87031
The lengths of the sides of a triangle are \(10+\) \(x^{2}, 10+x^{2}\) and \(20-2 x^{2}\). If for \(x=k\), the area of the triangle is maximum, then \(3 \mathrm{k}^{2}\) is equal to :
1 5
2 8
3 10
4 12
Explanation:
(C) : Given that, Sides of triangle are - \(a=20-2 x^{2}\) \(b=10+x^{2}\) \(c=10+x^{2}\) Perimeter, \(\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}\) \(=\frac{20-2 x^{2}+10+x^{2}+10+x^{2}}{2}=\frac{40}{2}=20\) Area of triangle, \(\Delta=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}\) \(=\sqrt{20\left(20-20+2 x^{2}\right)\left(20-10-x^{2}\right)\left(20-10-x^{2}\right)}\) \(=\sqrt{20\left(2 x^{2}\right)\left(10-x^{2}\right)\left(10-x^{2}\right)}\) \(=\sqrt{40 x^{2}\left(10-x^{2}\right)^{2}}=2 x\left(10-x^{2}\right) \sqrt{10}\) \(=2 \sqrt{10}\left(10 x-x^{3}\right)\) \(\mathrm{s}=\left(10 \mathrm{x}-\mathrm{x}^{3}\right)\) \(\frac{\mathrm{ds}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(10 \mathrm{x}-\mathrm{x}^{3}\right)\) \(=\left(10-3 x^{2}\right)=0\) \(3 x^{2}=10\) Put, \(\quad \mathrm{x}=\mathrm{k}\) \(3 \mathrm{k}^{2}=10\)
Shift-I
Application of the Integrals
87032
Let \(\mathbf{R}\) be the point \((3,7)\) and let \(P\) and \(Q\) be two points on the line \(x+y=5\) such that \(P Q R\) is an equilateral triangle. Then the area of \(\triangle \mathrm{PQR}\) is:
1 \(\frac{25}{4 \sqrt{3}}\)
2 \(\frac{25 \sqrt{3}}{2}\)
3 \(\frac{25}{\sqrt{3}}\)
4 \(\frac{25}{2 \sqrt{3}}\)
Explanation:
(D) : Given that, \(\text { line, } x+y=5\) \(\text { Point } \mathrm{R}=(3,7)\) Side of triangle \(=\mathrm{a}\) \(\therefore\) This triangle represents the equilateral triangle So, the angle \(\mathrm{Q}\) PR is \(60^{\circ}\) \(\quad h=\left|\frac{\mathrm{ax}_1+\mathrm{by}_1+c}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\right| \Rightarrow \mathrm{h}=\left|\frac{3+7-5}{\sqrt{1+1}}\right|\) \(\therefore \text { Height }=\frac{5}{\sqrt{2}}\) \(\operatorname{Sin} 60^{\circ}=\frac{5 / \sqrt{2}}{\mathrm{a}} \Rightarrow \mathrm{a} \times \frac{\sqrt{3}}{2}=\frac{5}{\sqrt{2}} \Rightarrow \mathrm{a}=\frac{5 \sqrt{2}}{\sqrt{3}}\) Area of triangle \(A=\frac{\sqrt{3}}{4} a^{2}\) \(=\frac{\sqrt{3}}{4} \times\left(\frac{5 \sqrt{2}}{\sqrt{3}}\right)^{2}=\frac{25 \times 2 \sqrt{3}}{4 \times 3}=\frac{25}{2 \sqrt{3}}\)
Shift-I
Application of the Integrals
87035
The curve \(y=a x^{2}+b x\) passes through the point \((1,2)\) and lies above the \(\mathrm{X}\)-axis for \(0 \leq \mathrm{x} \leq 8\). If the area enclosed by this curve, the \(\mathrm{X}\)-axis the line \(x=6\) is 108 square units, then \(2 b-a=\)
1 2
2 0
3 1
4 -1
Explanation:
(B) : Given, Curve \(y=a^{2}+b x\) passes through point \((1,2)\) \(\therefore 2=\mathrm{a}(1)^{2}+\mathrm{b}(1)\) \(\mathrm{a}+\mathrm{b}=2\) Given, area under curve \(=\int_{0}^{6}\left(a^{2}+b x\right) d x=108\) \({\left[\frac{a x^{3}}{3}+\frac{b x^{2}}{2}\right]_{0}^{6}=108}\) \(72 a+18 b=108\) \(4 a+b=6 \tag{ii}\) By solving equation (i) and (ii), we get - \(\mathrm{a}=\frac{4}{3} \quad \text { and } \quad \mathrm{b}=\frac{2}{3}\) \(\therefore \quad 2 \mathrm{~b}-\mathrm{a}=2 \times \frac{2}{3}-\frac{4}{3}=0\)
