86869
The normal to a curve at \(P(x, y)\) meets the \(x\) axis at \(G\). If the distance of \(G\) from the origin is twice the abscissa of \(P\) then curve is
1 A parabola
2 A circle
3 A hyperbola
4 An ellipse
Explanation:
(C): Given, \(\mathrm{G}=2 \mathrm{x}\), abcissa \(=\mathrm{x}\) Equation of normal is \(Y-y=-\frac{d x}{d y}(X-x)\) When, \(\mathrm{y}=0\) \(0-y=-\frac{d x}{d y}(X-x)\) \(\text { slope of tangent } \frac{\mathrm{dy}}{\mathrm{dx}}\) \(\text { slope of normal } \frac{-\mathrm{dx}}{\mathrm{dy}}\) \(\because \text { theproduct of both }=-1\) \(X-x=\frac{d y}{d x} y \Rightarrow X=y \frac{d y}{d x}+x\) \(\mathrm{G}=\mathrm{x}+\mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}\) It satisfies the equation \(\left[G=\left(x+y \frac{d y}{d x}, 0\right)\right]\) \(y \frac{d y}{d x}=x \Rightarrow y d y=x d x\) By integrate it, we get \(\int y d y=\int x d x\) \(\frac{y^{2}}{2}=\frac{x^{2}}{2}+c\) Therefore the curve will be hyperbola.
WB JEE-2021
Application of the Integrals
86870
The area (in sq. units) bounded by the curve \(y\) \(=x^{2}+2 x+1\) and the tangent to it at \((1,4)\) and the \(\mathbf{Y}\)-axis is
1 \(\frac{1}{3}\)
2 \(\frac{2}{3}\)
3 1
4 \(\frac{7}{3}\)
Explanation:
(A) : Given the curve \(y=x^{2}+2 x+1\) \(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}+2\) \(\left.\therefore \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{1,4}=2 \times 1+2=4\) \(\therefore\) The equation of the tangent is \(y-4=4(x-1) \Rightarrow y=4 x+c\) Required area \(=\int_{0}^{1} \mathrm{ydx}-\frac{1}{2} \mathrm{OA} \times \mathrm{AP}\) \(=\int_{0}^{1}\left(x^{2}+2 x+1\right) d x-\frac{1}{2} \times 1 \times 4\) \(=\left[\frac{\mathrm{x}^{3}}{3}+\frac{2 \mathrm{x}^{2}}{2}+\mathrm{x}\right]_{0}^{1}-2=\frac{1}{3}+1+1-2=\frac{1}{3}\) sq unit.
AP EAMCET-2019-21.04.2019
Application of the Integrals
86871
The area of the region bounded by the parabola \(y=x^{2}-4 x+5\) and the straight line \(y\) \(=\mathbf{x}+1\) is
1 \(\frac{1}{2}\)
2 2
3 3
4 \(\frac{9}{2}\)
Explanation:
(D) : Solving \(y=x+1\) and \(y=x^{2}-4 x+5\), we get \(\mathrm{y}_{2}=\mathrm{x}+1\) and \(\mathrm{y}\) \(\mathrm{x}^{2}-4 \mathrm{x}+5=\mathrm{x}+1\) or \(x^{2}-5 x+4=0\) or \(x^{2}-4 x-x+4=0\) or \(\quad x(x-4)-1(x-4)=0\) or \(\quad(x-1)(x-4)=0\) \(\therefore \mathrm{x}=1,4\) ired area is \(\int_{1}^{4}\left[(x+1)-\left(x^{2}-4 x+5\right)\right] d x=\int_{1}^{4}\left[x+1-x^{2}+4 x-5\right] d x\) \(=\int_{1}^{4}\left(5 x-x^{2}-4\right) d x=5 \int_{1}^{4} x d x-\int_{1}^{4} x^{2} d x-4 \int_{1}^{4} d x\) \(=5\left[\frac{x^{2}}{2}\right]_{1}^{4}-\left[\frac{x^{3}}{3}\right]_{1}^{4}-4[x]_{1}^{4}\) \(=\frac{5}{2}[16-1]-\frac{1}{3}[64-1]-4[4-1]\) \(=\frac{5}{2} \times 15-\frac{1}{3} \times 63-4 \times 3=\frac{75}{2}-21-12\) \(=\frac{75}{2}-33=\frac{75-66}{2}=\frac{9}{2}\)
WB JEE-2013
Application of the Integrals
86872
The area bounded by the curves \(y=|x|-1\) and \(y\) \(=-|\mathbf{x}|+1\) is
1 1 sq unit
2 2 sq units
3 \(2 \sqrt{2}\) squnits
4 4 sq units
Explanation:
(B) : The region is clearly square with vertices at the points \((1,0),(0,1),(-1,0)\) and \((0,-1)\). So, its area \(=\) \(\sqrt{2} \times \sqrt{2}=2\) sq units.
