NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of the Integrals
86860
The graph \(y^{2}+2 x y+50|x|=625\) divides the plane into regions. Then the area of bounded region is
1 500 sq. units
2 1250 sq. units
3 2500 sq. units
4 800 sq. units
Explanation:
(B) : We have, \(y^{2}+2 x y+50|x|=625\) When, \(x \geq 0\) we have \(y^{2}+2 x y+50 x-625=0\) \(y^{2}-25^{2}+2 x(y+25)=0\) \((y+25)(y-25)+2 x(y+25)=0\) \((y+25)(y-25+2 x)=0\) \(y=-25 \text { or } y-25+2 x=0\) When, \(\mathrm{x}\lt 0\) we have \(y^{2}+2 x y-50 x-625=0\) \(y^{2}-25^{2}+2 x y-50 x=0\) \((y-25)(y+25)+2 x(y-25)=0\) \((y-25)(y+25+2 x)=0\) \(y=25 \text { or } y+25+2 x=0\) Now area bounded by region Area of parallelogram ABCD \(25 \times 50=1250\) square units
AMU-2018
Application of the Integrals
86861
The area enclosed by, \(y=3 x-5, y=0, x=3\) and \(x=5\) is
1 12 sq. units
2 13 sq. units
3 \(13 \frac{1}{2}\) sq. units
4 14 sq. units
Explanation:
(D) : Given the lines \(y=3 x-5, y=0, x=3\) and \(\mathrm{x}=5\) For plotting \(\mathrm{y}=3 \mathrm{x}-5\) At \(\mathrm{x}-\) axis, \(\mathrm{y}=0\) \(3 \mathrm{x}-5=0\) or \(\mathrm{x}=\frac{5}{3}\) At \(\mathrm{y}\)-axis,\(\quad \mathrm{x}=0 \Rightarrow \mathrm{y}=-5\) \(\therefore\) Area enclosed is \(\mathrm{ABCD}\), which is equal to- \(A=\int_{3}^{5} y \cdot d x=\int_{3}^{5}(3 x-5) d x\) \(=\int_{3}^{5} 3 x d x-5 \int_{3}^{5} d x=3\left[\frac{x^{2}}{2}\right]_{3}^{5}-5[x]_{3}^{5}\) \(=\frac{3}{2}[25-9]-5[5-3]=\frac{3}{2} \times 16-5 \times 2\) \(=24-10=14\) Square unit.
Application of the Integrals
86862
The area bounded by the curve \(y=\left\{\begin{array}{cc}x^{1 / n x},& x \neq 1 \\ e &x=1\end{array}\right.\) and \(y=|x-e|\) is
1 \(\mathrm{e}^{2} / 2\)
2 \(e^{2}\)
3 \(2 \mathrm{e}^{2}\)
4 1
Explanation:
(B) : We have, \(y=\left\{\begin{array}{cc} x^{1 / \ln x},& x \neq 1\\ e & x=1 \end{array}\right.\) And, \(\quad y=|x-e|\) Required area \(=2 \mathrm{e} \times \mathrm{e}-\frac{1}{2} \mathrm{e} \times \mathrm{e}-\frac{1}{2}(\mathrm{e} \times \mathrm{e})\) \(=2 \mathrm{e}^{2}-\frac{\mathrm{e}^{2}}{2}-\frac{\mathrm{e}^{2}}{2}=2 \mathrm{e}^{2}-\mathrm{e}^{2}=\mathrm{e}^{2}\)
AMU-2018
Application of the Integrals
86863
The area of the region bounded by the curves \(y=x^{3}, y=\frac{1}{x}, x=2\) is
