85637
The minimum value of \(f(x)=e^{\left(x^{4}-x^{3}+x^{2}\right)}\) i
1 e
2 \(-\mathrm{e}\)
3 1
4 -1
Explanation:
(C) : We have \(f(x)=e^{\left(x^{4}-x^{3}+x^{2}\right)}\) Differentiating with respect to \(x\) we get- \(f^{\prime}(x)=e^{\left(x^{4}-x^{3}+x^{2}\right)} \quad\left(4 x^{3}-3 x^{2}+2 x\right)\) For maxima or minima \(f^{\prime}(x)=0\) \(e^{\left(x^{4}-x^{3}+x^{2}\right)}\) \(x\left(4 x^{3}-3 x^{2}+2 x\right)=0\) \(x=0\) The function has minimum value at \(\mathrm{x}=0\) Now, minimum value of function \(\mathrm{f}(\mathrm{x})=\mathrm{e}^{(0-0+0)}=\mathrm{e}^{0}\) \(\mathrm{f}(\mathrm{x})=1\) Hence, minimum value of function \(f(x)=e^{\left(x^{4}-x^{3}+x^{2}\right)} \text { is } 1\)
WB JEE-2010
Application of Derivatives
85638
A wire \(34 \mathrm{~cm}\) long is to be bent in the form of a quadrilateral of which each angle is \(90^{\circ}\). What is the maximum area which can be enclosed inside the quadrilateral?
1 \(68 \mathrm{~cm}^{2}\)
2 \(70 \mathrm{~cm}^{2}\)
3 \(71.25 \mathrm{~cm}^{2}\)
4 \(72.25 \mathrm{~cm}^{2}\)
Explanation:
(D) : Let one side of quadrilateral be \(x\) and another side be y so, \(2(x+y)=34\) \(\Rightarrow \quad x+y=17 \tag{i}\) We know from the basic principle that for a given perimeter square has the maximum area So, \(\quad \mathrm{x}=\mathrm{y}\) and putting this value in equation (i) We have \(\mathrm{x}=\mathrm{y}=\frac{17}{2}\) Area \(x . y=\frac{17}{2} \times \frac{17}{2}=\frac{289}{4}\) Area \(=72.25 \mathrm{~cm}^{2}\)
VITEEE-2019
Application of Derivatives
85639
If at \(x=1\), the function \(x^{4}-62 x^{2}+a x+9\) attains its maximum value on the interval \([0,2]\), then the value of \(a\) is
1 110
2 10
3 55
4 None of these
Explanation:
(D) : Let \(f(x)=x^{4}-62 x^{2}+a x+9\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=4 \mathrm{x}^{3}-124 \mathrm{x}+\mathrm{a}\) It is given that function its maximum Value on the interval \([0,2]\) at \(x=1\) \(\therefore \mathrm{f}^{\prime}(1)=0\) \(\Rightarrow 4 \times(1)^{3}-124(1)+\mathrm{a}=0\) \(\Rightarrow 4-124+\mathrm{a}=0\) \(\Rightarrow-120+\mathrm{a}=0\) \(\mathrm{a}=120\) Hence, the value of a is 120
VITEEE-2018
Application of Derivatives
85640
If \(f(x)=x^{2}, g(x)=2 x, 0 \leq x \leq 2\) then the value of
85637
The minimum value of \(f(x)=e^{\left(x^{4}-x^{3}+x^{2}\right)}\) i
1 e
2 \(-\mathrm{e}\)
3 1
4 -1
Explanation:
(C) : We have \(f(x)=e^{\left(x^{4}-x^{3}+x^{2}\right)}\) Differentiating with respect to \(x\) we get- \(f^{\prime}(x)=e^{\left(x^{4}-x^{3}+x^{2}\right)} \quad\left(4 x^{3}-3 x^{2}+2 x\right)\) For maxima or minima \(f^{\prime}(x)=0\) \(e^{\left(x^{4}-x^{3}+x^{2}\right)}\) \(x\left(4 x^{3}-3 x^{2}+2 x\right)=0\) \(x=0\) The function has minimum value at \(\mathrm{x}=0\) Now, minimum value of function \(\mathrm{f}(\mathrm{x})=\mathrm{e}^{(0-0+0)}=\mathrm{e}^{0}\) \(\mathrm{f}(\mathrm{x})=1\) Hence, minimum value of function \(f(x)=e^{\left(x^{4}-x^{3}+x^{2}\right)} \text { is } 1\)
WB JEE-2010
Application of Derivatives
85638
A wire \(34 \mathrm{~cm}\) long is to be bent in the form of a quadrilateral of which each angle is \(90^{\circ}\). What is the maximum area which can be enclosed inside the quadrilateral?