86869
The normal to a curve at \(P(x, y)\) meets the \(x\) axis at \(G\). If the distance of \(G\) from the origin is twice the abscissa of \(P\) then curve is
1 A parabola
2 A circle
3 A hyperbola
4 An ellipse
Explanation:
(C): Given, \(\mathrm{G}=2 \mathrm{x}\), abcissa \(=\mathrm{x}\) Equation of normal is \(Y-y=-\frac{d x}{d y}(X-x)\) When, \(\mathrm{y}=0\) \(0-y=-\frac{d x}{d y}(X-x)\) \(\text { slope of tangent } \frac{\mathrm{dy}}{\mathrm{dx}}\) \(\text { slope of normal } \frac{-\mathrm{dx}}{\mathrm{dy}}\) \(\because \text { theproduct of both }=-1\) \(X-x=\frac{d y}{d x} y \Rightarrow X=y \frac{d y}{d x}+x\) \(\mathrm{G}=\mathrm{x}+\mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}\) It satisfies the equation \(\left[G=\left(x+y \frac{d y}{d x}, 0\right)\right]\) \(y \frac{d y}{d x}=x \Rightarrow y d y=x d x\) By integrate it, we get \(\int y d y=\int x d x\) \(\frac{y^{2}}{2}=\frac{x^{2}}{2}+c\) Therefore the curve will be hyperbola.
WB JEE-2021
Application of the Integrals
86870
The area (in sq. units) bounded by the curve \(y\) \(=x^{2}+2 x+1\) and the tangent to it at \((1,4)\) and the \(\mathbf{Y}\)-axis is
1 \(\frac{1}{3}\)
2 \(\frac{2}{3}\)
3 1
4 \(\frac{7}{3}\)
Explanation:
(A) : Given the curve \(y=x^{2}+2 x+1\) \(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}+2\) \(\left.\therefore \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{1,4}=2 \times 1+2=4\) \(\therefore\) The equation of the tangent is \(y-4=4(x-1) \Rightarrow y=4 x+c\) Required area \(=\int_{0}^{1} \mathrm{ydx}-\frac{1}{2} \mathrm{OA} \times \mathrm{AP}\) \(=\int_{0}^{1}\left(x^{2}+2 x+1\right) d x-\frac{1}{2} \times 1 \times 4\) \(=\left[\frac{\mathrm{x}^{3}}{3}+\frac{2 \mathrm{x}^{2}}{2}+\mathrm{x}\right]_{0}^{1}-2=\frac{1}{3}+1+1-2=\frac{1}{3}\) sq unit.