1 \(4-\log _{\mathrm{e}} 2\)
2 \(\frac{1}{4}+\log _{\mathrm{e}} 2\)
3 \(3-\log _{\mathrm{e}} 2\)
4 \(\frac{15}{4}-\log _{\mathrm{e}} 2\)
Explanation:
(D): The area of the region bounded by the curves \(y=x^{3}, \quad y=1 / x, \quad x=2\) Let us plot these curves. Solving \(\mathrm{y}=\mathrm{x}^{3}\) and \(y=\frac{1}{x}\) we get \(\frac{1}{\mathrm{x}}=\mathrm{x}^{3}\) or \(1=\mathrm{x}^{4}\) \(x=1 \quad y=1\) Point A to \((1,1)\) \(\therefore \mathrm{A}=\int_{1}^{2}\left(\mathrm{x}^{3}-\frac{1}{\mathrm{x}}\right) \mathrm{dx}\) \(=\int_{1}^{2} \mathrm{x}^{3} \mathrm{dx}-\int_{1}^{2} \frac{1}{\mathrm{x}} \cdot \mathrm{dx}=\left[\frac{\mathrm{x}^{4}}{4}\right]_{1}^{2}-[\log \mathrm{x}]_{1}^{2}\) \(=\frac{1}{4}[16-1]-[\log 2-\log 1]=\frac{15}{4}-\log _{\mathrm{e}} 2\)
86860
The graph \(y^{2}+2 x y+50|x|=625\) divides the plane into regions. Then the area of bounded region is
1 500 sq. units
2 1250 sq. units
3 2500 sq. units
4 800 sq. units
Explanation:
(B) : We have, \(y^{2}+2 x y+50|x|=625\) When, \(x \geq 0\) we have \(y^{2}+2 x y+50 x-625=0\) \(y^{2}-25^{2}+2 x(y+25)=0\) \((y+25)(y-25)+2 x(y+25)=0\) \((y+25)(y-25+2 x)=0\) \(y=-25 \text { or } y-25+2 x=0\) When, \(\mathrm{x}\lt 0\) we have \(y^{2}+2 x y-50 x-625=0\) \(y^{2}-25^{2}+2 x y-50 x=0\) \((y-25)(y+25)+2 x(y-25)=0\) \((y-25)(y+25+2 x)=0\) \(y=25 \text { or } y+25+2 x=0\) Now area bounded by region Area of parallelogram ABCD \(25 \times 50=1250\) square units
AMU-2018
Application of the Integrals
86861
The area enclosed by, \(y=3 x-5, y=0, x=3\) and \(x=5\) is
1 12 sq. units
2 13 sq. units
3 \(13 \frac{1}{2}\) sq. units
4 14 sq. units
Explanation:
(D) : Given the lines \(y=3 x-5, y=0, x=3\) and \(\mathrm{x}=5\) For plotting \(\mathrm{y}=3 \mathrm{x}-5\) At \(\mathrm{x}-\) axis, \(\mathrm{y}=0\) \(3 \mathrm{x}-5=0\) or \(\mathrm{x}=\frac{5}{3}\) At \(\mathrm{y}\)-axis,\(\quad \mathrm{x}=0 \Rightarrow \mathrm{y}=-5\) \(\therefore\) Area enclosed is \(\mathrm{ABCD}\), which is equal to- \(A=\int_{3}^{5} y \cdot d x=\int_{3}^{5}(3 x-5) d x\) \(=\int_{3}^{5} 3 x d x-5 \int_{3}^{5} d x=3\left[\frac{x^{2}}{2}\right]_{3}^{5}-5[x]_{3}^{5}\) \(=\frac{3}{2}[25-9]-5[5-3]=\frac{3}{2} \times 16-5 \times 2\) \(=24-10=14\) Square unit.
