85807
If \(a\) and \(b\) are non-zero roots of \(x^{2}+a x+b=0\) then the least value of \(x^{2}+a x+b\) is
1 \(\frac{2}{3}\)
2 \(-\frac{9}{4}\)
3 \(\frac{9}{4}\)
4 1
Explanation:
(B) : Given, Sum of roots \(a+b=-a\) \(\mathrm{b}=-2 \mathrm{a} \tag{i}\) Product of root \(a b=b\) If \(b=0\) then \(a=0\) \(\mathrm{b}(\mathrm{a}-1)=0\) If \(b \neq 0\) then \(a=1\) and \(b=-2\) So, expression will be, \(f(x)=x^{2}+x-2\) \(=x^{2}+2 \frac{1}{2} x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}-2\) \(f(x)=\left(x+\frac{1}{2}\right)^{2}-\frac{9}{4}\) For \(\mathrm{f}(\mathrm{x})\) is minimum, \(\left(x+\frac{1}{2}\right)^{2}=0\) When, \(\mathrm{x}=\frac{-1}{2}\) Minimum value of function \(=\frac{-9}{4}\)
BITSAT-2015
Application of Derivatives
85808
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 \(\mathrm{m}\) and another side be \(n\). Perimeter of quadrilateral, \(2(\mathrm{~m}+\mathrm{n})=34\) \(\mathrm{~m}+\mathrm{n}=17 \tag{i}\) We know from the basic principle that for a given perimeter square has the maximum are a, So, \(\mathrm{m}=\mathrm{n}\) and putting this value in equation (i) \(\mathrm{m}=\mathrm{n}=\frac{17}{2}\) Area \(=\mathrm{m} \cdot \mathrm{n}=\frac{17}{2} \cdot \frac{17}{2}=72.25 \mathrm{~cm}^{2}\)
BITSAT-2016
Application of Derivatives
85809
If \(e^{x}=y+\sqrt{1+y^{2}}\), then the value of \(y\) is
85810
The part of straight line \(y=x+1\) between \(x=2\) and \(x=3\) is revolved about \(x\)-axis, then the curved surface of the solid thus generated is
1 \(\frac{37 \pi}{3}\)
2 \(7 \pi \sqrt{2}\)
3 \(37 \pi\)
4 \(\frac{7 \pi}{\sqrt{2}}\)
Explanation:
(B) : Curved surface area obtained on revolving a function between \(\mathrm{x}=\mathrm{a}\) and \(\mathrm{x}=\mathrm{b}\) along \(\mathrm{x}-\) axis is, Curved surface area \((A)=\int_{a}^{b} 2 \pi y \sqrt{\left\{1+\left(\frac{d y}{d x}\right)^{2}\right\} d x}\) Given, \(y=x+1 , x=2 \text { and } x=3\) \(\frac{d y}{d x}=1\) \(A=\int_{2}^{3} 2 \pi(x+1) \sqrt{\left\{1+(1)^{2}\right\} d x}\) \(=2 \sqrt{2} \pi\left[\frac{x^{2}}{2}+x\right]_{2}^{3}=2 \sqrt{2} \pi\left[\frac{9}{2}+3-\frac{4}{2}-2\right]=2 \sqrt{2} \pi \times \frac{7}{2}\) \(=7 \sqrt{2} \pi=7 \pi \sqrt{2}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of Derivatives
85807
If \(a\) and \(b\) are non-zero roots of \(x^{2}+a x+b=0\) then the least value of \(x^{2}+a x+b\) is
1 \(\frac{2}{3}\)
2 \(-\frac{9}{4}\)
3 \(\frac{9}{4}\)
4 1
Explanation:
(B) : Given, Sum of roots \(a+b=-a\) \(\mathrm{b}=-2 \mathrm{a} \tag{i}\) Product of root \(a b=b\) If \(b=0\) then \(a=0\) \(\mathrm{b}(\mathrm{a}-1)=0\) If \(b \neq 0\) then \(a=1\) and \(b=-2\) So, expression will be, \(f(x)=x^{2}+x-2\) \(=x^{2}+2 \frac{1}{2} x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}-2\) \(f(x)=\left(x+\frac{1}{2}\right)^{2}-\frac{9}{4}\) For \(\mathrm{f}(\mathrm{x})\) is minimum, \(\left(x+\frac{1}{2}\right)^{2}=0\) When, \(\mathrm{x}=\frac{-1}{2}\) Minimum value of function \(=\frac{-9}{4}\)