AP EAMCET-2019-21.04.2019
Application of the Integrals
86871
The area of the region bounded by the parabola \(y=x^{2}-4 x+5\) and the straight line \(y\) \(=\mathbf{x}+1\) is
1 \(\frac{1}{2}\)
2 2
3 3
4 \(\frac{9}{2}\)
Explanation:
(D) : Solving \(y=x+1\) and \(y=x^{2}-4 x+5\), we get \(\mathrm{y}_{2}=\mathrm{x}+1\) and \(\mathrm{y}\) \(\mathrm{x}^{2}-4 \mathrm{x}+5=\mathrm{x}+1\) or \(x^{2}-5 x+4=0\) or \(x^{2}-4 x-x+4=0\) or \(\quad x(x-4)-1(x-4)=0\) or \(\quad(x-1)(x-4)=0\) \(\therefore \mathrm{x}=1,4\) ired area is \(\int_{1}^{4}\left[(x+1)-\left(x^{2}-4 x+5\right)\right] d x=\int_{1}^{4}\left[x+1-x^{2}+4 x-5\right] d x\) \(=\int_{1}^{4}\left(5 x-x^{2}-4\right) d x=5 \int_{1}^{4} x d x-\int_{1}^{4} x^{2} d x-4 \int_{1}^{4} d x\) \(=5\left[\frac{x^{2}}{2}\right]_{1}^{4}-\left[\frac{x^{3}}{3}\right]_{1}^{4}-4[x]_{1}^{4}\) \(=\frac{5}{2}[16-1]-\frac{1}{3}[64-1]-4[4-1]\) \(=\frac{5}{2} \times 15-\frac{1}{3} \times 63-4 \times 3=\frac{75}{2}-21-12\) \(=\frac{75}{2}-33=\frac{75-66}{2}=\frac{9}{2}\)
WB JEE-2013
Application of the Integrals
86872
The area bounded by the curves \(y=|x|-1\) and \(y\) \(=-|\mathbf{x}|+1\) is
1 1 sq unit
2 2 sq units
3 \(2 \sqrt{2}\) squnits
4 4 sq units
Explanation:
(B) : The region is clearly square with vertices at the points \((1,0),(0,1),(-1,0)\) and \((0,-1)\). So, its area \(=\) \(\sqrt{2} \times \sqrt{2}=2\) sq units.
86869
The normal to a curve at \(P(x, y)\) meets the \(x\) axis at \(G\). If the distance of \(G\) from the origin is twice the abscissa of \(P\) then curve is
1 A parabola
2 A circle
3 A hyperbola
4 An ellipse
Explanation:
(C): Given, \(\mathrm{G}=2 \mathrm{x}\), abcissa \(=\mathrm{x}\) Equation of normal is \(Y-y=-\frac{d x}{d y}(X-x)\) When, \(\mathrm{y}=0\) \(0-y=-\frac{d x}{d y}(X-x)\) \(\text { slope of tangent } \frac{\mathrm{dy}}{\mathrm{dx}}\) \(\text { slope of normal } \frac{-\mathrm{dx}}{\mathrm{dy}}\) \(\because \text { theproduct of both }=-1\) \(X-x=\frac{d y}{d x} y \Rightarrow X=y \frac{d y}{d x}+x\) \(\mathrm{G}=\mathrm{x}+\mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}\) It satisfies the equation \(\left[G=\left(x+y \frac{d y}{d x}, 0\right)\right]\) \(y \frac{d y}{d x}=x \Rightarrow y d y=x d x\) By integrate it, we get \(\int y d y=\int x d x\) \(\frac{y^{2}}{2}=\frac{x^{2}}{2}+c\) Therefore the curve will be hyperbola.