1 \(68 \mathrm{~cm}^{2}\)
2 \(70 \mathrm{~cm}^{2}\)
3 \(71.25 \mathrm{~cm}^{2}\)
4 \(72.25 \mathrm{~cm}^{2}\)
Explanation:
(D) : Let one side of quadrilateral be \(x\) and another side be y so, \(2(x+y)=34\) \(\Rightarrow \quad x+y=17 \tag{i}\) We know from the basic principle that for a given perimeter square has the maximum area So, \(\quad \mathrm{x}=\mathrm{y}\) and putting this value in equation (i) We have \(\mathrm{x}=\mathrm{y}=\frac{17}{2}\) Area \(x . y=\frac{17}{2} \times \frac{17}{2}=\frac{289}{4}\) Area \(=72.25 \mathrm{~cm}^{2}\)
VITEEE-2019
Application of Derivatives
85639
If at \(x=1\), the function \(x^{4}-62 x^{2}+a x+9\) attains its maximum value on the interval \([0,2]\), then the value of \(a\) is
1 110
2 10
3 55
4 None of these
Explanation:
(D) : Let \(f(x)=x^{4}-62 x^{2}+a x+9\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=4 \mathrm{x}^{3}-124 \mathrm{x}+\mathrm{a}\) It is given that function its maximum Value on the interval \([0,2]\) at \(x=1\) \(\therefore \mathrm{f}^{\prime}(1)=0\) \(\Rightarrow 4 \times(1)^{3}-124(1)+\mathrm{a}=0\) \(\Rightarrow 4-124+\mathrm{a}=0\) \(\Rightarrow-120+\mathrm{a}=0\) \(\mathrm{a}=120\) Hence, the value of a is 120
VITEEE-2018
Application of Derivatives
85640
If \(f(x)=x^{2}, g(x)=2 x, 0 \leq x \leq 2\) then the value of
85637
The minimum value of \(f(x)=e^{\left(x^{4}-x^{3}+x^{2}\right)}\) i
1 e
2 \(-\mathrm{e}\)
3 1
4 -1
Explanation:
(C) : We have \(f(x)=e^{\left(x^{4}-x^{3}+x^{2}\right)}\) Differentiating with respect to \(x\) we get- \(f^{\prime}(x)=e^{\left(x^{4}-x^{3}+x^{2}\right)} \quad\left(4 x^{3}-3 x^{2}+2 x\right)\) For maxima or minima \(f^{\prime}(x)=0\) \(e^{\left(x^{4}-x^{3}+x^{2}\right)}\) \(x\left(4 x^{3}-3 x^{2}+2 x\right)=0\) \(x=0\) The function has minimum value at \(\mathrm{x}=0\) Now, minimum value of function \(\mathrm{f}(\mathrm{x})=\mathrm{e}^{(0-0+0)}=\mathrm{e}^{0}\) \(\mathrm{f}(\mathrm{x})=1\) Hence, minimum value of function \(f(x)=e^{\left(x^{4}-x^{3}+x^{2}\right)} \text { is } 1\)
WB JEE-2010
Application of Derivatives
85638
A wire \(34 \mathrm{~cm}\) long is to be bent in the form of a quadrilateral of which each angle is \(90^{\circ}\). What is the maximum area which can be enclosed inside the quadrilateral?
