86833
The area of region bounded by the lines \(y=m x, x=1\) and \(x=2\) and the \(X\)-axis is \(7.5 \mathrm{sq}\) units, then \(m\) is
1 2
2 3
3 4
4 5
Explanation:
(D) : Given lines, \(\mathrm{y}=\mathrm{mx}\) and \(\mathrm{x}=1, \mathrm{x}=2\) And the \(x-\) axis area \((A)=7.5\) square units Now, area, \(A=\int_{1}^{2} \mathrm{y} \cdot \mathrm{dx} \Rightarrow 7.5=\int_{1}^{2} \mathrm{mx} \cdot \mathrm{dx}\) \(7.5=\mathrm{m}\left[\frac{\mathrm{x}^{2}}{2}\right]_{1}^{2} \Rightarrow 7.5=\mathrm{m}\left[\frac{2^{2}}{2}-\frac{1^{2}}{2}\right]\) \(7.5=\mathrm{m}\left[2-\frac{1}{2}\right] \Rightarrow 7.5=\mathrm{m} \times \frac{3}{2} \Rightarrow \mathrm{m}=\frac{15}{3} \Rightarrow \mathrm{m}=5\)
[JCECE-2017]
Application of the Integrals
86835
The area of the triangle formed by the lines \(x^{2}\) \(-4 y^{2}=0\) and \(x=a\), is
1 \(2 a^{2}\)
2 \(\frac{a^{2}}{2}\)
3 \(\frac{\sqrt{3} \mathrm{a}^{2}}{2}\)
4 \(\frac{2 \mathrm{a}^{2}}{\sqrt{3}}\)
Explanation:
(B) : Given, \(\mathrm{x}^{2}-4 \mathrm{y}^{2}=0\) \(\mathrm{x}=\mathrm{a}\) Putting the value of \(x\) into equation (i)- \(a^{2}=4 y^{2} \Rightarrow 2 y= \pm a\) \(\mathrm{y}=\mathrm{a} / 2\) and \(\mathrm{y}=-\mathrm{a} / 2\) And from equation (i) when \(\mathrm{x}=0\) then \(\mathrm{y}=0\) So, area of \(\triangle \mathrm{COB}=2 \times\) area of \(\triangle \mathrm{OAB}\) \(=2 \times \frac{1}{2} \times \mathrm{OA} \times \mathrm{AB}=2 \times \frac{1}{2} \times \mathrm{a} \times \frac{\mathrm{a}}{2}=\frac{\mathrm{a}^{2}}{2} \text { sq. units }\)
[JCECE-2014]
Application of the Integrals
86836
The point on the curve \(x^{2}=2 y\) which are closest to the point \((0,5)\) are
1 \((2,2)(-2,2)\)
2 \((2 \sqrt{2}, 4)(-2 \sqrt{2}, 4)\)
3 \((\sqrt{6}, 3),(-\sqrt{6}, 3)\)
4 \((2 \sqrt{3}, 6),(-2 \sqrt{3}, 6)\)
Explanation:
(B) : The curve, \(\mathrm{x}^{2}=2 \mathrm{y} \quad\).......(i) Let the point \(A\) on the curve \(x^{2}=2 y\) is \((x, y)\) which is closest to \((0,5)\). If distance between them is \(\mathrm{s}\) then- \(s=\sqrt{(x-0)^{2}+(y-5)^{2}}\) Squaring both sides- \(s^{2}=(x)^{2}+(y-5)^{2} \tag{ii}\) From equation (i) and (ii)- \(s^{2}=2 y+(y-5)^{2}\) Form maxima or minima- \(\frac{\mathrm{ds}}{\mathrm{dy}}=0\) \(\therefore \frac{2 d x}{d y}=2+2(y-5)\) \(\text { Now } 2+2(y-5)=0 \quad\left(\because \frac{d x}{d y}=0\right)\) \(2 y=10-2=8\) \(y=4\) Putting the value of \(y\). in equation no. (i) \(\mathrm{x}^{2}=2 \times 4=8\) So the point is \(( \pm 2 \sqrt{2}, 4)\)
[JCECE-2009]
Application of the Integrals
86837
Let the straight line \(\mathrm{x}=\mathrm{b}\) divide the area enclosed by \(y=(1-x)^{2}, y=0\) and \(x=0\) into two parts \(R_{1}(0 \leq x \leq b)\) and \(R_{2}(b \leq x \leq 1)\) such that \(R_{1}-R_{2}=\frac{1}{4}\). Then \(b\) equals.