Application of the Integrals
86862
The area bounded by the curve \(y=\left\{\begin{array}{cc}x^{1 / n x},& x \neq 1 \\ e &x=1\end{array}\right.\) and \(y=|x-e|\) is
1 \(\mathrm{e}^{2} / 2\)
2 \(e^{2}\)
3 \(2 \mathrm{e}^{2}\)
4 1
Explanation:
(B) : We have, \(y=\left\{\begin{array}{cc} x^{1 / \ln x},& x \neq 1\\ e & x=1 \end{array}\right.\) And, \(\quad y=|x-e|\) Required area \(=2 \mathrm{e} \times \mathrm{e}-\frac{1}{2} \mathrm{e} \times \mathrm{e}-\frac{1}{2}(\mathrm{e} \times \mathrm{e})\) \(=2 \mathrm{e}^{2}-\frac{\mathrm{e}^{2}}{2}-\frac{\mathrm{e}^{2}}{2}=2 \mathrm{e}^{2}-\mathrm{e}^{2}=\mathrm{e}^{2}\)
AMU-2018
Application of the Integrals
86863
The area of the region bounded by the curves \(y=x^{3}, y=\frac{1}{x}, x=2\) is
1 \(4-\log _{\mathrm{e}} 2\)
2 \(\frac{1}{4}+\log _{\mathrm{e}} 2\)
3 \(3-\log _{\mathrm{e}} 2\)
4 \(\frac{15}{4}-\log _{\mathrm{e}} 2\)
Explanation:
(D): The area of the region bounded by the curves \(y=x^{3}, \quad y=1 / x, \quad x=2\) Let us plot these curves. Solving \(\mathrm{y}=\mathrm{x}^{3}\) and \(y=\frac{1}{x}\) we get \(\frac{1}{\mathrm{x}}=\mathrm{x}^{3}\) or \(1=\mathrm{x}^{4}\) \(x=1 \quad y=1\) Point A to \((1,1)\) \(\therefore \mathrm{A}=\int_{1}^{2}\left(\mathrm{x}^{3}-\frac{1}{\mathrm{x}}\right) \mathrm{dx}\) \(=\int_{1}^{2} \mathrm{x}^{3} \mathrm{dx}-\int_{1}^{2} \frac{1}{\mathrm{x}} \cdot \mathrm{dx}=\left[\frac{\mathrm{x}^{4}}{4}\right]_{1}^{2}-[\log \mathrm{x}]_{1}^{2}\) \(=\frac{1}{4}[16-1]-[\log 2-\log 1]=\frac{15}{4}-\log _{\mathrm{e}} 2\)
86860
The graph \(y^{2}+2 x y+50|x|=625\) divides the plane into regions. Then the area of bounded region is
1 500 sq. units
2 1250 sq. units
3 2500 sq. units
4 800 sq. units
Explanation:
(B) : We have, \(y^{2}+2 x y+50|x|=625\) When, \(x \geq 0\) we have \(y^{2}+2 x y+50 x-625=0\) \(y^{2}-25^{2}+2 x(y+25)=0\) \((y+25)(y-25)+2 x(y+25)=0\) \((y+25)(y-25+2 x)=0\) \(y=-25 \text { or } y-25+2 x=0\) When, \(\mathrm{x}\lt 0\) we have \(y^{2}+2 x y-50 x-625=0\) \(y^{2}-25^{2}+2 x y-50 x=0\) \((y-25)(y+25)+2 x(y-25)=0\) \((y-25)(y+25+2 x)=0\) \(y=25 \text { or } y+25+2 x=0\) Now area bounded by region Area of parallelogram ABCD \(25 \times 50=1250\) square units
AMU-2018
Application of the Integrals
86861
The area enclosed by, \(y=3 x-5, y=0, x=3\) and \(x=5\) is
1 12 sq. units
2 13 sq. units
3 \(13 \frac{1}{2}\) sq. units
4 14 sq. units
Explanation:
(D) : Given the lines \(y=3 x-5, y=0, x=3\) and \(\mathrm{x}=5\) For plotting \(\mathrm{y}=3 \mathrm{x}-5\) At \(\mathrm{x}-\) axis, \(\mathrm{y}=0\) \(3 \mathrm{x}-5=0\) or \(\mathrm{x}=\frac{5}{3}\) At \(\mathrm{y}\)-axis,\(\quad \mathrm{x}=0 \Rightarrow \mathrm{y}=-5\) \(\therefore\) Area enclosed is \(\mathrm{ABCD}\), which is equal to- \(A=\int_{3}^{5} y \cdot d x=\int_{3}^{5}(3 x-5) d x\) \(=\int_{3}^{5} 3 x d x-5 \int_{3}^{5} d x=3\left[\frac{x^{2}}{2}\right]_{3}^{5}-5[x]_{3}^{5}\) \(=\frac{3}{2}[25-9]-5[5-3]=\frac{3}{2} \times 16-5 \times 2\) \(=24-10=14\) Square unit.
