{2}+27 \mathrm{x}-7\) \(f^{\prime}(x)=12 x^{2}-36 x+27\) For maxima and minima, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(12 x^{2}-36 x+27=0\) \(4 \mathrm{x}^{2}-12 \mathrm{x}+9=0\) \((2 \mathrm{x}-3)(2 \mathrm{x}-3)=0\) \(x=\frac{3}{2}\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=24 \mathrm{x}-36\) At \(\quad \begin{aligned} \mathrm{x}=\frac{3}{2}, \mathrm{f}^{\prime \prime}\left(\frac{3}{2}\right) =24 \times \frac{3}{2}-36 \\ =0\end{aligned}\) Hence, The function has neither a maximum nor minimum.
85667
The maximum value of the function \(y=x(x-1)^{2}\), is
1 0
2 \(\frac{4}{27}\)
3 -4
4 None of these
Explanation:
(B) : Given, \(y=x(x-1)^{2}\) \(=x\left(x^{2}+1-2 x\right)=x^{3}+x-2 x^{2}\) \(\frac{d y}{d x}=3 x^{2}+1-4 x\) For maxima or minima, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(3 \mathrm{x}^{2}-4 \mathrm{x}+1=0\) \((3 \mathrm{x}-1)(\mathrm{x}-1)=0\) \(\mathrm{x}=1, \frac{1}{3}\) Now, \(\quad \frac{d^{2} y}{d x^{2}}=6 x-4\) At \(x=1, \frac{d^{2} y}{d^{2}}=6-4=2\) At \(\mathrm{x}=\frac{1}{3}, \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=6 \cdot \frac{1}{3}-4=-2\) Function has maxima at point \(\mathrm{x}=\frac{1}{3}\), then maximum value, \(\mathrm{y}=\frac{1}{3}\left(\frac{1}{3}-1\right)^{2} \Rightarrow \mathrm{y}=\frac{4}{27}\)
[JCECE-2010]
Application of Derivatives
85668
The function \(\mathrm{x}^{5}-5 \mathrm{x}^{4}+5 \mathrm{x}^{3}-1\) is
1 Neither maximum nor minimum at \(x=0\)
2 maximum at \(\mathrm{x}=0\)
3 maximum at \(\mathrm{x}=1\) and minimum at \(\mathrm{x}=3\)
4 Minimum at \(\mathrm{x}=0\)
Explanation:
(C) : \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{5}-5 \mathrm{x}^{4}+5 \mathrm{x}^{3}-1\) Differentiating both side with respect to \(x\) \(f^{\prime}(\mathrm{x})=5 \mathrm{x}^{4}-20 \mathrm{x}^{3}+15 \mathrm{x}^{2}\) For maxima or minima, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(5 \mathrm{x}^{4}-20 \mathrm{x}^{3}+15 \mathrm{x}^{2}=0\) \(5 \mathrm{x}^{2}\left(\mathrm{x}^{2}-4 \mathrm{x}+3\right)=0\) \((\mathrm{x}-1)(\mathrm{x}-3)=0\) \(\mathrm{x}=1,3\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=20 \mathrm{x}^{3}-60 \mathrm{x}^{2}+30 \mathrm{x}\) At \(x=1, \quad f^{\prime \prime}(1)=20(1)^{3}-60(1)^{2}+30(1)\) At \(\mathrm{x}=3, \quad \mathrm{f} \quad \mathrm{f}(3)=20(3)^{3}-60(3)^{2}+30(3)\) Hence, The function is maximum at \(\mathrm{x}=1\) and minimum at \(\mathrm{x}=3\).
UPSEE-2007
Application of Derivatives
85669
The maximum value of \(x^{3}-3 x\) in the interval \([0,2]\) is
1 -2
2 0
3 2
4 1
Explanation:
(A) : \(f(x)=x^{3}-3 x \quad x \in[0,2]\) \(f^{\prime}(x)=3 x^{2}-3\) For critical point, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(3 x^{2}-3=0\) \(3\left(\mathrm{x}^{2}-1\right)=0\) \(\mathrm{x}= \pm 1\) \(x=1 \in[0,2]\) Or \(\quad \mathrm{x}=1 \in[0,2]\) Function has maximum value, at \(\mathrm{x}=1\) \(f(1)=(1)^{3}-3(1)\) \(=-2\)
[JCECE-2007]
Application of Derivatives
85670
The function \(f(\mathrm{x})=4 \mathrm{x}^{3}-18 \mathrm{x}^{2}+27 \mathrm{x}-7\) has
1 one maximum
2 one minimum
3 one maximum and one minimum
4 neither a maximum nor minimum
Explanation:
(D) : \(\mathrm{f}(\mathrm{x})=4 \mathrm{x}^{3}-18 \mathrm{x}^{2}+27 \mathrm{x}-7\) \(f^{\prime}(x)=12 x^{2}-36 x+27\) For maxima and minima, \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(12 x^{2}-36 x+27=0\) \(4 \mathrm{x}^{2}-12 \mathrm{x}+9=0\) \((2 \mathrm{x}-3)(2 \mathrm{x}-3)=0\) \(x=\frac{3}{2}\) Now, \(\mathrm{f}^{\prime \prime}(\mathrm{x})=24 \mathrm{x}-36\) At \(\quad \begin{aligned} \mathrm{x}=\frac{3}{2}, \mathrm{f}^{\prime \prime}\left(\frac{3}{2}\right) =24 \times \frac{3}{2}-36 \\ =0\end{aligned}\) Hence, The function has neither a maximum nor minimum.