1 \(\frac{3}{4}\)
2 \(\frac{1}{2}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{4}\)
Explanation:
(B) : Given line, \(\mathrm{x}=\mathrm{b}\) and \(\mathrm{y}=(1-\mathrm{x})^{2}\) \(\mathrm{y}=0\) and \(\mathrm{x}=0\) Here, \(\mathrm{R}_{1}=\int_{0}^{\mathrm{b}}(1-\mathrm{x})^{2} \cdot \mathrm{dx}\) Given that, \(\mathrm{R}_{1}=\mathrm{R}_{2}=\frac{1}{4}\) So, \(\frac{(b-1)^{3}+1}{3}+\frac{(b-1)^{3}}{3}=\frac{1}{4} \Rightarrow \frac{2(b-1)^{3}+1}{3}=\frac{1}{4}\) \(\Rightarrow 2(\mathrm{~b}-1)^{3}=\frac{3}{4}-1 \Rightarrow(\mathrm{b}-1)^{3}=\frac{-1}{4 \times 2}\) \(\Rightarrow(\mathrm{b}-1)^{3}=\frac{-1}{8} \Rightarrow(\mathrm{b}-1)=-\frac{1}{2} \Rightarrow \mathrm{b}=\frac{1}{2}\)
86833
The area of region bounded by the lines \(y=m x, x=1\) and \(x=2\) and the \(X\)-axis is \(7.5 \mathrm{sq}\) units, then \(m\) is
1 2
2 3
3 4
4 5
Explanation:
(D) : Given lines, \(\mathrm{y}=\mathrm{mx}\) and \(\mathrm{x}=1, \mathrm{x}=2\) And the \(x-\) axis area \((A)=7.5\) square units Now, area, \(A=\int_{1}^{2} \mathrm{y} \cdot \mathrm{dx} \Rightarrow 7.5=\int_{1}^{2} \mathrm{mx} \cdot \mathrm{dx}\) \(7.5=\mathrm{m}\left[\frac{\mathrm{x}^{2}}{2}\right]_{1}^{2} \Rightarrow 7.5=\mathrm{m}\left[\frac{2^{2}}{2}-\frac{1^{2}}{2}\right]\) \(7.5=\mathrm{m}\left[2-\frac{1}{2}\right] \Rightarrow 7.5=\mathrm{m} \times \frac{3}{2} \Rightarrow \mathrm{m}=\frac{15}{3} \Rightarrow \mathrm{m}=5\)
[JCECE-2017]
Application of the Integrals
86835
The area of the triangle formed by the lines \(x^{2}\) \(-4 y^{2}=0\) and \(x=a\), is
1 \(2 a^{2}\)
2 \(\frac{a^{2}}{2}\)
3 \(\frac{\sqrt{3} \mathrm{a}^{2}}{2}\)
4 \(\frac{2 \mathrm{a}^{2}}{\sqrt{3}}\)
Explanation:
(B) : Given, \(\mathrm{x}^{2}-4 \mathrm{y}^{2}=0\) \(\mathrm{x}=\mathrm{a}\) Putting the value of \(x\) into equation (i)- \(a^{2}=4 y^{2} \Rightarrow 2 y= \pm a\) \(\mathrm{y}=\mathrm{a} / 2\) and \(\mathrm{y}=-\mathrm{a} / 2\) And from equation (i) when \(\mathrm{x}=0\) then \(\mathrm{y}=0\) So, area of \(\triangle \mathrm{COB}=2 \times\) area of \(\triangle \mathrm{OAB}\) \(=2 \times \frac{1}{2} \times \mathrm{OA} \times \mathrm{AB}=2 \times \frac{1}{2} \times \mathrm{a} \times \frac{\mathrm{a}}{2}=\frac{\mathrm{a}^{2}}{2} \text { sq. units }\)
[JCECE-2014]
Application of the Integrals
86836
The point on the curve \(x^{2}=2 y\) which are closest to the point \((0,5)\) are
1 \((2,2)(-2,2)\)
2 \((2 \sqrt{2}, 4)(-2 \sqrt{2}, 4)\)
3 \((\sqrt{6}, 3),(-\sqrt{6}, 3)\)
4 \((2 \sqrt{3}, 6),(-2 \sqrt{3}, 6)\)
Explanation:
(B) : The curve, \(\mathrm{x}^{2}=2 \mathrm{y} \quad\).......(i) Let the point \(A\) on the curve \(x^{2}=2 y\) is \((x, y)\) which is closest to \((0,5)\). If distance between them is \(\mathrm{s}\) then- \(s=\sqrt{(x-0)^{2}+(y-5)^{2}}\) Squaring both sides- \(s^{2}=(x)^{2}+(y-5)^{2} \tag{ii}\) From equation (i) and (ii)- \(s^{2}=2 y+(y-5)^{2}\) Form maxima or minima- \(\frac{\mathrm{ds}}{\mathrm{dy}}=0\) \(\therefore \frac{2 d x}{d y}=2+2(y-5)\) \(\text { Now } 2+2(y-5)=0 \quad\left(\because \frac{d x}{d y}=0\right)\) \(2 y=10-2=8\) \(y=4\) Putting the value of \(y\). in equation no. (i) \(\mathrm{x}^{2}=2 \times 4=8\) So the point is \(( \pm 2 \sqrt{2}, 4)\)
[JCECE-2009]
Application of the Integrals
86837
Let the straight line \(\mathrm{x}=\mathrm{b}\) divide the area enclosed by \(y=(1-x)^{2}, y=0\) and \(x=0\) into two parts \(R_{1}(0 \leq x \leq b)\) and \(R_{2}(b \leq x \leq 1)\) such that \(R_{1}-R_{2}=\frac{1}{4}\). Then \(b\) equals.