87031
The lengths of the sides of a triangle are \(10+\) \(x^{2}, 10+x^{2}\) and \(20-2 x^{2}\). If for \(x=k\), the area of the triangle is maximum, then \(3 \mathrm{k}^{2}\) is equal to :
1 5
2 8
3 10
4 12
Explanation:
(C) : Given that, Sides of triangle are - \(a=20-2 x^{2}\) \(b=10+x^{2}\) \(c=10+x^{2}\) Perimeter, \(\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}\) \(=\frac{20-2 x^{2}+10+x^{2}+10+x^{2}}{2}=\frac{40}{2}=20\) Area of triangle, \(\Delta=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}\) \(=\sqrt{20\left(20-20+2 x^{2}\right)\left(20-10-x^{2}\right)\left(20-10-x^{2}\right)}\) \(=\sqrt{20\left(2 x^{2}\right)\left(10-x^{2}\right)\left(10-x^{2}\right)}\) \(=\sqrt{40 x^{2}\left(10-x^{2}\right)^{2}}=2 x\left(10-x^{2}\right) \sqrt{10}\) \(=2 \sqrt{10}\left(10 x-x^{3}\right)\) \(\mathrm{s}=\left(10 \mathrm{x}-\mathrm{x}^{3}\right)\) \(\frac{\mathrm{ds}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(10 \mathrm{x}-\mathrm{x}^{3}\right)\) \(=\left(10-3 x^{2}\right)=0\) \(3 x^{2}=10\) Put, \(\quad \mathrm{x}=\mathrm{k}\) \(3 \mathrm{k}^{2}=10\)
Shift-I
Application of the Integrals
87032
Let \(\mathbf{R}\) be the point \((3,7)\) and let \(P\) and \(Q\) be two points on the line \(x+y=5\) such that \(P Q R\) is an equilateral triangle. Then the area of \(\triangle \mathrm{PQR}\) is:
1 \(\frac{25}{4 \sqrt{3}}\)
2 \(\frac{25 \sqrt{3}}{2}\)
3 \(\frac{25}{\sqrt{3}}\)
4 \(\frac{25}{2 \sqrt{3}}\)
Explanation:
(D) : Given that, \(\text { line, } x+y=5\) \(\text { Point } \mathrm{R}=(3,7)\) Side of triangle \(=\mathrm{a}\) \(\therefore\) This triangle represents the equilateral triangle So, the angle \(\mathrm{Q}\) PR is \(60^{\circ}\) \(\quad h=\left|\frac{\mathrm{ax}_1+\mathrm{by}_1+c}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\right| \Rightarrow \mathrm{h}=\left|\frac{3+7-5}{\sqrt{1+1}}\right|\) \(\therefore \text { Height }=\frac{5}{\sqrt{2}}\) \(\operatorname{Sin} 60^{\circ}=\frac{5 / \sqrt{2}}{\mathrm{a}} \Rightarrow \mathrm{a} \times \frac{\sqrt{3}}{2}=\frac{5}{\sqrt{2}} \Rightarrow \mathrm{a}=\frac{5 \sqrt{2}}{\sqrt{3}}\) Area of triangle \(A=\frac{\sqrt{3}}{4} a^{2}\) \(=\frac{\sqrt{3}}{4} \times\left(\frac{5 \sqrt{2}}{\sqrt{3}}\right)^{2}=\frac{25 \times 2 \sqrt{3}}{4 \times 3}=\frac{25}{2 \sqrt{3}}\)
Shift-I
Application of the Integrals
87035
The curve \(y=a x^{2}+b x\) passes through the point \((1,2)\) and lies above the \(\mathrm{X}\)-axis for \(0 \leq \mathrm{x} \leq 8\). If the area enclosed by this curve, the \(\mathrm{X}\)-axis the line \(x=6\) is 108 square units, then \(2 b-a=\)
1 2
2 0
3 1
4 -1
Explanation:
(B) : Given, Curve \(y=a^{2}+b x\) passes through point \((1,2)\) \(\therefore 2=\mathrm{a}(1)^{2}+\mathrm{b}(1)\) \(\mathrm{a}+\mathrm{b}=2\) Given, area under curve \(=\int_{0}^{6}\left(a^{2}+b x\right) d x=108\) \({\left[\frac{a x^{3}}{3}+\frac{b x^{2}}{2}\right]_{0}^{6}=108}\) \(72 a+18 b=108\) \(4 a+b=6 \tag{ii}\) By solving equation (i) and (ii), we get - \(\mathrm{a}=\frac{4}{3} \quad \text { and } \quad \mathrm{b}=\frac{2}{3}\) \(\therefore \quad 2 \mathrm{~b}-\mathrm{a}=2 \times \frac{2}{3}-\frac{4}{3}=0\)
