NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of the Integrals
87065
The area bounded by the lines \(y-2 x=2, y=4\) and the \(\mathbf{Y}\)-axis is equal to (in square units)
1 1
2 4
3 0
4 3
5 2
Explanation:
(A) : Given, Area bounded by lines \(y-2 x=2, \quad y=4\) \(A=\int_{2}^{4} \frac{(y-2)}{2} d y\). \(=\int_{2}^{4}\left(\frac{y}{2}-1\right) d y=\left|\frac{y^{2}}{4}-y\right|_{2}^{4}=4-4-1+2=1\)
Kerala CEE-2016
Application of the Integrals
87066
Area of the region enclosed by the parabola \(y=\) \(x^{2}\) and the line \(y=x+2\) is :
1 \(\frac{9}{2}\)
2 \(\frac{11}{2}\)
3 \(\frac{5}{2}\)
4 \(\frac{7}{2}\)
Explanation:
(A) : Given, \(y=x^{2}\) \(y=x+2\) On solving equation (i) and (ii), we get - \(x^{2}-x-2=0\) \((x-2)(x+1)=0\) \(x=2 \text { or }-1\) When, \(x=2, y=2+2=4\) \(x=-1, y=(-1)+2=1\) Therefore the point of intersection of parabola and the line are \((-1,1)\) and \((2,4)\) Required Area \(=\int_{-1}^{2}(x+2) d x-\int_{-1}^{-2} x^{2} d x\) \(=\left|\frac{x^{2}}{2}+2 x\right|_{-1}^{2}-\left|\frac{x^{3}}{3}\right|_{-1}^{2}\) \(=\frac{(2)^{2}}{2}+2 \times 2-\left[\frac{(-1)^{2}}{2}+2(-1)\right]-\frac{1}{3}\left[(2)^{3}-(-1)^{3}\right]\) \(=6-\frac{1}{2}+2-3=\frac{9}{2}\) square units
GUJCET-2023
Application of the Integrals
87067
The area and perimeter of a rectangle are \(A\) and \(P\), respectively, Then, \(P\) and \(A\) satisfy the inequality
1 \(\mathrm{P}+\mathrm{A}>\mathrm{PA}\)
2 \(\mathrm{P}^{2} \leq \mathrm{A}\)
3 \(\mathrm{A}-\mathrm{P}\lt 2\)
4 \(\mathrm{P}^{2} \leq 4 \mathrm{~A}\)
5 \(\mathrm{P}^{2} \geq 16 \mathrm{~A}\)
Explanation:
(E) : Area and perimeter of rectangle of sides \(x\) and \(y\) be \(A\) and \(P\) Area \((A)=x y\) Perimeter \((\mathrm{P})=2(\mathrm{x}+\mathrm{y})\) So, \(\quad \mathrm{AM} \geq \mathrm{GM}\) \((\mathrm{x}+\mathrm{y})^{2}=4 \mathrm{xy}\) \(\left(\frac{\mathrm{P}}{2}\right)^{2} \geq 4 \mathrm{~A}\) \(\mathrm{P}^{2} \geq 16 \mathrm{~A}\)
Kerala CEE-2015
Application of the Integrals
87068
If the straight lines \(y=2 x, y=2 x+1, y=-7 x\) and \(y=-7 x+1\) form a parallelogram, then the area of the parallelogram (in sq, units) is
1 \(\frac{1}{3}\)
2 \(\frac{2}{9}\)
3 \(\frac{1}{9}\)
4 \(\frac{1}{4}\)
5 9
Explanation:
(C) : Given, \(y=2 x, \quad y=2 x+1, \quad y=-7 x\) and \(y=-7 x+1\) Co-ordinates are \(0 \rightarrow(0,0)\) \(\mathrm{B} \rightarrow(0,1)\) \(\mathrm{C} \rightarrow\left(\frac{1}{9}, \frac{2}{9}\right)\) \(\mathrm{A} \rightarrow\left(-\frac{1}{9}, \frac{7}{9}\right)\) Area of parallelogram \(=2 \times\) Area of \(\triangle \mathrm{OBC}\) \(=\frac{2 \times 1}{2}\left[\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right]\) \(=\left|0+0+\frac{1}{9}(0-1)\right|=\frac{1}{9}\)
