(C) : Given that, \(y=\tan x\) From equation (i) at \(\mathrm{x}=\pi / 4, \mathrm{y}=1\) Differentiating w.r.t. \(\mathrm{x}\) of the equation- At \(x=\frac{d y}{d x}=\sec ^{2} x\) At \(x=\pi / 4 \Rightarrow\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}}=\sec ^{2} \frac{\pi}{4}=2\) Now, the equation of tangent to the curve \(y=\tan x\) at \(x\) \(=\pi / 4\) is given by- \((\mathrm{y}-1)=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=\pi / 4}(\mathrm{x}-\pi / 4)\) \((\mathrm{y}-1)=2(\mathrm{x}-\pi / 4)\) \(\mathrm{y}=2(\mathrm{x}-\pi / 4)+1\) Let equation (ii) meet at point \(\mathrm{A}\) on \(\mathrm{X}\) - Axis than, \(\mathrm{y}=0\) \(0=2 \mathrm{x}+1-\pi / 2 \Rightarrow \mathrm{x}=\frac{\pi}{4}-\frac{1}{2}\) So, as per the figure required area- \(\mathrm{A}=\) Area of \(\mathrm{OAB}-\) Area of \(\triangle \mathrm{ACB}\) \(\mathrm{A}=\int_{0}^{\pi / 4} \tan \mathrm{xdx}-\left(\frac{1}{2} \times\left(\frac{\pi}{4}-\left(\frac{\pi}{4}-\frac{1}{2}\right) \times 1\right)\right.\) \(A=[\log \sec x]_{0}^{\pi / 4}-\left[\frac{1}{2} \times\left[\frac{1}{2}\right] \times 1\right]\) \(A=\log \sqrt{2}-0-\frac{1}{4} \Rightarrow A=\log \sqrt{2}-\frac{1}{4}\) sq. units
BCECE-2004
Application of the Integrals
86844
If the ordinate \(x=a\) divides the area bounded by the curve \(y=\left(1+\frac{8}{x^{2}}\right), x\)-axis and the ordinates \(x=2, x=4\), into two equal parts then the value of \(a\) is :
1 \(2 \mathrm{a}\)
2 \(2 \sqrt{2}\)
3 \(\frac{\mathrm{a}}{2}\)
4 none of these
Explanation:
(B) : Given that, The curve, \(y=1+\frac{8}{x^{2}}\) and \(x=a, x=2, x=4\) Since, \(\mathrm{x}=\mathrm{a}\) divides the area into two equal parts so, \(A=\int_{2}^{a}\left(1+\frac{8}{x^{2}}\right) d x=\int_{a}^{4}\left(1+\frac{8}{x^{2}}\right) d x\) \(\left[\mathrm{x}-\frac{8}{\mathrm{x}}\right]_{2}^{\mathrm{a}}=\left[\mathrm{x}-\frac{8}{\mathrm{x}}\right]_{\mathrm{a}}^{4}\) \(\left(\mathrm{a}-\frac{8}{\mathrm{a}}\right)-\left(2-\frac{8}{2}\right)=\left(4-\frac{8}{4}\right)-\left(\mathrm{a}-\frac{8}{\mathrm{a}}\right)\) \(\underline{\left(\mathrm{a}^{2}-8\right)}+2=2-\underline{\left(\mathrm{a}^{2}-8\right)}\)
BCECE-2003
Application of the Integrals
86845
The area of the region bounded by \(x^{2}+y^{2}-2 y\) \(-3=0\) and \(y=|x|+1\) is
(C) : Given that, \(y=\tan x\) From equation (i) at \(\mathrm{x}=\pi / 4, \mathrm{y}=1\) Differentiating w.r.t. \(\mathrm{x}\) of the equation- At \(x=\frac{d y}{d x}=\sec ^{2} x\) At \(x=\pi / 4 \Rightarrow\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}}=\sec ^{2} \frac{\pi}{4}=2\) Now, the equation of tangent to the curve \(y=\tan x\) at \(x\) \(=\pi / 4\) is given by- \((\mathrm{y}-1)=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=\pi / 4}(\mathrm{x}-\pi / 4)\) \((\mathrm{y}-1)=2(\mathrm{x}-\pi / 4)\) \(\mathrm{y}=2(\mathrm{x}-\pi / 4)+1\) Let