1 \(68 \mathrm{~cm}^{2}\)
2 \(70 \mathrm{~cm}^{2}\)
3 \(71.25 \mathrm{~cm}^{2}\)
4 \(72.25 \mathrm{~cm}^{2}\)
Explanation:
(D) : Let one side of quadrilateral be \(x\) and another side be y so, \(2(x+y)=34\) \(\Rightarrow \quad x+y=17 \tag{i}\) We know from the basic principle that for a given perimeter square has the maximum area So, \(\quad \mathrm{x}=\mathrm{y}\) and putting this value in equation (i) We have \(\mathrm{x}=\mathrm{y}=\frac{17}{2}\) Area \(x . y=\frac{17}{2} \times \frac{17}{2}=\frac{289}{4}\) Area \(=72.25 \mathrm{~cm}^{2}\)
VITEEE-2019
Application of Derivatives
85639
If at \(x=1\), the function \(x^{4}-62 x^{2}+a x+9\) attains its maximum value on the interval \([0,2]\), then the value of \(a\) is
1 110
2 10
3 55
4 None of these
Explanation:
(D) : Let \(f(x)=x^{4}-62 x^{2}+a x+9\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=4 \mathrm{x}^{3}-124 \mathrm{x}+\mathrm{a}\) It is given that function its maximum Value on the interval \([0,2]\) at \(x=1\) \(\therefore \mathrm{f}^{\prime}(1)=0\) \(\Rightarrow 4 \times(1)^{3}-124(1)+\mathrm{a}=0\) \(\Rightarrow 4-124+\mathrm{a}=0\) \(\Rightarrow-120+\mathrm{a}=0\) \(\mathrm{a}=120\) Hence, the value of a is 120
VITEEE-2018
Application of Derivatives
85640
If \(f(x)=x^{2}, g(x)=2 x, 0 \leq x \leq 2\) then the value of
85637
The minimum value of \(f(x)=e^{\left(x^{4}-x^{3}+x^{2}\right)}\) i
1 e
2 \(-\mathrm{e}\)
3 1
4 -1
Explanation:
(C) : We have \(f(x)=e^{\left(x^{4}-x^{3}+x^{2}\right)}\) Differentiating with respect to \(x\) we get- \(f^{\prime}(x)=e^{\left(x^{4}-x^{3}+x^{2}\right)} \quad\left(4 x^{3}-3 x^{2}+2 x\right)\) For maxima or minima \(f^{\prime}(x)=0\) \(e^{\left(x^{4}-x^{3}+x^{2}\right)}\) \(x\left(4 x^{3}-3 x^{2}+2 x\right)=0\) \(x=0\) The function has minimum value at \(\mathrm{x}=0\) Now, minimum value of function \(\mathrm{f}(\mathrm{x})=\mathrm{e}^{(0-0+0)}=\mathrm{e}^{0}\) \(\mathrm{f}(\mathrm{x})=1\) Hence, minimum value of function \(f(x)=e^{\left(x^{4}-x^{3}+x^{2}\right)} \text { is } 1\)
WB JEE-2010
Application of Derivatives
85638
A wire \(34 \mathrm{~cm}\) long is to be bent in the form of a quadrilateral of which each angle is \(90^{\circ}\). What is the maximum area which can be enclosed inside the quadrilateral?
1 \(68 \mathrm{~cm}^{2}\)
2 \(70 \mathrm{~cm}^{2}\)
3 \(71.25 \mathrm{~cm}^{2}\)
4 \(72.25 \mathrm{~cm}^{2}\)
Explanation:
(D) : Let one side of quadrilateral be \(x\) and another side be y so, \(2(x+y)=34\) \(\Rightarrow \quad x+y=17 \tag{i}\) We know from the basic principle that for a given perimeter square has the maximum area So, \(\quad \mathrm{x}=\mathrm{y}\) and putting this value in equation (i) We have \(\mathrm{x}=\mathrm{y}=\frac{17}{2}\) Area \(x . y=\frac{17}{2} \times \frac{17}{2}=\frac{289}{4}\) Area \(=72.25 \mathrm{~cm}^{2}\)
VITEEE-2019
Application of Derivatives
85639
If at \(x=1\), the function \(x^{4}-62 x^{2}+a x+9\) attains its maximum value on the interval \([0,2]\), then the value of \(a\) is
1 110
2 10
3 55
4 None of these
Explanation:
(D) : Let \(f(x)=x^{4}-62 x^{2}+a x+9\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=4 \mathrm{x}^{3}-124 \mathrm{x}+\mathrm{a}\) It is given that function its maximum Value on the interval \([0,2]\) at \(x=1\) \(\therefore \mathrm{f}^{\prime}(1)=0\) \(\Rightarrow 4 \times(1)^{3}-124(1)+\mathrm{a}=0\) \(\Rightarrow 4-124+\mathrm{a}=0\) \(\Rightarrow-120+\mathrm{a}=0\) \(\mathrm{a}=120\) Hence, the value of a is 120
VITEEE-2018
Application of Derivatives
85640
If \(f(x)=x^{2}, g(x)=2 x, 0 \leq x \leq 2\) then the value of