87031
The lengths of the sides of a triangle are \(10+\) \(x^{2}, 10+x^{2}\) and \(20-2 x^{2}\). If for \(x=k\), the area of the triangle is maximum, then \(3 \mathrm{k}^{2}\) is equal to :
1 5
2 8
3 10
4 12
Explanation:
(C) : Given that, Sides of triangle are - \(a=20-2 x^{2}\) \(b=10+x^{2}\) \(c=10+x^{2}\) Perimeter, \(\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}\) \(=\frac{20-2 x^{2}+10+x^{2}+10+x^{2}}{2}=\frac{40}{2}=20\) Area of triangle, \(\Delta=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}\) \(=\sqrt{20\left(20-20+2 x^{2}\right)\left(20-10-x^{2}\right)\left(20-10-x^{2}\right)}\) \(=\sqrt{20\left(2 x^{2}\right)\left(10-x^{2}\right)\left(10-x^{2}\right)}\) \(=\sqrt{40 x^{2}\left(10-x^{2}\right)^{2}}=2 x\left(10-x^{2}\right) \sqrt{10}\) \(=2 \sqrt{10}\left(10 x-x^{3}\right)\) \(\mathrm{s}=\left(10 \mathrm{x}-\mathrm{x}^{3}\right)\) \(\frac{\mathrm{ds}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(10 \mathrm{x}-\mathrm{x}^{3}\right)\) \(=\left(10-3 x^{2}\right)=0\) \(3 x^{2}=10\) Put, \(\quad \mathrm{x}=\mathrm{k}\) \(3 \mathrm{k}^{2}=10\)
Shift-I
Application of the Integrals
87032
Let \(\mathbf{R}\) be the point \((3,7)\) and let \(P\) and \(Q\) be two points on the line \(x+y=5\) such that \(P Q R\) is an equilateral triangle. Then the area of \(\triangle \mathrm{PQR}\) is:
1 \(\frac{25}{4 \sqrt{3}}\)
2 \(\frac{25 \sqrt{3}}{2}\)
3 \(\frac{25}{\sqrt{3}}\)
4 \(\frac{25}{2 \sqrt{3}}\)
Explanation:
(D) : Given that, \(\text { line, } x+y=5\) \(\text { Point } \mathrm{R}=(3,7)\) Side of triangle \(=\mathrm{a}\) \(\therefore\) This triangle represents the equilateral triangle So, the angle \(\mathrm{Q}\) PR is \(60^{\circ}\) \(\quad h=\left|\frac{\mathrm{ax}_1+\mathrm{by}_1+c}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\right| \Rightarrow \mathrm{h}=\left|\frac{3+7-5}{\sqrt{1+1}}\right|\) \(\therefore \text { Height }=\frac{5}{\sqrt{2}}\) \(\operatorname{Sin} 60^{\circ}=\frac{5 / \sqrt{2}}{\mathrm{a}} \Rightarrow \mathrm{a} \times \frac{\sqrt{3}}{2}=\frac{5}{\sqrt{2}} \Rightarrow \mathrm{a}=\frac{5 \sqrt{2}}{\sqrt{3}}\) Area of triangle \(A=\frac{\sqrt{3}}{4} a^{2}\) \(=\frac{\sqrt{3}}{4} \times\left(\frac{5 \sqrt{2}}{\sqrt{3}}\right)^{2}=\frac{25 \times 2 \sqrt{3}}{4 \times 3}=\frac{25}{2 \sqrt{3}}\)
Shift-I
Application of the Integrals
87035
The curve \(y=a x^{2}+b x\) passes through the point \((1,2)\) and lies above the \(\mathrm{X}\)-axis for \(0 \leq \mathrm{x} \leq 8\). If the area enclosed by this curve, the \(\mathrm{X}\)-axis the line \(x=6\) is 108 square units, then \(2 b-a=\)
1 2
2 0
3 1
4 -1
Explanation:
(B) : Given, Curve \(y=a^{2}+b x\) passes through point \((1,2)\) \(\therefore 2=\mathrm{a}(1)^{2}+\mathrm{b}(1)\) \(\mathrm{a}+\mathrm{b}=2\) Given, area under curve \(=\int_{0}^{6}\left(a^{2}+b x\right) d x=108\) \({\left[\frac{a x^{3}}{3}+\frac{b x^{2}}{2}\right]_{0}^{6}=108}\) \(72 a+18 b=108\) \(4 a+b=6 \tag{ii}\) By solving equation (i) and (ii), we get - \(\mathrm{a}=\frac{4}{3} \quad \text { and } \quad \mathrm{b}=\frac{2}{3}\) \(\therefore \quad 2 \mathrm{~b}-\mathrm{a}=2 \times \frac{2}{3}-\frac{4}{3}=0\)