1 \(\frac{3}{4}\)
2 \(\frac{1}{2}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{4}\)
Explanation:
(B) : Given line, \(\mathrm{x}=\mathrm{b}\) and \(\mathrm{y}=(1-\mathrm{x})^{2}\) \(\mathrm{y}=0\) and \(\mathrm{x}=0\) Here, \(\mathrm{R}_{1}=\int_{0}^{\mathrm{b}}(1-\mathrm{x})^{2} \cdot \mathrm{dx}\) Given that, \(\mathrm{R}_{1}=\mathrm{R}_{2}=\frac{1}{4}\) So, \(\frac{(b-1)^{3}+1}{3}+\frac{(b-1)^{3}}{3}=\frac{1}{4} \Rightarrow \frac{2(b-1)^{3}+1}{3}=\frac{1}{4}\) \(\Rightarrow 2(\mathrm{~b}-1)^{3}=\frac{3}{4}-1 \Rightarrow(\mathrm{b}-1)^{3}=\frac{-1}{4 \times 2}\) \(\Rightarrow(\mathrm{b}-1)^{3}=\frac{-1}{8} \Rightarrow(\mathrm{b}-1)=-\frac{1}{2} \Rightarrow \mathrm{b}=\frac{1}{2}\)
86833
The area of region bounded by the lines \(y=m x, x=1\) and \(x=2\) and the \(X\)-axis is \(7.5 \mathrm{sq}\) units, then \(m\) is
1 2
2 3
3 4
4 5
Explanation:
(D) : Given lines, \(\mathrm{y}=\mathrm{mx}\) and \(\mathrm{x}=1, \mathrm{x}=2\) And the \(x-\) axis area \((A)=7.5\) square units Now, area, \(A=\int_{1}^{2} \mathrm{y} \cdot \mathrm{dx} \Rightarrow 7.5=\int_{1}^{2} \mathrm{mx} \cdot \mathrm{dx}\) \(7.5=\mathrm{m}\left[\frac{\mathrm{x}^{2}}{2}\right]_{1}^{2} \Rightarrow 7.5=\mathrm{m}\left[\frac{2^{2}}{2}-\frac{1^{2}}{2}\right]\) \(7.5=\mathrm{m}\left[2-\frac{1}{2}\right] \Rightarrow 7.5=\mathrm{m} \times \frac{3}{2} \Rightarrow \mathrm{m}=\frac{15}{3} \Rightarrow \mathrm{m}=5\)
[JCECE-2017]
Application of the Integrals
86835
The area of the triangle formed by the lines \(x^{2}\) \(-4 y^{2}=0\) and \(x=a\), is
1 \(2 a^{2}\)
2 \(\frac{a^{2}}{2}\)
3 \(\frac{\sqrt{3} \mathrm{a}^{2}}{2}\)
4 \(\frac{2 \mathrm{a}^{2}}{\sqrt{3}}\)
Explanation:
(B) : Given, \(\mathrm{x}^{2}-4 \mathrm{y}^{2}=0\) \(\mathrm{x}=\mathrm{a}\) Putting the value of \(x\) into equation (i)- \(a^{2}=4 y^{2} \Rightarrow 2 y= \pm a\) \(\mathrm{y}=\mathrm{a} / 2\) and \(\mathrm{y}=-\mathrm{a} / 2\) And from equation (i) when \(\mathrm{x}=0\) then \(\mathrm{y}=0\) So, area of \(\triangle \mathrm{COB}=2 \times\) area of \(\triangle \mathrm{OAB}\) \(=2 \times \frac{1}{2} \times \mathrm{OA} \times \mathrm{AB}=2 \times \frac{1}{2} \times \mathrm{a} \times \frac{\mathrm{a}}{2}=\frac{\mathrm{a}^{2}}{2} \text { sq. units }\)
[JCECE-2014]
Application of the Integrals
86836
The point on the curve \(x^{2}=2 y\) which are closest to the point \((0,5)\) are
1 \((2,2)(-2,2)\)
2 \((2 \sqrt{2}, 4)(-2 \sqrt{2}, 4)\)
3 \((\sqrt{6}, 3),(-\sqrt{6}, 3)\)
4 \((2 \sqrt{3}, 6),(-2 \sqrt{3}, 6)\)
Explanation:
(B) : The curve, \(\mathrm{x}^{2}=2 \mathrm{y} \quad\).......(i) Let the point \(A\) on the curve \(x^{2}=2 y\) is \((x, y)\) which is closest to \((0,5)\). If distance between them is \(\mathrm{s}\) then- \(s=\sqrt{(x-0)^{2}+(y-5)^{2}}\) Squaring both sides- \(s^{2}=(x)^{2}+(y-5)^{2} \tag{ii}\) From equation (i) and (ii)- \(s^{2}=2 y+(y-5)^{2}\) Form maxima or minima- \(\frac{\mathrm{ds}}{\mathrm{dy}}=0\) \(\therefore \frac{2 d x}{d y}=2+2(y-5)\) \(\text { Now } 2+2(y-5)=0 \quad\left(\because \frac{d x}{d y}=0\right)\) \(2 y=10-2=8\) \(y=4\) Putting the value of \(y\). in equation no. (i) \(\mathrm{x}^{2}=2 \times 4=8\) So the point is \(( \pm 2 \sqrt{2}, 4)\)
[JCECE-2009]
Application of the Integrals
86837
Let the straight line \(\mathrm{x}=\mathrm{b}\) divide the area enclosed by \(y=(1-x)^{2}, y=0\) and \(x=0\) into two parts \(R_{1}(0 \leq x \leq b)\) and \(R_{2}(b \leq x \leq 1)\) such that \(R_{1}-R_{2}=\frac{1}{4}\). Then \(b\) equals.
1 \(\frac{3}{4}\)
2 \(\frac{1}{2}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{4}\)
Explanation:
(B) : Given line, \(\mathrm{x}=\mathrm{b}\) and \(\mathrm{y}=(1-\mathrm{x})^{2}\) \(\mathrm{y}=0\) and \(\mathrm{x}=0\) Here, \(\mathrm{R}_{1}=\int_{0}^{\mathrm{b}}(1-\mathrm{x})^{2} \cdot \mathrm{dx}\) Given that, \(\mathrm{R}_{1}=\mathrm{R}_{2}=\frac{1}{4}\) So, \(\frac{(b-1)^{3}+1}{3}+\frac{(b-1)^{3}}{3}=\frac{1}{4} \Rightarrow \frac{2(b-1)^{3}+1}{3}=\frac{1}{4}\) \(\Rightarrow 2(\mathrm{~b}-1)^{3}=\frac{3}{4}-1 \Rightarrow(\mathrm{b}-1)^{3}=\frac{-1}{4 \times 2}\) \(\Rightarrow(\mathrm{b}-1)^{3}=\frac{-1}{8} \Rightarrow(\mathrm{b}-1)=-\frac{1}{2} \Rightarrow \mathrm{b}=\frac{1}{2}\)
86833
The area of region bounded by the lines \(y=m x, x=1\) and \(x=2\) and the \(X\)-axis is \(7.5 \mathrm{sq}\) units, then \(m\) is
1 2
2 3
3 4
4 5
Explanation:
(D) : Given lines, \(\mathrm{y}=\mathrm{mx}\) and \(\mathrm{x}=1, \mathrm{x}=2\) And the \(x-\) axis area \((A)=7.5\) square units Now, area, \(A=\int_{1}^{2} \mathrm{y} \cdot \mathrm{dx} \Rightarrow 7.5=\int_{1}^{2} \mathrm{mx} \cdot \mathrm{dx}\) \(7.5=\mathrm{m}\left[\frac{\mathrm{x}^{2}}{2}\right]_{1}^{2} \Rightarrow 7.5=\mathrm{m}\left[\frac{2^{2}}{2}-\frac{1^{2}}{2}\right]\) \(7.5=\mathrm{m}\left[2-\frac{1}{2}\right] \Rightarrow 7.5=\mathrm{m} \times \frac{3}{2} \Rightarrow \mathrm{m}=\frac{15}{3} \Rightarrow \mathrm{m}=5\)
[JCECE-2017]
Application of the Integrals
86835
The area of the triangle formed by the lines \(x^{2}\) \(-4 y^{2}=0\) and \(x=a\), is
1 \(2 a^{2}\)
2 \(\frac{a^{2}}{2}\)
3 \(\frac{\sqrt{3} \mathrm{a}^{2}}{2}\)