equation (ii) meet at point \(\mathrm{A}\) on \(\mathrm{X}\) - Axis than, \(\mathrm{y}=0\) \(0=2 \mathrm{x}+1-\pi / 2 \Rightarrow \mathrm{x}=\frac{\pi}{4}-\frac{1}{2}\) So, as per the figure required area- \(\mathrm{A}=\) Area of \(\mathrm{OAB}-\) Area of \(\triangle \mathrm{ACB}\) \(\mathrm{A}=\int_{0}^{\pi / 4} \tan \mathrm{xdx}-\left(\frac{1}{2} \times\left(\frac{\pi}{4}-\left(\frac{\pi}{4}-\frac{1}{2}\right) \times 1\right)\right.\) \(A=[\log \sec x]_{0}^{\pi / 4}-\left[\frac{1}{2} \times\left[\frac{1}{2}\right] \times 1\right]\) \(A=\log \sqrt{2}-0-\frac{1}{4} \Rightarrow A=\log \sqrt{2}-\frac{1}{4}\) sq. units
BCECE-2004
Application of the Integrals
86844
If the ordinate \(x=a\) divides the area bounded by the curve \(y=\left(1+\frac{8}{x^{2}}\right), x\)-axis and the ordinates \(x=2, x=4\), into two equal parts then the value of \(a\) is :
1 \(2 \mathrm{a}\)
2 \(2 \sqrt{2}\)
3 \(\frac{\mathrm{a}}{2}\)
4 none of these
Explanation:
(B) : Given that, The curve, \(y=1+\frac{8}{x^{2}}\) and \(x=a, x=2, x=4\) Since, \(\mathrm{x}=\mathrm{a}\) divides the area into two equal parts so, \(A=\int_{2}^{a}\left(1+\frac{8}{x^{2}}\right) d x=\int_{a}^{4}\left(1+\frac{8}{x^{2}}\right) d x\) \(\left[\mathrm{x}-\frac{8}{\mathrm{x}}\right]_{2}^{\mathrm{a}}=\left[\mathrm{x}-\frac{8}{\mathrm{x}}\right]_{\mathrm{a}}^{4}\) \(\left(\mathrm{a}-\frac{8}{\mathrm{a}}\right)-\left(2-\frac{8}{2}\right)=\left(4-\frac{8}{4}\right)-\left(\mathrm{a}-\frac{8}{\mathrm{a}}\right)\) \(\underline{\left(\mathrm{a}^{2}-8\right)}+2=2-\underline{\left(\mathrm{a}^{2}-8\right)}\)
BCECE-2003
Application of the Integrals
86845
The area of the region bounded by \(x^{2}+y^{2}-2 y\) \(-3=0\) and \(y=|x|+1\) is
(C) : Given that, \(y=\tan x\) From equation (i) at \(\mathrm{x}=\pi / 4, \mathrm{y}=1\) Differentiating w.r.t. \(\mathrm{x}\) of the equation- At \(x=\frac{d y}{d x}=\sec ^{2} x\) At \(x=\pi / 4 \Rightarrow\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}}=\sec ^{2} \frac{\pi}{4}=2\) Now, the equation of tangent to the curve \(y=\tan x\) at \(x\) \(=\pi / 4\) is given by- \((\mathrm{y}-1)=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=\pi / 4}(\mathrm{x}-\pi / 4)\) \((\mathrm{y}-1)=2(\mathrm{x}-\pi / 4)\) \(\mathrm{y}=2(\mathrm{x}-\pi / 4)+1\) Let equation (ii) meet at point \(\mathrm{A}\) on \(\mathrm{X}\) - Axis than, \(\mathrm{y}=0\) \(0=2 \mathrm{x}+1-\pi / 2 \Rightarrow \mathrm{x}=\frac{\pi}{4}-\frac{1}{2}\) So, as per the figure required area- \(\mathrm{A}=\) Area of \(\mathrm{OAB}-\) Area of \(\triangle \mathrm{ACB}\) \(\mathrm{A}=\int_{0}^{\pi / 4} \tan \mathrm{xdx}-\left(\frac{1}{2} \times\left(\frac{\pi}{4}-\left(\frac{\pi}{4}-\frac{1}{2}\right) \times 1\right)\right.\) \(A=[\log \sec x]_{0}^{\pi / 4}-\left[\frac{1}{2} \times\left[\frac{1}{2}\right] \times 1\right]\) \(A=\log \sqrt{2}-0-\frac{1}{4} \Rightarrow A=\log \sqrt{2}-\frac{1}{4}\) sq. units