BITSAT-2015
Application of Derivatives
85808
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 \(\mathrm{m}\) and another side be \(n\). Perimeter of quadrilateral, \(2(\mathrm{~m}+\mathrm{n})=34\) \(\mathrm{~m}+\mathrm{n}=17 \tag{i}\) We know from the basic principle that for a given perimeter square has the maximum are a, So, \(\mathrm{m}=\mathrm{n}\) and putting this value in equation (i) \(\mathrm{m}=\mathrm{n}=\frac{17}{2}\) Area \(=\mathrm{m} \cdot \mathrm{n}=\frac{17}{2} \cdot \frac{17}{2}=72.25 \mathrm{~cm}^{2}\)
BITSAT-2016
Application of Derivatives
85809
If \(e^{x}=y+\sqrt{1+y^{2}}\), then the value of \(y\) is
85810
The part of straight line \(y=x+1\) between \(x=2\) and \(x=3\) is revolved about \(x\)-axis, then the curved surface of the solid thus generated is
1 \(\frac{37 \pi}{3}\)
2 \(7 \pi \sqrt{2}\)
3 \(37 \pi\)
4 \(\frac{7 \pi}{\sqrt{2}}\)
Explanation:
(B) : Curved surface area obtained on revolving a function between \(\mathrm{x}=\mathrm{a}\) and \(\mathrm{x}=\mathrm{b}\) along \(\mathrm{x}-\) axis is, Curved surface area \((A)=\int_{a}^{b} 2 \pi y \sqrt{\left\{1+\left(\frac{d y}{d x}\right)^{2}\right\} d x}\) Given, \(y=x+1 , x=2 \text { and } x=3\) \(\frac{d y}{d x}=1\) \(A=\int_{2}^{3} 2 \pi(x+1) \sqrt{\left\{1+(1)^{2}\right\} d x}\) \(=2 \sqrt{2} \pi\left[\frac{x^{2}}{2}+x\right]_{2}^{3}=2 \sqrt{2} \pi\left[\frac{9}{2}+3-\frac{4}{2}-2\right]=2 \sqrt{2} \pi \times \frac{7}{2}\) \(=7 \sqrt{2} \pi=7 \pi \sqrt{2}\)
85807
If \(a\) and \(b\) are non-zero roots of \(x^{2}+a x+b=0\) then the least value of \(x^{2}+a x+b\) is
1 \(\frac{2}{3}\)
2 \(-\frac{9}{4}\)
3 \(\frac{9}{4}\)
4 1
Explanation:
(B) : Given, Sum of roots \(a+b=-a\) \(\mathrm{b}=-2 \mathrm{a} \tag{i}\) Product of root \(a b=b\) If \(b=0\) then \(a=0\) \(\mathrm{b}(\mathrm{a}-1)=0\) If \(b \neq 0\) then \(a=1\) and \(b=-2\) So, expression will be, \(f(x)=x^{2}+x-2\) \(=x^{2}+2 \frac{1}{2} x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}-2\) \(f(x)=\left(x+\frac{1}{2}\right)^{2}-\frac{9}{4}\) For \(\mathrm{f}(\mathrm{x})\) is minimum, \(\left(x+\frac{1}{2}\right)^{2}=0\) When, \(\mathrm{x}=\frac{-1}{2}\) Minimum value of function \(=\frac{-9}{4}\)
BITSAT-2015
Application of Derivatives
85808
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 \(\mathrm{m}\) and another side be \(n\). Perimeter of quadrilateral, \(2(\mathrm{~m}+\mathrm{n})=34\) \(\mathrm{~m}+\mathrm{n}=17 \tag{i}\) We know from the basic principle that for a given perimeter square has the maximum are a, So, \(\mathrm{m}=\mathrm{n}\) and putting this value in equation (i) \(\mathrm{m}=\mathrm{n}=\frac{17}{2}\) Area \(=\mathrm{m} \cdot \mathrm{n}=\frac{17}{2} \cdot \frac{17}{2}=72.25 \mathrm{~cm}^{2}\)
BITSAT-2016
Application of Derivatives
85809
If \(e^{x}=y+\sqrt{1+y^{2}}\), then the value of \(y\) is
85810
The part of straight line \(y=x+1\) between \(x=2\) and \(x=3\) is revolved about \(x\)-axis, then the curved surface of the solid thus generated is