Application of the Integrals
86862
The area bounded by the curve \(y=\left\{\begin{array}{cc}x^{1 / n x},& x \neq 1 \\ e &x=1\end{array}\right.\) and \(y=|x-e|\) is
1 \(\mathrm{e}^{2} / 2\)
2 \(e^{2}\)
3 \(2 \mathrm{e}^{2}\)
4 1
Explanation:
(B) : We have, \(y=\left\{\begin{array}{cc} x^{1 / \ln x},& x \neq 1\\ e & x=1 \end{array}\right.\) And, \(\quad y=|x-e|\) Required area \(=2 \mathrm{e} \times \mathrm{e}-\frac{1}{2} \mathrm{e} \times \mathrm{e}-\frac{1}{2}(\mathrm{e} \times \mathrm{e})\) \(=2 \mathrm{e}^{2}-\frac{\mathrm{e}^{2}}{2}-\frac{\mathrm{e}^{2}}{2}=2 \mathrm{e}^{2}-\mathrm{e}^{2}=\mathrm{e}^{2}\)
AMU-2018
Application of the Integrals
86863
The area of the region bounded by the curves \(y=x^{3}, y=\frac{1}{x}, x=2\) is
1 \(4-\log _{\mathrm{e}} 2\)
2 \(\frac{1}{4}+\log _{\mathrm{e}} 2\)
3 \(3-\log _{\mathrm{e}} 2\)
4 \(\frac{15}{4}-\log _{\mathrm{e}} 2\)
Explanation:
(D): The area of the region bounded by the curves \(y=x^{3}, \quad y=1 / x, \quad x=2\) Let us plot these curves. Solving \(\mathrm{y}=\mathrm{x}^{3}\) and \(y=\frac{1}{x}\) we get \(\frac{1}{\mathrm{x}}=\mathrm{x}^{3}\) or \(1=\mathrm{x}^{4}\) \(x=1 \quad y=1\) Point A to \((1,1)\) \(\therefore \mathrm{A}=\int_{1}^{2}\left(\mathrm{x}^{3}-\frac{1}{\mathrm{x}}\right) \mathrm{dx}\) \(=\int_{1}^{2} \mathrm{x}^{3} \mathrm{dx}-\int_{1}^{2} \frac{1}{\mathrm{x}} \cdot \mathrm{dx}=\left[\frac{\mathrm{x}^{4}}{4}\right]_{1}^{2}-[\log \mathrm{x}]_{1}^{2}\) \(=\frac{1}{4}[16-1]-[\log 2-\log 1]=\frac{15}{4}-\log _{\mathrm{e}} 2\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Application of the Integrals
86860
The graph \(y^{2}+2 x y+50|x|=625\) divides the plane into regions. Then the area of bounded region is
1 500 sq. units
2 1250 sq. units
3 2500 sq. units
4 800 sq. units
Explanation:
(B) : We have, \(y^{2}+2 x y+50|x|=625\) When, \(x \geq 0\) we have \(y^{2}+2 x y+50 x-625=0\) \(y^{2}-25^{2}+2 x(y+25)=0\) \((y+25)(y-25)+2 x(y+25)=0\) \((y+25)(y-25+2 x)=0\) \(y=-25 \text { or } y-25+2 x=0\) When, \(\mathrm{x}\lt 0\) we have \(y^{2}+2 x y-50 x-625=0\) \(y^{2}-25^{2}+2 x y-50 x=0\) \((y-25)(y+25)+2 x(y-25)=0\) \((y-25)(y+25+2 x)=0\) \(y=25 \text { or } y+25+2 x=0\) Now area bounded by region Area of parallelogram ABCD \(25 \times 50=1250\) square units
AMU-2018