WB JEE-2021
Application of the Integrals
86870
The area (in sq. units) bounded by the curve \(y\) \(=x^{2}+2 x+1\) and the tangent to it at \((1,4)\) and the \(\mathbf{Y}\)-axis is
1 \(\frac{1}{3}\)
2 \(\frac{2}{3}\)
3 1
4 \(\frac{7}{3}\)
Explanation:
(A) : Given the curve \(y=x^{2}+2 x+1\) \(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}+2\) \(\left.\therefore \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{1,4}=2 \times 1+2=4\) \(\therefore\) The equation of the tangent is \(y-4=4(x-1) \Rightarrow y=4 x+c\) Required area \(=\int_{0}^{1} \mathrm{ydx}-\frac{1}{2} \mathrm{OA} \times \mathrm{AP}\) \(=\int_{0}^{1}\left(x^{2}+2 x+1\right) d x-\frac{1}{2} \times 1 \times 4\) \(=\left[\frac{\mathrm{x}^{3}}{3}+\frac{2 \mathrm{x}^{2}}{2}+\mathrm{x}\right]_{0}^{1}-2=\frac{1}{3}+1+1-2=\frac{1}{3}\) sq unit.
AP EAMCET-2019-21.04.2019
Application of the Integrals
86871
The area of the region bounded by the parabola \(y=x^{2}-4 x+5\) and the straight line \(y\) \(=\mathbf{x}+1\) is
1 \(\frac{1}{2}\)
2 2
3 3
4 \(\frac{9}{2}\)
Explanation:
(D) : Solving \(y=x+1\) and \(y=x^{2}-4 x+5\), we get \(\mathrm{y}_{2}=\mathrm{x}+1\) and \(\mathrm{y}\) \(\mathrm{x}^{2}-4 \mathrm{x}+5=\mathrm{x}+1\) or \(x^{2}-5 x+4=0\) or \(x^{2}-4 x-x+4=0\) or \(\quad x(x-4)-1(x-4)=0\) or \(\quad(x-1)(x-4)=0\) \(\therefore \mathrm{x}=1,4\) ired area is \(\int_{1}^{4}\left[(x+1)-\left(x^{2}-4 x+5\right)\right] d x=\int_{1}^{4}\left[x+1-x^{2}+4 x-5\right] d x\) \(=\int_{1}^{4}\left(5 x-x^{2}-4\right) d x=5 \int_{1}^{4} x d x-\int_{1}^{4} x^{2} d x-4 \int_{1}^{4} d x\) \(=5\left[\frac{x^{2}}{2}\right]_{1}^{4}-\left[\frac{x^{3}}{3}\right]_{1}^{4}-4[x]_{1}^{4}\) \(=\frac{5}{2}[16-1]-\frac{1}{3}[64-1]-4[4-1]\) \(=\frac{5}{2} \times 15-\frac{1}{3} \times 63-4 \times 3=\frac{75}{2}-21-12\) \(=\frac{75}{2}-33=\frac{75-66}{2}=\frac{9}{2}\)
WB JEE-2013
Application of the Integrals
86872
The area bounded by the curves \(y=|x|-1\) and \(y\) \(=-|\mathbf{x}|+1\) is
1 1 sq unit
2 2 sq units
3 \(2 \sqrt{2}\) squnits
4 4 sq units
Explanation:
(B) : The region is clearly square with vertices at the points \((1,0),(0,1),(-1,0)\) and \((0,-1)\). So, its area \(=\) \(\sqrt{2} \times \sqrt{2}=2\) sq units.
86869
The normal to a curve at \(P(x, y)\) meets the \(x\) axis at \(G\). If the distance of \(G\) from the origin is twice the abscissa of \(P\) then curve is
1 A parabola
2 A circle
3 A hyperbola
4 An ellipse
Explanation:
(C): Given, \(\mathrm{G}=2 \mathrm{x}\), abcissa \(=\mathrm{x}\) Equation of normal is \(Y-y=-\frac{d x}{d y}(X-x)\) When, \(\mathrm{y}=0\) \(0-y=-\frac{d x}{d y}(X-x)\) \(\text { slope of tangent } \frac{\mathrm{dy}}{\mathrm{dx}}\) \(\text { slope of normal } \frac{-\mathrm{dx}}{\mathrm{dy}}\) \(\because \text { theproduct of both }=-1\) \(X-x=\frac{d y}{d x} y \Rightarrow X=y \frac{d y}{d x}+x\) \(\mathrm{G}=\mathrm{x}+\mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}\) It satisfies the equation \(\left[G=\left(x+y \frac{d y}{d x}, 0\right)\right]\) \(y \frac{d y}{d x}=x \Rightarrow y d y=x d x\) By integrate it, we get \(\int y d y=\int x d x\) \(\frac{y^{2}}{2}=\frac{x^{2}}{2}+c\) Therefore the curve will be hyperbola.