4 \(\frac{2 \mathrm{a}^{2}}{\sqrt{3}}\)
Explanation:
(B) : Given, \(\mathrm{x}^{2}-4 \mathrm{y}^{2}=0\) \(\mathrm{x}=\mathrm{a}\) Putting the value of \(x\) into equation (i)- \(a^{2}=4 y^{2} \Rightarrow 2 y= \pm a\) \(\mathrm{y}=\mathrm{a} / 2\) and \(\mathrm{y}=-\mathrm{a} / 2\) And from equation (i) when \(\mathrm{x}=0\) then \(\mathrm{y}=0\) So, area of \(\triangle \mathrm{COB}=2 \times\) area of \(\triangle \mathrm{OAB}\) \(=2 \times \frac{1}{2} \times \mathrm{OA} \times \mathrm{AB}=2 \times \frac{1}{2} \times \mathrm{a} \times \frac{\mathrm{a}}{2}=\frac{\mathrm{a}^{2}}{2} \text { sq. units }\)
[JCECE-2014]
Application of the Integrals
86836
The point on the curve \(x^{2}=2 y\) which are closest to the point \((0,5)\) are
1 \((2,2)(-2,2)\)
2 \((2 \sqrt{2}, 4)(-2 \sqrt{2}, 4)\)
3 \((\sqrt{6}, 3),(-\sqrt{6}, 3)\)
4 \((2 \sqrt{3}, 6),(-2 \sqrt{3}, 6)\)
Explanation:
(B) : The curve, \(\mathrm{x}^{2}=2 \mathrm{y} \quad\).......(i) Let the point \(A\) on the curve \(x^{2}=2 y\) is \((x, y)\) which is closest to \((0,5)\). If distance between them is \(\mathrm{s}\) then- \(s=\sqrt{(x-0)^{2}+(y-5)^{2}}\) Squaring both sides- \(s^{2}=(x)^{2}+(y-5)^{2} \tag{ii}\) From equation (i) and (ii)- \(s^{2}=2 y+(y-5)^{2}\) Form maxima or minima- \(\frac{\mathrm{ds}}{\mathrm{dy}}=0\) \(\therefore \frac{2 d x}{d y}=2+2(y-5)\) \(\text { Now } 2+2(y-5)=0 \quad\left(\because \frac{d x}{d y}=0\right)\) \(2 y=10-2=8\) \(y=4\) Putting the value of \(y\). in equation no. (i) \(\mathrm{x}^{2}=2 \times 4=8\) So the point is \(( \pm 2 \sqrt{2}, 4)\)
[JCECE-2009]
Application of the Integrals
86837
Let the straight line \(\mathrm{x}=\mathrm{b}\) divide the area enclosed by \(y=(1-x)^{2}, y=0\) and \(x=0\) into two parts \(R_{1}(0 \leq x \leq b)\) and \(R_{2}(b \leq x \leq 1)\) such that \(R_{1}-R_{2}=\frac{1}{4}\). Then \(b\) equals.
1 \(\frac{3}{4}\)
2 \(\frac{1}{2}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{4}\)
Explanation:
(B) : Given line, \(\mathrm{x}=\mathrm{b}\) and \(\mathrm{y}=(1-\mathrm{x})^{2}\) \(\mathrm{y}=0\) and \(\mathrm{x}=0\) Here, \(\mathrm{R}_{1}=\int_{0}^{\mathrm{b}}(1-\mathrm{x})^{2} \cdot \mathrm{dx}\) Given that, \(\mathrm{R}_{1}=\mathrm{R}_{2}=\frac{1}{4}\) So, \(\frac{(b-1)^{3}+1}{3}+\frac{(b-1)^{3}}{3}=\frac{1}{4} \Rightarrow \frac{2(b-1)^{3}+1}{3}=\frac{1}{4}\) \(\Rightarrow 2(\mathrm{~b}-1)^{3}=\frac{3}{4}-1 \Rightarrow(\mathrm{b}-1)^{3}=\frac{-1}{4 \times 2}\) \(\Rightarrow(\mathrm{b}-1)^{3}=\frac{-1}{8} \Rightarrow(\mathrm{b}-1)=-\frac{1}{2} \Rightarrow \mathrm{b}=\frac{1}{2}\)