Application of the Integrals
86861
The area enclosed by, \(y=3 x-5, y=0, x=3\) and \(x=5\) is
1 12 sq. units
2 13 sq. units
3 \(13 \frac{1}{2}\) sq. units
4 14 sq. units
Explanation:
(D) : Given the lines \(y=3 x-5, y=0, x=3\) and \(\mathrm{x}=5\) For plotting \(\mathrm{y}=3 \mathrm{x}-5\) At \(\mathrm{x}-\) axis, \(\mathrm{y}=0\) \(3 \mathrm{x}-5=0\) or \(\mathrm{x}=\frac{5}{3}\) At \(\mathrm{y}\)-axis,\(\quad \mathrm{x}=0 \Rightarrow \mathrm{y}=-5\) \(\therefore\) Area enclosed is \(\mathrm{ABCD}\), which is equal to- \(A=\int_{3}^{5} y \cdot d x=\int_{3}^{5}(3 x-5) d x\) \(=\int_{3}^{5} 3 x d x-5 \int_{3}^{5} d x=3\left[\frac{x^{2}}{2}\right]_{3}^{5}-5[x]_{3}^{5}\) \(=\frac{3}{2}[25-9]-5[5-3]=\frac{3}{2} \times 16-5 \times 2\) \(=24-10=14\) Square unit.
Application of the Integrals
86862
The area bounded by the curve \(y=\left\{\begin{array}{cc}x^{1 / n x},& x \neq 1 \\ e &x=1\end{array}\right.\) and \(y=|x-e|\) is
1 \(\mathrm{e}^{2} / 2\)
2 \(e^{2}\)
3 \(2 \mathrm{e}^{2}\)
4 1
Explanation:
(B) : We have, \(y=\left\{\begin{array}{cc} x^{1 / \ln x},& x \neq 1\\ e & x=1 \end{array}\right.\) And, \(\quad y=|x-e|\) Required area \(=2 \mathrm{e} \times \mathrm{e}-\frac{1}{2} \mathrm{e} \times \mathrm{e}-\frac{1}{2}(\mathrm{e} \times \mathrm{e})\) \(=2 \mathrm{e}^{2}-\frac{\mathrm{e}^{2}}{2}-\frac{\mathrm{e}^{2}}{2}=2 \mathrm{e}^{2}-\mathrm{e}^{2}=\mathrm{e}^{2}\)
AMU-2018
Application of the Integrals
86863
The area of the region bounded by the curves \(y=x^{3}, y=\frac{1}{x}, x=2\) is
1 \(4-\log _{\mathrm{e}} 2\)
2 \(\frac{1}{4}+\log _{\mathrm{e}} 2\)
3 \(3-\log _{\mathrm{e}} 2\)
4 \(\frac{15}{4}-\log _{\mathrm{e}} 2\)
Explanation:
(D): The area of the region bounded by the curves \(y=x^{3}, \quad y=1 / x, \quad x=2\) Let us plot these curves. Solving \(\mathrm{y}=\mathrm{x}^{3}\) and \(y=\frac{1}{x}\) we get \(\frac{1}{\mathrm{x}}=\mathrm{x}^{3}\) or \(1=\mathrm{x}^{4}\) \(x=1 \quad y=1\) Point A to \((1,1)\) \(\therefore \mathrm{A}=\int_{1}^{2}\left(\mathrm{x}^{3}-\frac{1}{\mathrm{x}}\right) \mathrm{dx}\) \(=\int_{1}^{2} \mathrm{x}^{3} \mathrm{dx}-\int_{1}^{2} \frac{1}{\mathrm{x}} \cdot \mathrm{dx}=\left[\frac{\mathrm{x}^{4}}{4}\right]_{1}^{2}-[\log \mathrm{x}]_{1}^{2}\) \(=\frac{1}{4}[16-1]-[\log 2-\log 1]=\frac{15}{4}-\log _{\mathrm{e}} 2\)