1 \(\frac{37 \pi}{3}\)
2 \(7 \pi \sqrt{2}\)
3 \(37 \pi\)
4 \(\frac{7 \pi}{\sqrt{2}}\)
Explanation:
(B) : Curved surface area obtained on revolving a function between \(\mathrm{x}=\mathrm{a}\) and \(\mathrm{x}=\mathrm{b}\) along \(\mathrm{x}-\) axis is, Curved surface area \((A)=\int_{a}^{b} 2 \pi y \sqrt{\left\{1+\left(\frac{d y}{d x}\right)^{2}\right\} d x}\) Given, \(y=x+1 , x=2 \text { and } x=3\) \(\frac{d y}{d x}=1\) \(A=\int_{2}^{3} 2 \pi(x+1) \sqrt{\left\{1+(1)^{2}\right\} d x}\) \(=2 \sqrt{2} \pi\left[\frac{x^{2}}{2}+x\right]_{2}^{3}=2 \sqrt{2} \pi\left[\frac{9}{2}+3-\frac{4}{2}-2\right]=2 \sqrt{2} \pi \times \frac{7}{2}\) \(=7 \sqrt{2} \pi=7 \pi \sqrt{2}\)
85807
If \(a\) and \(b\) are non-zero roots of \(x^{2}+a x+b=0\) then the least value of \(x^{2}+a x+b\) is
1 \(\frac{2}{3}\)
2 \(-\frac{9}{4}\)
3 \(\frac{9}{4}\)
4 1
Explanation:
(B) : Given, Sum of roots \(a+b=-a\) \(\mathrm{b}=-2 \mathrm{a} \tag{i}\) Product of root \(a b=b\) If \(b=0\) then \(a=0\) \(\mathrm{b}(\mathrm{a}-1)=0\) If \(b \neq 0\) then \(a=1\) and \(b=-2\) So, expression will be, \(f(x)=x^{2}+x-2\) \(=x^{2}+2 \frac{1}{2} x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}-2\) \(f(x)=\left(x+\frac{1}{2}\right)^{2}-\frac{9}{4}\) For \(\mathrm{f}(\mathrm{x})\) is minimum, \(\left(x+\frac{1}{2}\right)^{2}=0\) When, \(\mathrm{x}=\frac{-1}{2}\) Minimum value of function \(=\frac{-9}{4}\)
BITSAT-2015
Application of Derivatives
85808
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 \(\mathrm{m}\) and another side be \(n\). Perimeter of quadrilateral, \(2(\mathrm{~m}+\mathrm{n})=34\) \(\mathrm{~m}+\mathrm{n}=17 \tag{i}\) We know from the basic principle that for a given perimeter square has the maximum are a, So, \(\mathrm{m}=\mathrm{n}\) and putting this value in equation (i) \(\mathrm{m}=\mathrm{n}=\frac{17}{2}\) Area \(=\mathrm{m} \cdot \mathrm{n}=\frac{17}{2} \cdot \frac{17}{2}=72.25 \mathrm{~cm}^{2}\)
BITSAT-2016
Application of Derivatives
85809
If \(e^{x}=y+\sqrt{1+y^{2}}\), then the value of \(y\) is
85810
The part of straight line \(y=x+1\) between \(x=2\) and \(x=3\) is revolved about \(x\)-axis, then the curved surface of the solid thus generated is
1 \(\frac{37 \pi}{3}\)
2 \(7 \pi \sqrt{2}\)
3 \(37 \pi\)
4 \(\frac{7 \pi}{\sqrt{2}}\)
Explanation:
(B) : Curved surface area obtained on revolving a function between \(\mathrm{x}=\mathrm{a}\) and \(\mathrm{x}=\mathrm{b}\) along \(\mathrm{x}-\) axis is, Curved surface area \((A)=\int_{a}^{b} 2 \pi y \sqrt{\left\{1+\left(\frac{d y}{d x}\right)^{2}\right\} d x}\) Given, \(y=x+1 , x=2 \text { and } x=3\) \(\frac{d y}{d x}=1\) \(A=\int_{2}^{3} 2 \pi(x+1) \sqrt{\left\{1+(1)^{2}\right\} d x}\) \(=2 \sqrt{2} \pi\left[\frac{x^{2}}{2}+x\right]_{2}^{3}=2 \sqrt{2} \pi\left[\frac{9}{2}+3-\frac{4}{2}-2\right]=2 \sqrt{2} \pi \times \frac{7}{2}\) \(=7 \sqrt{2} \pi=7 \pi \sqrt{2}\)