87031
The lengths of the sides of a triangle are \(10+\) \(x^{2}, 10+x^{2}\) and \(20-2 x^{2}\). If for \(x=k\), the area of the triangle is maximum, then \(3 \mathrm{k}^{2}\) is equal to :
1 5
2 8
3 10
4 12
Explanation:
(C) : Given that, Sides of triangle are - \(a=20-2 x^{2}\) \(b=10+x^{2}\) \(c=10+x^{2}\) Perimeter, \(\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}\) \(=\frac{20-2 x^{2}+10+x^{2}+10+x^{2}}{2}=\frac{40}{2}=20\) Area of triangle, \(\Delta=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}\) \(=\sqrt{20\left(20-20+2 x^{2}\right)\left(20-10-x^{2}\right)\left(20-10-x^{2}\right)}\) \(=\sqrt{20\left(2 x^{2}\right)\left(10-x^{2}\right)\left(10-x^{2}\right)}\) \(=\sqrt{40 x^{2}\left(10-x^{2}\right)^{2}}=2 x\left(10-x^{2}\right) \sqrt{10}\) \(=2 \sqrt{10}\left(10 x-x^{3}\right)\) \(\mathrm{s}=\left(10 \mathrm{x}-\mathrm{x}^{3}\right)\) \(\frac{\mathrm{ds}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(10 \mathrm{x}-\mathrm{x}^{3}\right)\) \(=\left(10-3 x^{2}\right)=0\) \(3 x^{2}=10\) Put, \(\quad \mathrm{x}=\mathrm{k}\) \(3 \mathrm{k}^{2}=10\)
Shift-I
Application of the Integrals
87032
Let \(\mathbf{R}\) be the point \((3,7)\) and let \(P\) and \(Q\) be two points on the line \(x+y=5\) such that \(P Q R\) is an equilateral triangle. Then the area of \(\triangle \mathrm{PQR}\) is:
1 \(\frac{25}{4 \sqrt{3}}\)
2 \(\frac{25 \sqrt{3}}{2}\)
3 \(\frac{25}{\sqrt{3}}\)
4 \(\frac{25}{2 \sqrt{3}}\)
Explanation:
(D) : Given that, \(\text { line, } x+y=5\) \(\text { Point } \mathrm{R}=(3,7)\) Side of triangle \(=\mathrm{a}\) \(\therefore\) This triangle represents the equilateral triangle So, the angle \(\mathrm{Q}\) PR is \(60^{\circ}\) \(\quad h=\left|\frac{\mathrm{ax}_1+\mathrm{by}_1+c}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\right| \Rightarrow \mathrm{h}=\left|\frac{3+7-5}{\sqrt{1+1}}\right|\) \(\therefore \text { Height }=\frac{5}{\sqrt{2}}\) \(\operatorname{Sin} 60^{\circ}=\frac{5 / \sqrt{2}}{\mathrm{a}} \Rightarrow \mathrm{a} \times \frac{\sqrt{3}}{2}=\frac{5}{\sqrt{2}} \Rightarrow \mathrm{a}=\frac{5 \sqrt{2}}{\sqrt{3}}\) Area of triangle \(A=\frac{\sqrt{3}}{4} a^{2}\) \(=\frac{\sqrt{3}}{4} \times\left(\frac{5 \sqrt{2}}{\sqrt{3}}\right)^{2}=\frac{25 \times 2 \sqrt{3}}{4 \times 3}=\frac{25}{2 \sqrt{3}}\)
Shift-I
Application of the Integrals
87035
The curve \(y=a x^{2}+b x\) passes through the point \((1,2)\) and lies above the \(\mathrm{X}\)-axis for \(0 \leq \mathrm{x} \leq 8\). If the area enclosed by this curve, the \(\mathrm{X}\)-axis the line \(x=6\) is 108 square units, then \(2 b-a=\)
1 2
2 0
3 1
4 -1
Explanation:
(B) : Given, Curve \(y=a^{2}+b x\) passes through point \((1,2)\) \(\therefore 2=\mathrm{a}(1)^{2}+\mathrm{b}(1)\) \(\mathrm{a}+\mathrm{b}=2\) Given, area under curve \(=\int_{0}^{6}\left(a^{2}+b x\right) d x=108\) \({\left[\frac{a x^{3}}{3}+\frac{b x^{2}}{2}\right]_{0}^{6}=108}\) \(72 a+18 b=108\) \(4 a+b=6 \tag{ii}\) By solving equation (i) and (ii), we get - \(\mathrm{a}=\frac{4}{3} \quad \text { and } \quad \mathrm{b}=\frac{2}{3}\) \(\therefore \quad 2 \mathrm{~b}-\mathrm{a}=2 \times \frac{2}{3}-\frac{4}{3}=0\)