87065
The area bounded by the lines \(y-2 x=2, y=4\) and the \(\mathbf{Y}\)-axis is equal to (in square units)
1 1
2 4
3 0
4 3
5 2
Explanation:
(A) : Given, Area bounded by lines \(y-2 x=2, \quad y=4\) \(A=\int_{2}^{4} \frac{(y-2)}{2} d y\). \(=\int_{2}^{4}\left(\frac{y}{2}-1\right) d y=\left|\frac{y^{2}}{4}-y\right|_{2}^{4}=4-4-1+2=1\)
Kerala CEE-2016
Application of the Integrals
87066
Area of the region enclosed by the parabola \(y=\) \(x^{2}\) and the line \(y=x+2\) is :
1 \(\frac{9}{2}\)
2 \(\frac{11}{2}\)
3 \(\frac{5}{2}\)
4 \(\frac{7}{2}\)
Explanation:
(A) : Given, \(y=x^{2}\) \(y=x+2\) On solving equation (i) and (ii), we get - \(x^{2}-x-2=0\) \((x-2)(x+1)=0\) \(x=2 \text { or }-1\) When, \(x=2, y=2+2=4\) \(x=-1, y=(-1)+2=1\) Therefore the point of intersection of parabola and the line are \((-1,1)\) and \((2,4)\) Required Area \(=\int_{-1}^{2}(x+2) d x-\int_{-1}^{-2} x^{2} d x\) \(=\left|\frac{x^{2}}{2}+2 x\right|_{-1}^{2}-\left|\frac{x^{3}}{3}\right|_{-1}^{2}\) \(=\frac{(2)^{2}}{2}+2 \times 2-\left[\frac{(-1)^{2}}{2}+2(-1)\right]-\frac{1}{3}\left[(2)^{3}-(-1)^{3}\right]\) \(=6-\frac{1}{2}+2-3=\frac{9}{2}\) square units
GUJCET-2023
Application of the Integrals
87067
The area and perimeter of a rectangle are \(A\) and \(P\), respectively, Then, \(P\) and \(A\) satisfy the inequality
1 \(\mathrm{P}+\mathrm{A}>\mathrm{PA}\)
2 \(\mathrm{P}^{2} \leq \mathrm{A}\)
3 \(\mathrm{A}-\mathrm{P}\lt 2\)
4 \(\mathrm{P}^{2} \leq 4 \mathrm{~A}\)
5 \(\mathrm{P}^{2} \geq 16 \mathrm{~A}\)
Explanation:
(E) : Area and perimeter of rectangle of sides \(x\) and \(y\) be \(A\) and \(P\) Area \((A)=x y\) Perimeter \((\mathrm{P})=2(\mathrm{x}+\mathrm{y})\) So, \(\quad \mathrm{AM} \geq \mathrm{GM}\) \((\mathrm{x}+\mathrm{y})^{2}=4 \mathrm{xy}\) \(\left(\frac{\mathrm{P}}{2}\right)^{2} \geq 4 \mathrm{~A}\) \(\mathrm{P}^{2} \geq 16 \mathrm{~A}\)
Kerala CEE-2015
Application of the Integrals
87068
If the straight lines \(y=2 x, y=2 x+1, y=-7 x\) and \(y=-7 x+1\) form a parallelogram, then the area of the parallelogram (in sq, units) is
1 \(\frac{1}{3}\)
2 \(\frac{2}{9}\)
3 \(\frac{1}{9}\)
4 \(\frac{1}{4}\)
5 9
Explanation:
(C) : Given, \(y=2 x, \quad y=2 x+1, \quad y=-7 x\) and \(y=-7 x+1\) Co-ordinates are \(0 \rightarrow(0,0)\) \(\mathrm{B} \rightarrow(0,1)\) \(\mathrm{C} \rightarrow\left(\frac{1}{9}, \frac{2}{9}\right)\) \(\mathrm{A} \rightarrow\left(-\frac{1}{9}, \frac{7}{9}\right)\) Area of parallelogram \(=2 \times\) Area of \(\triangle \mathrm{OBC}\) \(=\frac{2 \times 1}{2}\left[\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right]\) \(=\left|0+0+\frac{1}{9}(0-1)\right|=\frac{1}{9}\)