WB JEE-2021
Application of the Integrals
86870
The area (in sq. units) bounded by the curve \(y\) \(=x^{2}+2 x+1\) and the tangent to it at \((1,4)\) and the \(\mathbf{Y}\)-axis is
1 \(\frac{1}{3}\)
2 \(\frac{2}{3}\)
3 1
4 \(\frac{7}{3}\)
Explanation:
(A) : Given the curve \(y=x^{2}+2 x+1\) \(\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}+2\) \(\left.\therefore \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{1,4}=2 \times 1+2=4\) \(\therefore\) The equation of the tangent is \(y-4=4(x-1) \Rightarrow y=4 x+c\) Required area \(=\int_{0}^{1} \mathrm{ydx}-\frac{1}{2} \mathrm{OA} \times \mathrm{AP}\) \(=\int_{0}^{1}\left(x^{2}+2 x+1\right) d x-\frac{1}{2} \times 1 \times 4\) \(=\left[\frac{\mathrm{x}^{3}}{3}+\frac{2 \mathrm{x}^{2}}{2}+\mathrm{x}\right]_{0}^{1}-2=\frac{1}{3}+1+1-2=\frac{1}{3}\) sq unit.
AP EAMCET-2019-21.04.2019
Application of the Integrals
86871
The area of the region bounded by the parabola \(y=x^{2}-4 x+5\) and the straight line \(y\) \(=\mathbf{x}+1\) is
1 \(\frac{1}{2}\)
2 2
3 3
4 \(\frac{9}{2}\)
Explanation:
(D) : Solving \(y=x+1\) and \(y=x^{2}-4 x+5\), we get \(\mathrm{y}_{2}=\mathrm{x}+1\) and \(\mathrm{y}\) \(\mathrm{x}^{2}-4 \mathrm{x}+5=\mathrm{x}+1\) or \(x^{2}-5 x+4=0\) or \(x^{2}-4 x-x+4=0\) or \(\quad x(x-4)-1(x-4)=0\) or \(\quad(x-1)(x-4)=0\) \(\therefore \mathrm{x}=1,4\) ired area is \(\int_{1}^{4}\left[(x+1)-\left(x^{2}-4 x+5\right)\right] d x=\int_{1}^{4}\left[x+1-x^{2}+4 x-5\right] d x\) \(=\int_{1}^{4}\left(5 x-x^{2}-4\right) d x=5 \int_{1}^{4} x d x-\int_{1}^{4} x^{2} d x-4 \int_{1}^{4} d x\) \(=5\left[\frac{x^{2}}{2}\right]_{1}^{4}-\left[\frac{x^{3}}{3}\right]_{1}^{4}-4[x]_{1}^{4}\) \(=\frac{5}{2}[16-1]-\frac{1}{3}[64-1]-4[4-1]\) \(=\frac{5}{2} \times 15-\frac{1}{3} \times 63-4 \times 3=\frac{75}{2}-21-12\) \(=\frac{75}{2}-33=\frac{75-66}{2}=\frac{9}{2}\)
WB JEE-2013
Application of the Integrals
86872
The area bounded by the curves \(y=|x|-1\) and \(y\) \(=-|\mathbf{x}|+1\) is
1 1 sq unit
2 2 sq units
3 \(2 \sqrt{2}\) squnits
4 4 sq units
Explanation:
(B) : The region is clearly square with vertices at the points \((1,0),(0,1),(-1,0)\) and \((0,-1)\). So, its area \(=\) \(\sqrt{2} \times \sqrt{2}=2\) sq units.