BCECE-2004
Application of the Integrals
86844
If the ordinate \(x=a\) divides the area bounded by the curve \(y=\left(1+\frac{8}{x^{2}}\right), x\)-axis and the ordinates \(x=2, x=4\), into two equal parts then the value of \(a\) is :
1 \(2 \mathrm{a}\)
2 \(2 \sqrt{2}\)
3 \(\frac{\mathrm{a}}{2}\)
4 none of these
Explanation:
(B) : Given that, The curve, \(y=1+\frac{8}{x^{2}}\) and \(x=a, x=2, x=4\) Since, \(\mathrm{x}=\mathrm{a}\) divides the area into two equal parts so, \(A=\int_{2}^{a}\left(1+\frac{8}{x^{2}}\right) d x=\int_{a}^{4}\left(1+\frac{8}{x^{2}}\right) d x\) \(\left[\mathrm{x}-\frac{8}{\mathrm{x}}\right]_{2}^{\mathrm{a}}=\left[\mathrm{x}-\frac{8}{\mathrm{x}}\right]_{\mathrm{a}}^{4}\) \(\left(\mathrm{a}-\frac{8}{\mathrm{a}}\right)-\left(2-\frac{8}{2}\right)=\left(4-\frac{8}{4}\right)-\left(\mathrm{a}-\frac{8}{\mathrm{a}}\right)\) \(\underline{\left(\mathrm{a}^{2}-8\right)}+2=2-\underline{\left(\mathrm{a}^{2}-8\right)}\)
BCECE-2003
Application of the Integrals
86845
The area of the region bounded by \(x^{2}+y^{2}-2 y\) \(-3=0\) and \(y=|x|+1\) is
(C) : Given that, \(y=\tan x\) From equation (i) at \(\mathrm{x}=\pi / 4, \mathrm{y}=1\) Differentiating w.r.t. \(\mathrm{x}\) of the equation- At \(x=\frac{d y}{d x}=\sec ^{2} x\) At \(x=\pi / 4 \Rightarrow\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}}=\sec ^{2} \frac{\pi}{4}=2\) Now, the equation of tangent to the curve \(y=\tan x\) at \(x\) \(=\pi / 4\) is given by- \((\mathrm{y}-1)=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=\pi / 4}(\mathrm{x}-\pi / 4)\) \((\mathrm{y}-1)=2(\mathrm{x}-\pi / 4)\) \(\mathrm{y}=2(\mathrm{x}-\pi / 4)+1\) Let equation (ii) meet at point \(\mathrm{A}\) on \(\mathrm{X}\) - Axis than, \(\mathrm{y}=0\) \(0=2 \mathrm{x}+1-\pi / 2 \Rightarrow \mathrm{x}=\frac{\pi}{4}-\frac{1}{2}\) So, as per the figure required area- \(\mathrm{A}=\) Area of \(\mathrm{OAB}-\) Area of \(\triangle \mathrm{ACB}\) \(\mathrm{A}=\int_{0}^{\pi / 4} \tan \mathrm{xdx}-\left(\frac{1}{2} \times\left(\frac{\pi}{4}-\left(\frac{\pi}{4}-\frac{1}{2}\right) \times 1\right)\right.\) \(A=[\log \sec x]_{0}^{\pi / 4}-\left[\frac{1}{2} \times\left[\frac{1}{2}\right] \times 1\right]\) \(A=\log \sqrt{2}-0-\frac{1}{4} \Rightarrow A=\log \sqrt{2}-\frac{1}{4}\) sq. units
BCECE-2004
Application of the Integrals
86844
If the ordinate \(x=a\) divides the area bounded by the curve \(y=\left(1+\frac{8}{x^{2}}\right), x\)-axis and the ordinates \(x=2, x=4\), into two equal parts then the value of \(a\) is :
1 \(2 \mathrm{a}\)
2 \(2 \sqrt{2}\)
3 \(\frac{\mathrm{a}}{2}\)
4 none of these
Explanation:
(B) : Given that, The curve, \(y=1+\frac{8}{x^{2}}\) and \(x=a, x=2, x=4\) Since, \(\mathrm{x}=\mathrm{a}\) divides the area into two equal parts so, \(A=\int_{2}^{a}\left(1+\frac{8}{x^{2}}\right) d x=\int_{a}^{4}\left(1+\frac{8}{x^{2}}\right) d x\) \(\left[\mathrm{x}-\frac{8}{\mathrm{x}}\right]_{2}^{\mathrm{a}}=\left[\mathrm{x}-\frac{8}{\mathrm{x}}\right]_{\mathrm{a}}^{4}\) \(\left(\mathrm{a}-\frac{8}{\mathrm{a}}\right)-\left(2-\frac{8}{2}\right)=\left(4-\frac{8}{4}\right)-\left(\mathrm{a}-\frac{8}{\mathrm{a}}\right)\) \(\underline{\left(\mathrm{a}^{2}-8\right)}+2=2-\underline{\left(\mathrm{a}^{2}-8\right)}\)
BCECE-2003
Application of the Integrals
86845
The area of the region bounded by \(x^{2}+y^{2}-2 y\) \(-3=0\) and \(y=|x|+1\) is
(C) : Given that, \(y=\tan x\) From equation (i) at \(\mathrm{x}=\pi / 4, \mathrm{y}=1\) Differentiating w.r.t. \(\mathrm{x}\) of the equation- At \(x=\frac{d y}{d x}=\sec ^{2} x\) At \(x=\pi / 4 \Rightarrow\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}}=\sec ^{2} \frac{\pi}{4}=2\) Now, the equation of tangent to the curve \(y=\tan x\) at \(x\) \(=\pi / 4\) is given by- \((\mathrm{y}-1)=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=\pi / 4}(\mathrm{x}-\pi / 4)\) \((\mathrm{y}-1)=2(\mathrm{x}-\pi / 4)\) \(\mathrm{y}=2(\mathrm{x}-\pi / 4)+1\) Let equation (ii) meet at point \(\mathrm{A}\) on \(\mathrm{X}\) - Axis than, \(\mathrm{y}=0\) \(0=2 \mathrm{x}+1-\pi / 2 \Rightarrow \mathrm{x}=\frac{\pi}{4}-\frac{1}{2}\) So, as per the figure required area- \(\mathrm{A}=\) Area of \(\mathrm{OAB}-\) Area of \(\triangle \mathrm{ACB}\) \(\mathrm{A}=\int_{0}^{\pi / 4} \tan \mathrm{xdx}-\left(\frac{1}{2} \times\left(\frac{\pi}{4}-\left(\frac{\pi}{4}-\frac{1}{2}\right) \times 1\right)\right.\) \(A=[\log \sec x]_{0}^{\pi / 4}-\left[\frac{1}{2} \times\left[\frac{1}{2}\right] \times 1\right]\) \(A=\log \sqrt{2}-0-\frac{1}{4} \Rightarrow A=\log \sqrt{2}-\frac{1}{4}\) sq. units
BCECE-2004
Application of the Integrals
86844
If the ordinate \(x=a\) divides the area bounded by the curve \(y=\left(1+\frac{8}{x^{2}}\right), x\)-axis and the ordinates \(x=2, x=4\), into two equal parts then the value of \(a\) is :
1 \(2 \mathrm{a}\)
2 \(2 \sqrt{2}\)
3 \(\frac{\mathrm{a}}{2}\)
4 none of these
Explanation:
(B) : Given that, The curve, \(y=1+\frac{8}{x^{2}}\) and \(x=a, x=2, x=4\) Since, \(\mathrm{x}=\mathrm{a}\) divides the area into two equal parts so, \(A=\int_{2}^{a}\left(1+\frac{8}{x^{2}}\right) d x=\int_{a}^{4}\left(1+\frac{8}{x^{2}}\right) d x\) \(\left[\mathrm{x}-\frac{8}{\mathrm{x}}\right]_{2}^{\mathrm{a}}=\left[\mathrm{x}-\frac{8}{\mathrm{x}}\right]_{\mathrm{a}}^{4}\) \(\left(\mathrm{a}-\frac{8}{\mathrm{a}}\right)-\left(2-\frac{8}{2}\right)=\left(4-\frac{8}{4}\right)-\left(\mathrm{a}-\frac{8}{\mathrm{a}}\right)\) \(\underline{\left(\mathrm{a}^{2}-8\right)}+2=2-\underline{\left(\mathrm{a}^{2}-8\right)}\)
BCECE-2003
Application of the Integrals
86845
The area of the region bounded by \(x^{2}+y^{2}-2 y\) \(-3=0\) and \(y=|x|+1\) is