87065
The area bounded by the lines \(y-2 x=2, y=4\) and the \(\mathbf{Y}\)-axis is equal to (in square units)
1 1
2 4
3 0
4 3
5 2
Explanation:
(A) : Given, Area bounded by lines \(y-2 x=2, \quad y=4\) \(A=\int_{2}^{4} \frac{(y-2)}{2} d y\). \(=\int_{2}^{4}\left(\frac{y}{2}-1\right) d y=\left|\frac{y^{2}}{4}-y\right|_{2}^{4}=4-4-1+2=1\)
Kerala CEE-2016
Application of the Integrals
87066
Area of the region enclosed by the parabola \(y=\) \(x^{2}\) and the line \(y=x+2\) is :
1 \(\frac{9}{2}\)
2 \(\frac{11}{2}\)
3 \(\frac{5}{2}\)
4 \(\frac{7}{2}\)
Explanation:
(A) : Given, \(y=x^{2}\) \(y=x+2\) On solving equation (i) and (ii), we get - \(x^{2}-x-2=0\) \((x-2)(x+1)=0\) \(x=2 \text { or }-1\) When, \(x=2, y=2+2=4\) \(x=-1, y=(-1)+2=1\) Therefore the point of intersection of parabola and the line are \((-1,1)\) and \((2,4)\) Required Area \(=\int_{-1}^{2}(x+2) d x-\int_{-1}^{-2} x^{2} d x\) \(=\left|\frac{x^{2}}{2}+2 x\right|_{-1}^{2}-\left|\frac{x^{3}}{3}\right|_{-1}^{2}\) \(=\frac{(2)^{2}}{2}+2 \times 2-\left[\frac{(-1)^{2}}{2}+2(-1)\right]-\frac{1}{3}\left[(2)^{3}-(-1)^{3}\right]\) \(=6-\frac{1}{2}+2-3=\frac{9}{2}\) square units
GUJCET-2023
Application of the Integrals
87067
The area and perimeter of a rectangle are \(A\) and \(P\), respectively, Then, \(P\) and \(A\) satisfy the inequality
1 \(\mathrm{P}+\mathrm{A}>\mathrm{PA}\)
2 \(\mathrm{P}^{2} \leq \mathrm{A}\)
3 \(\mathrm{A}-\mathrm{P}\lt 2\)
4 \(\mathrm{P}^{2} \leq 4 \mathrm{~A}\)
5 \(\mathrm{P}^{2} \geq 16 \mathrm{~A}\)
Explanation:
(E) : Area and perimeter of rectangle of sides \(x\) and \(y\) be \(A\) and \(P\) Area \((A)=x y\) Perimeter \((\mathrm{P})=2(\mathrm{x}+\mathrm{y})\) So, \(\quad \mathrm{AM} \geq \mathrm{GM}\) \((\mathrm{x}+\mathrm{y})^{2}=4 \mathrm{xy}\) \(\left(\frac{\mathrm{P}}{2}\right)^{2} \geq 4 \mathrm{~A}\) \(\mathrm{P}^{2} \geq 16 \mathrm{~A}\)
Kerala CEE-2015
Application of the Integrals
87068
If the straight lines \(y=2 x, y=2 x+1, y=-7 x\) and \(y=-7 x+1\) form a parallelogram, then the area of the parallelogram (in sq, units) is
1 \(\frac{1}{3}\)
2 \(\frac{2}{9}\)
3 \(\frac{1}{9}\)
4 \(\frac{1}{4}\)
5 9
Explanation:
(C) : Given, \(y=2 x, \quad y=2 x+1, \quad y=-7 x\) and \(y=-7 x+1\) Co-ordinates are \(0 \rightarrow(0,0)\) \(\mathrm{B} \rightarrow(0,1)\) \(\mathrm{C} \rightarrow\left(\frac{1}{9}, \frac{2}{9}\right)\) \(\mathrm{A} \rightarrow\left(-\frac{1}{9}, \frac{7}{9}\right)\) Area of parallelogram \(=2 \times\) Area of \(\triangle \mathrm{OBC}\) \(=\frac{2 \times 1}{2}\left[\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right]\) \(=\left|0+0+\frac{1}{9}(0-1)\right|=\frac{1}{9}\)
87065
The area bounded by the lines \(y-2 x=2, y=4\) and the \(\mathbf{Y}\)-axis is equal to (in square units)
1 1
2 4
3 0
4 3
5 2
Explanation:
(A) : Given, Area bounded by lines \(y-2 x=2, \quad y=4\) \(A=\int_{2}^{4} \frac{(y-2)}{2} d y\). \(=\int_{2}^{4}\left(\frac{y}{2}-1\right) d y=\left|\frac{y^{2}}{4}-y\right|_{2}^{4}=4-4-1+2=1\)
Kerala CEE-2016
Application of the Integrals
87066
Area of the region enclosed by the parabola \(y=\) \(x^{2}\) and the line \(y=x+2\) is :
1 \(\frac{9}{2}\)
2 \(\frac{11}{2}\)
3 \(\frac{5}{2}\)
4 \(\frac{7}{2}\)
Explanation:
(A) : Given, \(y=x^{2}\) \(y=x+2\) On solving equation (i) and (ii), we get - \(x^{2}-x-2=0\) \((x-2)(x+1)=0\) \(x=2 \text { or }-1\) When, \(x=2, y=2+2=4\) \(x=-1, y=(-1)+2=1\) Therefore the point of intersection of parabola and the line are \((-1,1)\) and \((2,4)\) Required Area \(=\int_{-1}^{2}(x+2) d x-\int_{-1}^{-2} x^{2} d x\) \(=\left|\frac{x^{2}}{2}+2 x\right|_{-1}^{2}-\left|\frac{x^{3}}{3}\right|_{-1}^{2}\) \(=\frac{(2)^{2}}{2}+2 \times 2-\left[\frac{(-1)^{2}}{2}+2(-1)\right]-\frac{1}{3}\left[(2)^{3}-(-1)^{3}\right]\) \(=6-\frac{1}{2}+2-3=\frac{9}{2}\) square units
GUJCET-2023
Application of the Integrals
87067
The area and perimeter of a rectangle are \(A\) and \(P\), respectively, Then, \(P\) and \(A\) satisfy the inequality
1 \(\mathrm{P}+\mathrm{A}>\mathrm{PA}\)
2 \(\mathrm{P}^{2} \leq \mathrm{A}\)
3 \(\mathrm{A}-\mathrm{P}\lt 2\)
4 \(\mathrm{P}^{2} \leq 4 \mathrm{~A}\)
5 \(\mathrm{P}^{2} \geq 16 \mathrm{~A}\)
Explanation:
(E) : Area and perimeter of rectangle of sides \(x\) and \(y\) be \(A\) and \(P\) Area \((A)=x y\) Perimeter \((\mathrm{P})=2(\mathrm{x}+\mathrm{y})\) So, \(\quad \mathrm{AM} \geq \mathrm{GM}\) \((\mathrm{x}+\mathrm{y})^{2}=4 \mathrm{xy}\) \(\left(\frac{\mathrm{P}}{2}\right)^{2} \geq 4 \mathrm{~A}\) \(\mathrm{P}^{2} \geq 16 \mathrm{~A}\)
Kerala CEE-2015
Application of the Integrals
87068
If the straight lines \(y=2 x, y=2 x+1, y=-7 x\) and \(y=-7 x+1\) form a parallelogram, then the area of the parallelogram (in sq, units) is
1 \(\frac{1}{3}\)
2 \(\frac{2}{9}\)
3 \(\frac{1}{9}\)
4 \(\frac{1}{4}\)
5 9
Explanation:
(C) : Given, \(y=2 x, \quad y=2 x+1, \quad y=-7 x\) and \(y=-7 x+1\) Co-ordinates are \(0 \rightarrow(0,0)\) \(\mathrm{B} \rightarrow(0,1)\) \(\mathrm{C} \rightarrow\left(\frac{1}{9}, \frac{2}{9}\right)\) \(\mathrm{A} \rightarrow\left(-\frac{1}{9}, \frac{7}{9}\right)\) Area of parallelogram \(=2 \times\) Area of \(\triangle \mathrm{OBC}\) \(=\frac{2 \times 1}{2}\left[\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right]\) \(=\left|0+0+\frac{1}{9}(0-1)\right|=\frac{1}{9}\)
